RBSE Solutions for Class 12 Chemistry

RBSE Solutions for Class 12 Chemistry Chapter 17 Chemistry in Daily Life

June 11, 2019 by Prasanna Leave a Comment

Rajasthan Board RBSE Class 12 Chemistry Chapter 17 Chemistry in Daily Life

RBSE Class 12 Chemistry Chapter 17 Text Book Questions

RBSE Class 12 Chemistry Chapter 17 Very Short Answer Type Questions

Question 1.
What is saponification?
Answer:
Soaps are sodium or potassium salts of higher fatty acids (containing 15 – 18 carbon atoms) such as stearic acid and palmitic acid. The process of formation of soap is known as saponification. The process of saponification can be shown as

(Here R may be – C₁₅H₃₁,- C₁₇H₃₃ or – C₁₇H₃₅)

Question 2.
What are soft and hard soaps?
Answer:
Soft soaps: Soft soaps are potassium salts of fatty acids.
Hard soaps: Hard soaps are sodium salts of fatty acids.

Question 3.
What are detergents?
Answer:
The difficulty encountered with soaps with regard to cleaning with hard water is resolved by taking detergents. Detergents are chemicals that have all the properties of soaps but do not contain soap.
Example:

Question 4.
What are biodegradable and non-biodegradable detergents?
Answer:
Biodegradable detergents:
Detergents which can be decomposed by microorganisms present in water are called biodegradable detergents.lt is found that detergents that have straight chains of hydrocarbons in the molecules are biodegradable. For example, sodium lauryl sulphate.

Non-biodegradable detergents:
Detergents that are not degraded (decomposed) by the micro-organisms are called non-biodegradable detergents. An example of such a detergent is sodium 4 – (1, 3, 5, 7 tetramethyl octyl) benzenesulfonate. Such detergent causes environment pollution.

Question 5.
Give an example of cationic detergent.
Answer:

Question 6.
What are chromophores? Give an example.
Answer:
Chromophores are organic structures which give rise to colour.
Example:

Question 7.
What are auxochromes? Give an example.
Answer:
It is a group which itself does not act as chromophore but when attached to the chromophore, it shifts the adsorption towards longer wavelength along with an increase in the intensity of adsorption.
Examples:
– OH, – NH₂ , – NHR, – NR₂ are auxochromes.

Question 8.
What are mordant dyes? Give an Example.
Answer:
These are applied with the help of binding material (e.g. metal hydroxide) called mordant depending upon the metal ion used, the same dye can give different colours. Alizarin is an important example of such a dye.
Example:

Question 9.
What are triphenylmethane dyes? Give an example.
Answer:
These dyes are amino derivatives of triphenyl methane. For example, malachite green is an important dye of this group. It is used to colour wool and silk directly.

Question 10.
What are vat dyes? Give examples.
Answer:
These dyes are known from ancient times. Vat dyes are insoluble in water cannot dye fibres directly.

RBSE Class 12 Chemistry Chapter 17 Short Answer Type Questions

Question 1.
What are soaps? Give an example?
Answer:
The sodium and potassium salts of higher fatty acids such as stearic acid, oleic acid are called soaps.
Example:
Sodium palmitate (C₁₅H₃₁COONa).

Question 2.
Write the differences between soaps and detergents?
Answer:
Differences between soaps and detergents:

Soaps	Detergents
Soaps are sodium salts of long chain fatty acids.	Detergents are sodium salts of long chain alkyl sulphonates or long chain alkyl benzene sulphonates.
They cannot be used in acidic solutions.	They can be used even in acidic solutions.
They cannot be used in hard water as they produce precipitates with Ca⁺ and Mg²⁺ ions.	They can be used even in hard water as they do not form any precipitates with Ca²⁺ or Mg²⁺ ions.

Question 3.
Explain the cleansing action of soaps and detergents with the help of micelle formation.
Answer:
Soaps and detergents consist of a large hydrocarbon tail with negatively charged head. The hydrocarbon tail is hydrophobic (water repelling) and the negatively charged head is hydrophilic (water-attracting).
The dirt in the cloth is due to the presence of dust particles in fat or grease which stick to the cloth. When soap or detergent is dissolved in water the molecules gather together as clusters called micelles when the dirty cloth is dipped in soap or detergent solution. The soap and dirt particles come in contact with each other.
The non-polar tails of the soap begin to dissolve in non – polar oil or grease. While the polar head part remains in the water. As more and more soap particles enter the grease each fat or oil particle is surrounded by a number of negatively charged ends. Since the similar charges repel each other the oil or grease droplets break off into small globules of oil. These are still surrounded by the negatively charged polar heads of the soap molecules.
This prevents the small globules from coming together to form bigger particles. The rubbing by hands or mechanical stirring also helps to break the grease particles. In this manner, the grease particles can be completely broken up and it forms an emulsion of grease or oil contained in dirt and the droplets are washed away with water.

Question 4.
What are soapless soaps? Explain with example.
Answer:
Synthetic detergents have all the soap’s properties but actually does not contain any soap. So they are known as “soapless soaps”. They are also known as detergents. Detergents can be used in both soft and hard water.
Example:

Question 5.
Explain cationic, anionic and non-ionic detergents with example.
Answer:
Cationic detergents:
Cationic detergents are those which have a cationic hydrophilic group. These are mostly acetates, bromides, chlorides of quaternary ammonium salts.
Example:

Anionic detergents:
Anionic detergents are those which have an anionic hydrophilic group. These are manufactured from long chain alcohols.
Example:
Sodium lauryl sulphate [C₁₁H₂₃CH₂OSO₃Na] sodium p – dodecyl benzene sulphonate

Natural detergents:
These detergents are simply long chain organic compounds and are esters in nature. These contain polar groups which can form hydrogen bonds with water.
Example: Polyethylene glycol stearate
[CH₃ (CH₂)₁₆ COO(CH₂CH₂O)_n CH₂CH₂OH]

Question 6.
Which type of dye is phenolphthalein? Write its structure.
Answer:
The compounds formed by a combination of phthalic anhydride and phenolic compounds are known as phthalein. For example, phenolphthalein has phthalein as a ring structure.

Question 7.
Give the structure of the following dyes:
1. Methyl orange
2. Fluorescence
3. Alizarin
Answer:
1. Methyl orange

2. Fluorescence

3. Alizarin

Question 8.
Write the differences between dyes and pigments.
Answer:
Differences between dyes and pigments:

Dyes	Pigments
They are soluble in water.	Pigments are colourants that are insoluble in water and most of the solvents.
Lower dyes are very much vulnerable. Light destroy coloured objects by breaking open electronic bonding within the molecule.	Traditionally pigments have been found to be more lightfast than dyes.
Usually, the dye is organic compounds.	Pigments are normally inorganic compounds.

Question 9.
Write the general characteristics of dyes.
Answer:
A dye has the following characteristics:

It must have a suitable colour.
It can be fixed on the fabric either directly or with the help of mordant.
It must be resistant to the action of water, acid and alkalies. The groups responsible for colour are called chromophores.
These should be unaffected by light.

Question 10.
Write short notes on
(a) Direct Dyes
(b) Disperse Dyes
(c) Ingrain dyes
Answer:
(a) Direct Dyes:
The fibres are soaked in a hot solution of dye then dries after taking out. These dyes applied directly to fibre and more useful to the fabrics containing H – bonding like cotton, rayon, wool, silk nylon,
Example: Martius yellow, congo red, etc.

(b) Disperse Dyes:
Disperse dyes are water-insoluble. These dyes are finely grinding and are available as a paste or a powder that gets dispersed in water.

(c) Ingrain dyes:
These dyes are also known as ‘ice-colours’ because they react at low temperatures. Ingrain dyes are that which are synthesized directly on the fabric.
For Example Azo dyes.

RBSE Class 12 Chemistry Chapter 17 Long Answer Type Questions

Question 1.
What are soaps? How are they prepared? Explain its cleansing actions.
Answer:
Soaps are sodium or potassium salts of higher fatty acids (containing 15 – 18 carbon atoms) such as stearic acid and palmitic acid. The process of formation of soap is known as saponification. The process of saponification can be shown as

(Here R may be – C₁₅H₃₁,- C₁₇H₃₃ or – C₁₇H₃₅)
Soaps and detergents consist of a large hydrocarbon tail with negatively charged head. The hydrocarbon tail is hydrophobic (water repelling) and the negatively charged head is hydrophilic (water-attracting).
The dirt in the cloth is due to the presence of dust particles in fat or grease which stick to the cloth. When soap or detergent is dissolved in water the molecules gather together as clusters called micelles when the dirty cloth is dipped in soap or detergent solution. The soap and dirt particles come in contact with each other.
The non-polar tails of the soap begin to dissolve in non – polar oil or grease. While the polar head part remains in the water. As more and more soap particles enter the grease each fat or oil particle is surrounded by a number of negatively charged ends. Since the similar charges repel each other the oil or grease droplets break off into small globules of oil. These are still surrounded by the negatively charged polar heads of the soap molecules.
This prevents the small globules from coming together to form bigger particles. The rubbing by hands or mechanical stirring also helps to break the grease particles. In this manner, the grease particles can be completely broken up and it forms an emulsion of grease or oil contained in dirt and the droplets are washed away with water.

Question 2.
What are detergents? Give its classification and its cleansing action.
Answer:
The difficulty encountered with soaps with regard to cleaning with hard water is resolved by taking detergents. Detergents are chemicals that have all the properties of soaps but do not contain soap.
Detergents are mainly classified into three categories:
1. Anionic detergents
2. Cationic detergents
3. Non-ionic detergents
Cleaning Action:
Soaps and detergents consist of a large hydrocarbon tail with negatively charged head. The hydrocarbon tail is hydrophobic (water repelling) and the negatively charged head is hydrophilic (water-attracting).
The dirt in the cloth is due to the presence of dust particles in fat or grease which stick to the cloth. When soap or detergent is dissolved in water the molecules gather together as clusters called micelles when the dirty cloth is dipped in soap or detergent solution. The soap and dirt particles come in contact with each other.
The non-polar tails of the soap begin to dissolve in non – polar oil or grease. While the polar head part remains in the water. As more and more soap particles enter the grease each fat or oil particle is surrounded by a number of negatively charged ends. Since the similar charges repel each other the oil or grease droplets break off into small globules of oil. These are still surrounded by the negatively charged polar heads of the soap molecules.
This prevents the small globules from coming together to form bigger particles. The rubbing by hands or mechanical stirring also helps to break the grease particles. In this manner, the grease particles can be completely broken up and it forms an emulsion of grease or oil contained in dirt and the droplets are washed away with water.

Question 3.
Explain Witt theory for structural properties of dyes.
Answer:
In 1876, a German scientist Otto Witt put forward a theory known as “Chromophore Auxochrome theory” or Witt theory. Some important points of this theory are:
(i) The colour of a substance is mainly due to the presence of unsaturated groups known as chromophores. The important chromophores are:

(ii) The compounds possessing chromophores are referred to as chromogens. The higher the number of chromophores in a chromogen higher is the colour bearing capacity. The chromospheric groups are of the following two types:
1. When a single chromophore is sufficient to impart colour to the compound.
Example: NO, NO₂, N = N, = N = N,N = NO, p – quinonoid etc.
2. When more than one chromophores are required to impart the colour.
Example: > C = O, > C < etc.
3. Certain groups, which do not produce colour themselves when present along with a chromophore in an organic substance, intensify the colour. Such colour assisting groups are called auxochromes. They make the colour deep and fast.

Question 4.
Classify dyes on the basis of applications.
Answer:
Classification of dyes on the basis of applications:
(i) Direct Dyes:
The fibres are soaked in a hot solution of dye and then dries after taking out. These dyes applied directly to fibre and are more useful to the fabrics containing H-bonding like cotton, rayon, wool, silk and nylon.
Examples:
Martius yellow, Congo red etc.

(ii) Acidic Dyes:
These dyes are used in slightly acidic medium and are generally sulphonic acid and its derivatives. These are water soluble. These do not have an affinity for cotton but are used for silk, wood etc.

(iii) Basic Dyes:
These dyes contain basic group (like – NH₂ group) and react with anionic sites present on the fabric. These are used to dye nylons and polyester.
Example:
Aniline yellow.

(iv) Disperse Dyes:
Disperse dyes are water-insoluble. Their dyes are finely grinding and are available as a paste or a powder that gets dispersed in water. These particles are dissolved in the fibres and impart colour to them. These dyes were originally developed for the dyeing of cellulose acetate but now they are used to dye nylon, cellulose triacetate and acrylic fibres too.

(v) Reactive Dyes:
Reactive dyes react with fibres molecules to form a chemical compound. Reactive dyes were originally used for cellulose fibres only but now their various types are used for wool, silk, nylon, acrylics and their blends as well.
Example:
Prussian red.
(vi) Ingrain Dyes:
These dyes are also known as ‘ice-colours’ because they react at low temperatures. Ingrain dyes are those which are synthesized directly on the fabric.
Example:
Azo-dye.

Question 5.
Classify dyes on the basis of structure.
Answer:
On the basis or structure, dyes can be classified as:
(i) Nitro and Nitroso Dyes:
These are ancient known dyes. These dyes contain nitro or nitroso groups as the chromophores and -OH as Auxo chrome. A few Examples:

(ii) Diphenyl Methane Dyes:
These dyes have a basic structure of diphenylmethane.
Example:
Auramine – O is an important dye of this group which used to colour silk, wool, paper, jute etc.

(iii) Triphenyl Methane Dye:
These dyes are amino derivatives of triphenyl methane.
Example:
Malachite green which is used to colour woof and silk directly.

(iv) Phthalein Dyes:
The compounds formed by the combination of phthalic anhydride and phenolic compounds are known as phthalein. Xanthene ring structure can also be included in this group.
Example:
Phenolphthalein has phthalein as a ring structure and fluorescein is a xanthene derivative.

(v) Azo – Dyes:
It is the largest group of synthesized dyes which contain almost all the colours. Dyes characterised by the presence of an azo group (- N = N -) as the chromophore.

(vi) Indigo Dyes:
This is the most important group of natural dyes. The dyestuff is extracted from Indigofera tinctoria, a bush pea family.

(vii) Anthraquinone Dyes:
Dyes that have anthraquinone as their base and the carbonyl group (> C = O) as the chromophore. These dyes can be used with mordant dye where it gives different colours with different ions.

(viii) Heterocyclic dyes:
In these dyes at least one odd cyclic ring should be present. It is also a large group and new dyes are discovered.

RBSE Solutions for Class 12 Chemistry

RBSE Solutions for Class 12 Chemistry Chapter 16 Stereo Chemistry

June 10, 2019 by Prasanna Leave a Comment

Rajasthan Board RBSE Class 12 Chemistry Chapter 16 Stereo Chemistry

RBSE Class 12 Chemistry Chapter 16 Text Book Questions

RBSE Class 12 Chemistry Chapter 16 Multiple Choice Questions

Question 1.
Which of the following is not involved in a branch of stereochemistry?
(a) Geometrical isomerism
(b) Conformational isomerism
(c) Functional group isomerism
(d) Optical isomerism

Question 2.
Which of the following compounds do not show geometrical isomerism?

Question 3.
Which of the following statements is not true about optical isomerism?
(a) Axis of symmetry is present in the molecule
(b) Plane of symmetry is present in the molecule
(c) Centre of symmetry is present in the molecule
(d) None of the above

Question 4.
Meso tartaric acid does not show optical activity because
(a) It has two chiral centres
(b) It shows external compensation
(c) It has a plane of symmetry
(d) It has an erythro form

Question 5.
Which of the following do not show optical isomerism?
(a) Ethyl Alcohol
(b) 2 – Butanol
(c) 2 – Chloropropane
(d) Lactic acid

Question 6.
Which of the following statement is not true for writing Fischer Projection formula?
(a) Two perpendicular lines intersecting each other are drawn
(b) First number carbon is placed on the left side.
(c) Both the groups of horizontal surface are directed above.
(d) Plane of the molecule can be rotated by 180°.

Question 7.
Which molecule is considered as standard in respect to relative configuration?
(a) Lactic acid
(b) Tartaric acid
(c) Glyceraldehyde
(d) Sodium potassium titrate

Question 8.
Which of the following method is not employed in the resolution of the racemic mixture?
(a) Biochemical method
(b) Mechanical method
(c) Effective distillation
(d) Column chromatography

Question 9.
Which of the following statements is not true about conformational isomerism?
(a) These can be represented by Newman and Sawhorse projections.
(b) The number of conformational isomers is infinite.
(c) The eclipsed form is most stable.
(d) Conformational isomerism is also found in the cyclic compound.

Question 10.
Which of the following statements is not true about diastereomers?
(a) They do not show optical activity.
(b) These isomers show the difference in physical properties.
(c) They show internal compensation.
(d) They scatter plane of polarised light.

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

RBSE Class 12 Chemistry Chapter 16 Very Short Answer Type Questions

Question 1.
Define isomerism.
Answer:
Two or more than two compounds having the same molecular formula but differing in their physical and chemical properties are called isomers (Greek: iso = equal, meros = parts) and this phenomenon is known as isomerism.

Question 2.
What do you understand by stereochemistry?
Answer:
The branch of chemistry which deals with three-dimensional molecules and their effect on physical and chemical properties is known as stereochemistry.

Question 3.
What is an optical activity?
Answer:
Compounds which rotate the plane of polarised light are called optically active compounds and the property is known as an optical activity.

Question 4.
What are the essential conditions for optical isomerism?
Answer:
The compound must be asymmetric or having at least one chiral centre for optical isomerism.

Question 5.
What do you understand by symmetry?
Answer:
Elements of symmetry offer a simple device to decide whether a molecule (or object) is chiral or not i.e. whether it is superimposable on its mirror image or not.

Question 6.
Meso form is optically inactive, Why?
Answer:
Meso form is optically inactive because the molecules in meso form have a plane of symmetry due to which the optical rotations of upper and lower parts are equal and in the opposite direction which balanced internally and compound becomes optically inactive. This property is called internal compensation.

Question 7.
What is racemisation?
Answer:
The optical rotation of equimolar mixture of two enantiomers (d and l) will be zero because rotation produces by one isomer will cancel the rotation produced by other isomers. So, such types of the mixture are called racemic mixture and the phenomenon of converting optically active isomers (d (or) l) into the optically inactive mixture is called racemization.

RBSE Class 12 Chemistry Chapter 16 Short Answer Type Questions

Question 1.
Define optical activity. Explain with the example which type of molecules show optical activity.
Answer:
Optical Activity:
Compounds which rotate the plane of polarized light are called optically active compounds and this property is known as optical

and is optically active compounds. They have one chiral centre each.

Question 2.
1 – Butanol does not show optical isomerism whereas 2 – Butanol shows, Why?
Answer:
1 – Butanol does not show optical isomerism whereas 2 – butanol shows because 1 – butanol has no asymmetric carbon and 2 – butanol has one asymmetric carbon.

Question 3.
What are the necessary conditions for optical activity? Give an example of optically active molecules.
Answer:
Necessary conditions for optical activity:

The compound must contain an asymmetrical carbon atom.
The molecule must contain a chiral axis.
The molecule must possess a chiral plane.
The molecule does not possess any elements of symmetry.

Examples of optically active compounds:

Question 4.
Interpret two optical isomers of Lactic acid?
Answer:
Lactic acid has only one asymmetric carbon atom. d – Lactic acid rotates the plane of polarised light to right (clockwise) and l – lactic acid rotates to left (anticlockwise).
Two optical isomers of lactic acid are as follows:

Question 5.
Describe racemisation giving suitable examples.
Answer:
Racemisation:
When equal amounts of dextrorotatory and laevorotatory isomers are mixed then the resulting mixture becomes optically inactive because optical activities of each isomer cancel each other. Such a mixture is called a racemic mixture and this phenomenon is called racemization. Racemisation can be done by the following methods:
1. By simple heating:
It is also called thermal racemisation. The racemic mixture of tartaric acid, lactic acid etc. can be obtained by this method.
2. Auto racemisation:
Racemisation of some compounds takes place at room temperature. It is called auto racemisation. For example, auto racemisation of dimethyl succinate takes place at room temperature.
3. Chemical racemisation:
When a racemic mixture is obtained by mixing a chemical then it is called chemical racemisation. For example, 2-butyl phenyl ketone gives a racemic mixture on adding an acid. We can represent the thermal racemisation of lactic acid as follows:

Question 6.
Explain erythro and three isomers giving suitable examples.
Answer:
When two similar groups at non-identical chiral carbons are on the same side of the vertical line of the Fischer projection, the isomer is called erythro isomer, if the groups are on opposite sides, the isomers are called three isomers.

Question 7.
What do you understand by absolute configuration?
Answer:
In 1950, when X-ray crystallographic techniques were developed then Bijvoet found that the form of glyceraldehyde which is given D-configuration is actually D – form. In other words, the configuration which is considered as standard for other compounds is actually its absolute configuration.

Question 8.
What are the differences between conformation and configuration?
Answer:
Differences between conformation and configuration:

Confirmation	Configuration
Conformations have low energy barrier, which varies from 4.2 to 46 kJ per mole.	The difference in energy between the two configurations is more than 84 kJ/ mole.
Conformers are non-ionisable.	Configurations are ionisable.
Conformers are easily interconvertible.	Configurations are not easily interconvertible.
A molecule could have an infinite number of conformations.	A molecule can have only one configuration.

Question 9.
What do you understand by geometrical isomerism? Write essential conditions for geometrical – isomerism.
Answer:
Geometrical isomerism:
The isomerism which arises due to frozen rotation about to bond in a molecule is known as geometrical or cis-trans isomerism. Conditions for geometrical isomerism: A compound will show geometrical isomerism if it fulfils the following two conditions:
1. There should be restricted (frozen) rotation about a bond in the molecule.
2. Both substituents on each carbon about which rotation is frozen (restricted) should be different.
Example:

Question 10.
Explain the stereochemistry of oximes.
Answer:
The reaction of aldehydes or ketones with hydroxylamines gives aldoximes and ketoximes as:

It is clear that in CN both C and N are in sp²hybridisation. Due to the presence of a double bond, free rotation is not possible. If OH group on N atoms is present on hydrogen attached to carbon or to a small group, then this is called ‘syn’ and if opposite then it is called ‘anti’.
For example:
Acetaldoxime (CH₃CH = NOH)

Benzaldoxime

Similarly, Acetophenone oxime

Question 11.
Draw the Newman projection formula for different conformations of ethane.
Answer:
Conformations:
The different spatial arrangements of atoms in a molecule which are readily interconvertible by rotation about single bonds are called conformations. Conformations represent conformers which are readily interconvertible and thus nonseparable.
Newman projection formula for different conformations of ethane:
When an ethane molecule rotates about its carbon – Carbon single bond, two extreme conformations can result in the staggered conformation and the eclipsed conformation.

Question 12.
Describe the salt formation method for the separation of enantiomers.
Answer:
The process of separation of a racemic mixture into the enantiomers.
Chemical method (salt formation method):
Consider a racemic mixture of enantiomeric organic acid (I) – HA. It can be separated into pure enantiomeric acid by reaction with an optically active base, say a laevorotatory base (-) – B. The mixture of enantiomeric acids on reaction with an optically active base result in the formation of two diastereomeric salts as shown in the figure. These diastereomeric salts have, of course, different physical properties, including solubility in a given solvent. They can, therefore, be separated, by a simple technique i.e., fractional crystallization.

RBSE Class 12 Chemistry Chapter 16 Long Answer Type Questions

Question 1.
Discuss optical isomerism in tartaric acid.
Answer:
Tartaric acid can be taken as an example.

It has two asymmetric carbon atoms so using the 2ⁿ formula there should be a maximum of 2² = 4 stereoisomers. These can be represented as:

I and II i.e., threo forms are mirror images of each other and are enantiomers. But in erythro form, the plane of symmetry is present. Thus, it does not show optical isomerism. Therefore, tartaric acid exists only in three isomeric forms of which one is meso (optically inactive) and the other two are optically active, i. e., d (or) (+) and l or (-).

Question 2.
Explain diastereoisomers with example.
Answer:
The optical isomers which do not mirror images of each other are called diastereoisomers and this phenomenon is called diastereoisomerism.
Their characteristics are given below:
1. These isomers can not superimposable to each other.
2. Diastereoisomerism is found only, in those compounds in which at least two chiral carbon centres are present.
3. The physical properties like melting point, boiling point, solubility, spectre rotation etc of diastereoisomers of a compound are different. So, these can be separated by fractional distillation, fractional crystallisation and chromatographic methods easily.
Example:
Cinnamic acid dibromide

Here, I & II and III & IV one enantiomers as there are mirror images of each other. However, I & II, I & IV, II & III and II & IV are diastereomers.

Question 3.
What is stereoisomerism? Discuss stereoisomerism in tartaric acid and write the structural formula of all its isomers. Explain when tartaric acid is synthesised then it is optically inactive.
Answer:
Compounds having the same constitutions but the different spatial arrangement of their atoms are known as stereoisomers and this phenomenon is called stereoisomerism.
Tartaric acid can be taken as an example:

It has two asymmetric carbon atoms so using the 2ⁿ formula there should be a maximum of 2² = 4 stereoisomers. These can be represented as:

I and II i.e. the forms are mirror images of each other and are called enantiomers but in erythro form i.e., meso form, the plane of symmetry is present thus it does not show optical activity. Therefore, tartaric acid exists only in three isomeric forms of which one is meso (optically inactive) and the other two are optical active i.e., d (or) (+) and l or (-).

Question 4.
What do you understand by optical isomerism? Discuss optical isomerism in compounds having two asymmetric carbon atoms.
Answer:
Optical isomerism:
Compounds which have the same molecular formula but differ in optical activity are known as optical isomers and the isomerism exhibited by them is called optical isomerism.
Optical isomerism in compounds having two asymmetric carbon atoms:
(i) – 2, 3 dichlorobutane acids (asymmetric molecule):

The structure I and II and structure III and IV are mirror images. These are known as enantiomers.
(ii) 2, 3 dihydroxy butanoic acids:

Structures I and II and structures III and IV are mirror images. These are known as enantiomers.

Question 5.
Explain stereoisomerism in tartaric acid. How many optical isomers are possible for tartaric acid? What are the differences between meso tartaric acid and racemic mixture?
Answer:
Compounds having the same constitutions but the different spatial arrangement of their atoms are known as stereoisomers and this phenomenon is called stereoisomerism.
Tartaric acid can be taken as an example:

It has two asymmetric carbon atoms so using the 2ⁿ formula there should be a maximum of 2² = 4 stereoisomers. These can be represented as:

I and II i.e. the forms are mirror images of each other and are called enantiomers but in erythro form i.e., meso form, the plane of symmetry is present thus it does not show optical activity. Therefore, tartaric acid exists only in three isomeric forms of which one is meso (optically inactive) and the other two are optical active i.e., d (or) (+) and l or (-).
Differences between meso and racemic tartaric acid:

Meso Tartaric Acid	Racemic Mixture
The angle of optical rotation of meso tartaric acid is 0°.	Its angle of optical rotation is also 0°.
It is melting point is 140°C.	It is melting point is 260°.
It can not be separated into two forms.	It can be separated into two forms.
It is optically inactive due to internal compensation.	It is optically inactive due I to external compensation.

Question 6.
The physical properties of geometrical isomers are different whereas physical properties of optical isomers are the same. Explain with reason.
Answer:
Although geometrical isomers have completely different physical and chemical properties (for example cis and trans 2 – butene, have different boiling points and densities), optical isomers (also called enantiomers) differ in only one characteristic – their interaction with plane polarized light.
Physical properties of geometrical isomers depend on
1. The polarity of the molecules.
2. The shape or symmetry of molecules.
The polarity strongly influences the relative boiling points as it determines the strength of the intermolecular forces.
For example, cis – 1, 2 dichloroethene has a net dipole moment and dipole-dipole attractions between its molecules in addition to the van der Waal’s forces, whereas trans 1, 2 dichloroethene which is non-polar has only van der Waal’s forces. The boiling point of the cis-isomers is therefore higher. Melting point is generally, more influenced by the symmetry of the molecules as this affects the packing in the solid state. The trans – isomers are able to pack more closely due to their greater symmetry, so the intermolecular forces are more effective than in the ch – isomer. The melting point of the trans-isomer is therefore higher.

RBSE Solutions for Class 12 Chemistry

RBSE Solutions for Class 12 Chemistry Chapter 15 Polymers

June 10, 2019 by Prasanna Leave a Comment

Rajasthan Board RBSE Class 12 Chemistry Chapter 15 Polymers

RBSE Class 12 Chemistry Chapter 15 Text Book Questions

RBSE Class 12 Chemistry Chapter 15 Multiple Choice Questions

Question 1.
Which of the following is not true about polymers?
(a) Polymers have low molecular weight.
(b) Polymers have no charge.
(c) Polymers have high viscosity.
(d) Polymers scatter light.

Question 2.
On the basis of the method of polymerisation, polymers are classified as:
(a) Only addition polymers
(b) Only condensation polymers
(c) Only addition and condensation polymers
(d) Copolymers

Question 3.
Teflon, styrene and neoprene are
(a) Copolymer
(b) Homopolymer
(c) Monomer
(d) Condensation polymer

Question 4.
Which of the following is condensation polymer?
(a) Dacron
(b) Teflon
(c) PVC
(d) Polythene

Question 5.
Which of the following is not a true polymer?
(a) Ethene
(b) Styrene
(c) Isoprene
(d) Butadiene

Question 6.
Which of the following has a ester bond?
(a) Nylon
(b) Bakelite
(c) Terylene
(d) Rubber

Question 7.
Terylene is a condensation polymer of ethylene glycol and:
(a) Terephthalic acid
(b) Phthalic acid
(c) Benzoic acid
(d) Acetic acid

Question 8.
The monomer of nylon – 6 is
(a) Adipic acid
(b) Caprolactum
(c) 1, 3 butadiene
(d) Chloroprene

Question 9.
Teflon is a polymer of:
(a) Tetrafluoroethylene
(b) Tetraiodoethylene
(c) Tetrachloroethylene
(d) Tetrabromoethylene

Question 10.
Which of the following is a bio-degradable polymer?
(a) PVC
(b) Cellulose
(c) Polythene
(d) Nylon – 6

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

RBSE Class 12 Chemistry Chapter 15 Very Short Answer Type Questions

Question 1.
Define polymers?
Answer:
Polymers are high molecular mass (10³ – 10⁷u) substances consisting of a very large number of simple repeating structural units
joined together through covalent bonds in a linear fashion.
Example: Polythene, Nylon – 6,6, Bakelite etc.

Question 2.
Give an example of natural and synthetic polymers.
Answer:
Natural polymers – Natural rubber

Synthetic polymer – Nylon 6, 6

Question 3.
Give one difference between homopolymer and copolymer.
Answer:
Homopolymer:
A polymer formed from only one kind of repeating units or monomer is called homopolymer.
Example: PVC is a homopolymer.
Copolymer:
A polymer formed from more than one kind of monomer units is called copolymer or mixed polymer.
Example: – 1, 3 – butadiene and styrene are examples of the mixed polymer.

Question 4.
How many methods of chain termination are there in the free radical mechanism?
Answer:
There are three methods of chain termination in the free radical mechanism.

Chain initiating.
Chain propagation.
Chain termination.

Question 5.
Which monomers are polymerised by cationic polymerisation?
Answer:
A vinyl derivative is monomers polymerised by cationic polymerisation.

Question 6.
Write a formula for number average molecular mass.
Answer:

Question 7.
Give the name of substances used in the vulcanisation of rubber.
Answer:
S (Sulphur).

Question 8.
Write the name of the monomer of neoprene.
Answer:
2 – Chlorobuta – 1, 3 diene is the monomer of neoprene.

Question 9.
Write the chemical formula of Nylon – 6, 6.
Answer:

Question 10.
Give two examples of synthetic biopolymers.
Answer:

PHBV (Poly β – hydroxybutyrate – co – β hydroxy valerate)
Nylon – 2, 6

RBSE Class 12 Chemistry Chapter 15 Short Answer Type Questions

Question 1.
On the basis of the source of how polymers are classified? Give one example of each?
Answer:
On the basis of source polymers are classified as:

Natural polymers (Example: Starch, Cellulose, Portions etc.)
Synthetic polymers (Example: PVC, -Nylon Teflon etc.)
Semi-synthetic polymers (Example.- Rayon, Cellulose nitrate)

Question 2.
Write differences between thermosetting and thermoplastic polymers.
Answer:
Thermoplastic polymers:
These are linear or slightly branched long chain molecules capable of repeatedly softening on heating and hardening on cooling. These polymers possess intermolecular forces of attraction intermediate between those in elastomers and fibres. Polythene and polystyrene are examples of thermoplastic polymers.

Thermosetting polymers:
These polymers are cross-linked or heavily branched molecules which on heating undergo extensive cross-linking in moulds and become infusible. These cannot be reused. Bakelite, urea formaldehyde, resins are examples of thermosetting polymers.

Question 3.
In free radical polymerisation, where do we use hydroquinone? Explain.
Answer:
Benzoquinone combines with free radical intermediate to form of non-reactive radical, which is highly stabilized by resonance.

Question 4.
Give the monomer of dacron and process of polymerisation of dacron.
Answer:
The monomer of dacron is:

It is also known as terylene and is the condensation product of ethylene glycol and terephthalic acid polymerisation is carried out at 420 to 460 K in the presence of zinc acetate and antimony trioxide.

Question 5.
What is the polydispersity index? Explain.
Answer:
Polydispersity Index:
In order to have an idea about the homogeneity of a polymer, we define the term polydispersity index. It is the ratio of mass average molecular mass to the number average molecular mass.

For natural polymers, PDI is usually equal to one which means that they are monodisperse. Such polymers are more homogeneous.
Synthetic polymers on the contrary generally have PDI > 1 which means that they were less homogeneous.

Question 6.
Why rubber is vulcanised? Explain.
Answer:
Natural rubber is soft and sticky at high temperatures and brittle at low temperatures. Therefore rubber is generally used in the narrow
temperature range (283 – 335 K) where its elasticity maintained. So to improve the property of natural rubber, it is vulcanised by heating it with about 5% sulphur at 373 K – 415 K.

Question 7.
What do LDPE and HOPE indicate? How are they formed?
Answer:
Low-Density polyethene (LDPE):
It is obtained by heating very pure ethylene at 350 K – 570 K under a pressure of about 1500 atm in the presence of traces of oxygen. It is a highly branched polymer. It is chemically inert, tough but flexible. It is used for packing, cable insulation and in the manufacture of pipes, toys etc.

High-Density polyethene (HDPE):
It is obtained by polymerisation of ethylene at about 333 – 343 K under 6 to 7 atmospheric pressure in the pressure of catalysts such as triethyl aluminium and titanium tetrachloride. It is chemically inert having greater toughness and hardness.

Question 8.
Write the names of monomers and structures of the following polymers:

Teflon
Terylene
Nylon – 6, 6
Bakelite

Answer:
1. Teflon – Monomer – CF₂= CF₂ (Tetrafluoroethylene)

2. Terylene – Monomers

3. Nylon – 6, 6
Monomers – HOOC – (CH₂)₄ – COOH (Adipic acid) and H₂N – (CH₂)₆ – NH₉ (Hexamethylene diamine)

4. Bakelite

Question 9.
What is PHBV?
Answer:
Polyhydroxybutyrate – Co – β – hydroxy valerate PHBV is used in especially packing. Orthopaedic devices and in controlled release of drugs. PHBV undergoes bacterial degradation in the environment.

Question 10.
Why the PDI of proteins is equal to one?
Answer:
For a protein, the weight average molecular weight is equal to the number average molecular mass. Therefore the value of PDI for a protein is equal to one.

RBSE Class 12 Chemistry Chapter 15 Long Answer Type Questions

Question 1.
Classify polymers on the following basis:
(i) On the basis of monomer units.
(ii) Type of polymerisation.
(iii) Molecular forces.
Answer:
(i) On the basis of monomer units:
Polymers can be classified into two types:
1. Homopolymers:
A polymer formed from only one kind of repeating units or monomers is called homopolymer.
Example: Polythene and PVC.

2. Copolymers:
A polymer formed from more than one kind of monomer units is called copolymer or mixed polymer. For example, Buna – S rubber which formed from 1, 3 – butadiene and styrene are examples of the copolymer.

(ii) Type of polymers:
Carothers in 1928 classified polymers on the basis of polymerisation process into two groups:
1. Addition polymers
2. Condensation polymers.
1. Addition polymers:
A polymer formed by direct addition of repeated monomer molecules possessing double or triple bond, without the elimination of any small molecule is called addition polymer.
Example: PVC, Teflon, Neoprene etc.
2. Condensation polymers:
These polymers are formed through a series of independent reactions. These reactions involve the bond formation condensation between two difunctional monomer molecules to form dimers, which in turn, produce tetramers and so on (with the loss of simple molecules like NH₃, H₂O etc.).
Example: Bakelite, Nylon – 6, 6 etc.
(iii) Type of polymers on the basis of molecular forces operating between different polymer chains, the polymers are of the
following types:

Elastomers: (Example: vulcanised rubber, natural rubber etc)
Fibres: (Example: nylon, Orlon, polyesters)
Thermoplastics Polymers: (Example: polypropylene, polythene etc)
Thermosetting polymers: (e.g. Bakelite, phenol-formaldehyde resin etc)

Question 2.
What is an addition polymerisation? Explain free radical addition polymerisation mechanism by taking an example.
Answer:
Addition polymerization:
A polymer formed by direct addition of repeated monomer molecules possessing double or triple bond, without the elimination of any small molecule is called addition polymerisation.
Example: polythene, PVC etc.
Mechanism of free radical addition is completed in three parts:
1. Chain initiating step:

2. Chain propagating step:

3. Chain terminating step:

Question 3.
How natural rubber is obtained? Write its composition and structure.
Answer:
To obtain natural rubber first of all latex of rubber plant is obtained. The latex contains, about 25 – 40% of rubber hydrocarbons dispersed in water along with stabilizer proteins and some fatty acids are mixed due to which rubber coagulates and precipitated out. It is separated by filtration. This rubber is known as crepe rubber.
Structure and composition:
Natural rubber may be considered as a linear polymer of isoprene (2 – methyl – 1, 3 butadiene) and also called as cis – 1,4 polyisoprene.

Question 4.
Why the molecular mass of polymers is taken as average molecular mass? Explain how many types of average molecular mass can be represented? Explain each type.
Answer:
A polymer sample contains chains of varying lengths and therefore its molecular mass. is always expressed as an average. On the other hand, natural polymers such as proteins contain chains of identical length and therefore they have definite molecular mass. The molecular mass of polymers can be expressed in two ways:
(i) Number average molecular mass (Mn):
If N₁, N₂, N₃ … are the number of molecules with molecular masses M₁, M₂, M₃ …. respectively then the number average molecular mass is

This may be expressed as

where N_i is the number of molecules with molecular mass M_i.

(ii) Weight average molecular mass (Mw):
If m₁, m₂, m₃…are the masses of species with molecular masses M₁, M₂, M₃… respectively, then the weight average molecular mass is:

where N_i is the number of molecules of mass M_i.

Question 5.
Write a short note on biopolymers and biodegradable polymers.
Answer:
1. Biopolymers:
Many polymers which are present in plants and animals such as polysaccharides (starch, cellulose), proteins and nucleic acid etc. which control various life processes in plants and animals are called biopolymers.
Examples: Protein, Carbohydrates, Nucleic acid, Lipids etc.

2. Biodegradable:
The polymers which are degraded by microorganisms within a suitable time period so that the polymers and their degraded products do not cause any serious effects on the environment are called biodegradable polymers.
Example: polylactic acid (PLA), Polyglycolic acid (PGA) and PHB etc.

Question 6.
Write method of polymerisations and uses of the following polymers:

Bakelite
PVC
Polyester
Nylon – 66

Answer:
1. Bakelite:
It is condensation polymer and is obtained from phenol and formaldehyde in the presence of either an acid or a base catalyst.
Uses:
Soft Bakelite with low degree polymerisation is used for making glue for binding laminated wooden planks and varnishes, High degree polymerisation gives hard bakelite which is making combs, fountain pens, barrels, electrical switches etc.

2. PVC:
It is an addition polymer. A polymer formed by direct addition of repeated monomers without the elimination of byproduct molecules is called addition polymer.
Uses:
PVC plastic is used to manufacture bottles that have application in a wide range of industries.

3. Polyesters:
Polyesters are condensation polymers.
Uses:
It is used

In blending with cotton or wool fibres because it increases resistant.
As glass reinforcing material in safety helmets.

4. Nylon – 66:
Nylon 66 is a condensation polymer.
It is used:

In making bristles for brushes.
In the manufacture of cords and climbing ropes.

RBSE Solutions for Class 12 Chemistry

RBSE Solutions for Class 12 Chemistry Chapter 14 Bio-Molecules

June 4, 2019 by Prasanna Leave a Comment

Rajasthan Board RBSE Class 12 Chemistry Chapter 14 Bio-Molecules

RBSE Class 12 Chemistry Chapter 14 Text Book Questions

RBSE Class 12 Chemistry Chapter 14 Multiple Choice Questions

Question 1.
Which is known as the powerhouse of the cell?
(a) Golgi body
(b) Mitochondria
(c) Cytosome
(d) Ribosome

Question 2.
Which of the following is a disaccharide?
(a) Starch
(b) Fructose
(c) Lactose
(d) Cellulose

Question 3.
The product of hydrolysis of starch is
(a) Fructose
(b) Sucrose
(c) Maltose
(d) Glucose

Question 4.
The molecular formula of most common disaccharide is:
(a) (C₆H₁₂O₆)₆
(b) (C₁₂H₂₂O₁₁)
(c) (C₁₀H₂₂O₁₁)
(d) (C₁₈H₂₂O₁₁)

Question 5.
Which of the following is not a reducing sugar?
(a) Glucose
(b) Fructose
(c) Sucrose
(d) Maltose

Question 6.
The product obtained after hydrolysis of protein in the presence of an enzyme is
(a) Amino acids
(b) Hydroxy acid
(c) Aromatic acids
(d) Dicarboxylic acid

Question 7.
The example of globular protein is
(a) Collagen
(b) Insulin
(c) Myosin
(d) Keratin

Question 8.
Alanine is an example of
(a) Alpha-amino acid
(b) B-fructose
(c) Gamma-lactose
(d) Delta-cellulose

Question 9.
Which of the following is a basic amino acid?
(a) Glycine
(b) Aspartic acid
(c) Lysine
(d) Glutamine

Question 10.
Enzymes are
(a) Carbohydrates
(b) Proteins
(c) Fats
(d) Salts

Question 11.
The conversion of protein to amino acids takes place by which of the following enzyme?
(a) Lipase
(b) Maltase
(c) Trypsin
(d) Renin

Question 12.
Chemical messengers are
(a) Hormones
(b) Enzymes
(c) Vitamins
(d) Nucleic acids

Question 13.
Number of the thyroid gland in humans is
(a) One
(b) Two
(c) Three
(d) Four

Question 14.
Growth hormone is secreted by
(a) Thyroid gland
(b) Pituitary gland
(c) Thymus gland
(d) Sex organs

Question 15.
Vitamin A deficiency causes
(a) Night blindness
(b) Scurvy
(c) Beriberi
(d) Anaemia

Question 16.
In nucleic acids, nucleotides are attached to one another by
(a) Hydrogen bond
(b) Peptide bond
(c) Phosphorous group
(d) Glycosidic bond

Question 17.
How many nucleotides sequence forms one codon in mRNA for an amino acid?
(a) One
(b) Two
(c) Three
(d) Four

Question 18.
DNA and RNA have chiral asymmetric carbon atom. The reason for their chirality is
(a) Asymmetric base
(b) D-sugar component
(c) L-sugar component
(d) Asymmetric phosphate ester units

Question 19.
In RNA, nitrogenous bases are
(a) Adenine and guanine and cytosine and thymine
(b) Adenine and guanine and thymine and cytosine
(c) Adenine and thymine and guanine and cytosine
(d) Adenine and guanine and uracil and cytosine

Question 20.
The sequence in nucleic acids is
(a) Base – sugar – phosphate
(b) Sugar – base – phosphate
(c) Phosphate – base – sugar
(d) Base – phosphate-sugar

Answers:
1. (b)
2. (c)
3. (d)
4. (b)
5. (c)
6. (a)
7. (b)
8. (a)
9. (c)
10. (b)
11. (c)
12. (a)
13. (a)
14. (b)
15. (a)
16. (c)
17. (c)
18. (d)
19. (a)
20. (a)

RBSE Class 12 Chemistry Chapter 14 Very Short Answer Type Questions

Question 1.
Write the chemical composition of the cell.
Answer:
Chemically a cell composed of carbohydrates, proteins, fats, lipids, minerals, water and various inorganic and organic substances.

Question 2.
What are monosaccharides?
Answer:
Monosaccharides:
A carbohydrate that cannot be hydrolysed further to give a simpler unit of polyhydroxy aldehyde and ketone is called a monosaccharide.
Example: glucose and fructose etc.

Question 3.
What are non- sugars?
Answer:
The carbohydrates having no sweet taste are called non-sugar.
For example starch, cellulose etc.

Question 4.
What is the main structural difference between starch and cellulose?
Answer:
The basic difference in starch and cellulose lies in the nature of the glucose molecules present. Starch consists of amylase and amylopectin. In both of them α – D (+) glucose molecules linked in C₁– C₄ manner. In amylopectin, these linear chains are further linked in C₁– C₆ manner. In cellulose β – D (+) glucose molecules are linked to one another in a C₁ – C₄ manner. Thus, starch and cellulose have a different arrangement of the constituents.

Question 5.
Define essential and non-essential amino acids with examples.
Answer:
(i) Essential Amino acids:
Those amino acids which are to be supplied through diet are called essential amino acids. They cannot be synthesized by the body.
For example arginine, valine, histidine, etc.
(ii) Non-essential Amino Acids:
Those amino acids that can be synthesized by the body are called non-essential amino acids.
For example glycine, alanine, serine, cysteine etc.

Question 6.
What is the main function of enzymes?
Answer:
Enzymes catalyse biological reactions take place within the body.

Question 7.
Why are hormones known as ‘glandular juice’?
Answer:
Hormones are also known as ‘Glandular Juices’ because they are secreted by endocrine glands.

Question 8.
Which vitamin, is soluble in water?
Answer:
Vitamin B, C is soluble in water.

Question 9.
Write the name of nitrogenous bases found in DNA.
Answer;
Purines: Adenine (A), Guanine (G)
Pyrimidines: Cytosine (C), Thymine (T)

Question 10.
What are the main functions of nucleic acids?
Answer:
1. Nucleic acids play a vital role in the transmission of heredity characters.
2. Nucleic acids help in the biosynthesis of proteins.

RBSE Class 12 Chemistry Chapter 14 Short Answer Type Questions

Question 1.
Write the functions of carbohydrates.
Answer:
1. Carbohydrates are the main source of energy for a living being.
2. Carbohydrates are the major constituent of the plant body.
3. Deoxyribose and ribose sugars are the constituents of genetic material i.e., DNA & RNA.
4. Starch is used as food. Cellulose acts as the source of shelter and clothing.

Question 2.
Write any two methods for the preparation of glucose.
Answer:
(i) Preparation of glucose from sucrose:
On boiling an aqueous solution of sucrose with dilute HCl or dilute H₂SO₄, glucose and fructose are formed in equimolar proportions.

(ii) Preparation of glucose from starch:
Starch, when boiled with dilute H₂SO₄ at 393 K under 2 to 3-atmosphere pressure, gives glucose.

Question 3.
What is Sickle Cell Anaemia?
Answer:
Sickle Cell Anaemia:
Haemoglobin, the molecule in the blood that carries oxygen consists of 574 amino acid units. Changing just one specific amino acid in the sequence results in defective haemoglobin found in patients suffering from sickle cell anaemia. In these patients the defective haemoglobin in RBCs precipitates causing the cells to sickle ie., the cells acquire a bent shape similar to a sickle, and some times cell bursts.
Normal Haemoglobin:
Val – His – Leu – Thr – Pro – Glu – Glu – Lys –
Sickle Cell Haemoglobin:
– Val – His – Leu – Thr – Pro – Val – Glu – Lys
The sickle cell haemoglobin contains valine (Val) in place of glutamic acid (Glu) present in normal haemoglobin.

Question 4.
Write the reactions of glucose with Fehling’s solution and Tollen’s reagent.
Answer:
(i) Reaction with Fehling’s Solution:
A deep blue solution is obtained by mixing Fehling A (aqueous CuSO₄ solution) and Fehling B (aqueous NaOH solution containing a small amount of Roschelle’s salt) in equal volumes. When heated with glucose then Fehling solution gives a red precipitate.

(ii) Reaction with Tollen’s Reagent:
Tollen’s reagent is ammonical silver nitrate solution. When heated with glucose in a clean test tube, a shining mirror gets deposited at the bottom of the tube.
AgNO3 + NH₄OH → AgOH + NH₄NO3

Question 5.
Why are hormones known as “Chemical messengers”?
Answer:
Hormones are known as chemical messengers because they are secreted by glands and they work in the cells and the tissues away from their origin.

Question 6.
What is the isoelectric point of amino acids? Define.
Answer:
α-Amino acids behave as dipolar ion or as Zwitterion due to the transfer of a proton (H⁺) from carboxylic group to the amino group in the same molecule.

In acidic medium, the cation formed migrates towards cathode while in the basic medium anion migrates towards anode on passing an electric current. At a certain pH of the medium called an isoelectric point of an α-amino acid, the Zwitterion behaves as neutral and does not migrate to any electrode on passing current. The iso-electric points are in the pH range of 5.5 to 6.3.

Question 7.
Write one difference and one similarity between enzymes and hormones.
Answer;
Similarity:
Hormones and enzymes act as biocatalyst.
Difference:
Hormones act on glands or organs which are different from their origin organs or glands. These are released into the blood before use. Enzymes are not utilized in the reaction.

Question 8.
What do you understand by denaturation of protein? Explain.
Answer:
Denaturation of Proteins:
A protein present in the biological system with a unique three-dimensional structure and biological activity is known as the native protein. When a native protein is subjected to either some physical change or chemical change by taking part in certain chemical reactions which bring about a change in pH, this leads to denaturation of the native protein. The denaturation is due to the breaking of the H-bonds as a result of which the globules unfold and helix get uncoiled. The protein loses its biological activity on account of denaturation. Chemically denaturation does not change the primary structure of the protein and only brings about change in secondary and tertiary structures.

Question 9.
What is the genetic code?
Answer:
The sequence of amino acids in the protein synthesised is predetermined from the nucleotide base sequence of mRNA which in turn is correspondingly derived from the base sequence of DNA. The DNA sequence that codes for a specific protein or a polypeptide are called a gene. Thus every protein in a cell has a corresponding gene. The relation between the nucleotide triplets and the amino acid is called the genetic code.

Question 10.
Differentiate between primary and secondary structure of a protein.

Primary Structure	Secondary Structure
Proteins may have one or more polypeptide chains. Each polypeptide in a protein has amino acids linked with each other in a specific sequence and this sequence of amino acids is said to be the primary structure of that protein.	The secondary structure of a protein refers to the shape in which a long polypeptide chain can exit. These structures arise due to the regular folding to the backbone of the polypeptide chain due to H-bonding between -COOH and -NH- groups of the peptide bond.

Question 11.
Explain mutarotation.
Answer:
Mutarotation:
The change in the specific rotation when the two anomeric forms are dissolved in water is known as mutarotation. Both the α and β forms of D (+) glucose are interconvertible through open chain structure. When α – form with specific rotation +112° is dissolved in water it slowly changes to β – form through the open chain structure. Similarly, when β – form with a specific rotation of +19° is dissolved in water it slowly changes to α – form in a similar manner. As a result of its specific rotation increase. Thus we conclude that both the forms in aqueous solution undergo a change in specific rotation and the change continues till the equilibrium, mixture acquires a specific rotation of about +52.7°. It has been established that the equilibrium contains about 36% of a-form and 63.5% of β – form with about 0.5% of open chain structure.

Question 12.
Write the name of the disease caused by the deficiency of Vitamin-B₁₂ and Vitamin A. Also, write the source of these vitamins.
Answer:

Vitamin	Deficiency Disease	Sources
B₁₂	Pernicious anaemia, inflammation of tongue and mouth	Milk, egg and animal tissues
A	Night blindness, stunted growth, xerosis (drying of the skin, xerophthalmia(cornea becomes opaque))	Milk, butter, egg, fish and fish oil. It can also be synthesized in the body from carotenoids present in carrots, tomatoes, ripe mangoes etc. Carotenoids are the precursors of vitamin A.

Question 13.
Write four differences between RNA and DNA.
Answer:
Differences between RNA and DNA

S.No	RNA	DNA
1.	The pentose sugar present in RNA is D-ribose.	The pentose sugar present in DNA is D-2-deoxyribose.
2.	RNA contains cytosine and uracil pyrimidine bases and guanine and adenine as purine bases.	DNA contains cytosine, and thymine as pyrimidine bases and guanine and adenine as purine bases.
3.	It is a single chain of a polynucleotide.	It is a double chain of a polynucleotide.
4.	It regulates protein synthesis.	It controls structure, metabolism, differentiation. Transfer the characters from our generation to the other.

Question 14.
Write the Haworth structure of glucose and fructose.
Answer:
(i) Haworth Structure of Glucose:
The two anomeric forms (α and β) of D (+) glucose have been assigned pyranose ring structures by Haworth. Pyran is a six-membered heterocyclic ring containing an oxygen atom. The two cyclic structures for D (+) glucose are shown in the pyranose form

(ii) Haworth Structure of Fructose:
Haworth has purposed furanose ring structures for the two anomeric forms of D (-) fructose. Furan is a five-membered heterocyclic ring with an oxygen atom. This forms the basis of the two cyclic structures of D (-) fructose known as α – D (-) fructofuranose and β – D (-) fructofuranose.

Question 15.
Define proteins and give their classification.
Answer:
Proteins:
A polypeptide with more than hundred amino acid residue having molecular mass higher than 10,000 u is called a protein.

RBSE Class 12 Chemistry Chapter 14 Long Answer Type Questions

Question 1.
Write the general chemical reactions of glucose.
Answer:
General chemical reactions of glucose can be grouped under two categories:
(a) Reactions due to alcoholic groups:
There are four secondary and one primary alcoholic groups present in a molecule of glucose. Some of the reactions of alcoholic groups are as follows:
(i) Acetylation:

(ii) Methylation:
Glucose reacts with methyl alcohol in the presence of dry HCl gas forming methyl glucoside which is an ether.

(b) Reactions due to aldehydic (-CHO) group
(i) Reduction:
When reduced with Na-Hg and water, the – CHO group in glucose is reduced to a primary alcoholic group to give sorbitol.

When reduction is carried with HI and red P, the product is n-hexane. This suggests that all six carbon atoms in glucose are linked to form a straight chain.

(ii) Oxidation:
With mild oxidising agent like Br2 water, glucose is oxidised to gluconic acid which is a carboxylic acid. This shows that the carbonyl group present in glucose is an aldehydic group.

In case the oxidation is carried out with conc. HNO3 the primary alcoholic group is also oxidised to carboxyl group to form a dicarboxylic acid which is saccharic acid.

(iii) Rection with hydrogen cyanide:
Glucose forms a cyanohydrin derivative with hydrogen cyanide.

(iv) Reaction with hydroxylamine:
Glucose reacts with hydroxylamine to form an oxime.

(v) Action with Tollen’s Reagent:
When Tollen’s reagent is heated with glucose in a clean test tube, a shining mirror gets deposited at the bottom of the test tube.
AgNO3 +NH4OH → AgOH+NH4NO3
AgOH + 2NH4OH → Ag(NH3)2OH + 2H2O

(vi) Action with Fehling’s Solution:
When heated with glucose, Fehling solution gives a red precipitate.

(vii) Reaction with Phenyl Hydrazine:
Glucose contains in it an aldehydic group and is expected to react with phenylhydrazine to form phenylhydrazone. However when the excess of phenylhydrazine is used the product formed is an osazone i.e., glucosamine.

Question 2.
What are the main sources of cellulose and starch? Explain their structure in brief.
Answer:
(i) Cellulose (C6H10O5)n
Sources:
Cellulose occurs exclusively in plants and it is the most abundant organic substance in the plant kingdom. It is a predominant constituent of the cell wall of the plant cell. It is present in wood, cotton clothes, jute, cotton etc. Inwood, it is present 50%, in dry grasses, it is 40 – 45%, injure 60 – 65%, in cotton 90 – 95%, in cotton clothes 90% cellulose and rest in fats and waxes.
Structure:
Cellulose is composed of β-D-glucose units linked by (1 → 4)glycosidic bonds. The X-ray analysis has shown that there are large linear chains of β – D(+) glucose molecules lying side by side in the form of bundles held together by H^–bonding in the neighbouring hydroxyl groups. In each linear chain, the D(+) glucose unit are attached to each other by C1 to C4 bonds through β – glycosidic linkages as shown below.

(ii) Starch (C6H10O5)n
Sources:
Starch is a major source of carbohydrates which are very much essential to the human body since they supply energy to the body. It occurs as granules mainly in seeds, fruits, tubers and also in the roots of plants. The chief commercial sources of starch are wheat, maize, rice, potatoes etc.
Structure:
Starch consists of two polysaccharide components. They are
(a) Amylose 15% – 20% soluble in water.
(b) Amylopectin (80% – 90%) insoluble in water.

Question 3.
What is the final product obtained by hydrolysis of the following:
(1) Fructose
(2) Lactose
(3) Sucrose
(4) Starch
(5) Maltose
(6) Cellulose
Answer:
(1) Fructose:
Being a monosaccharide it does not hydrolyse to is give simpler units at normal conditions.
(2) Lactose:
Upon hydrolysis with dilute mineral acid or with enzyme lactose gives equivalent amounts of β – D glucose and β – D galactose which is isomeric in nature.

(3) Sucrose:
Upon hydrolysis with dilute mineral acid or by enzyme sucrose form an equimolar mixture of D (+) glucose and D (-) fructose.

(4) Starch:
Upon hydrolysis with diastase enzyme starch gives maltose.

(5) Maltose:
Upon hydrolysis in presence of acid maltose gives two molecules of D – glucose.

(6) Cellulose:
Upon hydrolysis at high pressure in the presence of dilute acid cellulose gives β – D-glucose.

Question 4.
Define proteins. Explain its hydrolysis. Explain the primary and secondary structures of proteins.
Answer:
The name “Protein” is given by Berzelius in 1938. The word protein is derived from the Greek word, “proteios” which means primary or of prime importance because proteins are the most important chemical substances which are necessary for growth, repair and development of life.
Proteins are found in all living cells. Main sources of proteins are milk, cheese, pulses, peanuts, fish, meat, etc. They occur in every part of the body and form the fundamental basis of structure and functions of life. In humans, hair, skin, haemoglobin, nails, enzymes and cells etc., are made of proteins.

Composition of Proteins:
Proteins are the high molecular mass complex biopolymers of amino acids and organic compounds that are most important for life. Chemically, all proteins are nitrogen-containing carbon compounds. All proteins contain the elements carbon, hydrogen, oxygen, nitrogen and sulphur some of these may also contain phosphorus, iodine and traces of metals like iron, copper, zinc and manganese. Amino acids are liberated when proteins are hydrolyzed. Proteins are the polymers of α – amino acids.

On partial hydrolysis of proteins, peptides of different molecular weight are obtained which on complete hydrolysis gives α – amino acids. This means that proteins are polypeptides of α – amino acids. Their molecular weight is more than 10,000. These are monodisperse natural polymers.

Question 5.
Write the functions of enzymes. Also, give their classification.
Answer:
Functions of enzymes:

Where it is made?	Where does it work?	Enzyme	Substrate	Products
Salivary glands	Mouth	Amylase	Starch	Amino acids
Stomach cells	Stomach	Protease	Protein	Amino acids
Liver	Small intestine	Bile salts	Fat	Fat droplets
Pancreas	Small intestine	Amylase, Protease Lipase	Starch Protein Fat	Maltose amino acids glycerol and fatty acids
Small intestine	Small intestine	Maltase Protein	Maltase Protein	Glucose amino acids

IUB (International Union of Biochemistry) in 1965 classified enzymes into six groups:

Classification of Enzymes

S.No.	Group of enzyme	Reaction catalysed	Examples
1.	Oxidoreductases	Transfer of hydrogen and oxygen atoms or electrons from one substrate to another.	Dehydrogenases
2.	Transferases	Transfer of a specific group (a phosphate or methyl etc.) from one substrate to another.	Transaminase Kinases
3.	Hydrolases	Hydrolysis of a substrate.	Esterases Digestive enzymes
4.	Isomerases	Change of the molecular form of the substrate.	Phosphohexose isomerase, Fumarase
5.	Lyases	Nonhydrolytic removal of a group or addition of a group to a substrate.	Decarboxylases Aldolases
6.	Ligases(Synthetases)	Joining of two molecules by the formation of new bonds.	Citric acid synthetase

Question 6.
Write the names and biological functions of hormones secreted by the pituitary gland and thyroid gland.
Answer:

S.No.	Gland	Position	Hormone and their nature	Biological Functions
1.	Pituitary	Beneath Brain	(i) Follicle stimulating hormone (FSH) (Protein)	Stimulates spermatogenesis in male, growth of ovarian follicle in a female.
	(a) Anterior Lobe		(ii) Luteinising hormone (LH) (Protein)	Induces testosterone secretion in male. Ovulation and secretion of estrogen and progesterone in females.
			(iii) Somatotrophin (STH) (Protein)	Regulates protein metabolism and growth of bones. Deficiency causes dwarfism, excesses produce gigantism in young and acromegaly in adults.
			(iv) Prolactin (Protein)	Controls development of milk glands.
			(v) Adrenocorticotrophin (ACTH) (Peptide)	Stimulates secretion of hormones of the adrenal cortex.
			(vi) Thyrotrophin (TSH) (Protein)	Stimulates secretion of thyroid hormone.
	(b) Intermediate lobe		(i) Melanocyte-stimulating hormone (MSH) (Peptide)	Controls skin colour.
	(c) Posterior Lobe		(ii) Oxytocin (Peptide)	Regulate uterine contraction and release of milk.
2.	Thyroid	Front of larynx	(i) Thyroxine	Regulate oxidation of food, excess causes exophthalmic goitre.
			(ii) Triiodothyronine	Deficiency produces cretinism in children, myxoedema in adults. Also, control the working of kidneys. Deficiency also reduces urine output.
			(iii) calcitonin (CT) (Peptide)	Controls calcium and phosphorus balance.

Question 7.
What is vitamin B complex? Write the name of the disease caused by a deficiency of these vitamins.
Answer:
Vitamin B complex is a group of at least 13 vitamins usually named as B1, B2, B3 …etc. But to prevent confusion, their chemical names are now frequently used. The various members of the vitamin B complex are not related either chemically or physiologically yet they have many things in common. Diseases caused by the various vitamins of this group summarizes below:

S.no	Vitamin complex	Deficiency diseases
1.	Vitamin B1	Beri-beri (paralysis of legs, normal weakness) and lack of appetite.
2.	Vitamin B2 or riboflavin	Stunted growth, ulcers in mouth, skin dryness in corners of mouth and lips.
3.	Vitamin B3 (it is made up of three substance pyridoxal, pyridoxine and pyridoxine)	Anaemia (lack of blood), nervous system and normal weakness body.
4.	Vitamin B12 or Cyano-cobalamine	Pernicious Anaemia, loss of consciousness and swelling in the face.

Question 8.
Explain the molecular structure of DNA.
Answer:
Molecular structure of DNA:
DNA molecule is made up of three units. These are nitrogenous bases:
1. Sugar and
2. Phosphoric acid
Nitrogenous bases:
DNA molecule contains two types of nitrogenous bases i.e., purine bases and pyrimidine bases.

1. Sugar:
2 – Deoxyribose sugar (pentose sugar) is present in the DNA molecule. It contains a five-membered heterocyclic ring and is generally called pentose sugar.

Arrangement of constituents in DNA molecule
A combination of base and Sugar is known as a nucleoside. Similarly, base, sugar phosphates form nucleotides. DNA molecules are the polymer of nucleotides.
Nitrogenous base + Sugar = Nucleoside
Nucleoside + Phosphoric acid = Nucleotide
Nucleotides = Polynucleotide

DNA is a macromolecule consisting of two polydeoxyribonucleotide chains. Each polydeoxyribonucleotide is formed by linkage of many deoxyribonucleotides through phosphodiester bonds. These bonds are present between two adjacent nucleotide components.
In DNA molecules Adenine base joined with Thymine base by a double bond and cytosine with guanine by triple bond ie., A=T, C=G.

Question 9.
How protein is synthesized by nucleic acid? Explain.
Answer:
Synthesis of protein:
Protein synthesis is another important function of nucleic acids. The process of protein synthesis is a very complex process. As stated earlier, 20 main acids combine in different ways to synthesise protein. Living cells have above 200 enzymes and more than 70 RNA take part in protein synthesis. Protein synthesis is done by different RNA molecules. DNA provides code for protein synthesis.
The synthesis of protein takes place in two steps:
1. Transcription:
Transmission of heredity information to RNA.
2. Translation:
Transmission of information from RNA to protein, the synthesis of protein from DNA can be represented as:
1. Transcription:
The process which leads to the synthesis of RNA is called transcription. It can also be defined as the process of copying genetic information from one strand of the DNA into RNA is termed as transcription. Three types of RNA takes part in protein synthesis (m-RNA, r-RNA and t-RNA) Which are synthesised by DNA.
Transcription involves the binding of RNA-polymerase at the promoter site on DNA. As it moves along (through structural gene), the DNA unwinds and one of the two strands acts as a template to synthesize a meaningful RNA and other strand act as non-coding. A complementary RNA strand is synthesized with A, U, C and G as bases. RNA synthesis is terminated when the RNA-polymerase falls off a terminator sequence on the DNA.

2. Translation:
This process is very complex. The formation of protein from m-RNA is known as translation. More than 100 macromolecules are used in this process such as m-RNA, t-RNA, ribosomes etc. It refers to the process of polymerisation of amino acids to form a polypeptide. The order and sequence of amino acids are defined by the sequence of bases in the m-RNA. The amino acids are joined by a bond which is known as a peptide bond. Protein synthesis takes place in the cytoplasm.
After transcription, m-RNA moves to a ribosome in cytoplasm form nucleus. r-RNA components of ribosomes. The base sequence in RNA is studied in three groups. This group of three nucleotide triplet is known as a codon. Each codon represents a particular amino acid. t-RNA has a specific sequence of bases at the ends which are complementary to the base sequence of m-RNA. At the other end of t-RNA, a specific amino acid is attached.
In ribosomes, amino acids retransferred from t-RNA to m-RNA. Those amino acids whose base sequence in m-RNA are bonded by m-RNA through peptide linkage. In this way, the polypeptide chain is formed and protein synthesis takes place. In proteins, a sequence of amino acids is determined by m-RNA and in m-RNA, a base sequence is determined by DNA. This means that the translation of amino acids takes place by DNA on ribosomes.

RBSE Solutions for Class 12 Chemistry

RBSE Solutions for Class 12 Chemistry Chapter 13 Organic Compounds with Functional Group-Containing Nitrogen

June 1, 2019 by Prasanna Leave a Comment

Rajasthan Board RBSE Class 12 Chemistry Chapter 13 Organic Compounds with Functional Group-Containing Nitrogen

RBSE Class 12 Chemistry Chapter 13 Text Book Questions

RBSE Class 12 Chemistry Chapter 13 Multiple Choice Questions

Question 1.
Which of the following is most basic?
(a) CH₃NH₂
(b) (CH₃)₂NH
(c) (CH₃ )₃N
(d) C₆H₅NH₂

Question 2.
Hinsberg’s reagent is
(a) Benzene sulphonyl chloride
(b) Benzene sulphonic acid
(c) Benzene sulphonamide
(d) Phenyl isocyanide

Question 3.
C₃H₉N does not show
(a) Primary amine
(b) Quaternary salt
(c) Tertiary amine
(d) Secondary amine

Question 4.
The hybridisation of N atom in alkyl amine is
(a) sp²(b) sp³
(c) sp
(d) sp³d

Question 5.
The molecular formula of a compound having mustard oil smell is
(a) RCN
(b) RNC
(c) RNCO
(d) RNCS

Question 6.
The formula for chloropicrin is
(a) C(NO₂)Cl₃
(b) CCl(NO₂)₃
(c) C(NO₂)₂Cl₂
(d) None of these

Question 7.
Nitration of benzene gives nitrobenzene when HNO₃ and H₂SO₄ take part in the reaction. Here HNO₃ works as
(a) as a base
(b) as an acid
(c) a reducing agent
(d) as a catalyst

Question 8.
Benzene diazonium chloride reacts with X gives dye, here reactant X is
(a) C₂H₅OH
(b) C₆H₆
(c) C₆H₅NH₂
(d) H₂O

Question 9.
The formula of acetonitrile is
(a) CH₃CN
(b) CH₃COCN
(c) CH₃CH₂CN
(d) CNCH₂COOH

Question 10.
The major product of the reaction of methylamine with Tilden reagent is
(a) CH₃OH
(b) CH₃CHO
(c) CH₃Cl
(d) CH₃COOH

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

RBSE Class 12 Chemistry Chapter 13 Very Short Answer Type Questions

Question 1.
What is the reason that aromatic diazonium salt is more stable than aliphatic diazonium salt?
Answer:
Aromatic diazonium salt is more stable than aliphatic diazonium salt due to the resonance stability of aromatic diazonium salt.

Question 2.
Alkylamines are strong base than ammonia. Explain.
Answer:
Alkylamine is strongly basic than ammonia due to the electron releasing nature of the alkyl group.
Basicity α +I effect
Alkylamine having more + I effect than NH₃.

Question 3.
Identify X and Y in the following reactions:
R R – CO NH₂ \(\underrightarrow { { Br }_{ 2 }/{ NaOH } } \) X \(\underrightarrow { { CHCl }_{ 3 }/{ KOH } } \) Y
Answer:
X= R – NH₂ (Alkylamine)
Y = RNC N (Alkyl isocyanide)

Question 4.
Identify A and B in the following reactions.

Answer:

Question 5.
Dimethylamine is a stronger base than methylamine. Give reason.
Answer:
Dimethylamine is a stronger base than methylamine. Due to the more + I effect. + I effect for dimethyl group is more than the methyl group. So dimethylamine is the stronger base.

Question 6.
Give IUPAC name and structural formula of vinyl cyanide.
Answer:
Structural formula – CH₂ = CH – CN
IUPAC Name – Prop – 2 – en – nitrile.

Question 7.
Write the equation of the Mendius reduction.
Answer:

Question 8.
Write the equation for the conversion of aniline to phenyl isocyanide.
Answer:

Question 9.
Write the reaction to obtain ethanol from Ethylamine.
Answer:

Question 10.
Give the structural formula of urea and write its IUPAC name.
Answer:

IUPAC name – Carbonyl diamide.

Question 11.
Write the reaction for the reduction of nitrobenzene in the presence of Zn + NH₄Cl.
Answer:

Question 12.
Complete the following reaction NH₄CNO → K?
Answer:

Question 13.
Give one equation to show the basic nature of ethanolamine.
Answer:
It is clear from the above equations that ethylamine is basic in nature.

Question 14.
The boiling point of primary amine is greater than tertiary amine. Why?
Answer:
The intermolecular association is maximum in primary amine. Tertiary amine does not have an intermolecular association because of the absence of hydrogen atom for hydrogen bond formation.

RBSE Class 12 Chemistry Chapter 13 Short Answer Type Questions

Question 1.
What is the biuret test for urea? Explain with the chemical equation.
Answer:
When urea is heated at 155°C, two molecules of urea react to release of NH₃ and gives biuret, a white crystalline solid.

Above reaction is a test of urea because on adding an alkaline solution of copper sulphate to white solid biuret, violet colour is obtained.

Question 2.
Give the reaction of urea with:

Formaldehyde
Hydrazine
Malonic acid

Answer:
1. Reaction with Formaldehyde: Urea reacts with formaldehyde and forms dimethyl urea.

2. Reaction with Hydrazine: Urea forms semicarbazide with hydrazine.

3. Reaction with Malonic Acid: Urea reacts with malonic acid to form barbituric urea or malonic urea.

Question 3.
Complete the following reactions:
R-X + KCN → ?
R-X + AgCN → ?
Answer:

KCN is an ionic compound. On ionisation, it gives CN~ ion in which C and N both are nucleophiles but due to more nucleophilicity of negatively charged carbon, cyanide forms with KCN. Ag-C = N is a covalent compound. A lone pair of electrons is present on N atom of Ag-C=N which acts as a nucleophile. Hence, Ag- C N forms isocyanide.

Question 4.
Write the balanced chemical equation for the reduction of nitrobenzene in

Basic medium
Neutral medium

Answer:
1. Reduction of nitrobenzene in basic medium:

2. Reduction of nitrobenzene in neutral medium:

Question 5.
Write the aliphatic amines in increasing order of their basicity and give a short note on basicity.
Answer:
Due to the presence of unshared pair of electrons on nitrogen, amines are basic in nature. Aliphatic amines are a stronger base than ammonia due to +I effect of alkyl groups. Aromatic amines are weaker bases than ammonia due to the electron withdrawing nature of the aryl group.
The order of basicity of amines in the gaseous phase:
3° Amines > 2° Amines > 1° Amines > NH₃
In aqueous solution basicity of amines:

Question 6.
Write the structures of A, B and C in the following reactions:

Answer:
(i)

(ii)

(iii)

Question 7.
Write the reaction for hydrolysis of urea in different mediums.
Answer:
Hydrolysis of urea takes place with dilute acid, base and at high temperatures.
Basic medium:
NH₂CONH₂ + 2NaOH + 2HOH → 2NH₄OH +Na₂CO₃
Acidic medium:
NH₂CONH₂ + 2HCl + HOH → 2NH₄Cl + CO₂
At High temperature:
NH₂CONH₂ + 2HNO₂ → CO₂ + 2N₂+ 3H₂O

RBSE Solutions for Class 12 Chemistry

Rajasthan Board RBSE Class 12 Chemistry Chapter 17 Chemistry in Daily Life

RBSE Class 12 Chemistry Chapter 17 Text Book Questions

RBSE Class 12 Chemistry Chapter 17 Very Short Answer Type Questions

RBSE Class 12 Chemistry Chapter 17 Short Answer Type Questions

RBSE Class 12 Chemistry Chapter 17 Long Answer Type Questions

Rajasthan Board RBSE Class 12 Chemistry Chapter 16 Stereo Chemistry

RBSE Class 12 Chemistry Chapter 16 Text Book Questions

RBSE Class 12 Chemistry Chapter 16 Multiple Choice Questions

RBSE Class 12 Chemistry Chapter 16 Very Short Answer Type Questions

RBSE Class 12 Chemistry Chapter 16 Short Answer Type Questions

RBSE Class 12 Chemistry Chapter 16 Long Answer Type Questions

Rajasthan Board RBSE Class 12 Chemistry Chapter 15 Polymers

RBSE Class 12 Chemistry Chapter 15 Text Book Questions

RBSE Class 12 Chemistry Chapter 15 Multiple Choice Questions

RBSE Class 12 Chemistry Chapter 15 Very Short Answer Type Questions

RBSE Class 12 Chemistry Chapter 15 Short Answer Type Questions

RBSE Class 12 Chemistry Chapter 15 Long Answer Type Questions

Rajasthan Board RBSE Class 12 Chemistry Chapter 14 Bio-Molecules

RBSE Class 12 Chemistry Chapter 14 Text Book Questions

RBSE Class 12 Chemistry Chapter 14 Multiple Choice Questions

RBSE Class 12 Chemistry Chapter 14 Very Short Answer Type Questions

RBSE Class 12 Chemistry Chapter 14 Short Answer Type Questions

RBSE Class 12 Chemistry Chapter 14 Long Answer Type Questions

Rajasthan Board RBSE Class 12 Chemistry Chapter 13 Organic Compounds with Functional Group-Containing Nitrogen

RBSE Class 12 Chemistry Chapter 13 Text Book Questions

RBSE Class 12 Chemistry Chapter 13 Multiple Choice Questions

RBSE Class 12 Chemistry Chapter 13 Very Short Answer Type Questions

RBSE Class 12 Chemistry Chapter 13 Short Answer Type Questions

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