NCERT Solution of Class 12 Chemistry – Chapter 9: Amines

NCERT Solutions for Class 12 Chemistry Chapter 9: The Amines is a very important organic compound that contains nitrogen, discusses the structure, properties, and reactions of amines. Clearly mentioned is how amines are classified as primary, secondary, and tertiary according to the IUPAC nomenclature. Solutions give the structure of the Amines. The focus is given on the effect of the lone pair of electrons on the Nitrogen atom, that makes them basic in nature. Class 12 Chemistry Chapter 9 PDF describes how the electronic environment around Nitrogen is important for physical and chemical properties of themines. The various methods of preparation of amines, such as reduction of nitro compounds, reduction of amides, Hoffmann bromamide degradation, and Gabriel phthalimide synthesis, are also discussed in the NCERT Solutions for Class 12 Amines chapter. All of these techniques are explained with steps and reaction mechanisms so that students may have a clear idea of the synthesis of amines. There are more sectional methods involved in the Class 12 Chemistry Chapter 9 PDF, which has to be practiced. The solutions are also helpful during board and competitive exams like JEE and NEET. The solutions provided are useful in inculcating the fundamental concepts as well as enabling easy remembering and reproducing of the same.

Download PDF For NCERT Solutions for Chemistry Amines

The NCERT Solution of Class 12 Chemistry – Chapter 9: Amines are tailored to help the students master the concepts that are key to success in their classrooms. The solutions given in the PDF are developed by experts and correlate with the CBSE syllabus of 2023-2024. These solutions provide thorough explanations with a step-by-step approach to solving problems. Students can easily get a hold of the subject and learn the basics with a deeper understanding. Additionally, they can practice better, be confident, and perform well in their examinations with the support of this PDF.

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Access Answers to NCERT Solution of Class 12 Chemistry – Chapter 9: Amines

Students can access the NCERT Solution of Class 12 Chemistry – Chapter 9: Amines. Curated by experts according to the CBSE syllabus for 2023–2024, these step-by-step solutions make Chemistry much easier to understand and learn for the students. These solutions can be used in practice by students to attain skills in solving problems, reinforce important learning objectives, and be well-prepared for tests.

Amines

Question 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

Answer :

 (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°)

 


Question 2 :

Describe the method for the identification of primary, secondary and tertiary amines. Also write chemical equations of the reactions involved.

Answer :

The distinction in the three types of amines can be done by the following methods :
(i) Hinsberg’s Test:
This is a very useful test for the distinction of primary, secondary and tertiary amines. An amine is shaken with
Hinsberg’s reagent (benzene suiphonyl chloride) in the presence of excess of aqueous KOH solution. The reactions taking
place are given on the next page.

  1.  A primary amine forms N – alkyl benzene suiphonamide which dissolves in aqueous KOH solution to form potassium salt and upon acidification with dilute HCI regenerates the insoluble suiphonamide.
    NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Exercises Q6

  2.  A secondary amine forms N, N – dialkylbenzene suiphonamide which remains insoluble in aqueous KOH and even after acidification with dilute HCl
    NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Exercises Q6.1

  3. A tertiary amine does not react with benzene suiphonyl chloride and remains insoluble in aqueous KOH.
    However, on acidification with dilute HCI it gives a clear solution due to the formation of the ammonium salt.
    NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Exercises Q6.2

(ii) Reaction with nitrous acid:
All the three types of amines, aliphatic as well as aromatic, react with nitrous acid under different conditions to form variety of products. Since nitrous acid is highly unstable, it is prepared in situ by the action of dilute hydrochloric acid on sodium nitrite.

(a) Primary aliphatic amines react with nitrous acid at low temperature (cold conditions) to form primary alcohol and
nitrogen gas accompanied by brisk effervescence. Nitrous acid is unstable in nature and is prepared in situ by reacting sodium
nitrite with dilute hydrochloric acid. For example,
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Exercises Q6.3
The reaction is used as a tesijôr primary aliphadc amines as no other amine evolves nitrogen with nurous acid.

(b) Primary aromatic amines such as aniline react with nitrous acid under ice cold conditions (273 – 278 K) to form benzen diazonium salt. The reaction is known as diazotisation reaction.
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Exercises Q6.4
in case, the temperature is allowed to rise above 278 K, benzene diazortium chloride is decomposed by water to form phenol.
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Exercises Q6.5
Aliphatic primary amines also react with nitrous acid to form alkyl diazonium salts in a similar manner. But these are
quite unstable and decompose to form a mixture of alcohols, alkenes and alkyl halides along with the evolution of N2 gas.

NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Exercises Q6.6

(c) Secondary amines (both aliphatic and aromatic) react with nitrous acid to form nitrosoamines which separate as
Yellow oily liquids.

NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Exercises Q6.7

(d) Tertiary aliphatic amines dissolve in a cold solution of nitrous acid to form salts which decompose on warming to
give nitrosoamine and alcohol. For example,
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Exercises Q6.8
(e) Tertiary aromatic amines react with nitrous acid to give a coloured nitrosoderivative. This reaction is called
nitrosation and as a result, a hydrogen atom in the para position gets replaced by a nitroso (-NO) group. For example,
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Exercises Q6.9


Question 3 :

Classify the following amines as primary, secondary and tertiary:
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Intext Questions Q1

Answer :

 (i) 1° (ii) -3° (iii) 1° (iv) 2°


Question 4 :

Write the structures of different isomeric amines corresponding to the molecular formula, C4H11N.
(i) Write the IUPAC names of all the isomers
(ii) What type of isomerism is exhibited by different types of amines?

Answer :

 Eight isomeric amines are possible
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Intext Questions Q2
Isomerism exhibited by different amines

  • Chain isomers: (i) and (ii) ; (iii) and (iv) ; (i) and (iv)

  • Position isomers: (ii) and (iii) ; (ii) and (iv)

  • Metamers: (v) and (vi) ; (vii) and (viii)

Functional isomers: All the three types of amines are the functional isomers of each other.


Question 5 :

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

Answer :


NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Intext Questions Q3


Question 6 :

 Arrange the following in increasing order of their basic strength :
(i) C2H5NH2, C6H5NH2, NH3, C6H5CH2NH2, (C2H5)2NH
(ii) C2H5NH2, (C2H5)2NH, (C2H5)3N, C6H5NH2
(iii) CH3NH2, (CH3)2NH, (CH3)3N, C6HsNH2, C6H5CH2NH2

Answer :

In general, the basic character of ammonia (NH3) and the amines is linked with the availability of the lone electrons pair on the nitrogen atom. In other words, these are all Lewis bases.

 

Amines act as Lewis bases due to the presence of lone electron paîr on the nitrogen atom. Since the nitrogen atom is sp³ hybridised, its electron attracting tendency is considerably reduced. It can readily lose its electron pair and acts as a base. For example, amines form hydroxides with water.

NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Intext Questions Q4

Here Kt is called dissociation constant for the base. Greater the Kb value stronger will be the base. The basic strength

of amines can also be expressed as pKb value which is related to Kb as :

NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Intext Questions Q4.1

The Kb values are :

NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Intext Questions Q4.2

 


Question 7 :

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

Answer :


NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Intext Questions Q5


Question 8 :

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

Answer :


NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Intext Questions Q6


Question 9 :

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

Answer :


NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Intext Questions Q7


Question 10 :

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.

Answer :

 In ‘all, four structural isomers are possible. These are:
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Intext Questions Q8


Question 11 :

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

Answer :


NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Intext Questions Q9


Question 12 :

 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.

Answer :


NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Exercises Q2

NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Exercises Q2.1

 


Question 13 :

 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.

 

Answer :

 (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.

NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Exercises Q3

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

NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Exercises Q3.1

(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.

NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Exercises Q3.2

 


Question 14 :

 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

Answer :

(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:
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Exercises Q4
(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.
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Exercises Q4.1
(iii) (a) The presence of electron donating -CH3 group increases while the presence of electron withdrawing -NO2 group decreases the basic strength of amines.
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Exercises Q4.2
(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.
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Exercises Q4.3
(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,
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Exercises Q4.4
(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.
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Exercises Q4.5
(vi) Solubility decreases with increase in molecular mass of amines due to increase in the size of the hydrophobic hydrocarbon part and with decrease is the number of H-atoms on the N-atom which undergo H-bonding.
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Exercises Q4.6


Question 15 :

How will 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?

Answer :


NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Exercises Q5

NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Exercises Q5.1

 


Question 16 :

Write short notes on the following:
(i) Carbylamine reaction
(ii) Diazotisation
(iii) ‘Hofmann’s bromamide reaction
(iv) Coupling reaction
(v) Ammonolysis
(vi) Acetylation
(vii) Gabriel phthalimide synthesis

Answer :

(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.
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Exercises Q7
(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.
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Exercises Q7.1
(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.
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Exercises Q7.2
(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.
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Exercises Q7.3
(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.
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Exercises Q7.4
(vi) Acetylation: The process of introducing an acetyl (CH3CO-) group into molecule using acetyl chloride or acetic anhydride is called acetylation.
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Exercises Q7.5
(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.
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Exercises Q7.6


Question 17 :

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.

Answer :

NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Exercises Q8.2


Question 18 :

 Give the structures of A,B and C in the following reaction:
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Exercises Q9
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Exercises Q9.1

Answer :


NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Exercises Q9.2


Question 19 :

 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.

Answer :

From the available information, we find that ‘B’ upon heating with Br2 and KOH forms a compound ‘C’. The compound ‘B’ is
expected to be an acid amide. Since ‘B’ has been formed upon heating compound ‘A’ with aqueous ammonia, the compound ‘A’ is an aromatic acid.
It is benzoic acid. The reactions involved are given as follows:
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Exercises Q10


Question 20 :

Complete the following reactions:
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Exercises Q11

Answer :


NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Exercises Q11.1

NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Exercises Q11.2

 


Question 21 :

Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis?

Answer :

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.
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Exercises Q12


Question 22 :

 Write the reactions of (i) aromatic and (ii) aliphatic primary amines with nitrous acid.

Answer :

 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
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines Exercises Q13


Question 23 :

 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?

Answer :

(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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