RBSE Solutions for Class 12

RBSE Solutions for Class 12 Biology Chapter 9 Enzymes

July 29, 2019 by Prasanna Leave a Comment

Rajasthan Board RBSE Class 12 Biology Chapter 9 Enzymes

RBSE Class 12 Biology Chapter 9 Multiple Choice Questions

Question 1.
In what respect enzymes are different from catalysts?
(a) High diffusion rate
(b) Active at high temperature
(c) Protein nature
(d) Used in the reaction
Answer:
(c) Protein nature

Question 2.
The non-protein part of holoenzyme is called?
(a) Apoenzyme
(b) Co-factor
(c) Conjugated enzyme
(d) All the above
Answer:
(b) Co-factor

Question 3.
Which of the following statement is absolutely correct?
(a) All proteins are enzymes
(b) All enzymes are proteins
(c) Most enzymes are proteins
(d) None of the above
Answer:
(c) Most enzymes are proteins

Question 4.
Which enzyme was discovered first of all?
(a) Zymase
(b) Lipase
(c) Pepsin
(d) Isomerase
Answer:
(a) Zymase

Question 5.
Enzyme activity is affected by –
(a) pH only
(b) Substrate concentration
(c) only temperature
(d) All the above
Answer:
(d) All the above

Question 6.
Non-Competitive inhibitors are the substances which?
(a) Combine to the active sites of the enzyme
(b) Decrease the active sites of the enzyme
(c) Change the structural organisation of the enzyme
(d) Do not cause any change in the properties of an enzyme
Answer:
(c) Change the structural organisation of the enzyme

Question 7.
Which enzyme was obtained in a crystalline form first of all?
(a) Urease
(b) Catalase
(c) Amylase
(d) Aldolase
Answer:
(a) Urease

RBSE Class 12 Biology Chapter 9 Very Short Answer Questions

Question 1.
Who proposed lock and key theory and when was it proposed.
Answer:
Emil Fisher (1898).

Question 2.
The protein part and non-protein part of holoenzyme are called as?
Answer:
Protein part is named as apoenzyme and the non-protein part is called as co-factor.

Question 3.
Name a nonprotein enzyme.
Answer:
Ribozyme enzyme is made up of RNA and is the only non-proteinaceous enzyme known so far.

Question 4.
Define the prosthetic group.
Answer:
When the non-protein part of an enzyme is an organic substance and is firmly attached to the protein and can not be separated from it easily, it is called the prosthetic group.

RBSE Class 12 Biology Chapter 9 Short Answer Questions

Question 1.
What is co-enzyme? Give one example.
Answer:
When the non-protein part of the enzyme is an organic substance and it is closely attached to the protein part, which can dissociate from the protein part easily and can get attached again, it is called coenzyme.
Example: Co-A, NAD, NADP, FAD etc.

Question 2.
What do you understand by competitive inhibition? How this can be stopped.
Answer:
Some chemical substances which have structure closely similar to the substrate molecule, complete with the substrate in attaching with the active sites of the enzyme. Presence of such substances lowers the enzymatic activity. These substances are called competitive inhibitors and this inhibition is called competitive inhibition This type of inhibition can be stopped by increasing the concentration of substrate molecules.

Question 3.
Briefly explain the method of nomenclature of enzymes.
Answer:
There are several methods of naming enzymes. These are as follows:

Most enzymes are named by adding suffix-case at the end of the name of substrate catalyzed by them. Example: Maltase, Sucrase, Urease.
Many enzymes are named on the basis of the type of catalytic reaction governed by the enzyme. Example: Oxidase, Dehydrogenase etc.
In the modern method of nomenclature, enzymes are named by adding suffix-case after both the name of the substrate on which they act and the type of catalytic reaction governed by them. Example: Succinate dehydrogenase.
In some cases, traditional names of enzymes have been retained and are in practice. Example: Pepsin, Trypsin, Chymotrypsin.
International Union of biochemistry has proposed systematic nomenclature in which some code number is given to each enzyme. In this nomenclature, substrate name, type of reaction catalyzed and some other information is also included.

Question 4.
How the rate of enzyme activity is accelerated.
Answer:
Enzymes are macromolecules and have high molecular weight. There are present numerous active sites on the surface of enzymes where the substrate molecules attach to form an unstable complex. This complex is called an enzyme-substrate complex. It breaks to release the product and enzyme.

By increasing the amount of enzyme, the rate of reaction increases, until the concentration of the substrate becomes a limiting factor. At this stage, the rate of reaction can be increased by increasing the concentration of substrate.

RBSE Class 12 Biology Chapter 9 Essay Type Questions

Question 1.
Describe the structure of the enzyme and explain their specific characteristics.
Answer:

All enzymes are made up of proteins but all proteins are not enzymes.
Many enzymes are made up exclusively of protein and such enzymes are called simple enzymes.
In many enzymes, there is found a non-protein part associated with protein. Such enzymes are known as conjugated enzymes or holoenzymes.
The protein part of the conjugated enzyme is called apoenzyme.
The nonprotein part of the conjugated enzyme is called co-factor.
The co-factors are of three types:

- Prosthetic group
- Co-enzyme
- Activator

Conjugated enzyme or Holoenzyme = Apoenzyme + Co-factor
1. Prosthetic Group:

When the non-protein part attached with the apoenzyme is an organic compound and is firmly bonded with the protein part, it is called a prosthetic group.
The prosthetic group can not be separated from the protein part (the apoenzyme) without denaturation of the protein.
Example: Cytochrome, Flavoprotein.

2. Co-enzyme:

When the non-protein part is an organic compound and is only loosely attached with the protein part (the apoenzyme), it is called as co-enzyme.
The co-enzyme can be easily separated from the protein part and can be easily attached again with it.
Example: Co-enzyme A, NAD, FAD, NADP etc.

3. Activator:

When the non-protein part is an inorganic substance such as some metal ion or mineral ion, it is called as an activator.
The main role of the activator is to form a bond between the enzyme and the substrate molecule Example: K, Cu, Mn, Fe Zn, Ca etc.

Note: The main part of the holoenzyme or conjugated enzyme is made up of protein and it is called apoenzyme. The size (dimension) of the protein molecule and the sequence of amino acids varies in different enzymes.
As the proteins are colloidal in nature, they have large surface area per unit volume.
The protein part of the enzyme contains one or more specific sites called active sites. During the enzymatic activity, the substrate molecules attach at these active sites.
In the holoenzyme of the conjugated enzyme, the enzymatic activity is brought about by the holoenzyme.
The apoenzyme alone or the nonprotein part alone can not act as an enzyme.

Question 2.
Describe mechanisms of enzymatic action.
Answer:
All enzymes are macromolecules and have numerous specific areas called active sites on their surface.

During the enzymatic activity, substrate molecules attach at these active sites and an unstable enzyme-substrate complex is formed.
This enzyme-substrate complex dissociates to set free the product and the enzyme.

The mechanism of formation of the enzyme-substrate complex has been explained by two theories.

Lock and key hypothesis
Induced fit model theory

Formation of enzyme-substrate complex results in to lowering of the activation energy and hence the rate of reaction is accelerated.

Mode of Enzyme Action:
The mode of enzyme action depends upon the nature of the enzyme and the substrate molecule, and it can be understood by the following:
(1) Formation of Enzyme Substrate complex (ESC):
(2) Lowering of Activation energy.
(1) Formation of Enzyme Substrate Complex (ESC):

In all enzyme governed reactions the enzyme and the substrate combine to form an unstable complex called enzyme-substrate complex. (ESC). This complex dissociates into the product and the enzyme.
Enzyme + Substrate → Enzyme substrate complex.
Enzyme substrate complex → Enzyme + Product(s).
On the surface of each enzyme, several specific areas are found. These are called active sites.
The substrate molecules combine with the enzyme at these sites.
When the substrate molecules combine at the active sites, they can easily react with each other forming the product.
The product formed by formation of new bonds dissociates from the enzyme.
The enzyme becomes free to bind with other molecules of the substrate and new enzyme-substrate molecule complex is formed.

The mechanism of formation of the enzyme-substrate complex and the property of specificity of the enzyme has been explained by the following theories.
1. Lock and Key Theory:

This theory was postulated by Emil Fisher (1898).
According to this theory, like a lock can be operated by it’s key only, similarly, a specific substrate having a specific structure only can bind with the specific active site present on the surface of a specific enzyme.
The enzyme remains unchanged after the product is released.

2. Induced Fit Model Theory:

This theory has been proposed by Daniel Koshland (1966).
According to this theory, the shape (structure) of active sites found on the surface of enzymes is not rigid but flexible.
These active sites are initially not complementary to the substrate molecule, but as a specific substrate molecule comes in close proximity of the enzyme, it induces a change in the organisation of the active site and binds with it.
In other words, it can be said that the configuration of active sites is not fixed or rigid as has been stipulated in the lock and key theory but is flexible and when a specific substance comes in contact, the active site undergoes a minor change and becomes complementary to the substrate molecule.
Thus enzyme-substrate complex is formed between a specific substrate with a specific enzyme only.
This theory stipulates that there are two operative groups in the active sites one helps in binding of the substrate molecule and the other acts to convert the substrate into
the product.

(2) Lowering of Activation Energy:

All chemical compounds require an input of a certain amount of energy for initiating a chemical reaction. This energy is called activation energy.
Enzymes have the ability to reduce the energy required for the activation of the molecule. • Hence in the presence of an enzyme, the reactant substrates are converted into a product with much lesser input of energy.
On average the enzymes reduce the activation energy by approximately 65%.
It is because of this that a reaction which takes place at a high temperature outside the cell takes place in the cell at atmospheric temperature with the help of enzyme.

Question 3.
Explain in details, classficiation of enzymes.
Answer:
International Union of Biochemistry (IUB) suggested the use of all the reactions and equations for the purpose of classification of enzymes and the Enzyme commission has classified all enzymes into six main divisions (categories).
1. Oxidoreductases
2. Transferases.
3. Hydrolases
4. Lyases
5. Isomerases
6. Ligases or Synthetases.
The six division are as follows:
1. Oxidoreductases:
All those enzymes which catalyze oxidation-reduction reactions are included in this category. They catalyze the removal or addition of hydrogen, oxygen or electrons from the substrate or to the substrate and thus bring about oxidation or reduction.
Note: On the basis of their activity these can be a divide into three subgroups:
(a) Oxidases: These bring about oxidation of the substrate by transferring hydrogen from the substrate molecule to oxygen.
Example: Cytochrome oxidase, Ascorbic acid oxidase etc.
RBSE Solutions for Class 12 Biology Chapter 9 Enzymes 6
(b) Dehydrogenases: These enzymes bring about oxidation of substrate molecule by removal of hydrogen from the substrate as follows:
Example – Lactate dehydrogenase, Alcohol dehydrogenase.

(c)Reductases: These enzymes cause the addition of hydrogen or electron to the substrate and removal of oxygen.

2. Transferases:
These enzymes catalyze the transfer of a group (amino, phosphate, methyl, ketone etc.) from one molecule to another molecule.
Example: Transphosphatase, Transaminase.

3. Hydrolases:
These enzymes catalyze the addition or removal of water. They hydrolyse a variety of compounds and generally bring about cleavage of the substrate. This group includes digestive enzymes. Which breakdown macromolecules into smaller molecules.
Example: Carbohydrase, Amylase, Nuclease, Esterase etc.
Most hydrolytic reactions are reversible. These types of enzymes may also condense small molecules to form macromolecules. In such reactions, the water molecule is released. Example: Fumarase, Enolase etc.

4. Lyases:
These enzymes catalyze the removal of a group from their substrate molecule by breaking a special type of covalent bonds without hydrolysis.
Example: Aldolase.

5. Isomerases:
These enzymes catalyze intramolecular reorganisation in the substrate molecule and convert the substrate into its isomer.
Example: Phosphohexoisomerase.

6. Ligases or Synthetases:
These enzymes catalyze such reactions in which two compounds are linked together by forming covalent bonds between the two.
Example: Pyruvate carboxylase, Citrate synthetase.

Question 4.
What is enzyme inhibition? How many types of enzyme inhibition is known? Explain how the effect of inhibitors can be stopped.
Answer:
When an enzymatic activity is reduced or stopped by the effect of some chemicals others than the substrate molecules, the act is called enzyme inhibition. The chemicals or substances which bring about inhibition are called enzyme inhibitors.
Enzyme inhibition is of two types:

Competitive inhibition
Non-competitive inhibition

Enzyme inhibitors:
Some chemical substances inhibit enzyme activity. These substances are called enzyme
inhibitors and are of two types.
1. Competitive inhibitors:
These substances have structure closely similar to the structure of the substrate molecules. Hence these substances compete with the substrate in binding with the active site of an enzyme. This results in a decrease in enzyme activity.
Example: Malonic acid is a competitive inhibitor of succinic acid. The effect of this kind of inhibition can be overcome by increasing the concentration of substrate.

2. Non-competitive inhibitors:
Some substances do not show resemblance in structure with substrate molecule but attach with the active site of the enzyme permanently and bring about structural changes in their configuration. These inhibitors permanently block or destroy the active site and thus act as a poison. These are called non-competitive inhibitors or enzyme poisons. Example: [latex]{ Pb }^{ ++ }[/latex], [latex]{ Hg }^{ ++ }[/latex], [latex]{ Ag }^{ ++ }[/latex] etc.
Note: Cyanide causes denaturation of cytochrome oxidase enzyme which is used in respiration reactions and thus acts as a poison. The effect of non-competitive inhibitors can be overcome by increasing the concentration of the enzyme.

RBSE Solutions for Class 12 Biology

RBSE Solutions for Class 12 Biology Chapter 5 Plant Water Relations

July 27, 2019 by Prasanna Leave a Comment

Rajasthan Board RBSE Class 12 Biology Chapter 5 Plant Water Relations

RBSE Class 12 Biology Chapter 5 Multiple Choice Questions

Question 1.
The process of exchange of [latex]{ CO }_{ 2 }[/latex] and [latex]{ O }_{ 2 }[/latex] from the atmosphere by the leaves is called?
(a) Osmosis
(b) Diffusion
(c) Imbibition
(d) Endosmosis
Answer:
(a) Osmosis

Question 2.
Which of the following is permeable?
(a) Plasma membrane
(b) Tonoplast
(c) Cuticle
(d) Cell wall
Answer:
(d) Cell wall

Question 3.
The value of DPD in the flaccid state will be?
(a) Equal to OP
(b) More than OP
(c) Zero
(d) Less than OP
Answer:
(a) Equal to OP

Question 4.
In the flaccid state the value of which pressure is zero?
(a) Suction force
(b) Diffusion pressure
(c) Wall pressure
(d) Osmotic pressure
Answer:
(c) Wall pressure

Question 5.
The process of swelling of hydrophilic substances by adsorption of water or moisture is called –
(a) Imbibition
(b) Osmosis
(c) Diffusion
(d) De plasmolysis
Answer:
(a) Imbibition

RBSE Class 12 Biology Chapter 5 Very Short Answer Questions

Question 1.
Write an example of a semi-permeable membrane.
Answer:
The lipo-protein cell membranes are semipermeable in nature.

Question 2.
Write one importance of osmosis.
Answer:
Absorption of water from the soil by root hairs takes place by osmosis. Similarly, the movement of water from one cell to another in the plant body is also by the process of osmosis.

Question 3.
Define diffusion.
Answer:
The movement of molecules of a substance from the region of their higher concentration to the region of their lower concentration is called diffusion.

Question 4.
Define imbibition.
Answer:
Adsorption of water by the colloidal and solid substance without forming solution is called imbibition.

Question 5.
Explain T.P., W.P.
Answer:
The pressure developed on the cell wall by stretching of the cell membrane or plasma membrane of a turgid cell is called turgor pressure. The pressure developed by the cell wall on the cytoplasm in opposite direction against the turgor pressure is called wall pressure.

RBSE Class 12 Biology Chapter 5 Short Answer Questions

Question 1.
Explain the water potential.
Answer:
The difference between the free energy of water molecules in pure water and the free energy of water molecules present in a solution is called water potential.
This is expressed by Greek symbol ψ(Psi).
The unit of water potential is pascal (Pa).

Water potential (ψ) = [Free energy of molecules of pure water] – [Free energy of water molecules of a solution]

According to the modern concept, D.P.D. is expressed as water potential ([latex]{ \Psi }_{ w }[/latex]), osmotic pressure (O.P.) is [latex]{ \Psi }_{ s }[/latex]: expressed as osmotic potential ([latex]{ \Psi }_{ s }[/latex]) and turgor pressure (T.P.) is expressed as pressure potential ([latex]{ \Psi }_{ p }[/latex]).

Accordingly relation between water potential, osmotic potential and pressure potential is expressed as follows:
Water potential = Osmotic potential + Pressure potential
[latex]{ \Psi }_{ w }[/latex] = [latex]{ \Psi }_{ s }[/latex] + [latex]{ \Psi }_{ p }[/latex]

The value of water potential ([latex]{ \Psi }_{ w }[/latex]) and osmotic potential ([latex]{ \Psi }_{ s }[/latex]) is negative whereas the value of pressure potential [latex]{ \Psi }_{ p }[/latex] is positive. The maximum value of water potential zero.

Question 2.
Differentiate between endosmosis and exosmosis.
Answer:
RBSE Solutions for Class 12 Biology Chapter 5 Plant Water Relations 1

Question 3.
Explain the difference between plasmolysis and de-plasmolysis.
Answer:
RBSE Solutions for Class 12 Biology Chapter 5 Plant Water Relations 2

Question 4.
Differentiate between hypertonic and hypotonic solution.
Answer:
RBSE Solutions for Class 12 Biology Chapter 5 Plant Water Relations 3

Question 5.
Explain the diffusion pressure gradient.
Answer:
DPD of a certain cell is the measure of the capacity of the cell to draw water from the surrounding. When the DPD of a cell [A] is higher than the DPD of an adjacent cell [B], Cell (A) will draw water from cell (B) till the DPD of the two cells becomes equal i.e. the state of equilibrium is reached This difference in the value of DPD between two cells results in to a deficiency gradient, which is called as diffusion pressure gradient.

Question 6.
Explain the diffusion pressure deficit.
Answer:
At normal temperature and atmospheric pressure, the difference between the diffusion pressure of a solution and the diffusion pressure of pure water is called DPD. In fact, DPD of a cell is the measure of the force by which the cell can draw water, hence it is also called suction pressure.

Question 7.
Giving example, explain inhibition.
Answer:
Imbibition is a physical process. It is a process in which solid or semisolid organic compounds draw water from their surrounding and swell up but do not dissolve in water. The chemicals which show this property are called inhibitants. These substances are generally hydrophilic in nature. Due to the attraction between the molecules of hydrophilic substances and water molecules, plant cell wall, seeds and wood etc. draw water and swell up. The pressure developed due to imbibition is called imbibition pressure (IP). Imbibition pressure is equal to osmotic pressure. It is an exothermic process.

Examples:

The wooden doors and windows swell during rainy season due to this process.
Seeds swell up due to imbibition of water.
In the epiphytes, the hygroscopic nature of roots draws water by this process.
In many dry fruits and sporangia, dehiscence is by this process.
The imbibition pressure has been used to break large-sized rocks.

RBSE Class 12 Biology Chapter 5 Essay Type Questions

Question 1.
Describe osmotic potential, pressure potential and water potential and explain their mutual relationship.
Answer:

The hypothesis of water potential was proposed by Slatyer and Taylor (1960). According to laws of thermodynamics, each component of any system has some free energy which allows it to work.
During osmosis water molecules move across the semipermeable membrane and for this free energy is required. When a cell is kept in pure water, there exists a difference of free energy between molecules of pure water and the free energy of water molecules present in the solution.
This difference of free energy between the pure water and water molecules of the solution is called water potential.
In other words, the value of water potential of a solution is equal to the difference in the value of free energy of solvent molecules of the solution and free energy of molecules of the pure solvent. It is expressed by Greek symbol ψ(psi) and its unit is Bar or atm.
It is not possible to determine the absolute value of water potential and hence its value is taken as zero. Free energy of pure water (solvent) is maximum and when some solute is added to it, the free energy of solvent (water) decreases.
Hence free energy of any solution is less than that of pure water meaning by always negative.

According to laws of thermodynamics:

DPD of a solution is equal to the water potential but the value of DPD is positive and that of water potential is negative.
OP is called solute potential or osmotic potential and it is expressed as [latex]{ \Psi }_{ s }[/latex]. The value of osmotic pressure and the osmotic potential is equal but the only difference is that the value of osmotic pressure is positive and that of osmotic potential is negative. Its value is also negative.
TP is called pressure potential and it is expressed as [latex]{ \Psi }_{ p }[/latex]. Value of pressure potential is positive.

According to this hypothesis relationship between water potential, osmotic potential and pressure potential is as follows:
[latex]{ \Psi }_{ w }[/latex] = [latex]{ \Psi }_{ s }[/latex] + [latex]{ \Psi }_{ p }[/latex].

Where
[latex]{ \Psi }_{ w }[/latex] = Water potential
[latex]{ \Psi }_{ s }[/latex] = Osmotic Potential
[latex]{ \Psi }_{ p }[/latex] = Pressure potential

here value of
[latex]{ \Psi }_{ w }[/latex] is negative
[latex]{ \Psi }_{ s }[/latex] is negative
[latex]{ \Psi }_{ p }[/latex] is positive
so
[latex]{ \Psi }_{ w }[/latex] = [latex]{ \Psi }_{ s }[/latex] + [latex]{ \Psi }_{ p }[/latex].

Concept of an Osmotic System:

The concentrated sugar solution is filled in a bag made up of the semipermeable membrane.
It is kept in a beaker containing dilute sugar solution.
Water molecules from beaker begin to move to the concentrated solution in the bag through the semipermeable membrane.
After some time, the bag stretches due to endosmosis. When the bag becomes fully stretched it creates pressure on the solution to prevent further entry of water. This pressure is called turgor pressure.
The effect of turgor pressure on water potential is called pressure potential. In a plant cell, the cell wall is referred to as a matrix.
In a fully turgid plant cell, the pressure developed by the cell wall is called matrix potential. It is represented as [latex]{ \Psi }_{ m }[/latex].
Although in the soil the matrix potential is important it is negligible in the cellular osmotic system.

Hence
Water potential [latex]{ \Psi }_{ w }[/latex] = [latex]{ \Psi }_{ m }[/latex] + [latex]{ \Psi }_{ s }[/latex] + [latex]{ \Psi }_{ p }[/latex]
As the value of [latex]{ \Psi }_{ m }[/latex] is negligible in a cellular system,
[latex]{ \Psi }_{ w }[/latex] = [latex]{ \Psi }_{ s }[/latex] + [latex]{ \Psi }_{ p }[/latex]

Since the value of [latex]{ \Psi }_{ s }[/latex] is negative and that of [latex]{ \Psi }_{ p }[/latex] is positive. So in a fully turgid cell when the value of [latex]{ \Psi }_{ p }[/latex] becomes equal to [latex]{ \Psi }_{ s }[/latex], further endosmosis stops and ψw becomes zero.

Question 2.
Defining the process of osmosis and describe an experiment by which it may be demonstrated.
Answer:
Osmosis:
Diffusion of only solvent molecules through the semipermeable membrane is called osmosis. Osmosis can be defined as “movement of solvent molecules from the region of their higher concentration to the region of their lower concentration across a semipermeable or selectively permeable membrane is called osmosis”.

Osmosis is a special type of diffusion where a semipermeable membrane is necessarily present between two systems.
The process can be understood and demonstrated with the help of a thistle funnel experiment.
Parchment paper is properly tied over the broader face of the funnel.
The tube of the funnel is then filled with a sugar solution and its level is made.
The face of the funnel covered with parchment paper is kept in a beaker filled with water.
After some time it is observed that the level of sugar solution in the tube of the funnel rises up.
According to the process of osmosis, water molecules diffuse from the region of their higher concentrate (from beaker) to the region of their lower concentration (sugar solution in the funnel) whereas sugar molecules do not show diffusion because of the semipermeable nature of the parchment paper.

RBSE Solutions for Class 12 Biology Chapter 5 Plant Water Relations 4

Types of Osmosis:
1. Endosmosis:
When a normal plant cell is dipped in a hypotonic solution or pure water, it is observed that water molecules enter the plant cell. This is called endosmosis. Inward movement of water in any system through osmosis is called endosmosis.
Example:

Swelling of raisins dipped in water.
The entry of soil water in root hair cells.

2. Exosmosis:
When a plant cell is dipped in a hypertonic solution (concentrated sugar solution) it is observed that water molecules begin to come out of cells. This is called exosmosis.
Example:

Shrinking of grapes dipped in concentrated sugar solution.
Adverse effect on crop plants of the addition of excessive chemical fertilizer in the soil.

Importance of Osmosis:

Root hair absorbs water from the soil through osmosis. Plant Water Relations.
Absorption of some dissolved minerals also takes place partly through osmosis.
Turgidity of cells and growth of the young cells depend on osmosis.
Transportation of water from one part of the plant to another and cell to cell conduction occurs through osmosis.
Opening and closing of stomata are regulated by osmosis.
Turgid state of the cells is responsible for specific shape and form of leaf, flowers and fruits etc.
This process makes the plant resistant against freezing and desiccation.
It is helpful in the germination of seeds.

Question 3.
Describing briefly the process of osmosis, diffusion and imbibition, explain their importance in plant physiology.
Answer:
Osmosis:
Diffusion of only solvent molecules through the semipermeable membrane is called osmosis. Osmosis can be defined as “movement of solvent molecules from the region of their higher concentration to the region of their lower concentration across a semipermeable or selectively permeable membrane is called osmosis”.

Osmosis is a special type of diffusion where a semipermeable membrane is necessarily present between two systems.
The process can be understood and demonstrated with the help of a thistle funnel experiment.
Parchment paper is properly tied over the broader face of the funnel.
The tube of the funnel is then filled with a sugar solution and its level is made.
The face of the funnel covered with parchment paper is kept in a beaker filled with water.
After some time it is observed that the level of sugar solution in the tube of the funnel rises up.
According to the process of osmosis, water molecules diffuse from the region of their higher concentrate (from beaker) to the region of their lower concentration (sugar solution in the funnel) whereas sugar molecules do not show diffusion because of the semipermeable nature of the parchment paper.

RBSE Solutions for Class 12 Biology Chapter 5 Plant Water Relations 4

Swelling of raisins dipped in water.
The entry of soil water in root hair cells.

2. Exosmosis:
When a plant cell is dipped in a hypertonic solution (concentrated sugar solution) it is observed that water molecules begin to come out of cells. This is called exosmosis.
Example:

Shrinking of grapes dipped in concentrated sugar solution.
Adverse effect on crop plants of the addition of excessive chemical fertilizer in the soil.

Importance of Osmosis:

Root hair absorbs water from the soil through osmosis. Plant Water Relations.
Absorption of some dissolved minerals also takes place partly through osmosis.
Turgidity of cells and growth of the young cells depend on osmosis.
Transportation of water from one part of the plant to another and cell to cell conduction occurs through osmosis.
Opening and closing of stomata are regulated by osmosis.
Turgid state of the cells is responsible for specific shape and form of leaf, flowers and fruits etc.
This process makes a plant resistant against freezing and desiccation.
It is helpful in the germination of seeds.

Diffusion:

The movement of molecules of any substance (solid, liquid and gas) from the region of their higher concentration to the region of their lower concentration is called diffusion.
The property of diffusion is found in the molecules, particles and ions of all solids, liquids and gases, but the rate of their diffusion is different.
The diffusion of molecules or ions takes place continuously in all directions and they tend to spread evenly or uniformly in the available space.
The property of movement of molecules of any substance is due to their kinetic energy.
When a bottle of ammonia is opened in a room it is due to diffusion only that the smell of ammonia spreads in the entire room and when a crystal of copper sulphate is put in a glass of water entire water becomes blue coloured.
In a system of two or more substances, the direction and the rate of diffusion of molecules of the substances does not depend upon each other. For example, when concentrated sugar solution and pure water are brought in contact, sugar molecules move towards the water and water molecules move towards sugаr solution.
Diffusion of molecules of different substances present in a system depends on their concentration and is not influenced by the presence of other substance, this is called independent diffusion.
Example: In plants diffusion of [latex]{ O }_{ 2 }[/latex], [latex]{ CO }_{ 2 }[/latex] and water vapour is not influenced by the presence and concentration of the molecules of one another but is rather influenced by the concentration of the molecules of that particular substance in the atmosphere.

Factors Affecting Diffusion:
(1) Temperature:
Rate of diffusion is directly proportional to temperature, i.e. increase in temperature increases the kinetic energy of molecules, resulting in to increase in the rate of diffusion.

(2) Density:
Rate of diffusion of a substance is inversely proportional to the square root of the density of the diffusing substance i.e. increase in the size of molecules reduces the rate of their diffusion. This means lesser is the size of molecules higher is the rate of their diffusion.
RBSE Solutions for Class 12 Biology Chapter 5 Plant Water Relations 5
(3) Pressure:
Rate of diffusion is directly proportional to the pressure. Movement of molecules of any substance is always from the region of their higher diffusion pressure to the region of their lower diffusion pressure. In other words movement of molecules is from the region of their higher concentration to the region of their lower concentration.

(4) Medium:
Rate of diffusion is slower through a dense medium as compared to a sparse medium.

Importance of Diffusion:

During photosynthesis and respiration exchange of gases ([latex]{ CO }_{ 2 }[/latex] and [latex]{ O }_{ 2 }[/latex]) between plants and atmosphere takes place through diffusion.
During transpiration loss of water vapour from plant to the atmosphere is through diffusion.
Passive absorption of minerals to some extent occurs through diffusion.
Diffusion is also helpful in translocation of food material (from leaves to roots) in the plant body.
Osmosis and inhibition also involve diffusion.
Plant hormones are distributed in the plant body by diffusion.

Inhibition:

Absorption of water molecules by solid or colloidal substances without forming solution is called Inhibition. In fact, this kind of absorption is called adsorption.
In Inhibition colloidal substances and hydrophilic solid substances draw water. The substances which inhibit water are called Inhibition.
Inhibitants are hydrophilic colloids and can inhibit a large amount of water by surface attraction.
The primary and secondary wall layers of plant cells are made up of cellulose, pectin and lignin etc. These are hydrophilic in nature and can absorb a large amount of water.
The rate of inhibition is affected by temperature, nature of inhibitant, and a difference in water potential.
Inhibition can occur only when there is an affinity between the inhibitant and water.
A considerable force may develop due to inhibition within the plant body. This is called inhibition pressure.
In inhibition water always moves with some force from saturated region to drier region.

Importance of Inhibition:

In germinating seeds, seed coat burst due to inhibition pressure.
It also plays an important role (together with osmosis) in the intake of soil water by root-hairs.
It is believed to be an important force involved in the ascent of sap. In many plants, resurrection is on account of the presence of hydrophilic colloids.

RBSE Solutions for Class 12 Biology

RBSE Solutions for Class 12 Chemistry Chapter 11 Organic Compounds with Functional Group Containing Oxygen (Part-1)

May 31, 2019 by Prasanna Leave a Comment

Rajasthan Board RBSE Class 12 Chemistry Chapter 11 Organic Compounds with Functional Group-Containing Oxygen (Part-1)

RBSE Class 12 Chemistry Chapter 11 Text Book Questions

RBSE Class 12 Chemistry Chapter 11 Multiple Choice Questions

Question 1.
Which of the following is not obtained in any condition by the reaction of ethanol and conc.H₂SO₄?
(a) CH₃CHO
(b) CH₃CH₂HSO₄
(c) C₂H₄
(d) CH₃CH₂OCH₂CH₃

Question 2.
General formula of alcohol is
(a) C_nH_2n+2O
(b) C_nH_2n+1O
(c) C_n+1H_2nO
(d) C_n+2H_nO

Question 3.
R-MgX + HCHO [latex]\xrightarrow [ Ether ]{ Dry } [/latex] [P], here [P] is

Question 4.
The final product obtained by the reaction between alcohol and phosphorus pentachloride is
(a) Chloro alkene
(b) Dichloro alkene
(c) Chloroalkane
(d) Dichloro alkane

Question 5.
Which of the following phenol is strongest?
(a) o – nitrophenol
(b) m – nitrophenol
(c) p – nitrophenol
(d) p – chlorophenol

Question 6.
Victor Mayer test is not given by
(a) C₂H₅OH
(b) (CH₃)₃COH
(c) (CH₃)₂CHOH
(d) CH₃CH₂CH₂OH

Question 7.
Which one of the following is the strongest acid?
(a) Phenol
(b) m – chlorophenol
(c) Benzyl alcohol
(d) Cyclohexanol

Question 8.
The product formed when salicylic acidic heated with soda lime is
(a) Methyl Alcohol
(b) Ether
(c) Ethyl Alcohol
(d) phenol

Question 9
Phenol + Chloroform + Base → Main product. Main product is
(a) Salicylaldehyde
(b) Formaldehyde
(c) ketone
(d) Acetaldehyde

Question 10.
The product formed when ether is passed on alumina at 653 K is
(a) Alkene
(b) Alkane
(c) Alcohol
(d) Phenol

Answers:
1. (a)
2. (b)
3. (a)
4. (c)
5. (c)
6. (b)
7. (b)
8. (d)
9. (a)
10. (a)

RBSE Class 12 Chemistry Chapter 11 Very Short Answer Type Questions

Question 1.
Write the general formula of alcohol.
Answer:
C_nH_2n+1OH or R – OH.

Question 2.
Write the IUPAC name of ethyl alcohol.
Answer:
Ethanol.

Question 3.
Write the name of the product formed when Grignard’s reagent reacts with formaldehyde.
Answer:
When Grignard’s reagent reacts with formaldehyde primary alcohol forms.

Question 4.
Write the order of acidity for primary, secondary and tertiary alcohols.
Answer:
Primary (1°) > Secondary (2°) > Tertiary (3°).

Question 5.
Write Fries rearrangement.
Answer:
Fries Rearrangement:
When phenol is treated with acetyl chloride, phenyl esters are formed. When phenyl ester is heated in nitrobenzene in the presence of anhydrous AlCl₃ it undergoes rearrangement in which acyl group [latex s=1]\overset { { – }C{ -{ R } } }{ \underset { O }{ \parallel } } [/latex] migrates from the phenolic oxygen to an ortho and para position. This reaction is called Fries Rearrangement.

Question 6.
What happens when phenol is kept open in the air?
Answer:
Phenol gets slowly oxidised to a pink coloured compound p – benzoquinone when exposed to air.

Question 7.
What is the effect of electron withdrawing groups on the acidity of phenol?
Answer:
When an electron withdrawing group like the nitro group is present on the benzene ring, the acidic strength of phenol increases. When these groups are present at ortho and para positions then their effect is more because delocalisation of negative charge is more effective in peroxide ion.

Question 8.
Write the general formula of ether.
Answer:
R – O – R’ where R = R’ in simple ethers while R ≠ R’ in mixed ethers.

RBSE Class 12 Chemistry Chapter 11 Short Answer Type Questions

Question 1.
Explain Hydroboration-Oxidation reaction.
Answer:
Hydroboration-Oxidation Reaction:
Indirect hydration of alkenes can also be done by hydroboration-oxidation which is completed in two steps. In the first step, alkenes react with diborane (B₂H₆) as boron hydride (BH₃) to form an alkyl borane. In fact, the boron atom along with the hydrogen atom gets attached to the double bonded carbon atom with more number of hydrogen atoms. One hydrogen atom is then transferred to the other carbon atom. In this manner, all the three hydrogen atoms of boron are transferred to alkene molecule to form trialkyl borane as the product. In the next step, the alkyl borane is oxidized by alkaline H₂O₂ to form an alcohol. The indirect hydration process according to Antimarkownikov’s rule.
For example:

Question 2.
Write a method for preparation of primary alcohol from Grignard’s reagent.
Answer:
(i) The Grignard reagent reacts with formaldehyde to form primary alcohol.

(ii) When Grignard reagent reacts with oxirane an adduct forms which on hydrolysis gives primary alcohol.

Question 3.
Alcohols are soluble in water whereas diethyl ether is not, explain with reason.
Answer:
The solubility of alcohol in water is due to excessive intermolecular hydrogen bonding in their molecules.

The extent of hydrogen bonding in diethyl ether is less as compared to alcohols. Therefore diethyl ether is soluble in water only to a small extent.

Question 4.
Explain two commercial methods for the preparation of phenol.
Answer:
(i) From Cumene:
Phenol is prepared commercially from cumene. Cumene is isopropyl benzene formed by the alkylation of benzene with propane in the presence of H₃PO₄ (Friedel Craft reaction) at 523 K).

Oxygen is bubbled through the above solution to form cumene hydroperoxide which is decomposed with aqueous acid to form phenol and acetone as follows:

(ii) From Coaltar:
Phenol can also be prepared from the middle oil fraction (440 to 503 K) of coal tar distillation which mainly consists of phenol and naphthalene. Upon cooling the fraction, crystals of naphthalene separate out. The remaining oily solution is washed with dilute H2S04 to dissolve the basic impurities. From the solution, left phenol is extracted with aqueous NaOH to form soluble sodium phenate. CO₂ is then blown through the solution to liberating phenol.

Question 5.
Phenols show less acidity in comparison to carboxylic acids. Explain with reason.
Answer:
Carboxylic acids are stronger acids than phenols. It can be understood by comparing the hybrid structures of carboxylate ion and phenoxide ion. In carboxylate ion, the negative charge is equally distributed over two negatively charged oxygen atoms while in phenoxide ion it is present only on one oxygen atom. Thus carboxylate ion is more stabilised as compared to phenoxide ion. Hence carboxylic acids ionize to the greater extent than phenols furnishing higher concentration of H⁺ions. Therefore carboxylic acids behave as stronger acids than phenols.

Question 6.
Write the following reactions:
1. Gattermann Reaction.
2. Reimer-Tiemann reaction
3. Duff reaction
Answer:
1. Gattermann reaction:
When phenol reacts with the mixture of HCN and HCl in the presence of catalyst ZnCl₂, aldimine formed as intermediate, on hydrolysis aldimine forms p-hydroxybenzaldehyde.

This reaction is called Gattermann reaction.

2. Reimer-Tiemann Reaction:
When phenol reacts with chloroform in the presence of bases (NaOH/KOH), an aldehyde (-CHO) group gets introduced in the ring at a position ortho to the phenolic group and salicylaldehyde is formed.

This reaction is called the Reimer-Tiemann reaction.

3. Duff Reaction:
When phenol is heated with hexamethylene tetramine[(CH₂)₆N₄] and boric acid (H₃BO₃) in the presence of glycerol salicylaldehyde is formed. This reaction is called the Duff reaction.

Question 7.
Explain the halogenation reactions of diethyl ether.
Answer:
Halogenation reactions of diethyl ether:
When diethyl ether is heated with chlorine or bromine then hydrogen atoms present on a-carbon of ether get substituted by a halogen atom.
(i) Halogenation in dark:

(ii) Halogenation in the presence of sunlight:
Diethyl ether reacts with chlorine in the presence of sunlight to form a per-chloro-diethyl ether.

RBSE Class 12 Chemistry Chapter 11 Long Answer Type Questions

Question 1.
What does alcohol form by reaction with the following:
1. PCl₃2. SOCl₂Answer:
1. Reaction with PCl₃:
Alcohols react with PCl₃ to form alkyl halides.

For example:

2. Reaction with SOCl₂:
Alcohols react with thionyl chloride in the presence of pyridine to form alkyl chlorides.

For example:

Question 2.
What does phenol form by reaction with the following:
1. HCN and HCl
2. In the presence of NaOH or KOH

Answer:
1. When phenol reacts with the mixture of HCN and HCl in the presence of ZnCl₂ catalyst aldimine formed as intermediate. On hydrolysis, aldimine forms p-hydroxybenzaldehyde. This reaction is known as Gattermann reaction.

2. When phenol reacts with chloroform in the presence of bases (NaOH/ KOH) an aldehyde group (-CHO) gets introduced in the ring at a position ortho to the phenol group and salicylaldehyde is formed. This reaction is called the Reimer-Tiemann reaction.

Question 3.
Write the substitution reactions of diethyl ether.
Answer:
Substitution reactions of diethyl ether. When diethyl ether is heated with chlorine or bromine then hydrogen atoms present or a – carbon of diethyl ether gets substituted by halogen atoms.
(i) Halogenation in dark:

(ii) Halogenation in presence of sunlight:
Diethyl ether reacts with chlorine in the presence of sunlight to form a per-chloro-diethyl ether.

RBSE Solutions for Class 12 Chemistry

RBSE Solutions for Class 12 Chemistry Chapter 10 Halogen Derivatives

May 6, 2019 by Prasanna Leave a Comment

RBSE Solutions for Class 12 Chemistry Chapter 10 Halogen Derivatives are part of RBSE Solutions for Class 12 Chemistry. Here we have given Rajasthan Board RBSE Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives.

Rajasthan Board RBSE Class 12 Chemistry Chapter 10 Halogen Derivatives

RBSE Class 12 Chemistry Chapter 10 Text Book Questions

RBSE Class 12 Chemistry Chapter 10 Multiple Choice Questions

Question 1.
Out of the following compounds which compound will give Haloform reaction?
(a) Methanol
(b) Ethanol
(c) 1 – Propanol
(d) Butanol

Question 2.
What happens in Finkelstein reactions?
(a) Dehydrohalogenation
(b) Hydrogenation
(c) Halogen exchange
(d) Oxidation

Question 3.
Example of haloarene is:
(a) CH₃Cl
(b) C₆H – CH₂Cl
(c) C₆H₆Cl₆
(d) C₆H₅Cl

Question 4.
Which compound will form a yellow precipitate with AgNO₃?
(a) CHI₃
(b) CH₃l
(c) CHCl₃
(d) CH₃ – CH₂I

Question 5.
The intermediate formed in carbonyl amine reaction is:
(a) CN^–(b) N = C^–(c) CCl₂(d) Cl^–

Question 6.
What forms in SN²mechanism?
(a) Transition state
(b) Carbonium ion
(c) Carbanion
(d) Free radical

Question 7.
Out of the following which compound has zero dipole moment?
(a) CH₃Cl
(b) CHCl₃
(c) CCl₄(d) CHI₃

Answers:
1. (b)
2. (c)
3. (d)
4. (a)
5. (c)
6. (a)
7. (c)

RBSE Class 12 Chemistry Chapter 10 Very Short Answer Type Questions

Question 1.
Write full form or name of DDT and BHC.
Answer:
DDT = Dichloro diphenyl trichloroethane
BHC = Benzene hexachloride

Question 2.
Write name and formula of anyone tertiary alkyl halide.
Answer:

Question 3.
Write the name and formula of anyone alcohol and one ketone which give haloform reaction:
Answer:

Question 4.
Which reagent is used for the formation of methanol from methyl chloride?
Answer:
Methanol can be prepared by hydrolysis of methyl chloride with aqueous KOH solution.

Question 5.

Write the IUPAC name of
Answer:

Question 6.
Give any three examples of electrophiles and nucleophiles?
Answer:

Question 7.
Which compound is used in a fire extinguisher?
Answer:
Carbon tetrachloride (CCl₄) or tetrachloromethane is used in the fire extinguisher.

Question 8.
Write the formula of DDT and BHC.
Answer:

Question 9.
Write the possible dichloro derivatives of propane.
Answer:

Question 10.
Write Hunsdiecker reaction.
Answer:
Hunsdiecker Reaction – Silver salt of fatty acid can be converted into bromo alkane by refluxing with bromine in the presence of carbon tetrachloride. This reaction is known as Borodine Hunsdiecker reaction or simply Hunsdiecker reaction.

Question 11.
Write formula and uses of chloropicrin and chlorine.
Answer:
(i)
Uses:
Chloropicrin was manufactured for use as a poison gas in world war. In agriculture, chloropicrin is injected into the soil prior to planting a crop in order to fumigate soil.
(ii)
Uses:
It acts as a hypnotic and nervous sedative.

Question 12.
Which is the best reagent to obtain pure chloroform?
Answer:
Chloral hydrate is the best reagent to obtain pure chloroform.

Question 13.
Which gas is formed when chloroform is exposed to air?
Answer:
Poisonous gas phosgene is formed.

Question 14.
Which one is more reactive from methyl chloride and methyl iodide?
Answer:
CH₃ – I will be more reactive than CH₃ – Cl because I^–ion is a better leaving group than Cl^–ion. Alternatively, bond dissociation enthalpy of C – I bond is less than that of the C – Cl bond.

Question 15.
Write the structure of C₅H₁₂which forms only one mono chloro derivative.
Answer:
It must be of symmetrical nature and is neopentane (CH₃)₄C.

Question 16.
What is the use of DDT?
Answer:
It is used as an insecticide for killing or controlling mosquitoes, lice etc.

Question 17.
Write two examples of 2° alkyl halide.
Answer:

Question 18.
Arrange the following in order of reactivity towards SN1 reaction.
Answer:

RBSE Class 12 Chemistry Chapter 10 Short Answer Type Questions

Question 1.
C₆H₅Cl is less reactive towards nucleophilic substitution reactions than C₂H₅Cl. Explain.
Answer:
In chloroethane, the C-atom of C – Cl bond is sp³ hybridised while in C₆H₅Cl, the carbon atom is in sp²hybridisation state. The sp²hybridised C – an atom with a greater s-character is more electronegative and can hold the electron pair of the C – Cl bond more tightly than sp³ – hybridised C-atom in chloroethane with less s-character. Thus the bond cleavage in chlorobenzene is more difficult and is, therefore, less reactive as compared to chloroethane.

Question 2.
How ethyl bromide is prepared from Grignard reagent?
Answer:

Question 3.
Write the chemical reaction for the formation of BHC.
Answer:

Question 4.
How will you prepare the following compounds from chlorobenzene:
(a) Phenol
(b) Diphenyl
(c) Toluene
Answer:
(a) Phenol from chlorobenzene

(b) Diphenyl from chlorobenzene

(c) Toluene from chlorobenzene

Question 5.
Explain β -elimination.
Answer:
When haloalkane or alkyl halide with a β – hydrogen atom is heated with alcoholic solution of potassium hydroxide (alc. KOH), the OH^–ion acts as a base and eliminate an electron from β – carbon atom then this elimination is known as β – elimination. There is also a simultaneous elimination of halogen atom from a – carbon. As a result, an alkene is formed.

For Example:

This reaction is also called dehydrohalogenation.

Question 6.
Write the mechanism of Carbyl amine reaction.
Answer:
Carbylamine Reaction or Isocyanide Reaction – When chloroform is heated with a primary amine and alcoholic KOH then isocyanide is formed with an extremely unpleasent smell which is also known as a carbyl amine. This reaction is used as a test for primary amine and chloroform, therefore, it is also known as isocyanide test.

Mechanism:

Question 7.
How will you prepare the following from chloroform:
(a) Acetylene
(b) CCl₄
(c) Salicylaldehyde
Answer:
(a) Acetylene from Chloroform:
On warming with Ag powder, chloroform is converted into acetylene.

(b) CCl₄ from Chloroform:
Chloroform on chlorination gives carbon tetrachloride.

(c) Salicylaldehyde from chloroform:
Chloroform and phenol are heated with alcoholic KOH to give salicylaldehyde.

Question 8.
Write four uses of carbon tetrachloride.
Answer:
(i) It is used as a fire extinguisher.
(ii) It is used in dry cleaning.
(iii) It is used for the manufacture of freon and salicylic acid.
(iv) It is used as a solvent.

Question 9.
How will you prepare the following from aniline?
(a) Chlorobenzene
(b) Bromobenzene
(c) Iodobenzene
Answer:

Question 10.
Write the formula of following:
(a) Freon – 11
(b) Freon – 12
(c) Freon – 111
Answer:
(a) Freon – 11 = CFCl₃(b) Freon – 12 = CF₂Cl₂
(c) Freon – 111 = C₂FCl₅
Question 11.
What happens when:
(i) Does ethyl bromide react with silver cyanide?
(ii) Iodoform is heated with silver power?
Answer:

Question 12.
Benzyl chloride is more reactive than chlorobenzene. Why?
Answer:
Benzyl chloride is more reactive than chlorobenzene towards SN¹reaction because it readily ionises to give benzyl carbonation which is resonance stabilized.

In chlorobenzene, ionisation leading to carbonation is not possible. Therefore it is less reactive.

RBSE Class 12 Chemistry Chapter 10 Long Answer Type Questions

Question 1.
Explain the following:
(a) Classification of halogen derivatives
(b) Nature of C-X bond in halogen derivatives.
(c) Directive influence of halogen atom in haloarene.
Answer:
(a) Classification of Halogen Derivatives:
In the IUPAC system of nomenclature, halogen derivatives are considered as halo-substituted derivatives of the respective alkane. The number of carbon atoms is used to consider the position of the halogen atom to which it is attached. The carbon atom carrying the halogen atom gets the lowest number. If two different halogen atoms present in halogen derivative then preference is given in both numbering and writing name to that halogen atom which comes first in the alphabetical order.
For example 2-Bromo- 3- chloro butane.

(i) These may be classified as mono, di, tri, tetra etc compounds depending on the number of halogen atoms present in haloalkane. For example,

Formula	CH₃ – Cl	CH₂Cl₂	CHCl₃	CCl₄
Common Names	Methyl chloride	Methylene chloride	Chloroform	Carbon tetrachloride
IUPAC Name	Chloromethane	Dichloromethane	Trichloromethane	Tetrachloromethane
Series	Mono haloalkane	Dihalo alkane	Trihalo alkane	Tetra haloalkane

(ii) Monohaloalkanes are classified into three types on the basis of nature of carbon atom to which halogen is attached. ‘
Primary Alkyl Halides:
In these compounds, the halogen atom is attached with a primary carbon atom.
For example:

Formula	CH₃ – Cl	CH₃ – CH₂ – Cl	CH₃ – CH₂– CH₂ – Cl
Common Name	Methyl chloride	Ethyl chloride	Propyl Chloride
IUPAC Name	Chloromethane	Chloroethane	1 – Chloropropane

Secondary Alkyl Halides:
In these halogen atoms is attached with a secondary carbon atom.
For example:

Tertiary Alkyl Halide:
In these compounds, the halogen atom is attached with the tertiary carbon atom.

(iii) Allylic Halides:
In these compounds, the halogen atom is bonded to an sp³hybridised carbon atom next to the C = C bond. The sp³ hybridised carbon is known as allylic carbon.
For example:

(iv) Benzylic Halides:
In these compounds, the halogen atom is bonded to an sp³ hybridised carbon atom next to an aromatic ring. This carbon is known as benzylic carbon.

(v) Vinylic Halides:
In these compounds, the halogen atom is bonded to the sp² – hybridised C – atom of a carbon-carbon double bond.

(vi) Aryl Halides or Haloarenes:
In these compounds, halogen is bonded to the sp² – hybridised carbon atom of an aromatic ring.

(b) Nature of C – X bond in halogen derivatives:
In haloalkanes, the formation of C – X bond takes place by the overlapping of sp² – hybridised orbital of carbon and 3p (half filled orbital of chlorine (sp³– overlapping). This overlapping can be shown as:

Although the C-X bond is the covalent bond value of electronegativity of C – atom is 2.6 which is less than electronegativity of the halogen atom. So due to the difference in electronegativities, the nature of C – X bond is polar. Therefore the halogen atom bears a partial negative charge whereas the carbon atom bears a partial positive charge.

CH3 – Cl CHg – Br CH3 – I
H = 1.86D 1.79 D 1.62 D

(c) Directive Influence of Halogen atom in Haloarenes:
In haloarenes, the halogen atom present in the ring is ortho and para directing in nature and therefore in a substitution reaction, a mixture of isomeric products is formed, in which the para – isomer is the major product since the para position is less hindered in the molecule of haloarene compared to ortho position.
In haloarenes, the halogen atom has also a strong –I (inductive) effect which is likely to deactivate the ring and the electrophilic attack will become rather difficult. However, +M effect dominates over the –I effect when the electrophile is directed to ortho and para position in the ring. Therefore, the electrophilic substitution takes place at these positions though with difficulty. The electrophilic attack is not feasible at the meta position in the ring because the +M effect is rather weak and –I effect of the halogen atom is expressed to have greater magnitude.

Question 2.
How will you prepare:
(a) Alkyl halide from alcohol?
(b) Alkyl halide from halogen exchange?
(c) Chloroform from the acetone?
(d) Salicylic acid from carbon tetrachloride.
Answer:
(a) Alkyl halide from alcohol:

(i) By the reaction of alcohol and halogen acid.
R – OH + HCl [latex]\underrightarrow { ZnCL_{ 2 } }[/latex] R – Cl + HCl
R – OH + HBr [latex]\underrightarrow { { H }_{ 2 }S{ O }_{ 4 } }[/latex] R – Br + H₂O

(ii) By the action of thionyl chloride on alcohol.
R – OH + SOCl₂ [latex]\underrightarrow { { C }_{ 5 }{ H }_{ 5 }N }[/latex] R – Cl + SO₂ + HCl

(iii) By the action of phosphorus halides on alcohol.
R – OH +PCl₅ [latex]\underrightarrow { \triangle } [/latex] R – Cl +POCl₃ + HCl
3R – OH +PCl₃ [latex]\underrightarrow { \triangle } [/latex] 3R – Cl + H₃PO₃
(b) Alkyl halide from halogen exchange:
In this reaction, alkyl halides are prepared from alkyl halides only.
For Example:
R – Cl + Nal [latex]\xrightarrow [ Acetone ]{ \triangle } [/latex] R – I + NaCl
R – Br + Nal [latex]\xrightarrow [ Acetone ]{ \triangle } [/latex] R – I + NaBr
This reaction is known as the Finkelstein reaction. The synthesis of alkyl fluoride is best accomplished by heating alkyl iodide in the presence of a metallic fluoride such as AgF. This reaction is called “Swarts reaction”.
R – I + AgF [latex]\rightarrow [/latex] R -F + Agl

(c) Chloroform from Acetone:

(d) Salicylic acid from carbon tetrachloride:

Question 3.
Write short notes on the following:
(a) Haloform Reaction
(b) Carbylamine Reaction
(c) Darzen’s Reaction
(d)Sandmeyer Reaction.
Answer:
(a) Haloform Reaction:
When a methyl ketone (even acetaldehyde) is reacted with halogen in aqueous sodium hydroxide, the ketone gets oxidised to the sodium salt of acid with one carbon less than ketone and at the same time haloform (CHX3) also gets formed.

When the reaction is carried with iodine, a yellow precipitate of iodoform is obtained.

The iodoform test is quite useful to make a distinction between certain pairs of compounds one of which respond to the test while the other does not. For example:
(i) Methanal and ethanal (Test is given by ethanol)
(ii) Ethanal and propanal (Test is given by ethanol)
(iii) Pentan-2-one and pentane-3-one (Test is given by pentane – 2 – one)

(b) Carbylamine Reaction:
Chloroform is heated with primary aliphatic or aromatic amine and alcoholic KOH then it forms isocyanide or carbylamine with an extremely unpleasent smell. The reaction is not given by secondary and tertiary amines and is, therefore, a test for the primary amines.

(c) Darzen’s Reaction:
It involves the reaction of straight chain primary alcohols with thionyl chloride to form chloro derivatives without rearrangement.
R – OH + SOCl₂[latex]\underrightarrow { Pyridine } [/latex] RCl + SO₂+ HCl

(d) Sandmeyer Reaction:
In this reaction, benzene diazonium chloride is reacted with CuCl or CuBr in the presence of corresponding halogen acid to form haloarene.

Question 4.
Explain the mechanism of S_N¹ and S_N²Answer:
Unimolecular Nucleophilic Substitution or S_N¹(i) These reactions proceed in two steps:
(ii) In step I:
Heterolytic cleavage of C-X forms a carbocation (carbonium ion) (R^⊕) in the form of intermediate and a halide ion X^–. Step I is the slowest step, therefore, it is rate determining step.
R-X [latex]\xrightarrow [ slow ]{ step } [/latex] R^⊕+ X^–(iii) In the first step, only alkyl halide is used therefore the rate of reaction depends only upon the concentration of alkyl halide.
Rate α [ R-X ]
Therefore, this reaction is known as unimolecular nucleophilic substitution reaction (S_N¹).
(iv) In the second step, carbocation is attacked by nucleophile to form a product. Rate of the second step is much more than the first step.

(v) Like the stability of carbocation formed in step I increases, the reaction will easily proceed. Order of stability of carbocation is as follows:

Order of reactivity of S_N¹ mechanism if halogen atom is the same.

The intermediate carbocation is planar, therefore attack of the nucleophile may be accomplished from either side resulting in a mixture product. Therefore if the reactant is optical isomer then the product is a racemic mixture. Mechanism of S_N¹ reaction between tertiary butyl chloride and aqueous KOH can be explained in the following ways:

The overall reaction can be represented as follows:

(b) Bimolecular Nucleophilic Substitution or S_N²
(i) This type of reaction occurs in one step.
(ii) In this type of reaction the only formation of transition state occurs and no intermediate is formed.
(iii) For the formation of the transition state, the attacking nucleophile attacks from the opposite directions at from 180° angle of leaving the group (X^®). This is known as backside attack or rear attack.
(iv) In the transition, state attacking nucleophile (Nu^–) and leaving the group (X^–) both are partially attacked with the central carbon atom.
(v) The rate of reaction depends upon the concentration of both the alkyl halide and the nucleophile. Therefore, these reactions are known as bimolecular nucleophilic substitution reaction (S_N²).
Rate α [ R-X ] [ Nu^– ]
(vi) In these reactions, the configuration of products is opposite to that reactants, therefore in these reactions inversion of configuration occurs. This is’ known as “Walden inversion”.
(vii) In these reactions, the presence of bulky alkyl group on the carbon of C-X bond produces steric hindrance for the attack of the nucleophile. Therefore with the increase in the number of alkyl groups on carbon, the reactivity of R-X decreases. If the halogen atom is same then the order of reactivity of alkyl halides towards S_N² mechanism is as:

Therefore, primary alkyl halides react more rapidly by mechanism whereas tertiary alkyl halide reacts by the S_N² mechanism. Secondary alkyl halides will react with S_N¹and S_N² mechanism. This depends upon the nature of nucleophile and solvent.

Question 5.
Write a short note on the following?
(i) Freons
(ii) DDT
(iii) BHC
Answer:
(i) Freons:
Chlorofluoro derivatives of methane and ethane are called freons. They can be prepared by the reaction of HF with carbon tetrachloride (CCl₄) or hexachloroethane in the presence of SbCl₅.
For example:

Properties of Freons:
Freons odourless, volatile liquids. They are highly inert and highly stable even at high temperature and pressure.
Uses of Freons
(i) They are used as inert solvents.
(ii) They are used as refrigerants in the refrigerator, air conditioners.
(iii) They are used in the form of propellant in aerosols.

(ii) DDT (Dichlorodiphenyltrichloroethane)
It is synthesized by treating a mixture of chloral with chlorobenzene in the presence of concentrated H₂S0₄.

DDT is a white crystalline solid. It is used as an insecticide for killing or controlling mosquitoes, lice etc. DDT is a harmful poisonous substance and does not easily decompose.

(iii) B.H.C. (Benzene Hexachloride)
It has many common names like gammexane, lindane etc. Its IUPAC name is 1,2,3,4,5,6 – hexachlorocyclohexane. It is prepared by chlorination of benzene in the presence of UV radiations.

It is a mixture of many isomers (α, β, γ, δ, θ, η, ε, σ). BHC is used in an agricultural field in the form of insecticide. Insecticidal activity is highest in gamma-isomer (γ – BHQ), In comparison to other isomers, γ -isomer is comparatively small so its penetrating power is more.

Question 6.
Explain the electrophilic and nucleophilic reaction of chlorobenzene.
Answer:
(i) Electrophilic substitution reactions of chlorobenzene:
In chlorobenzene, due to – I effect of a chlorine atom, electron density decreases on benzene ring and attack of electrophile occurs slowly. Due to this reason, chlorobenzene is less reactive towards electrophilic substitution reactions in comparison to benzene. Some electrophilic substitution reactions of chlorobenzene are as follows:

(ii) Nucleophilic Substitution Reaction of Chlorobenzene:
Due to resonance, there is a partial double bond character in the C-X bond of chlorobenzene, as a result, the bond cleavage becomes difficult. Due to this reason, chlorobenzene is less reactive towards nucleophilic substitution reaction than alkyl halides. However, nucleophilic substitution in chlorobenzene occurs under drastic conditions like high-temperature. Some nucleophilic substitution reactions of chlorobenzene are as follows:

Question 7.
How will you prepare the following from alkyl halide:
(i) Alkyl isocyanide
(ii) Alkyl cyanide
(iii) Nitroalkane
(iv) Alkyl nitrile
(v) Isopropylbenzene
(vi) Tetramethylammonium chloride
Answer:
(i) When alkyl halide is treated with AgCN, alkyl isocyanide is obtained.

For example:

(ii) Alkyl halides react with an alcoholic solution of KCN to give alkyl cyanides.

For example:

(iii) When alkyl halide is treated with silver nitrate (AgNO₂), nitroalkanes are obtained.

For example:

(iv) When an alkyl halide is treated with sodium or potassium nitrite, alkyl nitrite is obtained.

For example:

(v) When secondary alkyl halide (isopropyl chloride) is treated with benzene in the presence of anhydrous AlCl₃isopropylbenzene is obtained.

(vii) When alkyl halide (e.g. chloromethane) is heated with an alcoholic solution of ammonia in a sealed tube to about 373 K, a mixture of products containing primary, secondary, tertiary amines and quaternary ammonium salt is formed. This reaction is called Hoffmann ammonolysis reaction.