NCERT Solutions For Class 12 Chemistry Chapter 15 Polymers

NCERT INTEXT QUESTIONS

15.1 What are polymers?
Ans. Polymers are high molecular mass substances (10³ — 10⁷u) consisting of a very large number of simple repeating structural units joined together through covalent bonds in a linear fashion. They are also called macromolecules. Ex: polythene, nylon 6,6, bakelite, rubber, etc.

15.2 How are polymers classified on the basis of structure?
Ans. On the basis of structure, polymers are classified as :
(i)Linear polymers in which the monomers are joined together to form long straight chains of polymer molecules. Forex: HDPE, PVC, nylons, etc.
(ii)Branched chain polymers in which the monomers not only join in linear fashion but also form branches of different lenghts along the main chain. For ex : LDPE, glycogen, etc.
(iii)Cross-linked polymers in which the intially formed linear polymer chains join together to form 3D network structure. For ex : bakelite, Urea-formaldehyde resin, etc.

15.3 Write the names of the monomers of the following polymers:


Ans. (i) Hexamethylene diamine NH₂-(CH₂)6NH₂ and adipic acid HOOC – (CH₂)₄ – COOH
(ii)Caprolactum
(iii)Tetrafluoroethene F₂C = CF₂

15.4 Classify the following as addition and condensation polymers:
Terylene, Bakelite, Polyvinyl chloride,Polythene
Ans.Addition polymers: Polyvinyl chloride, Polythene
Condensation polymers : Terylene, bakelite.

15.5 Explain the difference between Buna- N and Buna-S.
Ans. Both are copolymers. Buna-N is a copolymer of 1,3-butadiene and acrylonitrile whereas Buna-S is a copolymer of 1,3-butadiene and styrene.

15.6 Arrange the following polymers in increasing order of their intermolecuiar forces.
(i)Nylon 6,6, Buna-S, Polythene
(ii)Nylon 6, Neoprene, Polyvinyl chloride
Ans. On the basis of intermolecuiar forces, polymers
are classified as elastomers, fibres and plastics. The increasing order of intermolecuiar forces is: Elastomer < Plastic < fibre.
Thus, we have
(i)Buns-S < Polythene < Nylon 6,6
(ii)Neoprene < Polyvinyl chloride < Nylon 6.

NCERT EXRECISES

15.1 Explain the terms polymer and monomer.
Ans. Polymers are high molecular mass substances consisting of a very large number of simple repeating structural units joined together through covalent bonds in a regular fashion. Polymers are also called macromolecules. Some examples are polythene, nylon-66, bakelite, rubber, etc. Monomers are the. simple and reactive molecules from which the polymers are prepared either by addition or condensation polymerisation. Some examples are ethene, vinyl chloride, acrylonitrile, phenol and formaldehyde etc.

15.2 What are natural and synthetic polymers? Give two examples of each type.
Ans.Natural polymers: Polymers which are found in nature,i.e., in animals and plants are called natural polymers, e.g., proteins, starch, cellulose, nucleic acids, resins and natural Sol. rubber.
Synthetic polymers: Man-made polymers are called synthetic polymers, e.g., plastics (polythene, PVC), synthetic fibres (polyester, 15.8 nylon-66) and synthetic rubber (neoprene, Buna-S).

15.3 Distinguish between the terms homopolymer and copolymer and give an example of each.
Ans.Polymers whose repeating structural units are derived from only one type of monomer units are called homopolymers, e.g., PVC polyethene, PAN, teflon, polystyrene, nylon- 6 etc.
Polymers whose repeating structural units are derived from two or more types of monomer molecules are copolymers, e.g., Buna-S, Buna-N, nylon-66, polyester, bakelite.

15.4 How do you explain the functionality of a monomer?
Ans.Functionality means the number of binding sites in a molecule. For example, the functionality of ethene, propene, styrene, acrylonitrile is one while that of 1,3-butadiene, adipic acid, terephthaliC’ acid, hexa methylenediamine is two.

15.5 Define the term polymerisation?
Ans.It is a process of formation of a high molecular Sol. mass polymer from one or more monomers by linking together a large number of repeating structural units through covalent bonds.

15.6 Is (-NH — CHR—CO-)_n a homopolymer or copolymer?
Ans.It is a homopolymer because the repeating structural unit has only one type of monomer, i.e., NH₂—CHR—COOH.

15.7 In which classes, the polymers are classified on the basis of molecular forces?
Ans.(a)Elastomers
(b)Fibres
(c)Thermoplastics
(d)Thermosetting plastics

15.8 How can you differentiate between addition and condensation polymerisatiop?
Ans.In addition polymerization, the molecules of the same or different monomers simply add on to one another leading to the formation of a macromolecules without elimination of small molecules like H₂O, NH₃ etc. Addition polymerization generally occurs among molecules containing double and triple bonds. For example, formation of polythene from ethene and neoprene from chloroprene, etc. In condensation polymerisation, two or more bifunctional trifimctional molecules undergo a series of independent condensation reactions usually with the elimination of simple molecules like water, alcohol, ammonia, carbon dioxide and hydrogen chloride to form a macromolecule. For example, nylon-6,6 is a condensation polymer of hexamethylenediamine and adipic acid formed by elimination of water molecules.

15.9 Explain the term copolymerisation and give two examples.
Ans.When two or more different monomers are allowed to polymerise together the product formed is called a copolymer, and the process in called copolymerisation. Example, Buna-S and Buna-N. Buna- S is a copolymer of 1, 3- butadiene and styrene while Buna-N is a copolymer of 1,3-butadiene and acrylonitrile.

15.10 Write the free radical mechanism for the polymerisation of ethene.
Ans.

15.11 Define thermoplastics and thermo setting polymers with two examples of each
Ans. Thermoplastics polymers are linear polymer which can be repeatedly melted and moulded again and again on heating without any change in chemical composition and mechanical strength. Examples are polythene and polypropylene.
Thermosetting polymers, on the other hand, are permanently setting polymers. Once on heating in a mould, they get hardened and set, and then cannot be softened again. This hardening on heating is due to cross- linking between different polymeric chains to give a three dimensional network solid. Examples are bakelite, melamine-foimaldehyde polymer etc.

15.12 Write the monomers used for gettingThe following polymers:
(i) Polyvinylchloride
(ii) Teflon (iii) Bakelite
Ans.

15.13 Write the name and structure of one of the common initiators used in free radical addition polymerisation.
Ans.

15.4 How does the presence of double bonds in rubber molecules influence their structure and reactivity?
Ans.Natural rubber is cis-polyisoprene and is obtained by 1, 4-polymerization of isoprene units. In this polymer, double bonds are located between C₂ and C₃ of each isoprene unit. These cis-double bonds do not allow the polymer chains to come closer for effective interactions and hence intermolecular forces are quite weak. As a result, natural rubber, i.e., cis-polyisoprene has a randomly coiled structure not the linear one and hence show elasticity.

15.5 Discuss the main purpose of vulcanisation of rubber.
Ans.Natural rubber has the following disadvantages:
(a) It is soft and sticky and becomes even more so at high temperatures and brittle at low temperatures. Therefore, rubber is generally used in a narrow temperature range (283-335 K) where its elasticity is maintained.
(b)It has large water absorption capacity, has low tensile strength and low resistance to abrasion.
(c)It is not resistant to the action of organic solvents.
(d)It is easily attacked by oxygen and other oxidising agents. .
To improve all these properties, natural rubber is vulcanised by heating it with about 5% sulphur at 373-415 K. The vulcanized rubber thus obtained has excellent elasticity over a larger range of temperature, has low water absorption tendency and is resistant to the action of organic solvents and oxidising agents.

15.16 What are the monomeric repeating units of Nylon-6 and Nylon 6,6?
Ans.

15.17 Write the names and structures of the monomers of the following polymers:
(i) Buna-S (ii) Buna-N (iii) Dacron (iv) Neoprene
Ans.

15.18 Identify the monomer in the following polymeric structures:

Ans.

15.19 How is dacron obtained from ethylene glycol and terephthalic acid?
Ans.Dacron is obtained by condensation polymerization of ethylene glycol and terephthalic acid with the elimination of water molecules. The reaction is carried out at 420 – 460 K in presence of a catalyst consisting of a mixture of zinc acetate and antimony trioxide.

15.20 What is a biodegradable polymer ? Give an example of a biodegradable aliphatic polyester.
Ans. Polymers which disintegrate by themselves over a period of time due to environment degradation by bacteria, etc., are called biodegradable polymers. Example is PHBV, i. e., Poly-β-Hydroxybutyrate-co-β- Hydroxyvalerate.

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NCERT Solutions For Class 12 Maths Chapter 2 Inverse Trigonometric Functions

Chapter: Chapter 2 – Inverse Trigonometric Functions

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NCERT Solutions for Class 12 Maths Chapter 4 Determinants

Subject: Maths
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NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids

NCERT INTEXT QUESTION

12.1 Write the structures of the following compounds:
(i)α-Methoxypropionaldehyde
(ii)3-Hydroxybutanal
(iii)2-Hydroxycyclopentane carbaldehyde
(iv)4-OxopentanaI
(v)Di-sec.butylketone
(vi)4-fluoroaeetophenone
Ans.

12.2 Write the structures of products of following reactions:

Ans.

12.3 Arrange the following compounds in increasing order of their boiling points:
CH₃CHO, CH₃CH₂OH ,CH₃OCH₃, CH₃CH₂CH₃
Ans.The order is : CH₃CH₂CH₃ < CH₃OCH₃ < CH₃CHO <CH₃CH₂OH
All these compounds have comparable molecular masses CH₃CH₂OH undergoes extensive intermolecular Il-bonding and thus its b.pt. is the highest. CH₃CHO is more pdlar than CH₃OCH₃ so that dipole-dipoie interactions in CH₃CHO are greater than in CH₃OCH₃. Thus, b.pt. of CH₃CHO > CH₃OCH₃. CH₃CH₂CH₃ has only weak van der waals forces between its molecules and hence has the lowest b.pt.

12.4 Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions
(i)Ehtanal, propanaL, propanone, butanone
(ii)Benzaldehyde,p-Tolualdehyde, p-Nitrobenzaldehyde, acetophenone.
Ans. (i) Butanone < Propanone < Propanal < Ethanal .This is because as the no. of alkyl groups attached to carbonyl carbon increases, +I-effect increases. As a result, e^– density

12.5.Predict the products of the following reactions:


Ans.

12.6 Give the 1UPAC names of the following compounds:
(i)PhCH₂CH₂COOH
(ii)(CH₃)₂ C=CHCOOH

Ans.(i) 3 – Phenylpropanoic acid
(ii) 3 – Methylbut-2-enoic acid
(iii)2-Methylcyclohexanecarboxylic acid
(iv)2,4,6 – Trinitrobenzoic acid

12.7 Show how each of the following compounds can be converted into benzoic acid.
(i)Ethylbenzene
(ii)Acetophenone
(iii)Bromobenzene
(iv)Phenylethene (styrene)
Ans.

12.8 Which acid of each pair would you expect to be stronger?
(i)CH₃CO₂H or FCH₂CO₂H
(ii)FCH₂CO₂H or ClCH₂CO₂H
(iii)FCH₂CH2CH₂CO₂H
or CH₃CH(F)CH₂CO₂H

Ans.

NCERT EXERCISES

12.1 What is meant by the following terms? Give an example of the reaction in each case.
(i) Cyanohydrin (ii) Acetal (iii) Semicarbazone
(iv) Aldol (v) Hemiacetal (vi) Oxime
(vii) Ketal (viii) Imine (ix) 2,4-DNP derivative
(x) Schiff’s base.
Ans. (i) Cyanohydrin: gem-Hydroxynitriles, i.e., compounds possessing hydroxyl and cyano groups on the same carbon atom are called cyanohydrins. These are produced by addition of HCN to aldehydes or ketones in a weakly basic medium.

(ii) gem – Dialkoxy compounds in which the two alkoxy groups are present on the terminal carbon atom are called acetals. These are produced by the action of an aldehyde with two equivalents of a monohydric alcohol in presence of dry HCl gas.

These are easily hydrolysed by dilute mineral acids to regenerate the original aldehydes. Therefore, these are used for the protection of aldehyde group in organic synthesis.
(iii) Semicarbazones are derivatives of aldehydes and ketones and are produced by action of semicarbazide on them in acidic medium.

(iv) Aldols are P-hydroxy aldehydes or ketones and are produced by the condensation of two molecules of the same or one molecule each of two different aldehydes or ketones in presence of a dilute aqueous base. For example,

(v)gem – Alkoxyalcohols are called hemiacetals. These are produced by addition of one molecule of a monohydric alcohol to an aldehyde in presence of dry HCl gas.
(vi)Oximes are produced when aldehydes or ketones react with hydroxyl amine in weakly acidic medium.

(vii) Ketals are produced when a ketone is heated with dihydric alcohols like ethylene glycol in presence of dry HCl gas or /3-toluene sulphonic acid (PTS).

These are easily hydrolysed by dilute mineral acids to regenerate the original ketones. Therefore, ketals are used for protecting keto groups in organic synthesis.
(viii) Compounds containing -C = N – group are called imines. These are produced when aldehydes and ketones react with ammonia derivatives.

(ix)2, 4-Dinitrophenyl hydrazone (i.e., 2,4-DNP derivatives) are produced when aldehydes or ketones react with 2,4-dinitrophenyl hydrazine in weakly acidic medium.

(x) Aldehydes and ketones react with primary aliphatic or aromatic amines to form azomethines or SchifFs bases.

12.2 Name the following compounds according to IUPAC system of nomenclature:
(i) CH₃CH (CH₃)—CH₂ CH₂—CHO (ii) CH₃CH₂COCH(C₂H₅)CH₂CH₂Cl (iii) CH₃CH=CHCHO (iv) CH₃COCH₂COCH₃
(v)CH₃CH(CH₃)CH₂C(CH3)2COCH₃
(vi)(CH₃)₃CCH₂COOH.
(vii)OHCC₆H₄CHO-p
Ans. (i)4-Methyl pentanal
(ii)6-Chloro-4-ethylhexan-3-one
(iii)But-2-en-l-al
(iv)Pentane-2,4-dione
(v)3,3,5-Trimethyl-hexan-2-one
(vi)3,3-Dimethyl butanoic acid
(vii)Benzene-1,4-dicarbaldehyde

12.3 Draw the structures of the following compounds:
(i) 3-Methylbutanal
(ii) p-Nitropropiophenone
(iii)p-Methylbenzaldehyde
(iv)4-Methylpent-3-en-2-one
(v)4-Chloropentan-2-one
(vi)3-Bromo-4-phenylpentanoic acid
(vii) pp’-Dihydroxybenzophenone
(viii)Hex-2-en-4-ynoic acid
Ans.

12.4 Write the IUPAC names of the following ketones and aldehydes. Wherever possible, give also common names.(i)CH₃CO(CH₂)₄CH₃ (ii) CH₃CH₂CH BrCH₂CH(CH₃)CHO (iii) CH₃(CH₂)₅CHO (iv) Ph—CH=CH—CHO

Ans.

12.5 Draw structures of the following derivatives:
(i)The 2,4-dinitrophenylhydrazone of benzaldehyde
(ii)Cydopropanone oxime
(iii)Acetaldehydedimethylacetal
(iv)The semicarbazone ofcyclobutanone
(v)The ethylene ketal of hexan-3-one
(vi)The methyl hemiacetal of formaldehyde
Ans.

12.6 Predict the products formed when cyclohexanecarbaldehyde reacts with following reagents. .
(i) PhMgBr and then H₃O⁺
(ii) Tollen reagent
(iii) Semicarbazide and weak acid
(iv)Excess ethanol and acid
(v)Zinc amalgam and dilute hydrochloric acid
Ans.

12.7Which of the following compounds would undergo aldol condensation, which the Cannizzaro reaction and which neither? Write the structures of the expected products of aldol condensation and Cannizzaro reaction.
(i)Methanal (ii) 2-Methylpentanal (iii) Benzaldehyde .
(iv) Benzophenone (v) Cyclohexanone (vi) 1-Phenylpropanone
(vii) Phenylacetaldehyde (viii) Butan-l-ol 1 (ix) 2,2-Dimethylbutanal
Ans. 2-Methylpertfanal, cyclohexanone, 1-phenylpropanone and phenylacetaldehyde contain one or more a-hydrogen and hence undergo aldol condensation. The reactions and the structures of the expected products are given below:

12.8 How will you convert ethanal into the following compounds?
(i)Butane-1,3-diol (ii)But-2-enal (iii)But-2-enoic acid
Ans.

12.9 Write structural formulas and names of four possible aldol condensation products from propanal and butanal. In each case, indicate which aldehyde acts as nucleophile and which as electrophile.
Ans.

12.10 An organic compound with the molecular formula C₉H₁₀O forms 2,4-DNP derivative, reduces Tollen’s reagent and undergoes Cannizzaro reaction. On vigorous oxidation, it gives 1,2-benzenedicarboxylic acid. Identify the compound. 
Ans. Since the given compound with molecular formula C₉H₁₀O forms a 2,4-DNP derivative and reduces Tollen’s reagent, it must be an aldehyde. Since it undergoes Cannizzaro reaction, therefore, CHO group is directly attached to die benzene ring.
Since on vigorous oxidation, it gives 1, 2-benzene dicarboxylic acid, therefore, it must be an ortho- substituted benzaldehyde. The only o-substituted aromatic aldehyde having molecular formula C₉H₁₀O is o-ethyl benzaldehyde. Ail the reactions can now be explained on the basis of this structure.

12.11 An organic compound (A) (molecular formula C₈H₁₆O₂) was hydrolysed with dilute sulphuric acid to give a carboxylic acid (B} and an alcohol (C). Oxidation of (C) with chromic acid produced (B). (Q on dehydration gives but-l-ene. Write equations for the reactions involved.
Ans.Since an ester A with molecular formula C₈H₁₆O₂ upon hydrolysis gives carboxylic acid B and the alcohol C and oxidation of C with chromic acid produces the acid B, therefore, both the carboxylic acid B and alcohol C must contain the same number of carbon atoms.
Further, since ester A contains eight carbon atoms, therefore, both the carboxylic acid B and the alcohol C must contain four carbon atoms each.
Since the alcohol C on dehydration gives but-l-ene, therefore, C must be a straight chain alcohol, i.e., butan-l-ol.
If C is butan-l-ol, then the acid B must be butanoic acid and the ester A must be butyl butanoate.The chemical equations are as follows:

12.12Arrange the following compounds in increasing order of their property as indicated:
(i) Acetaldehyde, Acetone, Di-tert butyl ketone, Methyl tert-butyl ketone (reactivity towards HCN) (ii) CH₃CH₂CH(Br)COOH, CH₃CH(Br)CH₂COOH, (CH₃)₂CH COOH,CH₃CH₂CH₂COOH (acid strength).
(iii) Benzoic acid, 4-Nitrobenzoic acid, 3,4-Dinitrobenzoic add, 4-Methoxybenzok acid (acid strength).
Ans. (i) The reactivity of aldehydes and ketones towards HCN addition decreases as the +1 – effect of the alkyl groups increases. Secondly it decreases with increase in steric hindrance to the nucleophilic attack byCN^– at the carbonyl carbon. Thus the decreasing order of reactivity towards HCN is,

(ii)We know that + I-effect decreases while -I-effect increases the acidic strength of carboxylic acids. Since + I-effect of isopropyl group is more than that of propyl group, therefore, (CH₃)₂CHCOOH is a weaker acid than CH₃CH₂CH₂COOH. Further since -I-effect decreases with distance, therefore CH₃CH₂CHBrCOOH is a stronger acid than CH₃CHBrCH₂COOH. Thus, the overall acid strength increases in the order:

(iii) Since electron-donating groups decreases the acidic strength, therefore, 4-methoxy benzoic acid is a weaker acid than benzoic acid. Further since electron withdrawing groups increase the acidic strength, therefore, both 4-nitrobenzoic acid and 3,4-dinitrobenzoic acid are stronger acids than benzoic acid. Further due to the presence of an additional -NO₂ group at /w-position with respect to -COOH group, 3,4-dinitrobenzoic acid is a stronger acid than 4-nitrobenzoic acid. Thus, the overall acidic strength increases in the order:4-methoxy benzoic acid < benzoic acid < 4-nitrobenzoic acid < 3,4-dinitrobenzoic acid.

12.13 Give simple chemical tests to distinguish between the following pairs of compounds.
(i)PropanalandPropanone (ii)Acetophenone and Benzophenone
(iii)Phenol and Benzoic acid (iv)Benzoic acid and Ethyl benzoate
(v)Pentan-2-one and Pentan-3-one (vi)Benzaldehyde and Acetophenone.
(vii)EthanalandPropanal
Ans.

12.14 Row will you prepare the following compounds from benzene? You may use any inorganic reagent and any organic reagent having not more than one carbon atom.
(i) Methyl benzoate (ii) m-nitrobenzoic acid (iii) p-nitrobenzoic acid
(iv) Phenylaceticacid (v) p-nitrobenzaldehyde
Ans.

12.15 How will you bring about the following conversions in not more than two steps?
(i) PropanonetoPropene
(ii) Benzoic acid to Benzaldehyde
(iii) Ethanol to 3-Hydroxybutanal
(iv) Benzene to m-Nitroacetophenone
(v)Benzaldehyde to Benzophenone –
(vi)Bromobenzeneto 1-PhenylethanoL
(vii) Benzaldehyde to 3-Phenylpropan-1-ol .
(viil) Benzaldehyde to α Hydroxyphenylacetk acid
(ix) Benzoic acid to m-Nitrobenzy 1 alcohol
Ans.

12.16 Describe the following:
(i) Acetylation (ii) Cannizzaro reaction
(iii) Cross aldol condensation (iv) Decarboxylation
Ans. (i) Acetylation refers to the process of introducing an acetyl group into a compound namely, the substitution of an acetyl group for an active hydrogen atom. Acetylation is usually carried out in presence of a base such as pyridine, dimethylanitine, etc.

(ii)Cannizzaro reaction : Aldehydes which do not contain an a-hydrogen atom, when treated with concentrated alkali solution undergo disproportionation, i.e., self oxidation reduction. As a result, one molecule of the aldehyde is reduced to the corresponding alcohol at the cost of the other which is oxidised to the corresponding carboxylic acid. This reaction is called Cannizzaro reaction.

(iii) Cross aldol condensation: Aldol condensation between two different aldehydes is called cross aldol condensation.If both aldehydes contain a-hydrogens, It gives a mixture of four products.

(iv)Decarboxylation: The process of removal of a molecule of CO₂ from a carboxylic acid is called decarboxylation. Sodium salts of carboxylic acids when heated with soda-lime undergoes decarboxylation to yield alkanes.

12.17 Complete each synthesis by giving missing starting material, reagent or products.

Ans.

12.18 Give plausible explanation for each of the following:
(i) Cyclohexanone forms cyanohydrin in good yield but 2,2, fctrimethylcyclohexanone does not (ii) There are two – NH₂ groups in semicarbazide. However, only one is involved in the formation of semicarbazones.
(iii)During the preparation of esters from a carboxylic acid and an alcohol in the presence of an acid catalyst, the water or the ester should be removed as soon as it is formed.
Ans.

The yield of second reaction is very low because of the presence of three methyl groups at ex-positions with respect to the C = O, the nucleophilic attack by the CN^– ion does not occur due to steric hinderance. Since there is no such steric hindrance in cyclohexanone, therefore, nucleophilic attack by the CN^– ion occurs readily and hence cyclohexanone cyanohydrin is obtained in good yield.

Although semicarbazide has two – NH₂ groups but one of them (i.e., which is directly attached to C = O) is involved in resonance as shown above. As a result, electron density on N of this -NH₂ group decreases and hence it does not act as a nucleophile. In contrast, the other -NH₂ group (i.e.. attached to NH) is not involved in resonance and hence lone pair of electrons present on N atom of this -NH₂ group is available for nucleophilic attack on the C = O group of aldehydes and ketones.’
(iii) The formation of esters from a carboxylic acid and an alcohol in presence of an acid catalyst is a reversible reaction.

Thus to shift the equilibrium in the forward direction, the water or the ester formed should be removed as fast as it is formed.

12.19 An organic compound contains 69-77% carbon, 11-63 % hydrogen and rest oxygen. The molecular mass of the compound is 86. It does not reduce Tottens’ reagent but forms an addition compound with sodium hydrogensulphite and give positive iodoform test. On vigorous oxidation, it gives ethanoic and propanoic acid. Write the possible structure of the compound.
Ans.

Since the compound form sodium hydrogen sulphite addition product, therefore, it must be either an – aldehyde or methyl/ cyclic ketone. Since the compound does not reduce Tollens’ reagent therefore, it cannot be an aldehyde. Since the compound gives positive iodoform test, therefore, the given compound is a methyl ketone. Since the given compound on vigorous oxidation gives a mixture ofethanoic acid and propanoic acid, therefore, the methyl ketone is pentan-2-one, i.e.,

12.20 Although phenoxide ion has more number of resonating structures than carboxylate ion, carboxylic acid is a stronger acid than on phenol. Why?
Ans. Consider the resonating structures of carboxylate ion and phenoxide ion.

In case of phenoxide ion, structures (V – VII) carry a negative charge on the less electronegative carbon atom.Therefore, their contribution towards the resonance stabilization of phenoxide ion is very small.
In structures I and II, (carboxylate ion), the negative charge is delocalized over two oxygen atoms while in structures III and IV, the negative charge on the oxygen atom remains localized only the electrons of the benzene ring are delocalized. Since delocalization of benzene electrons contributes little towards the stability of phenoxide ion therefore, carboxylate ion is much more resonance stabilized than phenoxide ion. Thus, the release of a proton from carboxylic acids is much easier than from phenols. In other words, carboxylic acids are stronger acids than phenols.

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NCERT Solutions For Class 12 Chemistry Chapter 13 Amines

NCERT INTEXT QUESTIONS

13.1.Classify the following amines as primary, secondary and tertiary:

Ans. (i) 1° (ii)-3° (iii) 1° (iv) 2°

13.2.(i)Write the structures of different isomeric amines corresponding to the molecular formula, C₄H₁₁N.
(ii)Write 1UPAC names of all the isomers.
(iii)What type of isomerism is exhibited by different pairs of amines?
Ans.

Chain isomers: (a) and (c); (a) and (d); (b)and(c); (b)and(d)
Metamers: (e) and (g); (f) and (g) Functional isomers: All 10 amines are functional isomers of 2° and 3° amines and vice-versa.

13.3 How will you convert:
(i)Benzene into aniline
(ii)Benzene into N,N-dimethylaniline
(iii)Cl-(CH₂)₄-Cl into Hexane -1,6- diamine
Ans.

13.4 Arrange the following in increasing order of their basic strength: ‘
(i) C₂H₅NH₂,C₆H₅NH₂,NH₃,C₆H₅CH₂NH₂ and(C₂H₅)₂ NH
(ii) C₂H₅NH₂,(C₂H₅)₂NH,(C₂H₅)₃N,C₆H₅NH₂
(iii) CH₃NH₂, (CH₃)₂NH, (CH₃)₃N, C₆H₅NH₂, C₆H₅CH₂NH₂
Ans.C₆H₅NH₂ < NH₃ < C₆H₅CH₂NH₂ <  C₂H₅NH₂<(C₂H₅)₂NH
(ii)C₆H₅NH₂ < C₂H₅NH₂ < (C₂H₅)₃N <(C₂H₅)₂NH
(iii) C₆H₅NH₂ < C₆H₅CH₂NH₂ < (CH₃)₃ N < CH₃NH₂<(CH₃)₂NH

13.5 Complete the following acid-base reactions and name the products:
(i)CH₃CH₂CH₂NH₂+HCl ——–>
(ii)(C₂H₅)₃ N+HCl ——–>
Ans.

13.6 Write reactions of the final alkylation product of aniline with excess of methyl iodide in the presence of sodium carbonate solution.
Ans.

13.7 Write chemical reaction of aniline with benzoyl chloride and write the name of the product obtained.
Ans.

13.8 Write structures of different isomers corresponding to the molecular formula, C₃H₉N. Write IUPAC names of the isomers which will liberated N₂ gas on treatment with nitrons acid.
Ans. In ‘all, four structural isomers are possible. These are:

13.9 Convert:
(i)3-Methylanilineinto3-nitrotoluene
(ii)Aniline into 1,3,5- Tribromo benzene
Ans.

NCERT EXRECISES

13.1 Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines. 
(i) (CH₃)₂ CHNH₂ (ii) CH₃(CH₂)₂NH₂ (iii) CH₃NHCH(CH₃)₂
(iv) (CH₃)₃ CNH₂ (v) C₆H₅NHCH₃(vi) (CH₃CH₂)₂NCH₃
(vii)m-BrC₆H₄NH₂
Ans. (i) Propan-2-amine(1°)
(ii)Propan-1-amine (1°) ,
(iii)N-Methylpropan-2-amine (2°) .
(iv)2-Methylpropan-2-amine(l°)
(v)N-MethylbenzenamineorN-methylaniline(2°)
(vi)N-Ethyl-N-methylethanamine (3°)
(vii)3-Bromobenzenamine or 3-bromoaniline (1°)

13.2 Give one chemical test to distinguish between the following pairs of compounds:
(i)Methylamine and dimethylamine (ii) Secondary and tertiary amines (iii) Ethylamine and aniline (iv) Aniline and benzylamine (v) Aniline and N-Methylaniline.
Ans.

13.3 Account for the following
(i)pKb of aniline is more than that of methylamine
(ii) Ethylamine is soluble in water whereas aniline is not.
(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.
(iv)Although amino group is o and p – directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline.
(v)Aniline does not undergo Friedel-Crafts reaction.
(vi)Diazonium salts of aromatic amines are more stable than those of aliphatic amines.
(vii)Gabriel phthalimide synthesis is preferred for synthesising primary amines.
Ans. (i) In aniline, the lone pair of electrons on the N-atom is delocalised over the benzene ring.
As a result, electron density on the nitrogen . atom decreases. Whereas in CH₃NH₂,+ I-effect of -CH₃ group increases the electron density on the N-atom. Therefore, aniline is a weaker base than methylamine and hence its pKb value is higher than that of methylamine.
(ii) Ethylamine dissolves in water due to intermolecular H-bonding. However, in case of aniline, due to the large hydrophobic part, i.e., hydrocarbon part, the extent of H-bonding is very less therefore aniline is insoluble in water.

(iii)Methylamine being more basic than water, accepts a proton from water liberating OH^– ions,

(iv)Nitration is usually carried out with a mixture of cone HNO₃ + cone H₂SO₄. In presence of these acids, most of aniline gets protonated to form ahilinium ion. Therefore, in presence of acids, the reaction mixture consist of aniline and anilinium ion. Now, -NH2 group in aniline is activating and o, p-directing while the -+NH₃ group in anilinium ion is deactivating and rw-directing: Nitration of aniline (due to steric hindrance at o-position) mainly gives p-nitroaniline, the nitration of anilinium ion gives m-nitroaniline. In actual practice, approx a 1:1 mixture of p-nitroaniline and m-nitroaniline is obtained. Thus, nitration of aniline gives a substantial amount of m-nitroaniline due to protonation of the amino group.

13.4 Arrange the following:
(i) In decreasing order of pKb values: .
C₂H₅NH₂,C₆H₅NHCH₃,(C₂H₅)₂NH and C₆H₅NH₂
(ii) In increasing order of basic strength:
C₆H₅NH₂, C₆H₅N(CH₃)₂, (C₂H₅)₂ NH and CH₃NH₂.
(iii) In increasing order of basic strength:
(а)Aniline,p-nitroaniline andp-toluidine
(b)C₆H₅NH₂, C₆H₅NHCH₃, C₆H₅CH₂NH₂
(iv) In decreasing order of basic strength in gas phase:
C₂H₅NH₂, (C₂H₅)₂NH, (C₂H₅)₃N and NH₃
(v) In increasing order of boiling point:
C₂H₅OH, (CH₃)₂NH, C₂H₅NH₂
(vi) In increasing order of solubility in water:
C₆H₅NH₂,(C₂H₅)₂NH,C₂H₅NH₂
Ans. (i) Due to delocalisation of lone pair of electrons of the N-atom over the benzene ring,C₆H₅NH₂ and C₆H₅NHCH₃ are far less basic than C₂H₅NH₂ and (C₂H,)₂NH. Due to +I-effect of the -CH₃ group, C₆H₅NHCH₃ is little more basic that C₆H₅NH₂. Among C₂H₅NH₂ and (C₂H₅)₂NH, (C₂H₅)₂NH is more basic than C₂H₅NH₂ due to greater+I-effect of two -C₂H₅ groups. Therefore correct order of decreasing pKb values is:

(ii) Among CH₃NH₂ and (C₂H₅)₂NH, primarily due to the greater +I-effect of the two -C₂H₅ groups over one -CH₃ group, (C₂H₅)₂NH is more basic than CH₃NH₂.In both C₆H₅NH2 and C₆H₅N(CH₃)₂ lone pair of electrons present on N-atom is delocalized over the benzene ring but C₆H₅N(CH₃)₂ is more basic due to +1 effect of two-CH₃ groups.

(iii) (a) The presence of electron donating -CH₃ group increases while the presence of electron withdrawing -NO₂ group decreases the basic strength of amines.

(b) In C₆H₅NH₂ and C₆H₅NHCH₃, N is directly attached to the benzene ring. As a result, the lone pair of electrons on the N-atom is delocalised over the benzene ring. Therefore, both C₆H₅NH₂ and C₆H₅NHCH₃ are weaker base in comparison to C₆H₅CH₂NH₂. Among C₆H₅NH₂ and C₆H₅NHCH₃, due to +1 effect of-CH₃ group C₆H₅NHCH₃ is more basic.

(iv) In gas phase or in non-aqueous solvents such as chlorobenzene etc, the solvation effects i. e., the stabilization of the conjugate acid due to H-bonding are absent. Therefore, basic strength depends only upon the +I-effect of the alkyl groups. The +I-effect increases with increase in number of alkyl groups.Thus correct order of decreasing basic strength in gas phase is,

(v)Since the electronegativity of O is higher than thalof N, therefore, alcohols form stronger H-bonds than amines. Also, the extent of H-bonding depends upon flie number of H-atoms on the N-atom, thus the extent of H-bonding is greater in primary amine than secondary amine.

(vi)Solubility decreases with increase in molecular mass of amines due to increase in the size of the hydrophobic hydrocarbon part and with decrease iirthe number of H-atoms on the N-atom which undergo H-bonding.

13.5 How wjll you convert:
(i) Ethanoic acid into methanamine (ii) Hexanenitrile into 1-aminopentane
(iii) Methanol to ethanoic acid. (iv) Ethanamine into methanamine
(v) Ethanoic acid into propanoic acid (vi) Methanamine into ethanamine
(vii) Nitromethane into dimethylamine (viii) Propanoic acid into ethanoic acid?
Ans.

13.6 Describe a method for the identification of primary, secondary and tertiary amines. Also write chemical equations of the reactions involved.
Ans. The three type of amines can be distinguished by Hinsberg test. In this test, the amine is shaken with benzenesulphonyt chloride (C₆H₅SO₂Cl) in the presence of excess of aqueous NaOH or KOH. A primary amine reacts to give a clear solution, which on acidification yields an insoluble compound.

13.7 Write short notes on the following:
(i) Carbylamine reaction (il) Diazotisation (iii) ‘Hofmann’s bromamide reaction
(iv) Coupling reaction (v) Ammonolysis (vi) Acetylation
(vii) Gabriel phthalimide synthesis
Ans.(i) Carbylamine reaction: Both aliphatic and aromatic primary amines when warmed with chloroform and an alcoholic solution of KOH, produces isocyanides or carbylamines which have very unpleasant odours. This reaction is called carbylamine reaction.

(ii)Diazotisation: The process of conversion of a primary aromatic amino compound into a diazonium salt, is known as diazotisation. This process is carried out by adding an aqueous solution of sodium nitrite to a solution of primary aromatic amine (e.g., aniline) in excess of HCl at a temperature below 5°C.

(iii)Hoffmann’s bromamide reaction: When an amide is treated with bromine in alkali solution, it is converted to a primary amine that has one carbon atom less than the starting amide. This reaction is known as Hoffinann’s bromamide degradation reaction.

(iv) Coupling reaction: In this reaction, arene diazonium salt reacts with aromatic amino compound (in acidic medium) or a phenol (in alkaline medium) to form brightly coloured azo compounds. The reaction generally takes place at para position to the hydroxy or amino group. If para position is blocked, it occurs at ortho position and if both ortho and para positions are occupied, than no coupling takes place.

(v) Ammonolysis: It is a process of replacement of either halogen atom in alkyl halides (or aryl halides) or hydroxyl group in alcohols (or phenols) by amino group. The reagent used for ammonolysis is alcoholic ammonia. Generally, a mixture of primary, secondary and tertiary amine is formed.

(vi) Acetylation: The process of introducing an acetyl (CH₃CO-) group into molecule using acetyl chloride or acetic anhydride is called acetylation.

(vii) Gabriel phthalimide synthesis: It is a method of preparation of pure aliphatic and aralkyl primary amines. Phthalimide on treatment with ethanolic KOH gives potassium phathalimide which on heating with a suitable alkyl Or aralkyl halides gives N-substituted phthalimides, which on hydrolysis with dil HCI or with alkali give primary amines.

13.8 Accomplish the following conversions:
(i) Nitrobenzene to benzoic acid (ii) Benzene to m-bromophenol
(iii) Benzoic acid to aniline (iv) Aniline to 2,4,6-tribromofluorobenzene
(v) Benzyl chloride to 2-phenylethanamine (vi) Chlorobenzene to p-Chloroaniline
(vii) Aniline to p-bromoaniIine (viii)Benzamide to toluene
(ix) Aniline to benzyl alcohol.
Ans.

13.9 Give the structures of A,B and C in the following reaction:


Ans.

13.10 An aromatic compound ‘A’on treatment with aqueous ammonia and heating forms compound ‘B’ which on heating with Br₂ and KOH forms a compound ‘C’ of molecular formula C6H7N. Write the structures and IUPAC names of compounds A,B and C.
Ans. Since the compound ‘C’ with molecular formula C₆H₇N is formed from compound ‘B’ on treatment with Br₂ KOH, therefore, compound ‘B’ must be an amide and ‘C’ must be an amine.
The only amine having the molecular formula C₆H₇N, i. e., C₆H₅NH₂ is aniline.
Since ‘C’ is aniline, therefore, die amide from which it is formed must be benzamide (C₆H₅CONH₂). Thus, compound‘B’is benzamide. Since compound ‘B’ is formed from compound ‘A’ with aqueous ammonia and heating, therefore, compound ‘A’ must be benzoic acid.

13.11 Complete the following reactions:

Ans.

13.12 Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis?
Ans. The success of Gabriel phthalimide reaction depends upon the nucleophilic attack by the phthalimide anion on the organic halogen compound.
Since aryl halides do not undergo nucleophilic substitution reactions easily, therefore, arylamines, i.e., aromatic, primary amines cannot be prepared by Gabriel phthalimide reaction.

13.13 Write the reactions of (i) aromatic and (ii) aliphatic primary amines with nitrous acid.
Ans. Both aromatic and aliphatic primary amines react with HNO₂ at 273-278 K to form aromatic and aliphatic diazonium salts respectively. But aliphatic diazonium salts are unstable even at this low temperature and thus decompose readily to form a mixture of compounds. Aromatic and aliphatic primary amines react with

13.14 Give plausible explanation for each of the following:
(i) Why are amines less acidic than alcohols of comparable molecular masses?
(ii) Why do primary amines have higher boiling point than tertiary amines?
(iii) Why are aliphatic amines stronger bases than aromatic amines?
Ans. (i) Loss of proton from an amine gives an amide ion while loss of a proton from alcohol give an alkoxide ion.
R—NH₂—>R—NH^– +H⁺
R—O —H—>R— O^– +H⁺ .
Since O is more electronegative than N, so it wijl attract positive species more strongly in comparison to N. Thus, RO~ is more stable than RNH®. Thus, alcohols are more acidic than amines. Conversely, amines are less acidic than alcohols.
(ii) Due to the presence of two H-atoms on N-atom of primary amines, they undergo extensive intermolecular H-bonding while tertiary amines due to the absence of H-atom on the N-atom do not undergo H-bonding. As a result, primary amines have higher boiling points than tertiary amines of comparable molecular mass.
(iii) Aromatic amines are far less basic than ammonia and aliphatic amines because of following reasons:
(a)Due to resonance in aniline and other aromatic amines, the lone pair of electrons on the nitrogen atom gets delocalised over the benzene ring and thus it is less easily available for protonation. Therefore, aromatic amines are weaker bases than ammonia and aliphatic amines.
(b)Aromatic amines arS more stable than corresponding protonated ion; Hence, they hag very less tendency to combine with a proton to form corresponding protonated ion, and thus they are less basic.

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NCERT Solutions For Class 12 Chemistry Chapter 16 Chemistry in Everyday Life

NCERT INTEXT QUESTIONS

16.1 Sleeping pills are recommended by doctors to the patients suffering from sleeplessness but it is not advisable to take its doses without consultation with the doctor. Why?
Ans. Most of drugs taken in doses higher than recommended may produce harmful effects and act as poison and cause even death. Therefore, a doctor must always be consulted before taking the drug.

16.2 With refrence to which classification has the statement “ranitidine is an antacid”, been given?
Ans. This statement refers to the classification of drugs according to pharmacological effect because any drug which will be used to neutralise the excess acid present in the stomach will be called an antacid.

16.3 Why do we require artificial sweetening agents?
Ans. To reduce calorie intake and to protect teeth from decaying, we need artificial sweeteners.

16.4 Write the chemical equation for preparing sodium soap from glyceryl oleate and glyceryl palmitate. Structures of these compounds are given below:
(i)(C₁₅H₃₁COO)₃C₃H₅-Glyceryl palmitate
(ii)(C₁₇H₃₂COO)₃C₃H₅-Glyceryl oleate
Ans.

16.5 Following type of non-ionic detergents are present in liquid detergents, emulsifying agents and wetting agents. Label the hydrophilic and hydrophobic part in the molecule. Identify the functional group (s) present in the molecule.

Ans.

Functional groups present in the detergent molecule are:
(i)ether
(ii)1°alcoholic group

NCERT EXRECISES

16.1 Why do we need to classify drugs in different ways?
Ans.Drugs are classified in following different ways:
(a)Based on pharmacological effect.
(b)Based on action on a particular biochemical process.
(c)Based on chemical structure.
(d)Based on molecular targets.
Each classification has its own usefulness.
(а)Classification based on pharmacological effect is useful for doctors because it provides them the whole range of drugs available for the treatment of a particular disease.
(b)Classification based on action on a particular biochemical proc*ess is useful for choosing the correct compound for designing the synthesis of a desired drug.
(c)Classification based on chemical structure helps us to design the synthesis of a number of structurally similar compounds having different substituents and then choosing the drug having least toxicity.
(d)Classification on the basis of molecular targets is useful for medical chemists so that they can design a drug which is most effective for a particular receptor site.

16.2 Explain the term, target molecules or drug targets as used in medicinal chemistry.
Ans. Drugs interact with macromolecules like proteins, carbohydrates, lipids and nucleic acids thus these macro molecules are called drug targets. These macromolecules perform various functions in the body for example, proteins perform several roles in the body. Proteins which act as biological catalysts are called enzymes, those which are involved in communication system are called receptors. Carrier proteins carry polar molecules across the cell membrane. Nucleic acids have coded genetic information in the cell whereas lipids and carbohydrates form structural part of cell membranes.

16.3 Name the macro molecules that are chosen as drug targets.
Ans. Proteins, carbohydrates, lipids and nucleic acids are chosen as drug targets.

16.4 Why should not medicines be taken without consulting doctors?
Ans. Some drugs can cause side effects when drug binds to more than one type of receptor. Therefore, doctor’s consultation is must to choose the right drug that has the maximum affinity for a particular receptor site to have desired effect. Dose of the drug taken at a time is also crucial because some drugs in higher doses act as poisons and may cause death.

16.5 Define the term chemotherapy.
Ans. It is the branch of chemistry that deals with the treatment of diseases by using chemicals as medicines.

16.6 Which forces are involved in holding the drugs to the active site of enzymes?
Ans.The following forces are involved in holding the drugs to the active site of enzymes:
(a)Hydrogen bonding
(b)Ionic bonding
(c)Dipole-dipole interactions
(d)van der Waals interactions

16.7 While antacids and antiallergic drugs interfere with the function of histamines, why do these not interfere with the function of each other?
Ans. Drugs are designed to cure some ailment in one organ of the body do not affect the other because they work on different receptors. For example, secretion of histamine causes allergy. It also causes acidity due to release of hydrochloric acid in the stomach. Since antiallergic and antacids drugs work on different receptors, therefore, antihistamines remove allergy while antacids remove acidity.

16.8 Low level of noradrenaline is the cause of depression. What type of drugs are needed to cure this problem? Name two drugs.
Ans.In case of low level of neurotransmitter, . noradrenaline, tranquilizer (antidepressant) drugs are required because low levels of noradrenaline leads to depression. These drugs inhibit the enzymes which catalyse the degradation of noradrenaline. If the enzyme is inhibited, noradrenaline is slowly metabolized and can activate its receptor for longer periods of time thereby reducing depression. Two important drugs are iproniazid and phenylzine.

16.9 What is meant by the term broad spectrum antibiotics? Explain.
Ans. Broad spectrum antibiotics are effective against several different types or wide range of harmful bacteria. For example, tetracycline, chloramphenicol and of loxacin. Chloramphenicol can be used in case of typhoid, dysentry, acute fever, urinary infections, meningitis and pneumonia.

16.10 How do antiseptics differ from disinfectants? Give one example of each.
Ans. Antiseptics are chemical substances which prevent the growth of micro-organisms and may even kill them but they are not harmful for human or animal tissues. For example, dettol and savlon. They are generally applied on wounds, cuts, ulcers and diseased skin surfaces. Furacin and soframycin are well known antiseptic creams.
Disinfectants are chemical substances which kill microorganisms but are not safe to be applied to the living tissues. These are generally used to kill microorganisms present in the drains toilets, floors, etc. Some common examples of disinfectants are phenol ( 1% solution) and chlorine (0.2 to 0.4 ppm).

16.11 Why are cimetidine and ranitidine better antacids than sodium hydrogencarbonate or magnesium or aluminium hydroxide?
Ans. If excess of NaHCO₃ or Mg(OH)₂ or Al(OH)₃ is used, it makes the stomach alkaline and thus triggers the release of even more HCl which may cause ulcer in the stomach. In contrast, cimetidine and ranitidine prevent the interaction of histamine with the receptor cells in the stomach wall and thus release of HCl will be less as histamine stimulates the secretion of acid.

16.12 Name a substance which can be used as an antiseptic as well as disinfectant.
Ans. 0.2% solution of phenol acts as antiseptic while 1% solution acts as a disinfectant.

16.13 What are the main constituents of dettol?
Ans. Chloroxylenol .and α-terpineol in a suitable solvent.

16.14 What is tincture of iodine? What is its use?
Ans. 2-3% solution of iodine in alcohol and water is called tincture of iodine. It is a powerful antiseptic. It is applied on wounds.

16.15 What are food preservatives?
Ans. Chemical substances which are used to protect food against bacteria, yeasts and moulds are called preservatives. For example, sodium benzoate and sodium metabisulphite.

16.16 Why is the use of aspartame limited to cold foods and drinks?
Ans. This is because it decomposes at baking or cooking temperatures and hence can be used only in cold foods and drinks as an artificial sweetener

16.17 What are artificial sweetening agents? Give two examples.
Ans. Artificial sweeteners are chemical substances which are sweet in taste but do not add any calories to our body. They are excreted as such through urine. For example, saccharin, aspartame, alitame etc.

16.18 Name the sweetening agent used in the preparation of sweets for a diabetic patient.
Ans. Saccharine, aspartame or alitame may be used in the preparation of sweets for a diabetic patient.

16.19 What problem arises in using alitame as artificial sweetener?
Ans. Alitame is a high potency artificial sweetener.Therefore, it is difficult to control the sweetness of the food to which it is added.

16.20 How are synthetic detergents better than soaps?
Ans. They can be used in hard water as well as in acidic solution. The reason being that sulphonic acids and their calcium and magnesium salts are soluble in water thus they do not form curdy white precipitate with hard water but the fatty acids and their calcium and magnesium salts of soaps are insoluble. Detergents also works in slightly acidic solution due to formation of soluble alkyl hydrogen sulphates. Soaps react with acidic solution to form insoluble fatty acids.

16.21 Explain the following terms with suitable examples:
(i) cationic detergents (ii) anionic detergents and (iii) non-ionic detergents
Ans. (i) Cationic detergents: These are quaternary ammonium salts, chlorides, acetates, bromides etc containing one or more long chain alkyl groups. For example, cetyltrimethyl ammonium chloride.
(ii) Anionic detergents are called so because a large part of their molecules are anions. ‘These are of two types:
(a)Sodium alkyl sulphates: For example, sodium lauryl sulphate, C₁₁H₂₃CH₂OSO₃ Na⁺.
(b)Sodium alkylbenzenesulphonates.Vor example, sodium 4-(l-dodecyl) benzenesu Iphphonate (SDS).

(iii)Neutral or non-ionic detergents: These are esters of high molecular mass alcohols with fatty acids. These can also be obtained by treatment of long chain alcohols by with excess of ethylene oxide in presence of a base. For example, polyethylene glycol stearate,CH₃(CH₂)₁₆COO (CH₂CH₂O)₁₁ CH₂CH₂OH Polyethylene glycol stearate.

16.22 What are biodegradable and non-biodegradable detergents? Give one example of each.
Ans. Detergents having straight chain hydrocarbons are easily degraded (or decomposed) by microorganisms and hence are called biodegradable detergents while detergents containing branched hydrocarbon chains are not easily degraded by the microorganisms find hence are called non-biodegradable detergents. Consequently, non-biodegradable detergents accumulate in rivers and water ways thereby causing severe water pollution. Examples of biodegradable detergents are sodium lauryl sulphate, sodium 4-(-l-dodecyl) benzenesulphonate and sodium 4-(2-dodecyl) benzenesulphonate.
Examples of non-biodegradable detergents is sodium 4-(1, 3,5,7 – tetramethyloctyl) benzene sulphonate.

16.23 Why do soaps not work in hard water?
Ans. Hard water contains calcium and magnesium salts. Therefore, in hard water soaps get precipitated as calcium and magnesium soaps which being insoluble stick to the clothes as gummy mass.

16.24 Can you use soaps and synthetic detergents to check the hardness of water?
Ans. Soaps get precipitated as insoluble calcium and magnesium soaps in hard water but detergents do not. Therefore, soaps but not synthetic detergents can be used to check the hardness of water.

16.25 Explain the cleansing action of soaps.
Ans. Cleansing action of soaps : Soaps contain two parts, a large hydrocarbon which is a hydrophobic (water repelling) and a negative charged head, which is hydrophillic (water attracting). In solution water molecules being polar in nature, surround the ions & not the organic part of the molecule. When a soap is dissolved in water the molecules gather together as clusters, called micelles. The tails stick inwards & the head outwards. The hydrocarbon tail attaches itself to oily dirt. When water is agitated, the oily dirt tends to lift off from the dirty surface & dissociates into fragments. The solution now contains small globules of oil surrounded detergent molecules. The negatively charged heads present in water prevent the small globules from coming together and form aggregates. Thus the oily dirt is removed from the object.

16.26 If water contains dissolved calcium hydrogencarbonate, out of soaps and synthetic detergents, which one will you use for cleaning clothes?
Ans. Calcium hydrogencarbonate makes water hard. Therefore, soap cannot be used because it gets precipitated in hard water. On the other hand, a synthetic detergent does not precipitate in hard water because its calcium salt is also soluble in water. Therefore, synthetic detergents can be used for cleaning clothes in hard water.

16.27 Label the hydrophilic and hydrophobic parts in the following compounds.
(i)CH₃(CH₂)₁₀CH₂OSO₃ ^–Na⁺
(ii)CH₃(CH₂)₁₅ -N+(CH₃)₃Br^–
(iii)CH₃(CH₂)₁₆C00(CH₂CH₂O)₁₁CH₂CH₂OH
Ans.

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NCERT Solutions For Class 12 Chemistry Chapter 14 Biomolecules

NCERT INTEXT QUESTIONS

14.1 Glucose or sucrose are soluble in water but cyclohexane and benzene (simple six membred ring compounds) are insoluble in water Explain.
Ans. The .solubility of a solute in a given solvent follows the rule ‘ Like dissolves like’.Glucose contains five and sucrose contains eight -OH groups. These -OH groups form H-bonds with water. As a result of this extensive intermoleeular H-bonding, glucose and sucrose are soluble in water.On the other hand, benzene and cyclohexane do not contain -OH bonds and hence do not form H-bonds with water. Moreover, they are non-polar molecules and hence do not dissolve in polar water molecules.

14.2 What are the expected products of hydrolysis of lactose?
Ans. Lactose being a disaccharide gives two molecules of monosaccharides Le. one molecule each of D-(+) – glucose and D-(+)-galactbse.

14.3 How do you explain the absence of aldehyde group in the pentaacetate of D-glucose?
Ans. The cyclic hemiacetal form of glucose contains an -OH group at C-l which gets hydrolysed in aqueous solution to produce open chain aldehydic form which then reacts with NH₂OH -to form corresponding oxime. Thus, glucose contains an aldehydic group. However, when glucose is reacted with acetic anhydride, the -OH group at C-l along with the other -OH groups at C-2, C-3, C-4 and C-6 form a pentaacetate.
Since the penta acetate of1 glucose does not contain a free -OH group at C-l, it cannot get hydrolysed in aqueous solution to produce open chain aldehydic form and hence glucose pentaacetate does not react with NH₂OH to form glucose oxime. The reactions are shown as:

14.4 The melting points and solubility of amino acids in water are generally higher than those of corresponding haloacids. Explain.
Ans. The amino acids exist as zwitter ions, H₃N⁺ — CHR-COO-. Due to this dipolar salt like character, they have strong dipole-dipole attractions. Therefore, their melting points are higher than corresponding haloacids which do not have salt like character.
Due to salt like character, amino acids intereact strongly with water. As a result, their solubility in water is higher than corresponding haloacids which do not have salt like character.

14.5 Where does the water present in the egg go after boiling the egg?
Ans. When egg is boiled, proteins first undergo denaturation and then coagulation and the water present in the egg gets absorbed in coagulated protein, probably through H- bonding

14.6 Why cannot Vitamin C be stored in our body?
Ans. Vitamin C cannot be stored in the body because it is water soluble and is, therefore, easily excreted in urine.

14.7 What products would be formed when a nucleotide from DNA containing thymine is hydrolysed?
Ans. The products obtained are 2-deoxy-D-ribose, . phosphoric acid and thymine.

14.8 When RNA is hydrolysed, there is no relationship among the quantities of different bases obtained. What does this fact suggest about the structure of RNA?
Ans. A DNA molecule has two strands in which the four complementary bases pair each other, i.e., cytosine (C) always pair with guanine (G) while thymine (T) always pairs with adenine (A). Thus, when a DNA molecule is hydrolysed, the molar amounts of cytosine is always equal to that of guanine and that of adenine is always equal to thymine.In RNA, there is no relationship between the quantities of four bases (C, G, A and U) obtained, therefore, the base pairing principle, i.e. A pairs with U and C pairs with G is not followed. Therefore, unlike DNA, RNA has a single strand.

NCERT EXRECISES

14.1 What are monosaccharides ?
Ans. Monosaccharides are carbohydrates Which cannot be hydrolysed to smaller molecules.Their general formula is (CH₂O)n Where n=3-7 These are of two types: Those which contain an aldehyde group (-CHO) are called aldoses and those which contain a keto (C=O) group are called ketoses.
They are further classified as trioses , tetroses ,pentoses , hexoses and heptoses according as they contain 3,4,5,6, and 7 carbon atoms respectively.For example.

14.2 What are reducing sugars?
Ans. Carbohydrates which reduces Fehling’s solution to red precipitate of Cu₂0 or Tollen’s reagent to metallic Ag are called reducing sugars. All monosaccharides (both aldoses and ketoses) and disaccharides except sucrose are reducing sugars. Thus, D – (+) – glucose, D-(-)-fructose, D – (+) – maltose and D – (+) – lactose are reducing sugars.

14.3 Write two main functions of carbohydrates in plants.
Ans. Two major functions of carbohydrates in plants are following
(a)Structural material for plant cell walls: The polysaccharide cellulose acts as the chief structural material of the plant cell walls.
(b)Reserve food material: The polysaccharide starch is the major reserve food material in the plants. It is stored in seeds and act as the reserve food material for the tiny plant till it is capable of making its own food by photosynthesis.

14.4 Classify the following into monosaccharides and disaccharides. Ribose, 2-deoxyribose, maltose, galactose, fructose and lactose.
Ans. Monosaccharides: Ribose, 2-deoxyribose, galactose and fructose. Disaccharides: Maltose and lactose.

14.5 What do you understand by the term glycosidic linkage?
Ans. The ethereal or oxide linkage through which two monosaccharide units are joined together by the loss of a water molecule to form a molecule of disaccharide is called the glycosidic linkage. The glycosidic linkage in maltpse molecule is shown below:

14.6 What is glycogen? How is it different from starch?
Ans. Glycogen is a condensation polymer of α-D glucose. Starch is not a single compound but is a mixture of two components—a water soluble component called amyldse (15- 20%) and water insoluble component amylopectin (80 – 85%). Amylose is a linear polymer of α – D – glucose. But both glycogen and amylopectin are branched polymers of α – D – glucose; father glycogen is more highly branched than amylopectin as amylopectin chains consists of 20 – 25 glucose units, glycogen chains consist of 10 – 14 glucose units.

14.7 What are the hydrolysis products of (i) sucrose, and (ii) lactose?
Ans. Both sucrose and lactose are disaccharides. Sucrose on hydrolysis gives one molecule each of glucose and fructose but lactose on hydrolysis gives one molecule each of glucose and galactose.

14.8 What is the basic structural difference between starch and cellulose?
Ans. Starch consists of amylose and amylopectin. Amylose is a linear polymer of α-D-glucose while cellulose is a linear polymer of β -D- glucose. In amylose, C -1 of one glucose unit is connected to C – 4 of the other through α-glycosidic linkage. However in cellulose, C – 1 of one glucose unit is connected to C-4 of the other through β – glycosidic linkage. Amylopectin on the other hand has highly branched structure.

14.9 What happens when D-glucose is treated with . the following reagents.
(i)HI (ii) Bromine water (iii) HNO₃
Ans.

14.10 Enumerate the reactions of D-glucose which cannot be explained by its open chain structure.
Ans. (a) D (+) – glucose does not undergo certain characteristic reactions of aldehydes, e.g., glucose does not form NaHSO₃ addition product.
(b)Glucose reacts with NH₂OH to form an oxime but glucose pentaacetate does not. This implies that the aldehydic group is absent in glucose pentaacetate.
(c)D – (+) – glucose exists in two stereoisomeric forms, i.e., α -glucose and β-glucose.
(d)Both α – D – glucose and β – D – glucose undergo mutarotation in aqueous solution. Although the crystalline forms of α-  and β -D (+) – glucose are quite stable in aqueous solution but each form slowly changes into an equilibrium mixture of both.
(e)D (+) – glucose forms two isomeric methyl glucosides. Aldehydes normally react with two moles of methanol per mole of the aldehyde to form an acetal but D (+) – glucose when treated with methanol in presence of dry HCl gas, reacts with only one mole of methanol per mole of glucose to form a mixture of two methyl D – glucosides i. e., methyl – α – D – glucoside (melting point 43 8 K, specific rotation +158°) and methyl – β – D – glucoside (melting point 308 K, specific rotation – 33°).

14.11 What are essential and non-essential amino acids? Give two examples of each type.
Ans. α-Amino acids which are needed for good health and proper growth of human beings but are not synthesized by the human body are called- essential amino acids. For example, valine, leucine, phenylalanine, etc. On the other hand, α-amino acids which are needed for health and growth of human beings and are synthesized by the human body are called non-essential amino acids. For example, glycine, alanine, aspartic acid etc.

14.12 Define the following as related to proteins:
(i)Peptide linkage
(ii)Primary structure
(iii)Denaturation
Ans. (i) Peptide bond: Proteins are condensation polymers of α-amino acids in which the same or different α-amino acids are joined by peptide bonds. Chemically, a peptide bond is an amide linkage formed between – COOH group of one α-amino acid and -NH-, group of the other α-amino acid by loss of a molecule of water. For example,

(ii) Primary structure: Proteins may contain one or more polypeptide chains. Each . polypeptide chain has a large number of α-amino acids which are linked to one another in a specific manner. The specific sequence in which the various amino acids present in a protein linked to one another is called its primary structure. Any change in the sequence of α-amino acids creates a different protein.

(iii) Denaturation: Each protein in the biological system has a unique three-dimensional structure and has specific biologicalactivity. This is called native form of a protein. When a protein in its native form is subjected to a physical change such as change in temperature or a chemical change like change in pH, etc., hydrogen bonds gets broken. As a result, soluble forms of proteins such as globular proteins undergo coagulation or precipitation to give fibrous proteins which are insoluble in water. This coagulation also results in loss of biological activity of the proteins and this loss in biological activity, is called denaturation. During denaturation, 2° and 3° structures of proteins are destroyed but 1° structure remains intact.
The most common example of denaturation of proteins is the coagulation of albumin present in the white of an egg. When the egg is boiled hard, the soluble globular protein present in it is denatured and is converted into insoluble fibrous protein.

14.13 What are the common types of secondary structure of proteins?
Ans. The conformation which the polypeptide chains assume as a result of hydrogen bonding is called secondary structure of the proteins. The two types of secondary structures are α-helix and β-pleated sheet structure.

14.14 What type of bonding helps in stabilising the α-helix structure of proteins?
Ans. The α-helix structure of proteins is stabilized by intramolecular H-bonding between C = O of one amino acid residue and the N – H of the fourth amino acid residue in the chain. This causes the polypeptide chain to coil up into a spiral structure called right handed α- helix structure.

14.15 Differentiate between globular and fibrous proteins.
Ans. (i) Fibrous proteins: These proteins consist of linear thread like molecules which tend to lie side by side (parallel) to form fibres. The polypeptide chains in them are held together usually at many points by hydrogen bonds and some disulphide bonds. As a result,intermolecular forces of attraction are very’ strong and hence fibrous proteins are insoluble in water. Further, these proteins are stable to moderate changes in temperature and pH. Fibrous proteins serve as the chief structural material of animal tissues.For example, keratin in skin, hair, nails and wool, collagen in tendons, fibrosis in silk and myosin in muscles.
(ii) Globular proteins: The polypeptide chain in these proteins is folded around itself in such a way so as to give the entire protein molecule an almost spheroidal shape. The folding takes place in such a manner that hydrophobic (non-polar) parts are pushed inwards and hydrophilic (polar) parts are pushed outwards. As a result, water molecules interact strongly with the polar groups and hence globular protein are water soluble. As compared to fibrous proteins, these are very sensitive to small changes of temperature and pH. This class of proteins include all enzymes, many hormones such as insulin from pancreas, thyroglobulin from thyroid gland, etc.

14.16 How do you explain the amphoteric behaviour of amino acids?
Ans. Amino acids contain an acidic (carboxyl group) and basic (amino group) group in the same molecule. In aqueous solution, they neutralize each other. The carboxyl group loses a proton while the amino group accepts it. As a result, a dipolar or zwitter ion is formed.

In zwitter ionjc form, a-amino acid show amphoteric behaviour as they react with both acids and bases.

14.17 What are enzymes?
Ans. Enzymes are biological catalyst. Each biological reaction requires a different enzyme. Thus, as compared to conventional catalyst enzymes are very specific and efficient in their action. Each type of enzyme has its own specific optimum conditions of concentration, pH and temperature at which it works best.

14.18 What is the effect of denaturation on the structure of proteins?
Ans. During denaturation, 2° and 3° structures of proteins are destroyed but 1° structure remains intact. As a result of denaturation, die globular proteins (soluble in H₂O) are converted into fibrous proteins (insoluble in H₂O) and their biological activity is lost. For example, boiled egg which contains coagulated proteins cannot be hatched.

14.19 How are vitamins classified? Name the vitamin responsible for the coagulation of blood.
Ans. Vitamins are classified into two groups depending upon their solubility in water or fat: (i) Water soluble vitamins: These include vitamin B-complex (B₁, B₂, B₅, i.e., nicotinic acid,B₆, B₁₂, pantothenic acid, biotin, i.e., vitamin H and folic acid) and vitamin C.
(ii) Fat soluble vitamins: These include vitamins A, D, E and K. They are stored in liver and adipose (fat storing) tissues. Vitamin K is responsible for coagulation of blood.

14.20 Why are vitamin A and vitamin C essential to us? Give their important sources.
Ans. Vitamin A is essential for us because its deficiency causes xerophthalmia (hardening of cornea of eye) and night blindness.
Sources: Fish liver oil, carrots, butter, milk, etc. Vitamin C is essential for us because its deficiency causes scurvy (bleeding of gums) and pyorrhea (loosening and bleeding of teeth). Sources: Citrous fruits, amla, green leafy vegetables etc.

14.21 What are nucleic acids ? Mention their two important functions.
Ans. Nucleic acids are biomolecules which are found in the nuclei of all living cell in form of nucleoproteins or chromosomes (proteins contains nucleic acids as the prosthetic group).

Nucleic acids are of two types: deoxyribonucleic acid (DNA) and ribonucleic acid.(RNA).
The two main functions of nucleic acids are:
(a) DNA is responsible for transmission of hereditary effects from one generation to another. This is due to its unique property of replication, during cell division and two identical DNA strands are transferred to the daughter cells.
(b) DNA and RNA are responsible for synthesis of all proteins needed for the growth and maintenance of our body. Actually the proteins are synthesized by various RNA molecules (r-RNA, m-RNA) and t-RNA) in the cell but the message for the synthesis of a particular protein is coded in DNA.

14.22 What is the difference between a nucleoside and a nucleotide?
Ans. A nucleoside contains only two basic components of nucleic acids i.e., a pentose sugar and a nitrogenous base. It is formed when 1- position of pyrimidine (cytosine, thiamine or uracil) or 9-position of purine (guanine or adenine) base is attached to C -1 of sugar (ribose or deoxyribose) by a β-linkage. Nucleic acids are also called polynucleotides since the repeating structural unit of nucleic acids is a nucleotide.
A nucleotide contains all the three basic . components of nucleic acids, i.e., a phosphoric acid group, a pentose sugar and a nitrogenous base. These are obtained by esterification of C₅, – OH group of the pentose sugar by phosphoric acid.

14.23 The two strands in DNA are not identical but are complementary. Explain.
Ans. The two strands in DNA molecule are held together by hydrogen bonds between purine base of one strand and pyrimidine base of the other and vice versa. Because of different sizes and geometries of the bases, the only possible pairing in DNA are G (guanine) and C (cytosine) through three H-bonds, (i.e.,C = G) and between A (adenine) and T (thiamine) through two H-bonds (i.e., A = T). Due to this base -pairing principle, the sequence of bases in one strand automatically fixes the sequence of bases in the other strand. Thus, the two strands are complimentary and not identical.

14.24 Write the important structural and functional differences between DNA and RNA.
Ans.

14.25 What are the different types of RNA found in the cell?
Ans. There are three types of RNA:
(a) Ribosomal RNA (r RNA) (b) Messenger RNA (m RNA) (c) Transfer RNA (t RNA)

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NCERT Solutions For Class 12 Chemistry Chapter 15 Polymers

NCERT INTEXT QUESTIONS

15.1 What are polymers?
Ans. Polymers are high molecular mass substances (10³ — 10⁷u) consisting of a very large number of simple repeating structural units joined together through covalent bonds in a linear fashion. They are also called macromolecules. Ex: polythene, nylon 6,6, bakelite, rubber, etc.

15.2 How are polymers classified on the basis of structure?
Ans. On the basis of structure, polymers are classified as :
(i)Linear polymers in which the monomers are joined together to form long straight chains of polymer molecules. Forex: HDPE, PVC, nylons, etc.
(ii)Branched chain polymers in which the monomers not only join in linear fashion but also form branches of different lenghts along the main chain. For ex : LDPE, glycogen, etc.
(iii)Cross-linked polymers in which the intially formed linear polymer chains join together to form 3D network structure. For ex : bakelite, Urea-formaldehyde resin, etc.

15.3 Write the names of the monomers of the following polymers:


Ans. (i) Hexamethylene diamine NH₂-(CH₂)6NH₂ and adipic acid HOOC – (CH₂)₄ – COOH
(ii)Caprolactum
(iii)Tetrafluoroethene F₂C = CF₂

15.4 Classify the following as addition and condensation polymers:
Terylene, Bakelite, Polyvinyl chloride,Polythene
Ans.Addition polymers: Polyvinyl chloride, Polythene
Condensation polymers : Terylene, bakelite.

15.5 Explain the difference between Buna- N and Buna-S.
Ans. Both are copolymers. Buna-N is a copolymer of 1,3-butadiene and acrylonitrile whereas Buna-S is a copolymer of 1,3-butadiene and styrene.

15.6 Arrange the following polymers in increasing order of their intermolecuiar forces.
(i)Nylon 6,6, Buna-S, Polythene
(ii)Nylon 6, Neoprene, Polyvinyl chloride
Ans. On the basis of intermolecuiar forces, polymers
are classified as elastomers, fibres and plastics. The increasing order of intermolecuiar forces is: Elastomer < Plastic < fibre.
Thus, we have
(i)Buns-S < Polythene < Nylon 6,6
(ii)Neoprene < Polyvinyl chloride < Nylon 6.

NCERT EXRECISES

15.1 Explain the terms polymer and monomer.
Ans. Polymers are high molecular mass substances consisting of a very large number of simple repeating structural units joined together through covalent bonds in a regular fashion. Polymers are also called macromolecules. Some examples are polythene, nylon-66, bakelite, rubber, etc. Monomers are the. simple and reactive molecules from which the polymers are prepared either by addition or condensation polymerisation. Some examples are ethene, vinyl chloride, acrylonitrile, phenol and formaldehyde etc.

15.2 What are natural and synthetic polymers? Give two examples of each type.
Ans.Natural polymers: Polymers which are found in nature,i.e., in animals and plants are called natural polymers, e.g., proteins, starch, cellulose, nucleic acids, resins and natural Sol. rubber.
Synthetic polymers: Man-made polymers are called synthetic polymers, e.g., plastics (polythene, PVC), synthetic fibres (polyester, 15.8 nylon-66) and synthetic rubber (neoprene, Buna-S).

15.3 Distinguish between the terms homopolymer and copolymer and give an example of each.
Ans.Polymers whose repeating structural units are derived from only one type of monomer units are called homopolymers, e.g., PVC polyethene, PAN, teflon, polystyrene, nylon- 6 etc.
Polymers whose repeating structural units are derived from two or more types of monomer molecules are copolymers, e.g., Buna-S, Buna-N, nylon-66, polyester, bakelite.

15.4 How do you explain the functionality of a monomer?
Ans.Functionality means the number of binding sites in a molecule. For example, the functionality of ethene, propene, styrene, acrylonitrile is one while that of 1,3-butadiene, adipic acid, terephthaliC’ acid, hexa methylenediamine is two.

15.5 Define the term polymerisation?
Ans.It is a process of formation of a high molecular Sol. mass polymer from one or more monomers by linking together a large number of repeating structural units through covalent bonds.

15.6 Is (-NH — CHR—CO-)_n a homopolymer or copolymer?
Ans.It is a homopolymer because the repeating structural unit has only one type of monomer, i.e., NH₂—CHR—COOH.

15.7 In which classes, the polymers are classified on the basis of molecular forces?
Ans.(a)Elastomers
(b)Fibres
(c)Thermoplastics
(d)Thermosetting plastics

15.8 How can you differentiate between addition and condensation polymerisatiop?
Ans.In addition polymerization, the molecules of the same or different monomers simply add on to one another leading to the formation of a macromolecules without elimination of small molecules like H₂O, NH₃ etc. Addition polymerization generally occurs among molecules containing double and triple bonds. For example, formation of polythene from ethene and neoprene from chloroprene, etc. In condensation polymerisation, two or more bifunctional trifimctional molecules undergo a series of independent condensation reactions usually with the elimination of simple molecules like water, alcohol, ammonia, carbon dioxide and hydrogen chloride to form a macromolecule. For example, nylon-6,6 is a condensation polymer of hexamethylenediamine and adipic acid formed by elimination of water molecules.

15.9 Explain the term copolymerisation and give two examples.
Ans.When two or more different monomers are allowed to polymerise together the product formed is called a copolymer, and the process in called copolymerisation. Example, Buna-S and Buna-N. Buna- S is a copolymer of 1, 3- butadiene and styrene while Buna-N is a copolymer of 1,3-butadiene and acrylonitrile.

15.10 Write the free radical mechanism for the polymerisation of ethene.
Ans.

15.11 Define thermoplastics and thermo setting polymers with two examples of each
Ans. Thermoplastics polymers are linear polymer which can be repeatedly melted and moulded again and again on heating without any change in chemical composition and mechanical strength. Examples are polythene and polypropylene.
Thermosetting polymers, on the other hand, are permanently setting polymers. Once on heating in a mould, they get hardened and set, and then cannot be softened again. This hardening on heating is due to cross- linking between different polymeric chains to give a three dimensional network solid. Examples are bakelite, melamine-foimaldehyde polymer etc.

15.12 Write the monomers used for gettingThe following polymers:
(i) Polyvinylchloride
(ii) Teflon (iii) Bakelite
Ans.

15.13 Write the name and structure of one of the common initiators used in free radical addition polymerisation.
Ans.

15.4 How does the presence of double bonds in rubber molecules influence their structure and reactivity?
Ans.Natural rubber is cis-polyisoprene and is obtained by 1, 4-polymerization of isoprene units. In this polymer, double bonds are located between C₂ and C₃ of each isoprene unit. These cis-double bonds do not allow the polymer chains to come closer for effective interactions and hence intermolecular forces are quite weak. As a result, natural rubber, i.e., cis-polyisoprene has a randomly coiled structure not the linear one and hence show elasticity.

15.5 Discuss the main purpose of vulcanisation of rubber.
Ans.Natural rubber has the following disadvantages:
(a) It is soft and sticky and becomes even more so at high temperatures and brittle at low temperatures. Therefore, rubber is generally used in a narrow temperature range (283-335 K) where its elasticity is maintained.
(b)It has large water absorption capacity, has low tensile strength and low resistance to abrasion.
(c)It is not resistant to the action of organic solvents.
(d)It is easily attacked by oxygen and other oxidising agents. .
To improve all these properties, natural rubber is vulcanised by heating it with about 5% sulphur at 373-415 K. The vulcanized rubber thus obtained has excellent elasticity over a larger range of temperature, has low water absorption tendency and is resistant to the action of organic solvents and oxidising agents.

15.16 What are the monomeric repeating units of Nylon-6 and Nylon 6,6?
Ans.

15.17 Write the names and structures of the monomers of the following polymers:
(i) Buna-S (ii) Buna-N (iii) Dacron (iv) Neoprene
Ans.

15.18 Identify the monomer in the following polymeric structures:

Ans.

15.19 How is dacron obtained from ethylene glycol and terephthalic acid?
Ans.Dacron is obtained by condensation polymerization of ethylene glycol and terephthalic acid with the elimination of water molecules. The reaction is carried out at 420 – 460 K in presence of a catalyst consisting of a mixture of zinc acetate and antimony trioxide.

15.20 What is a biodegradable polymer ? Give an example of a biodegradable aliphatic polyester.
Ans. Polymers which disintegrate by themselves over a period of time due to environment degradation by bacteria, etc., are called biodegradable polymers. Example is PHBV, i. e., Poly-β-Hydroxybutyrate-co-β- Hydroxyvalerate.

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NCERT Solutions For Class 12 Maths Chapter 2 Inverse Trigonometric Functions

Chapter: Chapter 2 – Inverse Trigonometric Functions

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NCERT Solutions for Class 12 Maths Chapter 4 Determinants

Subject: Maths
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