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Balbharti Maharashtra State Board 11th Biology Important Questions Chapter 8 Plant Tissues and Anatomy Important Questions and Answers.

Maharashtra State Board 11th Biology Important Questions Chapter 8 Plant Tissues and Anatomy

Question 1.
How plant tissues are classified on the basis of their ability to divide?
Answer:
Plant tissues are classified into meristematic tissues and permanent tissues based on their ability to divide.

Question 2.
Identify the labels i, ii and iii in the given figure of meristematic tissue and write its characteristics.

Answer:
1. Cell wall
2. Nucleus
3. Cytoplasm
Characteristics of meristematic tissue:

1. It is a group of young, immature cells.
2. These are living cells with ability to divide in the regions where they are present.
3. These are polyhedral or isodiametric in shape without intercellular spaces.
4. Cell wall is thin, elastic and mainly composed of cellulose.
5. Protoplasm is dense with distinct nucleus at the centre and vacuoles if present, are very small.
6. Cells show high rate of metabolism.

Question 3.
With the help of neat and labelled diagram explain the classification of meristematic tissue based on its position.
Answer:
Classification of meristematic tissue based on its position:
1. Apical meristem:
a. It is produced from promeristem and forms growing point of apices of root, shoot and their lateral branches.
b. It brings about increase in length of plant body and is called as apical initials.
c. Shoot apical meristem is terminal in position whereas in root it is subterminal i.e. located behind the root cap.

2. Intercalary meristem:
a. Intercalary meristematic tissue is present in the top or base area of node.
b. Their activity is mainly seen in monocots.
c. These are short lived.

3. Lateral meristem:
a. It is present along the sides of central axis of organs.
b. It takes part in increasing girth of stem or root, e.g. Intrafascicular cambium.
c. It is found in vascular bundles of gymnosperms and dicot angiosperms.

Question 4.
Complete the given table representing types of meristematic tissue based on its function.
Answer:

Types of meristematic tissue	Function
1. Protoderm	It is found in young growing region of a plant forming a protective covering like epidermis around the various organs.
2. Procambium	It is involved in developing primary vascular tissue.
3. Ground meristem	It forms structures like cortex, endodermis, pericycle, medullary rays, pith.

Question 5.
Which are the simple permanent tissues in plants?
Answer:
Parenchyma, Collenchyma and Sclerenchyma are the simple permanent tissues in plants.

Question 6.
Complete the given chart by giving characteristics of following tissues:
Answer:

Question 7.
Name the type of tissue in the given figure, identify labels ‘a’ and ‘b’ and write its characteristics.

Answer:
1. The given figure represents simple permanent tissue i.e. Parenchyma.
2. a: Vacuole, b: Intercellular air spaces.

Characteristics of parenchyma: Parenchyma:

1. It is a type of simple permanent tissue.
2. Cells in this tissue are thin walled, isodiametric, round, oval to polygonal or elongated in shape.
3. Cell wall is composed of cellulose.
4. Cells are living with prominent nucleus and cytoplasm with large vacuole.
5. Parenchyma has distinct intercellular spaces. Sometimes, cells may show compact arrangement.
6. The cytoplasm of adjacent cells is interconnected through plasmodesmata and thus forms a continuous tissue.
7. This is less specialized permanent tissue.
8. Occurrence:
   These cells are distributed in all the parts of a plant body viz. epidermis, cortex, pericycle, pith, mesophyll cells, endosperm, xylem and phloem.
9. Functions:
   These cells store food, water, help in gaseous exchange, increase buoyancy, perform photosynthesis and different functions in plant body.
10. Dedifferentiation in parenchyma cells develops vascular cambium and cork cambium at the time of
    secondary growth.

Question 8.
Identify the type of tissue shown in the given figure and write its characteristics.

Answer:
The given figure represents Collenchyma tissue.
Characteristics of Collenchyma:

1. It is a simple permanent tissue made up of living cells.
2. The cell wall is cellulosic but shows uneven deposition of cellulose and pectin especially at comers.
3. The walls may show presence of pits.
4. Cells are similar like parenchyma, containing cytoplasm, nucleus and vacuoles but small in size and without intercellular spaces. Thus, the cells appear to be compactly packed.
5. The cells are either circular, oval or angular in transverse section.

Function:
Collenchyma is a living mechanical tissue and serves different functions in plants.
a. It gives mechanical strength to young stem and parts like petiole of leaf.
b. It allows bending and pulling action in plant parts and also prevents tearing of leaf.
c. It also allows growth and elongation of organs.
d. Collenchyma is usually absent in monocots and roots of dicot plant.

Question 9.
With the help of neat and labelled diagrams explain the Sclerenchyma Tissue.

Answer:
Sclerenchyma Tissue:

1. It is simple permanent tissue made up of compactly arranged thick walled dead cells.
2. The cells are living at the time of production but at maturity they become dead.
3. Cells are devoid of cytoplasm.
4. Their walls are thickened due to uniform deposition of lignin.
5. Cells remain interconnected through several pits.

Types of Sclerenchyma:
Sclerenchyma cells are categorized into two types on the basis of their size and shape as
fibres and sclereids:
a. Fibres:
Fibres are thread-like, elongated and narrow structures with tapering and interlocking end walls. Fibres are mostly in bundles. Pits are narrow, unbranched and oblique. They provide mechanical strength.

b. Sclereids:
Sclereids are usually broad, with blunt end walls.
These occur singly or in loose groups and their pits are deep branched and straight.
These are developed due to secondary thickening of parenchyma cells and provides stiffness only.

Functions:
a. This tissue functions as the main mechanical tissue.
b. It permits bending, shearing and pulling.
c. It gives rigidity to leaves and prevents it from falling.
d. It also gives rigidity to epicarps and seeds.

Question 10.
Give a brief account of water-conducting tissues in higher plants.
Answer:

1. Xylem is the water-conducting tissue in higher plants. It is a dead complex tissue.
It also provides mechanical strength to the plant body.
Components of xylem are tracheids, vessels, xylem parenchyma and xylem fibres.

2. Tracheids:
a. These are elongated, tubular and dead cells (without protoplasm).
b. The ends are oblique and tapering.
c. The cell walls is unevenly thickened and lignified. This provides mechanical strength.
d. Tracheids contribute 95% of wood in gymnosperms and 5% in angiosperms.
e. The different types of thickening patterns are seen on their walls such as annular (in the form of rings), spiral (in the form of spring/helix), scalariform (ladder like) and pitted (small circular area). Pitted is the most advanced type of thickening which may be simple or bordered.

3. Vessels:
a. Vessels are longer than tracheids with perforated or dissolved ends and formed by union of several vessels end to end.
b. These are involved in conduction of water and minerals.
c. Their lumen is wider than tracheids and the thickening is due to lignin and similar to tracheids.
d. In monocots, vessels are rounded where as they are angular in dicot angiosperms.
e. The first formed xylem vessels (protoxylem) are small and have either annular or spiral thickenings while latter formed xylem vessels are larger (metaxylem) and have reticulate or pitted thickenings.
f. When protoxylem is arranged towards pith and metaxylem towards periphery it is called as endarch
e. g. in stem and when the position is reversed as in the roots is called as exarch.

4. Xylem parenchyma:
a. Xylem parenchyma cells are small associated with tracheids and vessels.
b. This is the only living tissue among this complex tissue.
c. The function is to store food (starch) and sometimes tannins.
d. Xylem parenchyma are involved in lateral or radial conduction of water or sap.

5. Xylem fibres:
a. Xylem fibres are sclerenchymatous cells and serve mainly mechanical support. These are called wood fibres.
b. These are also elongated, narrow and spindle shaped.
c. Cells are tapering at both the ends and their walls are lignified.

Question 11.
Draw neat and labelled diagram of xylem tissue and vascular bundle.
Answer:

Question 12.
Match the following.

Column I	Column II
1. Protoxylem	(a) Xylem with larger vessels
2. Endarch Xylem	(b) Protoxylem arranged towards pith
3. Metaxylem	(c) Metaxylem arranged towards pith
4. Exarch xylem	(d) First formed xylem vessels

Answer:

Column I	Column II
1. Protoxylem	(d) First formed xylem vessels
2. Endarch Xylem	(b) Protoxylem arranged towards pith
3. Metaxylem	(a) Xylem with larger vessels
4. Exarch xylem	(c) Metaxylem arranged towards pith

Question 13.
Describe the structure of phloem.

Answer:
Structure of phloem:
1. Phloem is a living tissue. It is also called as bast.
It is responsible for conduction of organic food material from source (generally leaf) to a sink (other plant parts).
On the basis of origin, it can be protophloem (first formed) and metaphloem (latterly formed).
It is composed of sieve elements (sieve cells and sieve tubes), companion cells, phloem parenchyma and phloem fibres.

2. Sieve elements:
a. Sieve tubes are long tubular conducting channel of phloem.
b. These are placed end to end with bulging at end walls.
c. The sieve tube has sieve plate formed by septa with small pores.
d. The sieve plates connect protoplast of adjacent sieve tube cells.
e. The sieve tube cell is a living cell with a thin layer of cytoplasm, but loses its nucleus at maturity.
f. The sieve tube cell is connected to companion cell through phloem parenchyma by plasmodesmata.
g. Sieve cells are found in lower plants like pteridophytes and gymnosperms and sieve tubes are found in angiosperms.
h. The cells are narrow, elongated with tapering ends and sieve area located laterally.

3. Companion cells:
a. These are narrow elongated and living.
b. Companion cells are laterally associated with sieve tube elements.
c. Companion cells have dense cytoplasm and prominent nucleus.
d. Nucleus of companion cell regulates functions of sieve tube cells through simple pits.
e. From origin point of view, sieve tube cells and companion cell are derived from same cell. Death of the one result in death of the other type.

4. Phloem parenchyma:
a. Cells of phloem parenchyma are living, elongated found associated with sieve tube and companion cells.
b. Their chief function is to store food, latex, resins, mucilage, etc.
c. The cells carry out lateral conduction of food material.
d. These cells are absent in most of the monocots.

5. Phloem fibres (Bast fibres):
a. Phloem fibres are the only dead tissue among this unit.
b. They are sclerenchymatous.
c. They are generally absent in primary phloem, but present in secondary phloem.
d. These cells have with lignified walls and provide mechanical support.
e. They are used in making ropes and rough clothes.

Question 14.
Draw a diagram of phloem.
Answer:

Question 15.
Name the types of tissue systems in plants.
Answer:
The types of tissue systems in plants are epidermal tissue system, ground tissue system and vascular tissue system.

Question 16.
Write a short note on Epidermis.
Answer:
Epidermis:

1. It is the outermost protective cell layer made up of compactly arranged cells without intercellular spaces.
2. Cells show presence of central large vacuole, thin cytoplasm and a nucleus.
3. The outer side of the epidermis is often covered with a waxy thick layer called the cuticle which prevents the loss of water.
4. Root epidermis (Epiblema) has root hairs. These are unicellular, elongated and involved in absorption of sap from the soil.
5. In stem, epidermal hairs are called trichomes. These are generally multicellular, branched or unbranched, stiff or soft or even secretory. These help in preventing water loss due to transpiration.

Question 17.
Draw a diagram representing epidermal tissue system.
Answer:

Question 18.
Write a short note on Structure of stomata.
Answer:
Structure of stomata:

1. Small gateways in the epidermal cells are called as stomata.
2. Stoma is controlled or guarded by specially modified cells called guard cells.
3. These guard cells may be kidney shaped (dicot) or dumbbell shaped (monocot), collectively called as stomata.
4. Guard cells have chloroplasts to carry out photosynthesis.
5. Change in turgor pressure of guard cells causes opening and closing of stomata, which enables exchange of gases and water vapour.
6. Stomata are further covered by subsidiary cells.
7. Stoma, guard cells and subsidiary cells form a unit called stomatal apparatus.

Question 19.
Write the information related to diagrams given below.
Answer:
1. The given diagram represents stoma in dicot leaf.
[Note: We have given additional label of ‘chloroplast ’for better understanding of students]
2. Structure of stomata:

1. Small gateways in the epidermal cells are called as stomata.
2. Stoma is controlled or guarded by specially modified cells called guard cells.
3. These guard cells may be kidney shaped (dicot) or dumbbell shaped (monocot), collectively called as stomata.
4. Guard cells have chloroplasts to carry out photosynthesis.
5. Change in turgor pressure of guard cells causes opening and closing of stomata, which enables exchange of gases and water vapour.
6. Stomata are further covered by subsidiary cells.
7. Stoma, guard cells and subsidiary cells form a unit called stomatal apparatus.

Question 20.
Explain the term ground tissue.
Answer:
Ground tissue:

1. All the plant tissues excluding epidermal and vascular tissue is ground tissue.
2. It is made up of simple permanent tissue e.g. parenchyma.
3. It is present in cortex, pericycle, pith and medullary rays in the primary stem and root.
4. Collenchyma and sclerenchyma in the hypodermis and chloroplasts containing mesophyll tissue in leaves is also ground tissue.

Question 21.
Describe various types of vascular bundles.
Answer:
Vascular bundles occur in the form of distinct patches of the complex tissue viz. Xylem and Phloem. On the basis of their arrangement in the plant body they are classified as follows:
1. Radial vascular bundles:
When the complex tissues (xylem and phloem) are situated separately on separate radius as separate bundle, vascular bundle is called Radial vascular bundle. This is a common feature of roots.

2. Conjoint vascular bundles:
When the complex tissue (xylem and phloem) is collectively present as neighbours of each other on the same radius, vascular bundle is called Conjoint vascular bundle.
They are of two types:
a. Collateral vascular bundle:
In this type of vascular bundle, xylem lies inwards and the phloem lies outwards.
These bundles may be further of open type (secondary growth takes place) containing cambium in between xylem and phloem and closed type if cambium is not present (secondary growth absent).
b. Bicollateral vascular bundle:
When phloem is present in a vascular bundle on both the sides of xylem and intervening cambium tissue, it is called bicollateral vascular bundle. It is a feature of family Cucurbitaceae.

3. Concentric vascular bundle:
a. When one vascular tissue is completely encircling the other, it is called as concentric vascular bundle.
b. When phloem is encircled by xylem, it is called as leptocentric vascular bundle, whereas when xylem is encircled by phloem, it is called as hadrocentric vascular bundle.
c. When xylem is encircled by phloem on both faces, it is called as amphicribral vascular bundle. When phloem is encircled by xylem on both faces it is called as amphivasal vascular bundle.

Question 22.
Explain how formation of cambial ring occurs in dicot stem.
Answer:

1. The cambium present between the primary xylem and primary phloem of a vascular bundle is called intrafascicular cambium.
2. With the onset of favourable season, meristematic cells of intrafascicular cambium become active.
3. Simultaneously, the ray parenchyma cells, both fusiform initials and ray initials become meristematic. This is known as dedifferentiation.
4. These form patch of cambial cells (meristematic cells) in between the adjacent bundles and produce interfascicular cambium.
5. Both intrafascicular and interfascicular cambium join and form a complete ring, known as the cambial ring. This is possible because they lie in one plane.

Question 23.
‘Secondary growth is observed in most of the dicot and gymnospermic root.’ State whether the given statement is true or false and justify your answer.
Answer:

1. The given statement is true.
2. Secondary growth is observed in most of the dicot and gymnospermic root by producing secondary vascular tissue and periderm.
3. Secondary growth is produced by vascular cambium and cork cambium respectively.
4. Conjunctive parenchyma cells present on the inner edges of primary phloem bundles become meristematic.
5. These cells add secondary xylem and secondary phloem on the inner and outer side respectively which results in secondary growth.

Question 24.
Differentiate between heartwood and sap wood.
Answer:

Heartwood	Sap wood
1. It is central region of secondary xylem (wood).	It is the peripheral region of secondary xylem (wood).
2. It is darker in colour due to deposition of oils, gums, resins, tannins, etc	It is lighter in colour and without any depositions.
3. It is non- functional part of secondary xylem.	It is functional part of secondary xylem.
4. It is resistant to pathogens.	It is more susceptible to pathogens
5. It is not involved in conduction of sap.	It is involved in conduction of sap.
6. It is also called as duramen.	It is also called as alburnum.

Question 25.
What are tyloses?
Answer:
Tyloses:
1. Tracheary elements of heartwood are plugged by in-growth of adjacent parenchyma cells are known as tyloses.
2. Tyloses are fdled by oils, gums, resins, tannins called as extractives.

Question 26.
Explain how periderm is formed?
Answer:
Formation of periderm:
As the stem increase in diameter due to activity of vascular cambium, the outer cortical and epidermal layer get ruptured. Thus, it becomes necessaiy to replace these cells by new cells.

1. Phellogen (cork cambium) develops in extrastelar region (cortex region) of the stem.
2. The outer cortical cells of cortex become meristematic and produce a layer of thin walled, rectangular cells. These cells cut off new cells on both sides.
3. The cells produced on outer side develop phellem (cork), whereas on the inner side produce phelloderm (secondary cortex).
4. The cork is impervious in nature and does not allow entry of water due to suberized walls. Secondary cortex is parenchymatous in nature.
5. Phellogen, phellem and phelloderm constitute periderm.

Question 27.
Explain the given terms:
1. Bark
2. Lenticels
3. Anomalous secondary growth
Answer:
1. Bark:
a. Bark is non-technical term referring to all cell types found external to vascular cambium including secondary phloem.
b. Bark of early season is soft and of the late season is hard.

2. Lenticels:
a. Lenticels are aerating pores present as raised scars on the surface of bark.
b. These are portions of periderm, where phellogen activity is more.
c. Lenticels are meant for gaseous and water vapour exchange.

3. Anomalous secondary growth:
a. Monocot stems lack cambium hence secondary growth does not take place.
b. However, accessory cambium development in plants like, Dracaena, Agave, Palms and root of sweet potato shows presence of secondary growth. This is called as anomalous secondary growth.

Question 28.
With the help of neat and labelled diagram explain the anatomy of dicot root.
Answer:

The transverse section of a typical dicotyledonous root shows following anatomical features:
1. Epiblema: It is the outermost single layer of cells without cuticle. Some epidermal cells prolong to form unicellular root hairs.
2. Cortex: It is made up of many layers of thin walled parenchyma cells. Cortical cells store food and water.
3. Exodermis: After the death of epiblema, outer layer of cortex become cutinized and is called Exodermis.

4. Endodermis:
The innermost layer of cortex is called Endodermis.
The cells are barrel-shaped and their radial walls bear Casparian strip or Casparian bands composed of suberin. Near the protoxylem, there are unthickened passage cells.

5. Stele: It consists of pericycle, vascular bundles and pith.
a. Pericycle: Next to the endodermis, there is a single layer of thin walled parenchyma cells called pericycle. It forms outermost layer of stele or vascular cylinder.
b. Vascular bundle: Vascular bundles are radial. Xylem and Phloem occur in separate patches arranged on alternate radii. Xylem is exarch in root that means protoxylem vessels are towards periphery and metaxylem elements are towards centre. Xylem bundles vary from two to six number, i.e. they may be diarch, triarch, tetrarch, etc.
Connective tissue: A parenchymatous tissue is present in between xylem and phloem.
c. Pith: The central part of stele is called pith. It is narrow and made up of parenchymatous cells, with or without intercellular spaces.
6. At a later stage cambium ring develops between the xylem and phloem causing secondary growth.

Question 29.
With the help of neat and labelled diagram explain the anatomy of monocot root.
Answer:
The transverse section of a typical monocotyledonous root shows following anatomical features:
1. Epiblema: It is the outermost single layer of cells without cuticle. Some epidermal cells prolong to form unicellular root hairs.
2. Cortex: It is made up of many layers of thin walled parenchyma cells. Cortical cells store food and water.
Endodermis: It is innermost layer of cortex. The cells of endodermis are thick walled except the passage cells which lie just opposite to the protoxylem.
Stele: It consists of pericycle, vascular bundles and pith.
a. Pericycle: Pericycle is present below the endodermis.
b. Vascular bundle: Vascular bundles are radial. Xylem and Phloem occur in separate patches arranged on alternate radii. Xylem is exarch in root that means protoxylem vessels are towards periphery and metaxylem elements are towards centre. Polyarch condition of xylem is observed, (xylem bundles are more than six).
Pith: Pith is large and well developed.
Secondary growth does not occur due to absence of cambium.

Question 30.
What is polyarch condition of root?
Answer:
Polyarch condition is the one in which roots possess more than six xylem bundles.

Question 31.
Explain in detail anatomy of sunflower stem.
Answer:
A transverse section of sunflower (dicot) stem shows the following structures:
1. Epidermis: It is a single, outermost layer with multicellular outgrowth called trichomes. A layer of cuticle
is usually present towards the outer surface of epidermis.

2. Cortex: Cortex is situated below the epidermis and is usually differentiated into three regions namely, hypodermis, general cortex and endodermis.
a. Hypodermis: It is situated just below the epidermis and is made of 3-5 layers of collenchymatous cells. Intercellular spaces are absent.
b. General cortex: It is made up of several layers of large parenchymatous cells with intercellular spaces.
c. Endodermis: It is an innermost layer of cortex which is made up of barrel shaped cells. It is also called starch sheath, as it is rich in starch grain.

3. Stele: It is differentiated into pericycle, vascular bundles and pith.
a. Pericycle: It is the outermost layer of vascular system situated between the endodermis and vascular bundles. In sunflower, it is multi-layered and also called hard bast.
b. Vascular bundles: Vascular bundles are conjoint, collateral, open, and are arranged in a ring. Each one is composed of xylem, phloem and cambium. Xylem is endarch. A strip of cambium is present between xylem and phloem.
c. Pith: It is situated in the centre of the young stem and is made up of large-sized parenchymatous cells with conspicuous intercellular spaces.

Question 32.
With the help of neat and labelled diagram explain the anatomy of maize stem.
Answer:

A transverse section of maize (monocot) stem shows the following structures:

1. Epidermis: It is single layered and without trichomes.
2. Hypodermis: It is sclerenchymatous.
3. Ground tissue: It consists of thin walled parenchyma cells. It extends from hypodermis to the centre. It is not differentiated into cortex, endodermis, pericycle and pith.
4. Vascular bundles: Vascular bundles are numerous and are scattered in ground tissue. Each vascular bundle is surrounded by a sclerenchymatous bundle sheath. Vascular bundles are conjoint, collateral and closed (without cambium). Xylem is endarch and shows lysigenous cavity.
5. Pith: Pith is absent.

Question 33.
With the help of a neat and labelled diagram, describe the internal structure of dorsiventral leaf.
Answer:
1. Structure of dorsiventral leaf: The mesophyll tissue is differentiated into palisade and spongy parenchyma in a dorsiventral leaf. This type is very common in dicot leaf. The different parts of this leaf are as follows:
2. Upper epidermis: It consists of a single layer of tightly packed rectangular, barrel shaped, parenchymatous cells which are devoid of chloroplast. A distinct layer of cuticle lies on the outside of the epidermis. Stomata are generally absent.
3. Mesophyll: Between upper and lower epidermis, there is chloroplast-containing photosynthetic tissue called mesophyll It is differentiated into Palisade parenchyma and Spongy parenchyma.
a. Palisade parenchyma:
Palisade parenchyma is present below upper epidermis and consists of closely packed elongated cells. The cells contain abundant chloroplasts and help in photosynthesis.
b. Spongy parenchyma:
Spongy parenchyma is present below palisade tissue and consists of loosely arranged irregularly shaped cells with intercellular spaces. The spongy parenchyma cells contain chloroplast and are in contact with the atmosphere through stomata.

4. Vascular system: It is made up of a number of vascular bundles of varying size depending upon the venation. Each one is surrounded by a thin layer of parenchymatous cells called bundle sheath. Vascular bundles are closed. Xylem lies towards upper epidermis and phloem towards lower epidermis. Cambium is absent, hence there is no secondary growth in the leaf.
5. Lower epidermis: It consists of a single layer of compactly arranged rectangular, parenchymatous cells. A thin layer of cuticle is also present. The lower epidermis contains a large number of microscopic pores called stomata. There is an air-space called substomatal chamber at each stoma.

Question 34.
With the help of a neat labelled diagram, describe the anatomy of isobilateral leaf.
Answer:
The parts of isobilateral leaf are as follows:

1. Epidermis:
It is single layered, present on both sides of the leaf.
It consists of compactly arranged rectangular transparent parenchymatous cells.
Both the surfaces contain stomata.
Both the surfaces have a distinct layer of cuticle.
2. Mesophyll:
Mesophyll is not differentiated into palisade and spongy tissue.
3. Vascular bundle:
These are conjoint, collateral and closed.

Question 35.
Compare between dorsiventral and isobilateral leaf.
Answer:

Dorsiventral leaf	Isobilateral leaf
1. Dorsiventral Leaf is very common in dicotyledonous plants.	Isobilateral leaf is very common in monocotyledonous plants.
2. In this mesophyll tissue is differentiated into palisade and spongy parenchyma.	In this mesophyll tissue is not differentiated into palisade and spongy parenchyma.
3. The leaves are commonly horizontal in orientation with distinct upper and lower surfaces. The upper surface which faces the sun is darker than the lower surface.	In this leaf both the surfaces are equally illuminated as both the surface can face the sun, and show similar structure. The two surfaces are equally green.
4. Stomata is absent on the upper surface of these leaves.	Stomata is present on both the upper and lower surfaces of these leaves.

Question 36.
Distinguish between anatomy of dicot root and monocot root.
Answer:

Anatomy of dicot root	Anatomy of monocot root
1. Pith is narrow.	Pith is large and well developed.
2. Diarch, triarch or tetrarch condition can be observed. (Xylem bundles vary from two to six number)	Polyarch condition is observed, (xylem bundles are more than six)
3. Cambium is formed in later stage between xylem and phloem which causes secondary growth.	Secondary growth is absent.

Question 37.
Distinguish between anatomy of dicot stem and monocot stem.
Answer:

Anatomy of Dicot stem	Anatomy of Monocot stem
1. Epidermis shows presence of multicellular trichomes.	Epidermis is without trichomes.
2. Hypodermis is made up of collenchymatous cells.	Hypodermis is made up of sclerenchymatous cells.
3. Medullary rays are present between vascular bundles.	Medullary rays are absent.
4. Vascular bundles are arranged in the form of a ring.	Vascular bundles are scattered in the ground tissue.
5. It is conjoint, collateral and open (Cambium present)	They are conjoint, collateral and closed (cambium is absent).
6. Vascular bundle is not surrounded by a sclerenchymatous bundle sheath.	Vascular bundle is surrounded by a sclerenchymatous bundle sheath.
7. Secondary growth takes place due to presence of cambium.	Secondary growth does not occur due to absence of cambium.
8. Pith is present.	Pith is absent.

Question 38.
Apply Your Knowledge

Question 1.
Which plant part would show the following:

1. Radial vascular bundles.
2. Large and well-developed pith.
3. Differentiation of mesophyll into palisade and spongy tissue.
4. Presence of stomata on both upper and lower epidermis.

Answer:

1. Root
2. Monocot root and Dicot stem,
3. Dicot leaf
4. Monocot leaf

Question 2.
When a tree is debarked, which tissues are removed?
Answer:
The bark is made up of tissues like cork, cork cambium and secondary cortex, which are removed when a tree is debarked.

Question 3.
While eating fruits like pear or guava, it feels gritty. What gives stiffness to these fruits?
Answer:
Sclereids are found in pulp of fruits like pear and guava which gives them stiffness and thus we feel gritty while eating these fruits.

Question 39.
Quick Review

Question 40.
Exercise

Question 1.
Define tissue.
Answer:
A group of cells having essentially a common function and origin is called as tissue.

Question 2.
Classify the meristematic tissue based on its origin.
Answer:
Classification of meristematic tissue on the basis of origin:
1. Promeristem / Primordial meristem:
a. It is also called as embryonic meristem.
b. It usually occupies very minute area at the tip of root and shoot.

2. Primary meristem:
a. It originates from the primordial meristem and occurs in the plant body from the beginning, at the root and shoot apices.
b. Cells are always in active state of division and give rise to permanent tissues.

3. Secondary meristem:
a. These tissues develop from living permanent tissues during later stages of plant growth hence are called as secondary meristems.
b. This tissue occurs in the mature regions of root and shoot of many plants.
c. Secondary meristem is always lateral (to the central axis) in position e.g. Fascicular cambium, inter fascicular cambium, cork cambium.

Question 3.
Explain in detail classification of meristematic tissue based on its position.
Answer:
Classification of meristematic tissue based on its position:
1. Apical meristem:
a. It is produced from promeristem and forms growing point of apices of root, shoot and their lateral branches.
b. It brings about increase in length of plant body and is called as apical initials.
c. Shoot apical meristem is terminal in position whereas in root it is subterminal i.e. located behind the root cap.

2. Intercalary meristem:
a. Intercalary meristematic tissue is present in the top or base area of node.
b. Their activity is mainly seen in monocots.
c. These are short lived.

3. Lateral meristem:
a. It is present along the sides of central axis of organs.
b. It takes part in increasing girth of stem or root, e.g. Intrafascicular cambium.
c. It is found in vascular bundles of gymnosperms and dicot angiosperms.

Question 4.
Give any two examples of secondary meristematic tissue.
Answer:
Secondary meristem is always lateral (to the central axis) in position e.g. Fascicular cambium, inter fascicular cambium, cork cambium.

Question 5.
Draw a diagram of meristematic cells.
Answer:
1. Cell wall
2. Nucleus
3. Cytoplasm
Characteristics of meristematic tissue:

1. It is a group of young, immature cells.
2. These are living cells with ability to divide in the regions where they are present.
3. These are polyhedral or isodiametric in shape without intercellular spaces.
4. Cell wall is thin, elastic and mainly composed of cellulose.
5. Protoplasm is dense with distinct nucleus at the centre and vacuoles if present, are very small.
6. Cells show high rate of metabolism.

Question 6.
Write a short note on tracheids.
Answer:
Tracheids:
a. These are elongated, tubular and dead cells (without protoplasm).
b. The ends are oblique and tapering.
c. The cell walls is unevenly thickened and lignified. This provides mechanical strength.
d. Tracheids contribute 95% of wood in gymnosperms and 5% in angiosperms.
e. The different types of thickening patterns are seen on their walls such as annular (in the form of rings), spiral (in the form of spring/helix), scalariform (ladder like) and pitted (small circular area). Pitted is the most advanced type of thickening which may be simple or bordered.

Question 7.
Describe parenchyma in detail.
Answer:
Cell is the component that brings about important processes in the living organisms.

Question 8.
Describe the structure of xylem in detail.
Answer:
1. Xylem is the water conducting tissue in higher plants. It is a dead complex tissue.
It also provides mechanical strength to the plant body.
Components of xylem are tracheids, vessels, xylem parenchyma and xylem fibres.

2. Tracheids:
a. These are elongated, tubular and dead cells (without protoplasm).
b. The ends are oblique and tapering.
c. The cell walls is unevenly thickened and lignified. This provides mechanical strength.
d. Tracheids contribute 95% of wood in gymnosperms and 5% in angiosperms.
e. The different types of thickening patterns are seen on their walls such as annular (in the form of rings), spiral (in the form of spring/helix), scalariform (ladder like) and pitted (small circular area). Pitted is the most advanced type of thickening which may be simple or bordered.

3. Vessels:
a. Vessels are longer than tracheids with perforated or dissolved ends and formed by union of several vessels end to end.
b. These are involved in conduction of water and minerals.
c. Their lumen is wider than tracheids and the thickening is due to lignin and similar to tracheids.
d. In monocots, vessels are rounded where as they are angular in dicot angiosperms.
e. The first formed xylem vessels (protoxylem) are small and have either annular or spiral thickenings while latter formed xylem vessels are larger (metaxylem) and have reticulate or pitted thickenings.
f. When protoxylem is arranged towards pith and metaxylem towards periphery it is called as endarch
e. g. in stem and when the position is reversed as in the roots is called as exarch.

4. Xylem parenchyma:
a. Xylem parenchyma cells are small associated with tracheids and vessels.
b. This is the only living tissue among this complex tissue.
c. The function is to store food (starch) and sometimes tannins.
d. Xylem parenchyma are involved in lateral or radial conduction of water or sap.

5. Xylem fibres:
a. Xylem fibres are sclerenchymatous cells and serve mainly mechanical support. These are called wood fibres.
b. These are also elongated, narrow and spindle shaped.
c. Cells are tapering at both the ends and their walls are lignified.

Question 9.
What are Sclerenchyma fibres?
Answer:
a. Fibres:
Fibres are thread-like, elongated and narrow structures with tapering and interlocking end walls. Fibres are mostly in bundles. Pits are narrow, unbranched and oblique. They provide mechanical strength.

Question 10.
Write the functions of parenchyma cells.
Answer:
Parenchyma, Collenchyma and Sclerenchyma are the simple permanent tissues in plants.

Question 11.
Write function of collenchyma tissue.
Answer:
Function:
Collenchyma is a living mechanical tissue and serves different functions in plants.
a. It gives mechanical strength to young stem and parts like petiole of leaf.
b. It allows bending and pulling action in plant parts and also prevents tearing of leaf.
c. It also allows growth and elongation of organs.

Question 12.
Which are the different types of tracheids based on the types of thickenings on their walls?
Answer:
The different types of thickening patterns are seen on their walls such as annular (in the form of rings), spiral (in the form of spring/helix), scalariform (ladder like) and pitted (small circular area). Pitted is the most advanced type of thickening which may be simple or bordered.

Question 13.
Death of companion cell causes death of sieve tube cells and vice versa. Justify.
Answer:
Companion cells:
a. These are narrow elongated and living.
b. Companion cells are laterally associated with sieve tube elements.
c. Companion cells have dense cytoplasm and prominent nucleus.
d. Nucleus of companion cell regulates functions of sieve tube cells through simple pits.
e. From origin point of view, sieve tube cells and companion cell are derived from same cell. Death of the one result in death of the other type.

Question 14.
Which are the three types of simple permanent tissues?
Answer:
Parenchyma, Collenchyma and Sclerenchyma are the simple permanent tissues in plants.

Question 15.
Write the functions of sclerenchyma tissue.
Answer:
Functions:
a. This tissue functions as the main mechanical tissue.
b. It permits bending, shearing and pulling.
c. It gives rigidity to leaves and prevents it from falling.
d. It also gives rigidity to epicarps and seeds.

Question 16.
What are the components of xylem?
Answer:
1. Xylem is the water conducting tissue in higher plants. It is a dead complex tissue.
It also provides mechanical strength to the plant body.
Components of xylem are tracheids, vessels, xylem parenchyma and xylem fibres.

Question 17.
Name the living component of xylem.
Answer:
This is the only living tissue among this complex tissue.

Question 18.
Name the dead component of phloem.
Answer:
Phloem fibres are the only dead tissue among this unit.

Question 19.
What is closed vascular bundle?
Answer:
When cambium is not present between xylem and phloem, it is known as closed vascular bundle.

Question 20.
Describe two types of conjoint vascular bundles.
Answer:
Conjoint vascular bundles:
When the complex tissue (xylem and phloem) is collectively present as neighbours of each other on the same radius, vascular bundle is called Conjoint vascular bundle.
They are of two types:
a. Collateral vascular bundle:
In this type of vascular bundle, xylem lies inwards and the phloem lies outwards.
These bundles may be further of open type (secondary growth takes place) containing cambium in between xylem and phloem and closed type if cambium is not present (secondary growth absent).
b. Bicollateral vascular bundle:
When phloem is present in a vascular bundle on both the sides of xylem and intervening cambium tissue, it is called bicollateral vascular bundle. It is a feature of family Cucurbitaceae.

Question 21.
Write the function of trichomes.
Answer:
In stem, epidermal hairs are called trichomes. These are generally multicellular, branched or unbranched, stiff or soft or even secretory. These help in preventing water loss due to transpiration.

Question 22.
What are bicollateral vascular bundle?
Answer:
Bicollateral vascular bundle:
When phloem is present in a vascular bundle on both the sides of xylem and intervening cambium tissue, it is called bicollateral vascular bundle. It is a feature of family Cucurbitaceae.

Question 23.
Name the tissue that are not included in ground tissue.
Answer:
All the plant tissues excluding epidermal and vascular tissue is ground tissue.

Question 24.
Which type of conjoint – vascular bundles are found in members of Cucurbitaceae family?
Answer:
Bicollateral vascular bundle:
When phloem is present in a vascular bundle on both the sides of xylem and intervening cambium tissue, it is called bicollateral vascular bundle. It is a feature of family Cucurbitaceae.

Question 25.
What is concentric vascular bundle?
Answer:
Concentric vascular bundle:
a. When one vascular tissue is completely encircling the other, it is called as concentric vascular bundle.
b. When phloem is encircled by xylem, it is called as leptocentric vascular bundle, whereas when xylem is encircled by phloem, it is called as hadrocentric vascular bundle.
c. When xylem is encircled by phloem on both faces, it is called as amphicribral vascular bundle. When phloem is encircled by xylem on both faces it is called as amphivasal vascular bundle.

Question 26.
Define intrafascicular cambium.
Answer:
The cambium present between the primary xylem and primary phloem of a vascular bundle is called intrafascicular cambium.

Question 27.
What is the difference between spring wood and autumn wood?
Answer:
During favourable conditions, spring wood (early wood) is formed which has broader xylem bands, lighter colour, tracheids with thin wall and wide lumen, fibres are less in number, low density. Whereas, during unfavourable conditions, autumn wood (late wood) is formed which has narrow xylem band, darker in colour, lumen is narrow and walls are thick with abundant fibres, high density.

Question 28.
Explain how growth rings are formed in trees?
Answer:
1. Growth rings are formed due cambial activity during favourable and non-favourable climatic conditions.
2. During favourable conditions, spring wood (early wood) is formed which has broader xylem bands, lighter colour, tracheids with thin wall and wide lumen, fibres are less in number, low density. Whereas, during unfavourable conditions, autumn wood (late wood) is formed which has narrow xylem band, darker in colour, lumen is narrow and walls are thick with abundant fibres, high density.
3. Spring wood and autumn wood that appear as alternate light and dark concentric rings, constitute an annual ring or growth ring.

Question 29.
Which tissues are together called as periderm?
Answer:
Phellogen, phellem and phelloderm constitute periderm.

Question 30.
What is bark?
Answer:
Bark:
a. Bark is non-technical term referring to all cell types found external to vascular cambium including secondary phloem.
b. Bark of early season is soft and of the late season is hard.

Question 31.
What is the function of lenticels?
Answer:
Lenticels are meant for gaseous and water vapour exchange.

Question 32.
Explain the term anomalous secondary growth.
Answer:
Anomalous secondary growth:
a. Monocot stems lack cambium hence secondary growth does not take place.
b. However, accessory cambium development in plants like, Dracaena, Agave, Palms and root of sweet potato shows presence of secondary growth. This is called as anomalous secondary growth.

Question 33.
Explain in detail anatomical structure of a dicot stem.
Answer:
A transverse section of sunflower (dicot) stem shows the following structures:
1. Epidermis: It is a single, outermost layer with multicellular outgrowth called trichomes. A layer of cuticle
is usually present towards the outer surface of epidermis.

2. Cortex: Cortex is situated below the epidermis and is usually differentiated into three regions namely, hypodermis, general cortex and endodermis.
a. Hypodermis: It is situated just below the epidermis and is made of 3-5 layers of collenchymatous cells. Intercellular spaces are absent.
b. General cortex: It is made up of several layers of large parenchymatous cells with intercellular spaces.
c. Endodermis: It is an innermost layer of cortex which is made up of barrel shaped cells. It is also called starch sheath, as it is rich in starch grain.

3. Stele: It is differentiated into pericycle, vascular bundles and pith.
a. Pericycle: It is the outermost layer of vascular system situated between the endodermis and vascular bundles. In sunflower, it is multi-layered and also called hard bast.
b. Vascular bundles: Vascular bundles are conjoint, collateral, open, and are arranged in a ring. Each one is composed of xylem, phloem and cambium. Xylem is endarch. A strip of cambium is present between xylem and phloem.
c. Pith: It is situated in the centre of the young stem and is made up of large-sized parenchymatous cells with conspicuous intercellular spaces.

Question 34.
Draw neat and labelled diagrams of dicot and monocot root and differentiate between their anatomical characters.
Answer:
The transverse section of a typical dicotyledonous root shows following anatomical features:
1. Epiblema: It is the outermost single layer of cells without cuticle. Some epidermal cells prolong to form unicellular root hairs.
2. Cortex: It is made up of many layers of thin walled parenchyma cells. Cortical cells store food and water.
3. Exodermis: After the death of epiblema, outer layer of cortex become cutinized and is called Exodermis.

4. Endodermis:
The innermost layer of cortex is called Endodermis.
The cells are barrel-shaped and their radial walls bear Casparian strip or Casparian bands composed of suberin. Near the protoxylem, there are unthickened passage cells.

5. Stele: It consists of pericycle, vascular bundles and pith.
a. Pericycle: Next to the endodermis, there is a single layer of thin walled parenchyma cells called pericycle. It forms outermost layer of stele or vascular cylinder.
b. Vascular bundle: Vascular bundles are radial. Xylem and Phloem occur in separate patches arranged on alternate radii. Xylem is exarch in root that means protoxylem vessels are towards periphery and metaxylem elements are towards centre. Xylem bundles vary from two to six number, i.e. they may be diarch, triarch, tetrarch, etc.
Connective tissue: A parenchymatous tissue is present in between xylem and phloem.
c. Pith: The central part of stele is called pith. It is narrow and made up of parenchymatous cells, with or without intercellular spaces.
6. At a later stage cambium ring develops between the xylem and phloem causing secondary growth.

Question 35.
Which type of vascular bundles are observed in isobilateral leaf?
Answer:
Vascular bundle:
These are conjoint, collateral and closed.

Question 36.
Describe the internal structure of a leaf in which mesophyll is differentiated in palisade and spongy parenchyma.
Answer:
1. Structure of dorsiventral leaf: The mesophyll tissue is differentiated into palisade and spongy parenchyma in a dorsiventral leaf. This type is very common in dicot leaf. The different parts of this leaf are as follows:
2. Upper epidermis: It consists of a single layer of tightly packed rectangular, barrel shaped, parenchymatous cells which are devoid of chloroplast. A distinct layer of cuticle lies on the outside of the epidermis. Stomata are generally absent.
3. Mesophyll: Between upper and lower epidermis, there is chloroplast-containing photosynthetic tissue called mesophyll It is differentiated into Palisade parenchyma and Spongy parenchyma.
a. Palisade parenchyma:
Palisade parenchyma is present below upper epidermis and consists of closely packed elongated cells. The cells contain abundant chloroplasts and help in photosynthesis.
b. Spongy parenchyma:
Spongy parenchyma is present below palisade tissue and consists of loosely arranged irregularly shaped cells with intercellular spaces. The spongy parenchyma cells contain chloroplast and are in contact with the atmosphere through stomata.
4. Vascular system: It is made up of a number of vascular bundles of varying size depending upon the venation. Each one is surrounded by a thin layer of parenchymatous cells called bundle sheath. Vascular bundles are closed. Xylem lies towards upper epidermis and phloem towards lower epidermis. Cambium is absent, hence there is no secondary growth in the leaf.
5. Lower epidermis: It consists of a single layer of compactly arranged rectangular, parenchymatous cells. A thin layer of cuticle is also present. The lower epidermis contains a large number of microscopic pores called stomata. There is an air-space called substomatal chamber at each stoma.

Question 37.
Multiple Choice Questions:

Question 1.
Meristematic tissues are found
(A) only in stems of the plants
(B) in both roots and stems
(C) in all growing tips of the plant body
(D) only in roots of the plants
Answer:
(C) in all growing tips of the plant body

Question 2.
The tissue responsible for translocation of food material is _________
(A) xylem
(B) cambium
(C) parenchyma
(D) phloem
Answer:
(D) phloem

Question 3.
________ are used in making ropes and rough clothes.
(A) Phloem parenchyma
(B) Trachieds
(C) Phloem fibres
(D) Sieve tube elements
Answer:
(C) Phloem fibres

Question 4.
_______ are the only dead tissue among the phloem.
(A) Phloem parenchyma
(B) Sieve tubes
(C) Companion cells
(D) Phloem fibres
Answer:
(D) Phloem fibres

Question 5.
Phloem was named as _______ by Haberlandt as similar to xylem.
(A) Bast
(B) Leptome
(C) Wood fibres
(D) Casparian
Answer:
(B) Leptome

Question 6.
The sieve tube cell is connected to companion cell through phloem parenchyma by
(A) Plasmodesmata
(B) Interfascicular cambium
(C) Pericycle
(D) Hypodermis
Answer:
(A) Plasmodesmata

Question 7.
Which of the following tissues is with dead thick-walled cells without intercellular spaces?
(A) parenchyma
(B) collenchyma
(C) sclerenchyma
(D) phloem
Answer:
(C) sclerenchyma

Question 8.
The tissue which is present in between xylem and phloem of stem is called
(A) apical meristem
(B) pericycle
(C) vascular cambium
(D) cork cambium
Answer:
(C) vascular cambium

Question 9.
In stem, epidermal hairs are called as
(A) Cuticles
(B) Casparian strip
(C) Trichomes
(D) Companion cells
Answer:
(C) Trichomes

Question 10.
_______ forms the outer covering of plant body and is derived from protodenn or dermatogen.
(A) Ground tissue system
(B) Interfascicular cambium
(C) Vascular tissue system
(D) Epidermal tissue system
Answer:
(D) Epidermal tissue system

Question 11.
_______ play a vital role in exchange of gases and water vapour.
(A) Vascular bundles
(B) Stomata
(C) Ground tissues
(D) Trichomes
Answer:
(B) Stomata

Question 12.
In which of the following leaf possesses dumbbell shaped guard cell’?
(A) Pisum sativum
(B) Wheat
(C) Datura
(D) Sunflower
Answer:
(B) Wheat

Question 13.
Which of the following is NOT a characteristic of spring wood?
(A) Tracheids with wide lumen
(B) Less number of fibres
(C) Narrow xylem band
(D) Lighter colour
Answer:
(C) Narrow xylem band

Question 14.
Periderm consists of
(A) Phellogen
(B) Phellem
(C) Phelloderm
(D) All of these
Answer:
(D) All of these

Question 15.
Which of the following is essential for secondary growth?
(A) Xylem
(B) Pith
(C) Phloem
(D) Cambium
Answer:
(D) Cambium

Question 16.
Vascular bundles of dicot root are
(A) radial exarch
(B) radial endarch
(C) conjoint exarch
(D) conjoint endarch
Answer:
(A) radial exarch

Question 17.
In which of the following characters, a monocot root differs from dicot root?
(A) Open vascular bundle
(B) Large pith
(C) Radial vascular bundles
(D) Scattered vascular bundles
Answer:
(B) Large pith

Question 18.
Which of the following plant shows isobilateral leaves?
(A) Hibiscus
(B) Maize
(C) Mangifera indica
(D) Sunflower
Answer:
(B) Maize

Question 19.
Secondary growth does not occur in
(A) Maize stem
(B) Mango leaf
(C) Carina root
(D) All of these
Answer:
(D) All of these

Question 20.
Stele of a dicot stem consists of all the given below, EXCEPT
(A) Pericycle
(B) Cortex
(C) Vascular bundles
(D) Pith
Answer:
(B) Cortex

Question 21.
Hypodermis is collenchymatous in
(A) monocot stem
(B) dicot stem
(C) monocot root
(D) both (A) and (B)
Answer:
(B) dicot stem

Question 22.
Lysigenous cavity filled with water is present in
(A) dicot stem
(B) monocot stem
(C) monocot root
(D) dicot root
Answer:
(B) monocot stem

Question 23.
The vascular bundles in a dicot stem are
(A) collateral and open
(B) radial
(C) bicollateral and open
(D) collateral and closed
Answer:
(A) collateral and open

Question 38.
Competitive Corner

Question 1.
Phloem in gymnosperms lacks:
(A) companion cells only
(B) both sieve tubes and companion cells
(C) albuminous cells and sieve cells
(D) sieve tubes only
Answer:
(B) both sieve tubes and companion cells

Question 2.
Grass leaves curl inwards during very dry weather. Select the most appropriate reason from the following:
(A) Shrinkage of air spaces in spongy mesophyll
(B) Tyloses in vessels
(C) Closure of stomata
(D) Flaccidity of bulliform cells
Hint: Grass leaves curl inwards to the minimize water loss.
Answer:
(D) Flaccidity of bulliform cells

Question 3.
Which of the statements given below is NOT true about formation of ‘annual rings’ in trees?
(A) Activity of cambium depends upon variation in climate.
(B) Annual rings are not prominent in trees of temperate region.
(C) Annual ring is a combination of spring wood and autumn wood produced in a year.
(D) Differential activity of cambium causes light and dark bands of tissue – early and late wood respectively.
Hint: Annual rings are formed due to activity of cambium. The activity of cambium is under the control of many physiological and environmental factors. In temperate regions, the climatic conditions are not uniform throughout the year due to this, annual rings are formed.
Answer:
(B) Annual rings are not prominent in trees of temperate region.

Question 4.
Regeneration of damaged growing grass following grazing is largely due to:
(A) secondary meristem
(B) lateral meristem
(C) apical meristem
(D) intercalary meristem
Hint: Intercalary meristems occur in grasses at the base of intemode, which regenerates the grass damaged due to grazing.
Answer:
(D) intercalary meristem

Question 5.
In the dicot root, the vascular cambium originates from:
(A) intrafascicular and interfascicular tissue in a ring
(B) tissue located below the phloem bundles and a portion of pericycle tissue above protoxylem
(C) cortical region
(D) parenchyma between endodermis and pericycle
Answer:
(B) tissue located below the phloem bundles and a portion of pericycle tissue above protoxylem

Question 6.
Casparian strips occur in
(A) Cortex
(B) Pericycle
(C) Epidermis
(D) Endodermis
Answer:
(D) Endodermis

Question 7.
Plants having little or no secondary growth are
(A) Conifers
(B) Deciduous angiosperms
(C) Grasses
(D) Cycads
Hint: Secondary growth takes place in stems and roots of dicotyledons and gymnosperms, but does not occur in monocotyledons.
Answer:
(C) Grasses

Question 8.
Secondary xylem and phloem in dicot stem are produced by
(A) Phellogen
(B) Vascular cambium
(C) Apical meristems
(D) Axillary meristems
Answer:
(B) Vascular cambium

Question 9.
The vascular cambium normally gives rise to
(A) Phelloderm
(B) Primary phloem
(C) Secondary xylem
(D) Periderm
Answer:
(C) Secondary xylem

Question 10.
Identify the wrong statement in context of heartwood.
(A) Organic compounds are deposited in it
(B) It is highly durable
(C) It conducts water and minerals efficiently
(D) It comprises dead elements with highly lignified walls
Hint: In old trees, secondary xylem (wood) becomes physiologically non active. It does not conduct water and becomes dark due to organic deposits (tannins, resins, oils, aromatic substances, etc.) It comprises of dead elements and called as heart wood. It is non conductive, hard, durable and resistant to microbes and insects.
Answer:
(C) It conducts water and minerals efficiently

Question 11.
Which of the following is made up of dead cell?
(A) Xylem parenchyma
(B) Collenchyma
(C) Phellem
(D) Phloem
Hint: Cork cambium (phellogen) cuts off cells on both the sides. The outer cells differentiate into cork or phellem. The cork is impervious to water due to suberin deposition in the cell wall.
Answer:
(C) Phellem

Balbharti Maharashtra State Board 12th Chemistry Important Questions Chapter 16 Green Chemistry and Nanochemistry Important Questions and Answers.

Maharashtra State Board 12th Chemistry Important Questions Chapter 16 Green Chemistry and Nanochemistry

Question 1.
Define the following:
a. Atom economy.
Answer:
Atom economy : Atom economy is a measure of the amount of atoms from the starting materials that are present
in the final product at the end of chemical process.

Question 2.
How will you prevent the generation of waste or by-products?
Answer:
To prevent generating waste, there is the need to develop the zero waste technology (ZWT). ZWT in a chemical synthesis should result in waste product being zero or minimum. To use the waste product of one system as the raw material for other system is also the aim of ZWT.

For example :

• Cement and brick industry can use the bottom ash of thermal power station as the raw material.
• Thermal power station can use the effluent coming out from cleansing of machinery parts as coolant water.

Question 3.
(1) Calculate the atom economy of the following:
(At mass of C = 12, 11 = 1 ,0 = 16)

Answer:

Formula weight of ethanol = 46
ethene 28
water= 18
% atom economy = \(\frac{28}{46}\) x 100 = 60.9%

(2) Calculate the atom economy of fermentation of sugar (glucose) to ethanol.

Answer:

Formula wt of glucose = 180
Formula Wi of ethanol =46
Relative massiwt of desired useful product in thc equation = 2 x 46 =92
% Atom economy = 92/180 x 100 = 51.1%

Question 4.
ExpLain less hazardous chemical synthesis with suitable example.
Answer:

1. To avoid formation of hazardous waste from chemical processes, the chemical reactions and synthesis routes should be designed to be as safe as possible.
2. Earlier used insecticide DDT (Dichlorodiphenyltrichloroethanc) was found to be harmful for human beings. So DDT has been replaced by benzenc hexachioride (BHC) as an insecticide, one of the y-isomer (gamma) of BHC is called gammexane or lindane.

Question 5.
How will you develop products that are less toxic or which require less toxic raw materials ?
Answer:

• There is a need to design safer chemicals to prevent the workers in chemical industries from being exposed to toxic environment.
• Adipic acid is extensively used in polymer industry. In synthesis of adipic acid, benzene is used as the starting material, but benzene is carcinogenic and being volatile organic compound (VOC) it pollutes the air and environment.
• To overcome this health hazard Green technology developed by Drath and Frost, adipic acid is enzymatically synthesised from glucose.

Question 6.
How to apply the principle of green chemistry to achieve energy efficiency?
Answer:

• Energy requirements during chemical synthesis is huge. To minimize the energy use it is better to carry out reactions at room temperature and pressure.
• This can be achieved by applying the principle of green chemistry i.e. use of catalyst, use of micro-organisms or biocatalyst and use of renewable materials, etc.
• The use of less energy can be achieved by improving the technology of heating system, use of microwave, etc.

Question 7.
Explain the use of renewable feed stocks.
Answer:

• Industries use a lot of non-renewable feed stocks like petroleum. These resources are depleting fast and the future generation will be deprived. The excessive use of these resources have also put a burden on the environment.
• If renewable resources like agricultural or biological products are used, this will ensure the sharing of resources by future generations. This practice will also not put a burden on the environment.
• The products and waste produced are generally biodegradable and environmental friendly hence leading to a sustainable future.

Question 8.
Explain the need to design degradable chemicals.
Answer:

• Environment protection is the prime concern which has lead to the need for designing chemicals that degrade and can be discarded easily. These chemicals and their degradation products should be non-toxic, non-bioaccumulative or should not be environmentally persistent.
• This principle aims at waste product being automatically degradable to clean the environment. Thus the preference for biodegradable polymers and pesticides.
• To make the separation and segregation easier for the consumer an international plastic recycle mark is printed on larger items.

Question 9.
Define the role of real time analysis in pollution prevention.
Answer:

• There is a dire need to develop improvised analytical methods to allow for real time, in process monitoring and control prior to the formation of hazardous substances.
• It is very much important for the chemical industries and nuclear reactors to develop or modify analytical
  methodologies so that continuous monitoring of the manufacturing and processing unit is possible.

Question 10.
Define the role of safer chemistry in accident prevention.
Answer:
(1) It is needed to develop chemical processes that are safer and minimize the risk of accidents. It is important to select chemical substances used in a chemical reaction in such a way that they can minimize the occurrence of chemical accidents, explosions, fire and emissions.

(2) For example : Chemical process that works with the gaseous substances can lead to relatively higher possibilities of accidents including explosion as compared to the system working with nonvolatile liquid and solid substances.

Question 11.
Green chemistry plays an important role in sustainable development. Explain.
Answer:
Sustainable development is a development that protects the environment and the world’s resources. We can achieve sustainable development by adapting the twelve principles of green chemistry.

Green chemistry designs safer chemicals which are less toxic. It normally leads to low cost, use of less energy, environmentally friendly solvents and less production of waste. Green chemistry works on the principle of atom economy and minimum or no waste production. It encourages the use of renewable feed stocks and reduces the use of toxic and hazardous chemicals. It eliminates majorly stoichiometry reactions and prefers to use catalysis. It preserves the environment and safety requirements with added benefit of cost reduction.

Question 12.
How are nanomaterials classified ?
Answer:
Nanoparticles, nanowires and nanotubes can be classified according to dimensions. The nano structured materials may be large organic molecules, inorganic cluster compounds and metallic or semiconductor particles.

Question 13.
What are zero, one and two dimensional nanoscale system ?
Answer:

• Zero-Dimensional Nanostructures : A zero-dimensional structure is one in which all three dimensions are in the nanoscale.
  For example : Nanoparticles.
• One-Dimensional Nanostructures : A one-dimensional nanostructure is one in which two dimensions are in the nanoscale. For example : Nanowires and Nano rods.
• Two-Dimensional Nanostructures : A two-dimensional nanostructure is one in which one dimension is in the nanoscale. For example : Thin films.

Question 14.
State the different characteristic features of nanoparticles.
Answer:
The nanoparticle science is special as at such a small scale, different laws dominate than what we experience in our everyday life.

The characteristic features like optical properties, catalytical activities, have huge surface area and good thermal properties mechanical strength electrical conductivity vary than that of bulk material.

(1) Colour : At nanoscale this optical property behaves differently. Elemental gold has nice shining yellow colour, but nanoparticles of gold show red colour.

(2) Catalytic activity : Since the surface area of nanoparticles is large they show increased catalytic activity. They are usually heterogenous catalyst that means catalysts are solid form and the reactions occur on the surface of the catalyst. These catalysts can be easily separated and recycled. For example : Pd, Pt metal nanoparticles used in hydrogenation reactions. Ti02, ZnO are used in photocatalysis. Gold in bulk is unreactive but the nanoparticles of gold behave as very good catalyst for organic reactions.

(3) Surface area : High surface-to-volume ratio is a very important characteristic of nanoparticles. Bulk material if subdivided into a group of individual nanoparticles, the total volume remains the same, but the collective surface area is largely increased. With large surface area for the same volume, these small particles react much faster because more surface area provides more number of reaction sites, leading to more chemical reactivity. Explanation of increase in surface area with decrease in particle size.

(4) Thermal strength : The melting point of nanomaterial changes drastically with size.

For example : Sodium clusters (Nan) of 1000 atoms melts at 288 K, 10000 atoms melt at 303 K and bulk sodium melts at 371 K.

(5) Mechanical strength : The mechanic al strength of nano clusters increase the hardness of the metal.

For example : nanoparticles of copper and palladium clusters with diameter in the range of 5-7 nm have hardness up to 500 r. greater than the bulk metal.

(6) Electrical conductivity : At nanoscale level the electrical conductivity changes. For example : Carbon nanotubes behave as a conductor or semiconductor whereas carbon is nonconductor.

Question 15.
Describe the two methods of synthesising nanomaterials (nanoparticles).
Answer:
The two methods of synthesising nanomaterials :
(1) Bottom-up and
(2) Top down methods :

(1) Bottom-up method : Synthesis of nanoparticles in the bottom-up approach molecular components arrange themselves into more complex assemblies atom by atom, molecule by molecule and cluster by cluster from the bottom. Example : synthesis of nanoparticles by colloidal dispersion.

(2) Top-down method : In the top-down approach, involves nanomaterials being synthesised from bulk material by breaking the material. The bulk solids are disassembled into finer pieces until they are constituted of only few atoms. Example : Nanoparticles are synthesised by colloidal dispersion.

Question 16.
Discuss the various analytical tools used for characterization of nanoparticles.
Answer:
The analytical tools used for characterization of nanoparticles are

• U.V visible spectroscopy – It gives the preliminary confirmation of formation of nanoparticles.
• X-ray Diffraction (XRD) – The information given by this tool is about particle size, crystal structure and geometry.
• Scanning electron microscopy (SEM) : This is used to study the structure of surface of material that is the morphology of the material.
• Transmission electron microscopy (TEM) gives information about the particles size.
• (FTIR) Fourier transform infrared spectroscopy gives information about absorption of functional groups and binding nature of the nanomaterial.

Question 17.
Give evidence of use of nanoparticles by humans in ancient times with appropriate examples.
Answer:
There is enough evidence that nanomaterials have been produced and used by humans in ancient times. For example :

• Gold and silver nanoparticles trapped in the glass matrix gives ruby red colour in some ancient glass paintings.
• The decorative glaze or metallic film known as lustre found on some medieval pottery is due to certain spherical metallic nanoparticles.
• Carbon black is a nanostructured material that is used in tyres of car to increase the life of tyre. (Discovery in 1900). Carbon nanotubes are made up of graphite sheets with nanosized diameter. They have highest strength.
• Fumed silica, a component of silicon rubber, coatings, sealants and adhesives is also a nanostructured material.

Question 18.
Explain the different applications of nanoparticles.
Answer:
The contribution of nanochemistry in number of innovative products in various disciplines due to their unique physical, chemical, optical, structural, catalytic properties. Few applications are as follows :

• Nanoparticles contribute to stronger, lighter, cleaner and smarter surfaces and systems. They are used in the manufacture of scratchproof eyeglasses, transport, sunscreen, crack resistant paints, etc.
• Used in electronic devices like Magnetoresistive Random Access Memory (MRAM).
• Nanotechnology plays an important role in water purification techniques. Silver nanoparticles are used in water purification system to get safe drinking water.
• Self cleansing materials : Lotus is an example of self cleansing. Nanostructures on lotus leaves repel water which carries dirt as it rolls off. Lotus effect is the basis of self cleaning windows.

Question 19.
State the advantages of nanoparticles and nanotechnology.
Answer:

• Nanotechnology has revolutionalized electronics and computing.
• Nanotechnology has benefited the energy sector by making solar power more economical and energy storage more efficient.
• Nanotechnology has transformed the medical field with the manufacture of smart dmgs which help cure the life threatening diseases like cancer and diabetes faster and without side effects.

Question 20.
State the disadvantages of nanoparticles and nanotechnology.
Answer:
Despite the benefits that nanotechnology offers to the world, it is accompanied by certain disadvantages and potential risks.

The standard of living has been raised by nanotechnology but at the same time it has increased the environmental pollution. The kind of pollution caused by nanotechnology is very dangerous for living organism, it is called nano pollution.

Nanoparticles can be potential health hazard depending on the size, chemical composition and shape. They can be inhaled and can be deposited in the human respiratory tract and in the lungs, causing lung damage.

Question 21.
Name the development that meets the needs of the present, without compromising the ability of future generation to meet their own need.
Answer:
Sustainable development

Question 22.
Give name of father of green chemistry.
Answer:
Paul T. Anastas

Question 23.
Environmentally safe chemistry is known as.
Answer:
Green chemistry

Question 24.
How many principles does green chemistry have ?
Answer:
Twelve

Question 25.
Which principle of green chemistry has its perspective largely towards petrochemicals?
Answer:
Use of renewable feedstocks.

Question 26.
Name the chemical which leachs out of plastic packaging materials.
Answer:
Phthalate

Question 27.
Name the materials having structural components with at least one dimension in the nanometer scale.
Answer:
Nanomaterials.

Question 28.
Name the class of nanomaterial i.e. nanotubes, fibres, nanowires belong to.
Answer:
Two dimensions are in the nanoscale.

Question 29.
Name the nanoparticles used in sunscreen.
Answer:
Zinc oxide (ZnO) and Titanium dioxide (TiO₂).

Question 30.
What is the colour of gold nanoparticles ?
Answer:
Red

Question 31.
Name the nanoparticles used as catalyst in hydrogenation reaction.
Answer:
Palladium and Platinum.

Question 32.
Name the two approaches used to synthesize nanomaterials.
Answer:
Bottom up and Top down.

Question 33.
Give the name of the wet chemical synthetic process for nanomaterials.
Answer:
Sol-gel process.

Question 34.
Give the steps involved in preparation of nanoparticle using sol-gel process.
Answer:
Hydrolysis, polycondensation, drying, thermal decomposition.

Question 35.
Name the analytical techniques used for characterisation of nanomaterials.
Answer:
u.v-visible spectroscopy, x-ray diffraction (XRD), scanning electron microscopy (SEM), Transmission electron microscopy (TEM), Fourier transform infrared spectroscopy (FTIR).

Question 36.
Name the technique used to analyse particle size, crystal structure and geometry of a nanoparticle.
Answer:
x-ray diffraction (XRD)

Question 37.
Name the analytical technique used to study the morphology (structure of surface) of a material.
Answer:
Scanning electron microscopy (SEM)

Question 38.
Which innovative material has been developed using the lotus effect ?
Answer:
Self cleansing material

Question 39.
Which are the sectors that are revolutionalized by nanoparticles ?
Answer:
Electronics, energy sector and medical fields.

Question 40.
What are the disadvantages of nanotechnology ?
Answer:
Nano pollution and lung damage.

Question 41.
Name the scientist who coined the word nanotechnology.
Answer:
Nario Taniguchi (University professor at Tokyo in 1974).

Question 42.
Select and write the most appropriate answer from the given alternatives for each subquestion:

1. The measure of the amount of atoms from the starting materials that are present in the useful product at the end of chemical process is known as
(a) catalyst
(b) atom economy
(c) design of safer chemicals
(d) design for efficient energy
Answer:
(b) atom economy

2. The atom economy of the following reaction is CH₃ – CH₂ – CH₂ – CH₂ – OH + NaBr + H₂SO₄ → CH₃ – CH₂ – CH₂ – CH₂ – Br + NaHSO₄ + H₂O
(a) 49.81%
(b) 49%
(c) 50%
(d) 100%
Answer:
(a) 49.81%

3. Green chemistry reduces risk by
(a) developing the process for reuse and recycle of solvents and chemicals
(b) inventing technologies to clean the environ-ment
(c) minimize the use of chemicals
(d) reducing or eliminating the use or generation of hazardous chemicals in chemical products and process
Answer:
(d) reducing or eliminating the use or generation of hazardous chemicals in chemical products and process

4. Chemical synthesis should be designed to mini-mizes the use of
(a) liquid fuels
(b) solid fuels
(c) gaseous fuels
(d) energy
Answer:
(d) energy

5. The chemistry that applies across the life cycle of a chemical product like design, manufacture and use is called
(a) eco-friendly chemistry
(b) green chemistry
(c) environmental chemistry
(d) inorganic chemistry
Answer:
(b) green chemistry

6. According to the principles of green chemistry the chemicals involved in the production must be
(a) non-hazardous
(b) toxic
(c) polluting
(d) highly toxic
Answer:
(a) non-hazardous

7. Which of the following is not one of the twelve principles of green chemistry ?
(a) using renewable feedstocks
(b) designing safer chemicals and products
(c) maximizing atom economy
(d) avoiding the use of catalysts
Answer:
(d) avoiding the use of catalysts

8. Chemical synthesis based on principle of green chemistry encourages the use of
(a) hazardous chemicals
(b) reactions with low atom efficiency
(c) catalyst
(d) high energy requirements
Answer:
(c) catalyst

9. The plastic bottles made of HDPE are used to store household cleaner and shampoo can be recycled to make
(a) carpets, furniture, new containers
(b) detergent bottles, fencing, floor tiles, pens
(c) custom-made products
(d) cables, mudflaps, panelling, roadway gutters
Answer:
(b) detergent bottles, fencing, floor tiles, pens

10. The plastic ketch-up bottles and syrup bottles made from polypropylene (pp) can be recycled to make
(a) battery cables, brooms, ice scrapers, rakes
(b) envelopes, floor tiles, lumber
(c) custom-made products
(d) carpet, furniture, new containers
Answer:
(a) battery cables, brooms, ice scrapers, rakes

11. The role of green chemistry aims to
(a) design chemical processes and products that maximize profits
(b) design safer chemicals and products by reduc¬ing or eliminating the use and generation of hazardous substances
(c) design processes and products that work efficiently
(d) utilize non-renewable feedstocks
Answer:
(b) design safer chemicals and products by reducing or eliminating the use and generation of hazardous substances

12. The study of phenomena and manipulation of materials of atomic, molecular and macromolecular scales where properties differ significantly from those at a large scale is called
(a) nanoscience
(b) nanochemistry
(c) nanotechnology
(d) nanomaterial
Answer:
(a) nanoscience

13. The term nanotechnology was first used by whom and when ?
(a) Richard Feynman, 1959
(b) Nario Taniguchi, 1974
(c) Eric Drexter, 1986
(d) Sumia Lijima, 1991
Answer:
(b) Nario Taniguchi, 1974

14. Which one of these statements is NOT true ?
(a) Gold at the nanoscale is red.
(b) A very highly useful application of nanochem¬istry is medicine.
(c) Sunscreen contains nanoparticles of zinc oxide (ZnO) and (SiO₂) silicon oxide.
(d) Silicon at nanoscale is not an insulator
Answer:
(c) Sunscreen contains nanoparticles of zinc oxide (ZnO) and (SiO₂) silicon oxide.

15. Which of the historical works mentioned below contain nanotechnology?
(a) Lycurgus cup
(b) Medieval stained glass windows in churches
(c) Damascus steel swords
(d) All of the above
Answer:
(d) All of the above

16. The nanometer scale is conventionally defined as
(a) 10 – 100nm
(b) 1 – 100nm
(c) 1 – 1000 nm
(d) 1 – 10000 nm
Answer:
(b) 1-100 nm

17. The material synthesized on the nanometer scale possess
(a) same bulk properties
(b) different bulk properties
(c) unique optical, magnetic, electrical properties
(d) no change in properties
Answer:
(c) unique optical, magnetic, electrical properties

18. Nanomaterials of zero dimension is
(a) one in which all three dimensions are in the nanoscale
(b) one in which two dimensions are in the nanoscale
(c) one in which one dimension is in the nanoscale
(d) None of the above
Answer:
(a) one in which all three dimensions are in the nanoscale

19. The science which deals with the design and synthesis of material on nanoscale with different size and shape is called
(a) nanoscience
(b) nanochemistry
(c) nanophysics
(d) nanotechnology
Answer:
(b) nanochemistry

20. Elemental has a shining yellow colour, but the colour of nanoparticles of gold is
(a) green
(b) yellow
(c) red
(d) blue
Answer:
(c) red

21. The surface area of nanoparticles
(a) is the same as in bulk
(b) increases with the same volume of the bulk
(c) decreases with the same volume of the bulk
(d) does not change with particle size
Answer:
(b) increases with the same volume of the bulk

22. The nanomaterial based catalyst are usually
(a) homogeneous catalyst
(b) heterogeneous catalyst
(c) good catalyst
(d) bad catalyst
Answer:
(b) heterogeneous catalyst

23. The catalyst used in photocatalysis is
(a) gold
(b) Raney Ni
(c) TiO₂
(d) AI₂O₃
Answer:
(c) TiO₂

24. Nanosized copper clusters have the mechanical strength of
(a) less than the bulk copper wire
(b) 100% more than the bulk metal
(c) 200% more than the bulk metal
(d) 500% more than the bulk metal
Answer:
(d) 500% more than the bulk metal

25. The most common method used for synthesis of nanomaterials is
(a) sol-gels method
(b) only sol method
(c) only gel method
(d) colloidal dispersion method
Answer:
(a) sol-gels method

26. What is the information obtained from uv-visible spectroscopy when used for nanomaterials ?
(a) morphology of structure
(b) preliminary conformation of formation of nanoparticle
(c) particle size
(d) functional group present
Answer:
(b) preliminary conformation of formation of nanoparticle

27. What information of the nanoparticles is obtained from transmission electron microscopy technique ?
(a) structure
(b) functional group
(c) particle size
(d) geometry
Answer:
(c) particle size

28. The analytical tool used to study the structure of surface of nanoparticle i.e. morphology is
(a) Absorption spectroscopy
(b) Scanning electron microscopy
(c) Emission spectroscopy
(d) Nuclear magnetic resonance spectroscopy
Answer:
(b) Scanning electron microscopy

29. The constituents of carbon nanotubes are
(a) nanosized graphite sheets
(b) nanosized carbon black
(c) nanosized coal black
(d) None of the above
Answer:
(a) nanosized graphite sheets

30. Self cleansing windows are example of the
(a) Nanoparticle effect
(b) Crompton effect
(c) Lotus effect
(d) Tyndal effect
Answer:
(c) Lotus effect

31. Which highly effective and cost effective nano-particles are used for water purification ?
(a) gold nanoparticle
(b) copper nanoparticle
(c) silver nanoparticle
(d) silica nanoparticle
Answer:
(c) silver nanoparticle

32. The sectors revolutionalized by nanotechnology are
(a) electronics and computing
(b) energy
(c) medicine
(d) All of the above
Answer:
(d) All of the above

33. Name the body part that gets affected by the hazardous nano pollution.
(a) heart
(b) brain
(c) lungs
(d) eyes
Answer:
(c) lungs

34. The pollution caused by nanotechnology is known as
(a) air pollution
(b) nano pollution
(c) ground pollution
(d) environmental pollution
Answer:
(b) nano pollution.

Balbharti Maharashtra State Board 12th Chemistry Important Questions Chapter 10 Halogen Derivatives Important Questions and Answers.

Maharashtra State Board 12th Chemistry Important Questions Chapter 10 Halogen Derivatives

Question 1.
What are halogen derivatives of hydrocarbons?
Answer:
The replacement of hydrogen atom/s in aliphatic or aromatic hydrocarbons by halogen atom/s results in the formation of halogen derivatives of hydrocarbons.

Question 2.
How are halogen derivatives of hydrocarbons classified?
Answer:
Halogen derivatives of alkane are classified as :

1. Monohalogen derivative (or alkyl halide) : It is a halogen derivative of an alkane in which one hydrogen atom is replaced by one halogen atom and it is also called alkyl halide. E.g. C₂H₅Br.
2. Poh halogen derivatives : These are halogen derivatives in which more than one hydrogen atoms of an alkane are substituted by corresponding number of halogen atoms.

They are classified as follows :
(i) Dihalogen derivatives : The compounds formed by the substitution of two hydrogen atoms of an alkane by two halogen atoms are called dihalogen derivatives.

They are further classified as :

• Vicinal dihalides (Two halogen atoms on vicinal or adjacent carbon atoms)
• Geminal dihalides : CH₃ – CHCI₂ (Two halogen atoms on the same carbon atom)

(ii) Trihalogen derivatives : The compounds formed by the substitution of three hydrogen atoms of an alkane by three halogen atoms are called trihalogen derivatives.

(iii) Tetrahalogen derivatives : The compounds formed by the substitution of four hydrogen atoms of an alkane by four halogen atoms are called tetrahalogen derivatives. E.g. CCI₄.

Question 3.
What are alkyl halides? How are they classified?
Answer:
The compound formed by the replacement of one hydrogen atom in an alkane by a halogen atom is called an alkyl halide. The halogen atom is bonded to sp³ hybridised carbon. Alkyl halides are classified into the following three classes depending on the type of the carbon to which halogen atom is bonded.

(1) Primary (1°) alkyl halide : Alkyl halide in which a halogen atom is bonded to a primary carbon atom is called primary alkyl halide.

[Primary (1°) carbon atom i.e., the carbon atom which is attached to only one carbon atom.]

They are represented by the general formula R – ^1°CH₂ – X.

(2) Secondary (2°) alkyl halide : Alkyl halide in which a halogen atom is bonded to a secondary carbon atom is called secondary alkyl halide. [Secondary (2°) carbon i.e., the carbon atom which is attached to two other carbon atoms.]

They are represented by the general formula (R and R’ can be same or different)

(3) Tertiary (3°) alkyl halide : Alkyl halide in which halogen atom is bonded to a tertiary carbon atom is called tertiary alkyl halide. [Tertiary (3°) carbon i.e., the carbon atom which is attached to three other carbon atoms.]

They are represented by the general formula  (R, R’ and R” may be same or different)

Question 4.
Explain the following :
(1) Alkyl halide or haloalkanes
(2) Allylic halides
(3) Benzylic halide
(4) Vinylic halide
(5) Haloalkyne
(6) Aryl halide or haloarenes.
Answer:
(1) Alkyl halide or haloalkanes : In alkyl halides or haloalkanes the halogen atom is bonded to sp3 hybridized carbon which is a part of saturated carbon chain.
Example :

(2) Allylic halides : In allylic halides, halogen atom is bonded to a sp³ hybridized carbon atom next to a carbon-carbon double bond.

Example :

(3) Benzylic halide : In benzylic halides, halogen atom is bonded to a sp³ hybridized carbon atom which is further bonded to an aromatic ring.
Example :

(4) Vinylic halides : In vinylic halides, halogen atom is bonded to a sp² hybridized carbon atom of aliphatic chain. Vinylic halide is a haloalkene.
Example :

(5) Haloalkyne : In haloalkynes, halogen atom is bonded to a sp hybridized carbon atom.
Example :

(6) Aryl halides or haloarenes : In aryl halides, halogen atom is directly bonded to the sp² hybridized carbon atom of aromatic ring.
Example :

Question 5.
Give the IUPAC names of the following :
Answer:

Question 6.
Draw the structures of the following compounds:
Answer:

Question 7.
Write the structure of-
(a) 3-chloro-3-ethylhex-l-ene
(b) 1-Iodo-2, 3-dimethylbutane
(c) 1, 3, 5-tribromobenzene
Answer:

Question 8.
Write structures of
(a) 2-iodo-3-methyl pentane
(b) 3-chiorolleNane
(c) 1-chloro-2, 2-dimethyl propane
(d) 1-chloro-4-ethyl cyclohexane.
Answer:

Question 9.
Write the possible isomers of monochloro derivatives of 2,3-Dimethylbutane and write their IUPAC names.
Answer:
The given parent hydrocarbon has molecular formula, C₆H₁₄. The monochloro derivative of this compound has molecular formula C₆H₁₃CI.
The parent hydrocarbon is,

Hence the structures of isomers of monochioroderivative are.

Question 10.
Write structures and IUPAC names of all possible isomers of C₅H₁₁Br and classify them as l°/2°/3°.
Answer:
C₅H₁₁Br is a monohalogen derivative.

Question 11.
How are following compounds obtained from alcohols :
(1) ethyl chloride C₂H₅CI
(2) isopropyl chloride (CH₃ CHCI – CH₃)
(3) tert-butyl chloride (CH₃)₃ – CI?
Answer:
Alcohols in the presence of Lucas reagent which is a solution of concentrated HCI and ZnCI₂ form alkyl halides. Hydrogen chloride is used with zinc chloride (Grooves’ process) for primary and secondary alcohols.

(3) Tertiary alcohols don’t need ZnCI₂ to react with HCI.

The order of reactivity of alcohols with a given halo acid is 3° >2°> 1°.

Question 12.
How are following compounds prepared from alcohols:
(1) ethyl bromide (C₂H₅Br)
(2) isopropyl bromide (CH₃ – CHBr – CH₃)
(3) tert-butyl bromide (CH₃)₃ C – Br?
Answer:
(1) Ethyl alcohol on heating with conc. hydrobromic acid (48%) forms ethyl bromide.
OR
When ethyl alcohol is treated with a mixture of NaBr and H₂SO₄, ethyl bromide is formed. Here HBr is generated in situ.

(2) Isopropyl alcohol, on reaction with NaBr and dil. H₂SO₄ forms isopropyl bromide.

(3) Tertiary alcohol on reaction with sodium bromide and dil. H₂SO₄ forms tert-butyl bromide.

Question 13.
How is ethyl iodide obtained from ethyl alcohol?
Answer:
When ethyl alcohol is treated with sodium or potassium iodide in 95 % phosphoric acid, ethyl iodide is formed. Here HI is generated in situ.

Question 14.
How will you prepare the following :

(1) Ethyl chloride (chloroethane) from ethyl alcohol using
(i) PCI₃
(ii) PCI₅ and
(iii) SOCI₂.
Answer:
(i) When ethyl alcohol is refluxed with phosphorus trichloride, ethyl chloride is formed.

(ii) When ethyl alcohol is refluxed with phosphorus pentachloride, ethyl chloride is formed.

(iii) When ethyl alcohol is refluxed with thionyl chloride, in the presence of pyridine, ethyl chloride is formed. The by-products obtained are gases. Therefore, this method is preferred for preparation of alkyl chloride.

(2) Isopropyl chloride (2-chloropropane) from isopropyl alcohol using
(i) PCI₃
(ii) PCI₅
(iii) SOCI₂.
Answer:
When isopropyl alcohol is refluxed with phosphorus trichloride, isopropyl chloride is formed.

When isopropyl alcohol is refluxed with phosphorus pentachloride, isopropyl chloride is formed.

When isopropyl alcohol is refluxed with thionyl chloride, in the presence of pyridine, isopropyl chloride is formed. The by-products obtained are gases. Therefore, this method is preferred for the preparation of alkyl chloride.

(3) Ethyl bromide (bromoethane) from ethyl alcohol.
Answer:
When ethyl alcohol is treated with a mixture of red phosphorus and bromine or hydrobromic acid (phosphorus tribromide is generated in situ), ethyl bromide is formed.

(4) Ethyl iodide (do ethane) from ethyl alcohol.
Answer:
When ethyl alcohol is heated with a mixture of red phosphorus and iodine, (phosphorus triiodide is generated in situ), ethyl iodide is formed.

Question 15.
Explain halogenation of methane.
Answer:
Halogenation : A reaction of alkanes with halogens (CI₂, Br₂, I₂) in the presence of appropriate conditions forming a mixture of alkyl halides.
(1) Chlorination:

(2) When excess of chlorine is used, tetrachioro methane, a major product is obtained. When excess of methane is used, chioromethane, a major product is obtained. The order of reactivity of halogens towards alkane is

(3) lodination:

However, iodination reaction is a reversible reaction. HI being a strongest reducing agent reduces methyl iodide back to methane.

(4) Fluorination: A reaction of alkane with fluorine is explosive and also hydrofluoric acid is poisonous and corrosive. Hence, alkyl fluorides are not prepared by halogenation of alkane.

Question 16.
Predict the possible products of the following reaction :
(1) Bromination of propane
(2) Bromination of n-butane
(3) Bromination of 2-Methyl propane.
Answer:
(1) Bromination of propane

(2) Bromination of n-butane

(3) Bromination of 2-Methyl propane

Question 17.
How are following compounds prepared by halogenation of ethane :
(1) Chloroethane
(2) Bromoethane
(3) Iodoethane?
Answer:
(1) Chlorination of ethane : When ethane (excess) is reacted with a limited quantity of chlorine in the presence of diffused sunlight or U.V. light or at high temperature, chloroethane is obtained.

(2) Bromination of ethane : When ethane is heated with Br₂ in the presence of anhydrous AlBr₃, bromoethane is obtained.

(3) Iodination : When ethane is reacted with I₂ in the presence of suitable oxidising agents like-HgO or HIO₃ or dilute HNO₃ iodoethane is obtained.

Question 18.
Direct iodination of alkanes is not possible.
Answer:
(1) Direct iodination of alkanes using iodine is highly reversible.
\(\mathrm{RH}+\mathrm{I}_{2} \rightleftharpoons \mathrm{RI}+\mathrm{HI}\)
(2) Hydroiodic acid HI being strong reducing agent, it reduces RI to alkane RH.
(3) The reaction takes place only in the presence of a suitable oxidizing agent like HgO, HIO₃ or dilute HNO₃ which decomposes HI. Hence, direct iodination of alkanes is not possible.

Question 20.
How are following compounds obtained from alkenes :

Answer:
(1) Ethene on reaction with hydrogen chloride forms

(2) Ethene on reaction with hydrogen bromide forms

(3) Propene on reaction with hydrogen iodide forms

(4) but-2-ene on reaction with hydrogen iodide forms 2-iodobutane.

Question 19.
State and explain Markovnikov’s rule.
Answer:
Markovnikov’s rule : When an unsymmetrical reagent is added to an unsymmetrical alkene, the negative part of the reagent gets attached to that carbon atom of the double bond which carries less number of hydrogen atoms.

Example : Addition of HBr’to unsymmetrical alkene like propene gives two products.

Isopropyl bromide is the major product, since the negative part (Br^–) of HBr is attached to carbon atom of a double bond with less number of hydrogen atoms.

Question 20.
Explain peroxide effect.
OR
Write a note on the Kharasch-Mayo effect.
OR
Explain the addition of HBr to (unsymmetrical alkene) propane in the presence of benzoyl peroxide.
Answer:
The addition of HBr to an unsymmetrical alkene (propane) in the presence of benzoyl peroxide takes place in the opposite orientation to that of Markovnikov’s rule and this is known as Kharasch-Mayo effect or peroxide effect or Anti-Markovnikov addition.

Question 21.
Write the structure of alkyl halide obtained by the action hydrogen bromide on 2-Methyiprop-1-ene in the presence of peroxide.
Answer:
In the presence of peroxide. HBr to 2-Methyl prop-I-cne forms l-Bromo-2-methylpropane.

Question 22.
How are alkyl iodides prepared from alkyl chlorides/bromides?
Answer:
Alkyl iodide is prepared by treating alkyl chloride or alkyl bromide with sodium iodide, in the presence of dry acetone, sodium chloride or sodium bromide precipitates from the solution and can be separated by filtration. This reaction is known as Finkelstein reaction.

Question 23.
How are alkyl fluorides prepared with alkyl chlorides/alkyl bromides?
Answer:
When alkyl chloride or alkyl bromide is heated with metallic fluorides like AgF, CaF₂, CoF₂ or Hg₂F₂, alkyl fluoride is formed.

This reaction is known as Swarts reaction.

Question 24.
Explain the preparation of haloarenes using electrophilic substitution.
Answer:
When arene is treated with chlorine or bromine in dark at ordinary temperature in the presence of lewis acid as a catalyst like Fe, FeCI₃ or anhydrous AlCI₃, aryl chloride or aryl bromide is formed.

When toluene is brominated in dark at ordinary temperature in the presence of iron, a mixture of ortho and para bromo tolerene is obtained.

Ortho and para isomers can be easily separated as there is large difference in melting points of ortho and para isomers.

Question 25.
Write a note on Sandmeyer’s reaction.
Answer:
Aryl halides are most commonly prepared by replacement of nitrogen of diazonium salt. The replacement of diazonium group by -Cl or -Br using cuprous salt is called Sandmeyer’s reaction. When a primary aromatic amine (like aniline) suspended in cold F1C1, is treated with sodium nitrite, a diazonium salt (benzene diazonium chloride) is formed. When diazonium salt is treated with cuprous chloride or cuprous bromide, aryl halide (chlorobenzene or bromobenzene) is formed.

When benzene diazonium salt is mixed with potassium iodide, iodobenzene is formed.

Question 26.
Define the following :
Answer:
(1) Monochromatic light : It consists of rays of single wavelength vibrating in different planes perpendicular to the direction of propagation of the light.

(2) Plane polarized light : A light having oscillations only in one plane perpendicular to direction of propagation of light is known as plane polarized light.

(3) Optical isomerism : The steroisomerism in which the isomers have different spatial arrangements of groups/atoms around a chiral atom is called optical isomerism.

(4) Optical activity : The property of a substance by which it rotates plane of polarization of incident plane polarized light is known as optical activity.

(5) Optically active compound : The compound which rotate the plane of plane polarized light is called optically active compound.

(6) Enantiomers : The optical isomers which are non-superimposable mirror images of each other are called enantiomers or enantiomorphs or optical antipodes.
Example : 2-chlorobutane, lactic acid

(7) Chiral carbon atom : Carbon atom in a molecule which carries four different groups/atoms is called chiral carbon atom.

Chiral atom in a molecule is marked with asterisk (*)

For example : C in lactic acid

(8) Chiral molecule : When a molecule contains one chiral atom, it acquires a unique property i.e. it is non- superimposable with its mirror image is said to be chiral molecule.

(9) Chirality : The relationship between a chiral molecule and its mirror image is similar to the relationship between left and right hands. Therefore it is called handedness or chirality.

(10) Dextrorotatory substance or r/-Isomer : An optically active substance (or isomer) which rotates the plane of a plane polarized light to the right hand side (RHS) is called dextrorotatory substance (or isomer) and denoted by d or (+) sign.

(11) Laevorotatory substance or /-Isomer : An optically active substance (or isomer) which rotates the plane of a plane polarized light to the left hand side (LHS) is called laevorotatory substance (or isomer) and denoted by / or (-) sign.

(12) Racemic mixture or Racemate : A mixture containing equimolar quantities of dextro (d) and laevo (/) optical
isomers which is optically inactive due to molecules of one enantiomer is cancelled by equal and opposite optical rotation due to molecules of the other enantiomer is called a racemic mixture or racemate. It is represented as (dl) or (+).

Question 27.
Calculate the number of isomers for 2-chlorobutane.
Answer:
The number of optical isomers possible for a compound is 2ⁿ where n = number of asymetric carbon atoms.
As n = 1 for 2-ehlorobutane, 2ⁿ = 2¹ = 2. Hence, it has two optical isomers.

Question 28.
How many optical isomers are possible for C₅H₁₁ CI?
Answer:
The number of optical isomers : 3.

Question 29.
How many optical isomers are possible for glucose?
Answer:
The number of optical isomers : 16.

Question 30.
Draw the structures and indicate the chiral carbon atoms in
(1) Lactic acid
(2) 2-Chlorobutane.
Answer:
(1) In lactic acid structure, the starred carbon atom is chiral carbon atom as it is attached to four different substituents, COOH, OH, CH₃ and H.
(2) In 2-chlorobutane structure, the starred carbon atom is chiral carbon atom as it is attached to four different substituents, -CH₂ – CH₃ (ethyl), CH₃ (methyl), Cl and H.

Question 31.
Identify chira! and achiral molecules.

Question 32.
Complete the following reactions and explain optical activity of the products formed:
(i) Pent-1-ene with HBr
Answer:

(ii) Pent-2-ene with HBr
Answer:

Question 33.
C₆H₁₂ (A) on treatment with HCI produced a compound Y. Which is optically active, what is structure A?
Answer:

Question 34.
A racemic mixture is optically inactive. Explain.
Answer:

• A racemic mixture contains equimolar (or equimolecular) quantities of the dextrorotatory (d-) and laevorotatory (l-) isomers (enantiomers) of a compound.
• The d-enantiomer rotates the plane of plane-polarized light to the right, while the l-enantiomer rotates the same to the left to the same extent.
• The quantities of the d- and l-enantiomers being the same, both the rotations are of the same magnitude, but of opposite directions. Hence, they cancel each other. Hence, a racemic mixture is optically inactive.
• It is represented as dl or ( + ). Example : ( ± ) lactic acid

Question 35.
Explain Fischer projection formula with illustration.
OR
Write a note on Fischer projection formula.
Answer:
Fischer projection formula or cross formula : The three dimensional (3-D) view of a molecule is presented on plane of paper. A Fischer projection formula can be drawn by visualizing the main carbon chain verical in the molecule. Each carbon on the vertical chain is represented by a cross.

Conventionally the horizontal lines of the cross represent bonds projecting up from the carbon and the vertical lines represent the bonds going below the carbon.

Question 36.
Explain Wedge formula with illustration
OR
Write a note on Wedge formula.
Answer:
Wedge formula : When a tetrahedral carbon is imagined to be present in the plane of paper all the four bonds at this carbon cannot lie in the same plane. The bonds in the plane of paper are represented by normal lines, the bonds projecting above the plane of paper are represented by solid wedges (or simply by bold lines) while bonds going below the plane of paper are represented by broken wedges (or simply by broken lines).

Question 37.
Give a laboratory test to confirm the presence of halogen in the original organic compound.
Answer:
Haloalkanes are of neutral type in aqueous medium. On warming with aqueous sodium or potassium hydroxide the covalently bonded halogen in haloalkane is converted to halide ion.

When this reaction mixture is acidified by adding dilute nitric acid and silver nitrate solution is added a precipitate of silver halide is formed which confirms presence of halogen in the original organic compound.

Question 38.
Define the following :
Answer:
(1) Mechanism of a reaction : It is a step by step description of exactly how the reactants are transformed into products in as much details as possible.
(2) Substitution reaction : When a group bonded to a carbon in a substrate is replaced by another group to get a product with no change in state of hybridization of that carbon, the reaction is called substitution reaction.

Question 39.
Describe the action of aqueous KOH (or NaOH) on :
(1) ethyl bromide
(2) isopropyl bromide
(3) tert-butyl chloride
(4) methyl bromide
(5) 2-chlorobutane.
Answer:
(1) Ethyl bromide : When ethyl bromide (bromoethane) is refluxed with aqueous potassium hydroxide, ethyl alcohol is formed. The reaction is called a hydrolysis reaction.

(2) Isopropyl bromide : When isopropyl bromide (2-bromopropane) is boiled with aqueous potassium hydroxide, isopropyl alcohol is formed.

(3) Tert-butyl chloride : When tert-butyl chloride is refluxed with aqueous potassium hydroxide, tert-butyl alcohol is formed.

(4) Methyl bromide : When methyl bromide (bromomethane) is heated with aq. KOH, it is hydrolysed to methyl alcohol (methanol).

(5) 2-chlorobutane : When 2-Chlorobutane is boiled with aqueous KOH, Butan-2-ol is obtained.

Question 40.
Describe the action of sodium ethoxide on
(1) ethyl bromide
(2) methyl bromide :
OR
Write a note on Williamson’s synthesis.
OR
How are ethers prepared from alkyl halides?
Answer:
Williamson’s synthesis : When an alkyl halide (R – X) is heated with sodium alkoxide (R – O – Na), an ether is obtained. In this reaction halide (-X) of alkyl halide is replaced by an alkoxy group (-OR). This reaction is known as Williamson’s synthesis. This method is used to prepare simple (or symmetrical) ethers and mixed (or unsymmetrical) ethers.

Sodium alkoxide is obtained by a reaction of sodium with an alcohol.

(1) Simple (symmetrical) ether : When an alkyl halide and sodium alkoxide having similar alkyl groups are heated, symmetrical ether is obtained.

e.g., When ethyl bromide is heated with sodium ethoxide, diethyl ether is formed.

(2) Mixed (unsymmetrical) ether : When an alkyl halide and sodium alkoxide having different alkyl groups are heated, unsymmetrical ether is obtained.

When methyl bromide is heated sodium ethoxide, ethyl methyl ether is formed.
When ethyl bromide is heated with sodium meihoxide, ethyl methyl ether is formed.

Question 41.
What is the action of silver salt of carboxylic acid on alkyl halide?
Answer:
When an alkyl halide (R – X) is heated with silver salt of carboxylic acid (R -COOAg). an ester is obtained.

Question 42.
Describe the action of alcoholic silver acetate on
(1) methyl bromide
(2) ethyl bromide.
Answer:
(1) Methyl bromide : When methyl bromide is heated with an alcoholic silver acetate, methyl acetate is formed.

(2) Ethyl bromide : When ethyl bromide is heated with an alcoholic silver acetate, ethyl acetate is formed.

Question 43.
What is the action of alcoholic silver propionate on ethyl bromide?
Answer:
When ethyl bromide is heated with an alcoholic silver propionate. ethyl propionate is formed.

Question 44.
Describe the action of excess of ammonia on (I) ethyl bromide (2) n.propyl bromide.
Answer:
(1) Ethyl bromide : When ethyl bromide is boiled under pressure with an excess of alcoholic ammonia, ethylamine (ethanamine) is formed. This is known as ‘ammonolysis of ethyl bromide.

(2) n-propyl bromide : When n-propy1 bromide is boiled under pressure with an excess of ammonia, n-propyl amine (propanamine) is formed.

Question 45.
What is ammonolysis? Give a suitable example for the reaction.
Answer:
When an alkyl halide is boiled under pressure with an excess of alcoholic solution of ammonia (NH₃), corresponding (primary amine) alkyl amine is formed. This reaction is known as ammonolysis of alkyl halide.

(1) Ethyl bromide : When ethyl bromide (bromoethane) is refluxed with aqueous potassium hydroxide, ethyl alcohol is formed. The reaction is called a hydrolysis reaction.

(2) Isopropyl bromide : When isopropyl bromide (2-bromopropane) is boiled with aqueous potassium hydroxide, isopropyl alcohol is formed.

(3) Tert-butyl chloride : When tert-butyl chloride is refluxed with aqueous potassium hydroxide, tert-butyl alcohol is formed.

(4) Methyl bromide : When methyl bromide (bromomethane) is heated with aq. KOH, it is hydrolysed to methyl alcohol (methanol).

(5) 2-chlorobutane : When 2-Chlorobutane is boiled with aqueous KOH, Butan-2-ol is obtained.

Question 46.
Describe the action of aqueous alcoholic potassium cyanide on
(1) ethyl bromide
(2) methyl iodide.
Answer:
Ethyl bromide : When ethyl bromide (bromoethane) is boiled with alcoholic solution of potassium cyanide in aqueous ethanol, ethyl cyanide (ethyl nitrile) is formed.

(2) Methyl iodide : When methyl iodide is boiled with alcoholic solution of potassium cyanide, methyl cyanide is formed.

Question 47.
Describe the action of alcoholic silver cyanide on
(1) ethyl bromide
(2) methyl chloride.
OR
Explain isocyanide reaction of
(1) ethyl bromide
(2) methyl chloride.
Answer:
(1) Ethyl bromide : When ethyl bromide is heated with alcoholic silver cyanide, ethyl isocyanide is formed.

(2) Methyl chloride : When .methyl chloride is heated with alcoholic silver cyanide, methyl isocyanide is formed.

The above reactions (1) and (2) are called isocyanide reaction.

Question 48.
Describe the action of potassium nitrite on
(i) ethyl bromide,
(ii) methyl chloride.
Answer:
(1) Ethyl bromide : When ethyl bromide is treated with potassium nitrite, ethyl nitrite is formed.

(2) Methyl chloride : When methyl chloride is treated with potassium nitrite, methyl nitrite is formed.

Question 49.
Describe the action of silver nitrite on (1) ethyl chloride (2) n-propyl bromide.
Answer:
(1) Ethy chloride : When ethyl chloride is treated with silver nitrite, nitroethane is obtained.

(2) n-Propyt bromide: When n-propyl bromide is treated with silver nitrate, nitropropane is obtained.

Question 50.
How will you convert (the following:

(1) Ethyl bromide to ethanol.
Answer:

(2) Ethyl bromide to propane nitrile.
Answer:

(3) Ethyl bromide to ethyl amine.
Answer:

(4) Ethyl bromide to ethyl acetate.
Answer:

(5) Ethyl bromide to ethyl isocyanide.
Answer:

(6) Ethyl bromide to ethyl methyl ether.
Answer:

(7) Ethyl bromide to n-butane.
Answer:

(8) Ethyl bromide to Ethyl magnesium bromide.
Answer:

Question 51.
Define the following :
Answer:
(1) Nucleophilic bimolecular reaction (SN²) : The substitution reaction in which a nucleophile reacts with the substrate and the rate of the reaction depends on the concentration of the substrate and the nucleophile is called a nucleophilic bimolecular reaction.

Example :

(2) SN¹ reaction : The substitution reaction in which a nucleophile reacts with the substrate and the rate of the reaction depends only on the concentration of the substrate is called nucleophilic unimolecular or first order reaction or SN¹ reaction.

Question 52.
Explain, the mechanism of alkaline hydrolysis (reaction with aqueous KOH) of tert-butyl bromide (2-Bromo-2-methylpropane) with energy profile diagram.
OR
Explain only reaction mechanism for alkaline hydrolysis of tert-butyl bromide.
Answer:
(i) Consider alkaline hydrolysis of tert-butyl bromide (2-Bromo-2 methylpropane) with aqueous NaOH or KOH.

(ii) Kinetics of the reaction : Due to steric hindrance of voluminous three methyl groups around carbon, nucleophile OH- cannot attack carbon atom directly. Hence, the reaction takes place in two steps.

Step I : This involves heterolytic fission of C – Br covalent bond in the substrate forming carbocation and Br^–. This is a slow process.

Step II : This step involves attack of nucleophile OH- or carbocation forming C – OH bond and product tert-butyl alcohol. Since it involves ionic charge neutralisation, it is a fast step.

Rate Determining Step (R.D.S.) : Since the first step is a slow step, it is R.D.S., and therefore the rate of the reaction depends on the concentration of only one reactant, (CH₃) C – Br.
Rate = R = k [(CH₃)₃ C – Br] where k is a rate constant of the reaction.

SN¹ reaction : The reaction between tert.butyl bromide and hydroxide ion to form tert.butyl alcohol follows a first-order kinetics. The rate of this reaction depends only on the concentration of one substance (tert-butyl bromide) and is independent of the concentration of alkali added. It is an unimolecular first (1st) order Nucleophilic Substitution reaction denoted as SN reaction.

Stereochemistry and mechanism of the reaction : The reaction takes place in two steps and both the steps involve formation of transition states (T.S.).

T.S. -1 for first step :

In this transition state, C – Br bond is partially broken, so that carbon atom carries partial positive charge (+δ) and Br carries partial negative charge (-δ) which further breaks forming carbocation and Br . Tert-butyl cation (carbocation) has a planar structure and the CH₃ – C – CH₃ bond angle is 120°. It is the intermediate of the reaction. It is unstable. In this step, hybridisation of carbon atom changes from sp³ (tetrahedral geometry) to sp (planar geometry).

T.S. – II for second step :

In this transition state, C – OH bond is partially fonned so that carbon atom carries partial positive charge (+ δ) and OH carries partial negative charge ( -δ) which further forms tert-butyl alcohol.

Formation of a racemic mixture : Since OH^– has equal probability of the attack on carbocation from frontside and from backside, the products obtained are equal. In case of optical active alkyl halide, a racemic mixture is obtained.

Question 53.
Discuss SN² mechanism of methyl bromide using aqueous KOH. Draw energy profile diagram.
OR
Discuss the mechanism of alkaline hydrolysis of methyl bromide or Bromomethane.
Answer:
(1) Consider alkaline hydrolysis of methyl bromide (Bromomethane). CH₃Br with aqueous NaOH or KOH.

(2) Stereochemistry and Kinetics of the reaction iR.D.S.) : This hydrolysis reaction takes place only in one step which is a rate determining step i.e. R.D.S. The rate of hydrolysis reaction depends on the Concentration of CH3Br
and 0H which are present in the R.D.S. of the reaction.
Rate = R = k [CH₃Br] (OH]
where k is rate constant of the reaction.

SN² reaction : The reaction between methyl bromide and hydroxide ion to form methanol follows a second order kinetics, since the rate of the reaction depends on the concentrations of two reacting species, namely methyl bromide and hydroxide ion it is bimolecular second order (2nd) Nucleophilic Substitution reaction denoted by SN².

(3) Mechanism of the reaction :
(i) It is a single step mechanism. The reaction takes place in the following steps :

(ii) Backside attack of the nucleophile : Nucleophile, OH^– attacks carbon atom of CH₃Br from back side i. e. from opposite side to that of the leaving group i.e. Br^– to experience minimum steric repulsion and electrostatic repulsion between the incoming nucleophile (OH^–) and leaving Br^–.

(iii) Transition state : When a nucleophile, OH^– approaches carbon atom of CH₃Br, the potential energy of the system increases until a transition state (T.S.) of maximum potential energy is formed in which C – Br bond is partially broken and C – OH bond is partially formed. The negative charge is equally shared by both incoming nucleophile- OH^– and outgoing, leaving group-Br^–. (Thus, the total negative charge is diffused.)

(iv) In CH₃Br, carbon atom is sp³ -hybridized and CH₃Br molecule is tetrahedral. The hybridisation of carbon atom changes to sp² hybridisation. The transition state contains pentacoordinate carbon having three δ (sigma) bonds in one plane making bond angles of 120° with each other i.e., H₁; H₂ and H₃ atoms lie in one plane while two partial covalent bonds containing Br and OH lie collinear and on opposite sides perpendicular to the plane.

(v) Inversion of configuration : The transition state decomposes fast by the complete breaking of the C-Br bond and the new C-OH bond is formed on the other side. The breaking of C-Br bond and the formation of C-OH bond take place simultaneously. The energy required to break the C-Br bond is partly obtained from the energy released when C-OH bond is formed. The formation of product CH₃OH is accompanied by complete or 100% inversion of configuration forming again sp³-hybridized carbon atom giving tetrahedral CH₃OH molecule. But in this structure the positions of H₂ and H₃ atoms in the reactant (CH₃Br) and in product are on the opposite side. This inversion of configuration is called Walden inversion.

Question 54.
Discuss the factors influencing S_N1 and S_N2 mechanism.
Answer:
(1) Nature of substrate : S_N2 : The transition state (T.S.) of S_N2 mechanism + is pentacoordinate, it is crowded. As a result S_N2 mechanism is favoured in primary halides and least favoured in tertiary halides.

S_N1 : A planar carbocation intermediate is formed in S_N1 reaction. Bulky alkyl groups can be easily accommodated in planar carbocation and it has no steric crowding. As a result SN1 mechanism is favoured in tertiary halides and least favoured in primary halides.

The carbocation intermediate is stabilized by + effect of alkyl substituents and also by hyperconjugation y effect of alkyl substituents containing a-hydrogens. As a result, S_N1 mechanism is favoured in tertiary halides and least favoured in primary halides.

Thus, tertiary alkyl halides undergo nucleophilic substitution by S_N1 mechanism while primary halides follow S_N2 mechanism.

(2) Nucleophilicity of the reagent : A strong nucleophile attacks the substrate faster and favours S_N2 mechanism. The rate of S_N1 mechanism is independent of the nature of nucleophile. Nucleophile does not react in the 1st step (slow step) of S_N1. Nucleophile reacts fast after the carbocation intermediate is formed.

(3) Solvent polarity : (1) S_N1 reaction proceeds more rapidly in polar protic solvents than in aprotic solvent. Polar protic solvent decreases the rate of S_N2 reaction. (2) In S_N2 mechanism, rate depends on substrate as well as nucleophile. A polar solvent stabilizes nucleophile by solvation. Thus solvent deactivates the nucleophile by stabilizing it. Hence, aprotic solvents or solvent of low polarity will favour S_N2 mechanism.

Question 55.
How does relative reactivity for alkaline hydrolysis with respect to S_N2 and S_N1 vary in the following alkyl halides :
(1) Bromomethane
(2) Bromoethane
(3) 2-Bromopropane
(4) 2-Bromo-2-methylpropane ?
Answer:
(A) Relative reactivity for S_N2 mechanism decreases in the order of :

(B) Relative reactivity for S_N1 mechanism decreases in the order of :
2-Bromo-2-methylpropane > 2-Bromopropane > Bromoethane > Bromomethane

Question 56.
Explain with reason the relative order of reactivity of l°/2°/3° alkyl halides by S_N1 mechanism.
Answer:
In alkaline hydrolysis of an alkyl halide by S_N1 mechanism, the formation of carbocation as an intermediate product is involved.

The increasing order of a stability of carbocation is,

The stability order for carbocation is 3° > 2° > 1°.
Therefore the increasing order of reactivity by S_N1 mechanism of alkyl halides is
(1°) primary < (2°) secondary < (3°) tertiary

Question 57.
Which one of the following is more easily hydrolysed in S_N1 and S_N2 reaction by aqueous KOH, C₆H₅ CHCIC₆H₅ and C₆H₅CH₂CI?
Answer:
In S_N1 reaction C₆H₅CHCI C₆H₅ will be more easily hydrolysed than C₆H₅CH₂CI
In S_N2 reaction C₆H₅ CH₂CI will be more easily hydrolysed than C₆H₅CHCIC₆H₅.

Question 58.
Choose the member that will react faster than the following pairs by S_N1 mechanism.
(1) l-bromo-2, 2-dimethyl propane or 2-bromopropane.

Answer:
The reactivity of S_N1 reaction depends on the steric hindrance, in 2-bromopropane, a-carbon atom is attached to two methyl groups suffers greater steric hindrance to nucleophilic attack than l-bromo-2, 2-dimethyl propane. Hence, 2-bromopropane react faster by S_N1 mechanism.

(2) 2-Iodo-2-methyl butane or 2-iodio-3-methyl butane.

Answer:
Since, 2-Iodo-2-methyl butane is a tertiary alkyl halide, it undergoes SN^-1 reaction faster than 2-iodo-3-methyl butane.

(3) 1-Chloro propane or 2-chloropropane.

Answer:
Since, 2-chloropropane is a secondary alkyl halide, it undergoes SN^-1 reaction faster than 1-chloropropane.

(4) 2-Iodo-2-methyl butane or tert-butyl chloride.

Answer:
Since, iodine is a better leaving group than chloride 2-iodo-2-methyl butane undergo SN^-1 reaction faster than tert-butyl chloride.

Question 59.
Write a note on elimination reaction.
OR
Explain dehydrohalogenation reaction.
Answer:
When alkyl halide having at least one β-hydrogen is boiled with alcoholic solution of potassium hydroxide (KOH), an alkene is formed due to elimination of hydrogen atom from β-carbon and halogen atom from α-carbon, is called dehydrohalogenation.

Tertiary butyl bromide when heated with alcoholic solution of potassium hydroxide forms isobutylene.

This reaction is called β-elimination (or 1,2-elimination) reaction as it involves elimination of halogen and a β-hydrogen atom.

As hydrogen and halogen is removed in this reaction it is also known as dehydrohalogenation reaction.

Question 60.
Describe the action of alcoholic potassium hydroxide (aic. KOH) on
(1) ethyl bromide
(2) n-propyl bromide
(3) isopropyl bromide
(4) tert-butyl chlorIde.
Answer:
(1) Ethyl bromide : When ethyl bromide (bromoethane) is heated with alcoholic potassium hydroxide (alcoholic alkali). ethene (gas) is formed by the dehydrobrominaion reaction.

(2) n-PropI bromide : When n-propyl bromide is heated with alcoholic potassium hydroxide, propene is formed.

(3) Isopropvl bromide : When isopropyl bromide (2-bromopropane) is boiled with alcoholic potassium hydroxide, propcne is formed.

(4) Tert-hutyl chloride: When ten-butyl chloride (2-chloro-2-methyl propanc) is hcatcd with alcoholic KOH.

Question 61.
Describe the action of alc.KOH on 2-bromobutane.
When 2-bromobutane is boiled with alc.KOH on 2-bromobutane, a mixture of but-l-ene and but-2-ene is formed.
Answer:

Question 62.
Explain Saytzelf’s rule with suitable example.
Answer:
Saytzcff’s rule : In dehydrohalogenation reaction the preferred product is that alkene which has the greater number of alkyl groups attached to the doubly bonded carbon atoms.

Hence the number of alkyl substituents on doubly bonded Carbon atoms increases, the stability of the alkene giving its major products.

Hence the increasing stability of alkenes is.

There are two types of fi hydrogens (β₁ and β₂) therefore two alkenes are expected.

Question 63.
What is a Grignard reagent ?
Answer:
Grignard reagent : An organometallic compound in which the divalent magnesium is directly linked to an alkyl group (R -) and a halogen atom (X), and has general formula R – Mg – X is called Grignard reagent. OR When alkyl halide is treated with magnesium in dry ether as solvent, it gives alkyl magnesium halide. It is known as Grignard reagent.

The carbon-magnesium bond is highly polar and magnesium-halogen bond is in ionic in nature. Grignard reagent is highly reactive. It is an important reagent and used in the preparation of a large number of organic compounds.

Question 64.
How is Grignard reagent prepared ?
Answer:
Grignard reagent is an alkyl magnesium halide, R – Mg – X obtained by the reaction of alkyl halide R – X with magnesium (Mg) in dry ether.

When an alkyl halide like CH₃I is added from a dropping funnel to a flask containing pieces of pure Mg in pure and dry ether (diethyl ether) and a trace of iodine, Grignard reagent, CH₃ – Mg – I is formed.

Ethyl iodide when treated with magnesium in presence of dry ether forms ethyl magnesium iodide.

Question 65.
Write a note on Grignard reagent.
Answer:
(1) Ethyl bromide : When ethyl bromide (bromoethane) is heated with alcoholic potassium hydroxide (alcoholic alkali). ethene (gas) is formed by the dehydrobrominaion reaction.

(2) n-PropI bromide : When n-propyl bromide is heated with alcoholic potassium hydroxide, propene is formed.

(3) Isopropvl bromide : When isopropyl bromide (2-bromopropane) is boiled with alcoholic potassium hydroxide, propcne is formed.

(4) Tert-hutyl chloride: When ten-butyl chloride (2-chloro-2-methyl propanc) is hcatcd with alcoholic KOH.

Question 66.
Describe the action of water on
(1) methyl magnesium iodide
(2) ethyl magnesium iodide.
Answer:
(1) Methyl magnesium iodide : When methyl magnesium iodide is treated with water, methane is obtained

(2) Ethyl magnesium iodide : When ethyl magnesium iodide is treated with water, ethane is obtained.

Question 67.
Describe the action of ammonia on
(1) ethyl magnesium bromide
(2) n-propyl magnesium chloride.
Answer:
(1) Ethyl magnesium bromide : When ethyl magnesium bromide is treated with ammonia, ethane is formed.

(2) n-Propyl magnesium chloride : When n-propyl magnesium chloride is treated with ammonia, propane is formed.

Question 68.
Explain Wurtz reaction. OR Explain the action of sodium with alkyl halides.
Answer:
(1) When an alkyl halide is treated with metallic sodium in dry ether, the corresponding higher alkane is formed. This is called Wurtz reaction or Wurtz coupling reaction.

(2) In this reaction the alkyl radicals from two molecules of the reacting alkyl halide combine or couple to form the higher alkane.

(3) Thus, methyl bromide reacts with sodium in ether to form ethane (C₂H₆), while ethyl bromide under the same conditions forms n-butane (C₄H₁₀).

(4) If a mixture of two different alkyl halides is treated with Na in dry ether, then a mixture of alkanes is obtained called self coupling products. For example, a mixture of CH₃Br and C₂H₅Br gives propane along with C₂H₆ and C₄H₁₀.

Question 69.
Explain the reaction of haloarene with alkyl halide and sodium metal.
Write a note on Wurtz-Fittig reaction.
Answer:
When an alkyl halide and an aryl halide is treated with sodium metal in dry ether the corresponding alkylarene (alkyl benzene) is formed. The reaction is known as Wurtz-Fittig reaction. This reaction allows alkylation of alkyl halides.

Question 70.
Describe the action of aryl halide on sodium metal.
Answer:
Aryl halide reacts with sodium metal in dry ether, biphenyl is formed. This reaction is known as Fittig reaction.

Question 71.
Identify the product A of following reaction.

Answer:

Question 72.
Explain the following substitution reactions of chlorobenzene :
(1) Halogenation
(2) Nitration
(3) Sulphonation.
Answer:
(1) Halogenation : When chlorobenzene is reacted with chlorine in presence of anhydrous ferric chloride, a mixture of ortho and para-dichlorobenzene (major product) is formed.

(2) Nitration : When chlorobenzene is heated with nitrating mixture (cone, nitric acid -I- cone, sulphuric acid) a mixture of l-chloro-4-nitro benzene (major product) and l-chloro-2-nitrobenzene is formed.

(3) Sulphonation : When chlorobenzene is heated with concentrated sulphuric acid, a mixture of 4-chlorobenzene sulphonic acid (major product) and 2-chlorobenzene sulphonic acid is formed.

Question 73.
Describe the action of the following on chlorobenzene :
(1) Methyl chloride in the presence of anhydrous AICI₃
(2) Acetyl chloride in the presence of anhydrous AICI₃.
Answer:
(1) Methyl chloride in the presence of anhydrous AICI₃ : When chlorobenzene is treated with methyl chloride in the presence of anhydrous AICI₃, a mixture of l-chloro-4-methyl benzene (major product) and l-chloro-2-methyl benzene is formed. Since, the alkyl group is introduced in the benzene ring, the reaction is termed as Friedel Craft’s alkylation.

(2) Acetyl chloride in the presence of anhydrous AICI₃ : When chlorobenzene is reacted with acetyl chloride in the presence of anhydrous AICI₃, a mixture of 2-chloro acetophenone and 4-chloro acetophenone (major product) is formed. Since, the acetyl group is introduced in the benzene ring, the reaction is termed as Friedel Craft’s acylation.

Question 74.
Write a note on Friedel Craft’s reaction.
Answer:

(1) Methyl chloride in the presence of anhydrous AICI₃ : When chlorobenzene is treated with methyl chloride in the presence of anhydrous AICI₃, a mixture of l-chloro-4-methyl benzene (major product) and l-chloro-2-methyl benzene is formed. Since, the alkyl group is introduced in the benzene ring, the reaction is termed as Friedel Craft’s alkylation.

(2) Acetyl chloride in the presence of anhydrous AICI₃ : When chlorobenzene is reacted with acetyl chloride in the presence of anhydrous AICI₃, a mixture of 2-chloro acetophenone and 4-chloro acetophenone (major product) is formed. Since, the acetyl group is introduced in the benzene ring, the reaction is termed as Friedel Craft’s acylation.

Question 75.
Convert 1-chlorobutane into the following compounds :

Answer:
(1) 1-Chlorobutane to butan-l-ol :

(2) 1-Chlorobutane to 1-iodobutane :

(3) 1-Chlorobutane to n-butyl cyanide (CH₃ – CH₂ – CH₂ – CH₂ – CN) :

(4) l-Chlorobutane to

Question 76.
Predict the expected product of substitution reactions :
(1) Isobutyl chloride + sodium ethoxide.
Answer:

(2) n-butyl chloride + sodium.
Answer:

(3) 1-chloropropane + aq. potassium hydroxide.
Answer:

(4) Aniline + NaNO₂/HCl.
Answer:

Question 77.
Write the products:

Question 78.
Identify A and B in the following :

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

Question 79.
State the uses of the following compounds :
(1) Dichloromethane (CH₂CI₂)
(2) Trichloromethane or Chloroform (CHCI₃)
(3) Tetrachloromethane or carbon tetrachloride (CCI₄)
(4) Iodoform (CHI₃)
(5) Freons
(6) DDT (p, p’-Dichlorodiphenyl trichloroethane).
Answer:
(1) Dichloromethane (CH₂CI₂) :

• Dichloromethane dissolves wide range of organic compounds, hence it is used as solvent for many chemical reactions.
• It is used as a solvent as a paint remover and degreaser.
• It is used as propellant in aerosols and as a fumigant pesticide for grains and strawberries.
• It is used to decaffinate tea or coffee.

(2) Trichloromethane or Chloroform (CHCI₃) :

• Chloroform in the production of chlorofluoromethane, freon refrigerant R-22.
• It is used as solvent in pharmaceuticals, pesticides, gums, fats, resins and dye industry.
• It is a good source of dichlorocarbene species.

(3) Tetrachloromethane or carbon tetrachloride (CCI₄) :

• Carbon tetrachloride is used in the manufacture of refrigerants.
• It is used as a dry cleaning agent and as a pesticide for stored grains.
• It is very useful solvent for oils, fats and resins. It serves as a source of chlorine.

(4) Iodoform (CHI₃) :

• Iodoform is used as antiseptic, dressing of wounds and sores.
• On small scale it is used as disinfectant.

(5) Freons :

• Freons are widely used as propellants in aerosol, products of food, cosmetics and pharmaceutical industries.
• Freons containing bromine in their molecules are used as fire extinguishers.
• They are used in aerosol insecticides, solvent for cleaning clothes and metallic surfaces.
• It is used as foaming agents in the preparation of foamed plastics and in production of certain fluorocarbons.
• It is used as refrigerants and air conditioning purposes.

Question 80.
State the environmental effects of the following compounds :
(1) Dichloromethane (CH₂CI₂)
(2) Trichloromethane or chloroform (CHCI₃)
(3) Tetrachloromethane or carbon tetrachloride (CCI₄)
(4) Iodoform (CHI₃)
(5) Freon.s (CCI₂F₂, CCI₃F, CHCIF₂)

Answer:
(1) Dichloromethane (CH₂CI₂) :

• Higher levels of dichloromethane in air causes nausea, numbness in fingers and toes, dizziness.
• Lower levels of dichloromethane causes impaired vision and hearing.
• Direct contact with eyes can damage cornea.

(2) Trichloromethane or chloroform (CHCI₃) :

• When chloroform is exposed to air in the presence of sunlight, it slowly oxidised to phosgene, a poisonous compound, therefore it is stored in dark, amber coloured bottles.
• Chloroform vapour when inhaled for a short time causes dizziness, headache and fatigue and if inhaled for a long time affects central nervous system.

(3) Tetrachloromethane or carbon tetrachloride (CCI₄) :

• Exposure to carbon tetrachloride causes eye irritation, damages nerve cells, vomiting sensation, dizziness, unconciousness or death. Long exposure to chloroform may affect liver.
• When mixed with air it causes depletion of the ozone layer, which affects human skin leading to cancer.

(4) Iodoform (CHI₃) : Iodoform has a strong smell. It causes irritation to skin and eyes. It may cause respiratory irritation or breathing difficulty, dizziness, nausea, depression of central nervous system, visual disturbance.

(5) Freons (CCI₂F₂, CCI₃F, CHCIF₂) :

• Freon as refrigerant causes ozone depletion.
• Freons have low toxicity and low biological activity.
• Freons from propane group are more toxic in nature.
• Regular large inhalation of freon results in breathing problems, organ damage, loss of consciousness.

(6) DDT :

• DDT is not readily metabolised by animals.
• It is deposited and stored in fatty tissues.
• Exposure to high doses of DDT may cause vomiting, tremors or shakiness.
• Laboratory animal studies showed adverse effect of DDT on liver and reproduction.
• DDT is a pressistent organic pollutant, readily absorbed in soils and tends to accumulate in the ecosystem.
• When dissolved in oil or other lipid, it is readily absorbed by the skin. It is resistant to metabolism.
• There is a ban on use of DDT due to all these adverse effects.

Question 81.
What is the chemical name of freon?
Answer:
The chemical name of freon is Dichlorodifluoromethane.

Question 82.
What is the chemical name of DDT ?
Answer: The chemical name of DDT is p, p’-Dichlorodiphenyltrichloroethane.

Activity :
(1) Collect detailed information about Freons and their uses.
(2) Collect information about DDT as a persistent pesticide.
Reference books :
(1) Organic chemistry by Morrison, Boyd, Bhattacharjee, 7th edition, Pearson.
(2) Organic chemistry by Finar, Vol 1, 6th edition, Pearson

Multiple Choise Questions

Question 83.
Select and write the most appropriate answer from the given alternatives for each sub-question :

Question 1.

Answer:
(b) CH₃ – CH₂ – CH₂ – I

Question 2.
The rate of S_N2 reaction depends on the concentra¬tion of
(a) only the substrate
(b) only the reagent
(c) both the substrate and the reagent
(d) neither the substrate nor the reagent
Answer:
(c) both the substrate and the reagent

Question 3.
In S_N2 reaction, the hydrolysis of alkyl halide shows
(a) the retention of configuration
(b) the inversion of configuration
(c) both retention and inversion of configuration
(d) no change in the configuration
Answer:
(b) the inversion of configuration

Question 4.
The one step exothermic reaction is
(a) S_N1
(b) S_N2
(C) S_N
(d) S2_N
Answer:
(b) S_N2

Question 5.
Which of the following is correct about S_N2 mechanism?
(a) Two step reaction
(b) Complete inversion of configuration
(c) Formation of carbonium ion
(d) Favoured by polar solvent
Answer:
(b) Complete inversion of configuration

Question 6.
Which of the following is not a nucleophile?
(a) Ammonia
(b) Ammonium ion
(c) Primary amine
(d) Secondary amine
Answer:
(d) Secondary amine

Question 7.
Which of the following undergoes nucleophilic substitution exclusively by S_N2 mechanism ?
(a) ethyl chloride
(b) isopropyl chloride
(c) chlorobenzene
(d) benzyl chloride
Answer:
(d) benzyl chloride

Question 8.
Which of the following is most reactive towards nucleophilic substitution reaction ?
(a) CH₂ = CH – CI
(b) CH₃CH = CHCI
(c) C₆H₅CI
(d) CICH₂ – CH = CH₂
Answer:
(d) CICH₂ – CH = CH₂

Question 9.
The stability order of carbocation is
(a) 2° > 3° > 1°
(b) 3° > 2° > 1°
(c) 3° > 1° > 2°
(d) 1° > 3° > 2°
Answer:
(b) 3° > 2° > 1°

Question 10.

(a) ethane
(b) propane
(c) n-butane
(d) n-pentane
Answer:
(c) n-butane

Question 11.
Which of the following characteristic properties of the enantiomers is correct?
(a) The enantiomers possess same physical and chemical properties
(b) The enantiomers are optically active compounds
(c) The enantiomers have different optical rotations
(d) All of these
Answer:
(d) All of these

Question 12.
The optically inactive compound is
(a) glucose
(b) lactic acid
(c) isopropyl alcohol
(d) 2-bromo butane
Answer:
(c) isopropyl alcohol

Question 13.
A compound with the molecular formula CH₂OH(CHOH)₃CH₂OH has optically active forms
(a) 3
(b) 4
(c) 6
(d) 8
Answer:
(d) 8

Question 14.
A racemic mixture consists of
(a) equal amount of d and l isomers
(b) unequal amounts of d and / isomers
(c) unknown amounts of d and / isomers
(d) only d isomers
Answer:
(a) equal amount of d and l isomers

Question 15.
Which of the following compounds is not optically active ?
(a) Lactic acid
(b) Secondary butyl chloride
(c) n-propyl iodide
(d) Glucose
Answer:
(c) n-propyl iodide

Question 16.
Which of the following compounds shows optical activity ?
(a) n-butyl chloride
(b) isobutyl chloride
(c) sec-butyl chloride
(d) t-butyl chloride
Answer:
(c) sec-butyl chloride

Question 17.
The major product of the following reaction is

Answer:
(c)

Question 18.
The above reaction is known as

(a) Wurtz-Fittig reaction
(b) Friedel Craft’s reaction
(c) Sandmeyer’s reaction
(d) Swarts reaction
Answer:
(b) Friedel Craft’s reaction

Question 19.
Iodoform is used as
(a) an anaesthetic
(b) an antiseptic
(c) an analgesic
(d) an antibiotic
Answer:
(b) an antiseptic

Question 20.
p, p’-dichlorodiphenyl trichloroethane is used as
(a) insecticide
(b) anaesthetic
(c) antiseptic
(d) refrigerant
Answer:
(a) insecticide

Question 21.
The order of reactivity in nucleophilic substitution reaction is
(a) CH₃F < CH₃C1 < CH₃I < CH₃Br
(b) CH₃F < CH₃C1 < CH₃Br < CH₃I
(c) CH₃F < CH₃Br < CH₃C1 < CH₃I
(d) CH₃I < CH₃Br < CH₃C1 < CH₃F
Answer:
(b) CH₃F < CH₃C1 < CH₃Br < CH₃I

Question 22.
Racemate is
(a) optically active
(b) optically dextro rotatory
(c) optically inactive
(d) optically laevorotatory
Answer:
(c) optically inactive

Question 23.
The number of asymmetric carbon atoms in glucose are
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(c) 4

Question 24.
The geometry of carbon lum ion is
(a) Tetrahedral
(b) planar
(c) linear
(d) pyramidal
Answer:
(b) planar

Question 25.
In its nucleophilic substitution reaction, aryl halide resembles
(a) Vinyl chloride
(b) allyl chloride
(e) Benzyl chloride
(d) ethyl chloride
Answer:
(a) Vinyl chloride

Question 26.
The weakest C-Cl bond is present in

Answer:
(d)

Question 27.
Which alkyl halide among the following com¬pounds has the highest boiling point ?
(a) (CH₃)₃CCI
(b) CH₃CH₂CH₂CH₂CI
(c) CH₃CH₂CH₂C1
(d) CH₃CH(CH₃)CH₂CI
Answer:
(b) CH₃CH₂CH₂CH₂CI

Question 28.
It is difficult to break C-Cl bond in CH₂ = CH – CI due to
(a) Hyper conjugation
(b) Resonance
(c) Electromeric effect
(d) Inductive effect
Answer:
(b) Resonance

Question 29.
Which one of the following when heated with metallic sodium will not give the corresponding alkane ?

Answer:
(c)

Question 30.
The most reactive alkyl halide towards SN2 reac¬tion is
(a) CH₃X
(b) R₃CX
(C) R₂CHX
(d) RCH₂X
Answer:
(a) CH₃X

Question 31.
The number of electrons surrounding the carbon- ium ion is
(a) 6
(b) 8
(c) 10
(d) 7
Answer:
(a) 6

Question 32.
The lowest stability of carbocation among the compounds

Answer:
(a)

Question 33.
Carbon atom in methyl carbocation contains how many pairs of electrons?
(a) 8
(b) 4
(c) 3
(d) 5
Answer:
(b) 4

Question 34.
The optically inactive compound is
(a) Glucose
(b) Lactic acid
(c) 2-Chlorobutane
(d) 2-Chloropropane
Answer:
(d) 2-Chloropropane

Question 35.
The hydrogen halide which does not obey Markownikv rule in presence of peroxide is
(a) HC1
(b) HBr
(c) HF
(d) HI
Answer:
(b) HBr

Question 36.
Which one of the following is NOT used to prepare alkyl halide from an alcohol ?
(a) SOCl₂
(b) PC1₃
(c) HC1 + ZnCl₂
(d) NaCl
Answer:
(d) NaCl

Question 37.
The total number of electrons present in the central carbon atom of a free radical is
(a) 7
(b) 8
(c) 9
(d) 6
Answer:
(a) 7

Question 38.
In which of the following pairs both are nucleophiles ?
(a) BF₃, AICI₃
(b) NO⁺₂, Cl^–
(c) CN^–, NH₃
(d) Br⁺, BC1₃
Answer:
(c) CN^–, NH₃

Question 39.
Which one of the following alkane is NOT formed in Wurtz reaction ?
(a) Methane
(b) Ethane
(c) Propane
(d) Butane
Answer:
(a) Methane

Question 40.
Which of the following groups has highest priority according to R, S convention?
(a) CH₂OH
(b) COOH
(c) COCH₃
(d) COOCH₃
Answer:
(d) COOCH₃

Question 41.
The halogen atom in aryl halides is
(a) o- and p-di reefing
(b) m-directing
(c) o, m and p-di reefing
(d) only m-directing
Answer:
(a) o- and p-di reefing

Question 42.
Chlorobenzene can be obtained by benzene diazonium chloride by
(a) Friedel Craft’s reaction
(b) Wurtz reaction
(c) Gatterman’s reaction
(d) Fittig reaction
Answer:
(c) Gatterman’s reaction

Question 43.
Which of the following carbocations is least stable ?


Answer:
(c)

Question 44.
But-l-ene on reaction with HCI in the presence of sodium peroxide yields
(a) n-butyl chloride
(b) isobutyl chloride
(c) secondary butyl chloride
(d) tertiary butyl chloride
Answer:
(c) secondary butyl chloride

Question 45.
Carbon tetrachloride is used as
(a) anaesthetic
(b) antiseptic
(c) dry cleaning agent
(d) fire extinguisher
Answer:
(c) dry cleaning agent

Question 46.
Identify the product D in the following sequence of reactions :

(a) 2, 2-dimethyl butane
(b) 2, 3-dimethyl butane
(C) hexane
(d) 2, 4-dimethylpentane
Answer:
(b) 2, 3-dimethyl butane

Question 47.
The preparation of alkyl fluoride from alkyl chlor ide, in presence of metallic fluorides is known as
(a) Williamson’s reaction
(b) Finkeistein reaction
(c) Swarts reaction
(d) Wurlz reaction
Answer:
(c) Swarts reaction

Question 48.
UPAC name of the following compound is

(a) 3-Bromo-3, 4-dimetbyiheptane
(b) 3,4-dimethyl-3-bromoheptane
(c) 5-Bromo-4,5-dimethylheptane
(d) 4,5-dimethyl-5-bromoheptane
Answer:
(a) 3-Bromo-3, 4-dimetbyiheptane