Class 12th Chemistry Amines Notes and Important Question.

www.free-education.in is a platform where you can get pdf notes from 6th to 12th class notes, General Knowledge post, Engineering post, Career Guidelines , English Speaking Trick , How to crack interview and lots more. Class 12 Chemistry Amines ( Amines Notes and Important Question for Board Exam )

Class 12th Amines Notes

Introduction

  • Amines constitute an important class of organic compounds derived by replacing one or more hydrogen atoms of ammonia molecule by alkyl/aryl group(s).
  • In nature, they occur among proteins, vitamins, alkaloids and hormones.
  • Amines have lot of importance in our day to day life.
  • For example:-
    • Polymers, dyestuffs and drugs contain amines.
    • Two biologically active compounds, namely adrenaline and ephedrine, both containing secondary amino group, are used to increase blood pressure.
    • Novocain, a synthetic amino compound, is used as an anaesthetic in dentistry.
    • Benadryl, a well-known antihistaminic drug also contains tertiary amino group.
    • Quaternary ammonium salts are used as surfactants. Diazonium salts are intermediates in the preparation of a variety of aromatic compounds including dyes.
Class_12_Chemistry_Amines_Diagram1

Structure of Amines

  • Amines can be considered as derivatives of ammonia, obtained by replacement of one, two or all the three hydrogen atoms by alkyl and/or aryl groups.
  • Nitrogen atom of amines is trivalent and carries an unshared pair of electrons.
Class_12_Chemistry_Amines_Structure_Of_Amines
  • Nitrogen orbitals of amines are sp3 hybridized and the geometry of amines is pyramidal.
  • Each of the three sp3 hybridised orbitals of nitrogen overlap with orbitals of hydrogen or carbon depending upon the composition of the amines.
  • The fourth orbital of nitrogen in all amines contains an unshared pair of electrons.
  • Due to the presence of unshared pair of electrons, the angle C–N–E, (where E can be either C or H). is less than 109.50.
Class_12_Chemistry_Amines_Structure_Of_Amines1
  • In case of trimethylamine the bond angle is 1080.

Classification of Amines as Primary, Secondary & Tertiary

  • Amines are classified as primary (1o), secondary (2o) and tertiary (3o) depending upon the number of hydrogen atoms replaced by alkyl or aryl groups in ammonia molecule.
  • If one hydrogen atom of ammonia is replaced by R or Ar, we get RNH2 or ArNH2, a primary amine (1o).
  • If two hydrogen atoms of ammonia or one hydrogen atom of R-NH2 are replaced by another alkyl/aryl (R’) group then it is called as secondary amine.
  • The second alkyl/aryl group may be same or different.
  • Replacement of another hydrogen atom by alkyl/aryl group leads to the formation of tertiary amine.
Class_12_Chemistry_Amines_Classification_Of_Amines
Class_12_Chemistry_Amines_Classification_Of_Amines1
  • Amines are said to be ‘simple’ when all the alkyl or aryl groups are the same, and ‘mixed’ when they are different.
Class_12_Chemistry_Amines_Classification_Of_Amines2

Problem:

Classify the following amines as primary, secondary or tertiary:

Class_12_Chemistry_Amines_Classification_Of_Amines3

Answer:-

  • It is a primary amine because only 1 hydrogen is replaced if we compare this structure with ammonia.
  • It is a tertiary amine as all the 3 hydrogen is replaced.
  • It is a primary amine as only 1 hydrogen is replaced.

 It is a secondary amine as 2 hydrogens are

Nomenclature of Amines by IUPAC System

Common Names of Amines

  • An aliphatic amine is named by prefixing alkyl group to amine, i.e. alkylamine as 1 word.
  • We are considering primary amines.
Class_12_Chemistry_Amines_Nomenclature_Of_Amines
  • In secondary and tertiary amines, when two or more groups are the same, the prefix di or tri is added to the before the name of alkyl group.
Class_12_Chemistry_Amines_Nomenclature_Of_Amines1
  • In case of mixed amine, alkyl group are written in alphabetical order.
Class_12_Chemistry_Amines_Nomenclature_Of_Amines2

IUPAC System

  • In IUPAC system, amines are named as alkanamines, derived by replacement by ‘e’ of alkane by the word amine.
  • For Example: – Methanamine (CH3NH2), Ethan amine (C2H5NH2).
  • For naming higher member hydrocarbon, longest chain containing amino group is selected. C atom to which amino group is attached is given to lower number.
Class_12_Chemistry_Amines_Nomenclature_Of_Amines3
  • In case, more than one amino group is present at different positions in the parent chain, their positions are specified by giving numbers to carbon atoms bearing –NH2 groups and suitable prefix such as di, tri etc. is attached to the amine. The letter ’e’ of the suffix of the hydrocarbon part is retained.
  • For Example:- (Ethane 1,2 diamine) H2N-CH2-CH2-NH2
  • Each alkyl group bonded to the N atom is named as N-alkyl group.

Aryl Amines

  • In aryl amines, -NH2 group is directly attached to the benzene ring.
  • C6H5NH2 is the simplest example of aryl amine. It is also known as aniline.
Class_12_Chemistry_Amines_Nomenclature_Of_Amines4
  • Naming aryl amines according to IUPAC system, suffix ‘e’ of arene is replaced by ‘amine’.
  • For Example: – C6H5-NH2 is named as benzenamine.

Preparation of Amines

Amines are prepared by the following methods:

Reduction of nitro compounds

Ammonolysis of alkyl halides

Reduction of nitriles

Reduction of amides

Gabriel phthalimide synthesis

Hoffmann bromamide degradation reaction

  1. Reduction of nitro compounds
  • Nitro compounds are reduced to amines by passing hydrogen gas in the presence of finely divided nickel, palladium or platinum and also by reduction with metals in acidic medium.

R- NO2 + 3H2 à R-NH2 + 2H2O   {In the presence of Ni, Pt or Pd}

Ar- NO + 3H2à Ar- NH2 + 2H2O

  • Nitro alkanes can also be similarly reduced to the corresponding alkanamines.
  • Reduction with iron scrap and hydrochloric acid is preferred because FeCl2 formed gets hydrolysed to release hydrochloric acid during the reaction.
  • Thus, only a small amount of hydrochloric acid is required to initiate the reaction.
  • Nitro compounds can also be reduced with active metals such as Fe, Sn, Zn etc. with conc. HCl

R- NO2 + 3H2 à R- NH2 + 2H2O   {In the presence of Sn/HCl}

Ar- NO2 + 3H2 à Ar- NH2 + 2H2O

  1. Ammonolysis of alkyl halides
  • The carbon – halogen bond in alkyl or benzyl halides can be easily cleaved by a nucleophile.
Class_12_Chemistry_Amines_Preparation_Of_Amines_Ammonylsis
  • Therefore, an alkyl or benzyl halide on reaction with an ethanolic solution of ammonia undergoes nucleophilic substitution reaction in which the halogen atom is replaced by an amino (–NH2) group.
  • This process of cleavage of the C–X bond by ammonia molecule is known as ammonolysis.
Class_12_Chemistry_Amines_Preparation_Of_Amines_Ammonylsis1
  • The reaction is carried out in a sealed tube at 373 K. The primary amine thus obtained behaves as a nucleophile and can further react with alkyl halide to form secondary and tertiary amines, and finally quaternary ammonium salt.
  • The free amine can be obtained from the ammonium salt by treatment with a strong base:
Class_12_Chemistry_Amines_Preparation_Of_Amines_Ammonylsis2
  • Disadvantages:-
  • When ammonia is taken in excess primary amine is formed as main Product
  • When alkyl halide is used in excess quaternary ammonium salt is formed as main product
  • Note: – This method is not suitable for preparation of aryl amines because aryl amines are relatively less reactive than alkyl halides towards nucleophilic substitution reactions.
  1. Reduction of nitriles
  • Nitriles on reduction with lithium aluminium hydride (LiAlH4) or catalytic hydrogenation produce primary amines.
  • This reaction is used for ascent of amine series, i.e., for preparation of amines containing one carbon atom more than the starting amine.
Class_12_Chemistry_Amines_Preparation_Of_Amines_Reaction1
  1. Reduction of amides
  • The amides on reduction with lithium aluminium hydride yield amines.
Class_12_Chemistry_Amines_Preparation_Of_Amines_Reaction2
  1. Gabriel phthalimide synthesis
  • Gabriel synthesis is used for the preparation of primary amines.
  • Phthalimide on treatment with ethanolic potassium hydroxide forms potassium salt of phthalimide which on heating with alkyl halide followed by alkaline hydrolysis produces the corresponding primary amine.
  • Aromatic primary amines cannot be prepared by this method because aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide.
Class_12_Chemistry_Amines_Preparation_Of_Amines_GabrielReaction
  1. Hoffmann bromamide degradation reaction
  • Hoffmann developed a method for preparation of primary amines by treating an amide with bromine in an aqueous or ethanolic solution of sodium hydroxide.
  • In this degradation reaction, migration of an alkyl or aryl group takes place from carbonyl carbon of the amide to the nitrogen atom.
  • The amine so formed contains one carbon less than that present in the amide.
Class_12_Chemistry_Amines_Preparation_Of_Amines_Hoffmann_BromamideReaction

Physical & Chemical Properties of Amines

General:-

  • The lower aliphatic amines are gases with fishy odour.
  • Primary amines with three or more carbon atoms are liquid and still higher ones are solid.
  • Aniline and other arylamines are usually colourless but get coloured on storage due to atmospheric oxidation.
  • Solubility:-
    1.  Lower aliphatic amines are soluble in water because they can form hydrogen bonds with water molecules.
    2. However, solubility decreases with increase in molar mass of amines due to increase in size of the hydrophobic alkyl part. Higher amines are essentially insoluble in water.
    3. Amines are soluble in organic solvents like alcohol, ether and benzene.

Boiling Points:-

Primary and secondary amines are engaged in intermolecular association due to hydrogen bonding between nitrogen of one and hydrogen of another molecule.

This intermolecular association is more in primary amines than in secondary amines as there are two hydrogen atoms available for hydrogen bond formation in it.

Tertiary amines do not have intermolecular association due to the absence of hydrogen atom available for hydrogen bond formation. Therefore, the order of boiling points of isomeric amines is as follows:

Primary > Secondary > Tertiary

  • Intermolecular hydrogen bonding in primary amines is as:-
Class_12_Chemistry_Amines_Properties_Of_Amines

Chemical Properties of Amines

  • Amines behave as nucleophiles due to the presence of unshared electron pair.
  • Difference in electronegativity between nitrogen and hydrogen atoms and the presence of unshared pair of electrons over the nitrogen atom makes amines reactive.
  • Primary, secondary and tertiary amines differ in reactions because of number of hydrogen attached to nitrogen.

Following are the chemical properties of Amines:-

  • Basicity of Amines
    • They are basic in nature and form salts when react with acids.
Class_12_Chemistry_Amines_Basicity_Of_Amines
  • Amine salts on treatment with a base like NaOH, regenerate the parent amine.
Class_12_Chemistry_Amines_Basicity_Of_Amines1
  • Amine salts are soluble in water but insoluble in organic solvents like ether.
  • They form ammonium salts when react with mineral acids.
  • Amines have an unshared pair of electrons on nitrogen atom due to which they behave as Lewis base.

Why Amines are basic in nature in terms of Kb and pKb values?

  • Larger the value of Kb or smaller the value of pKb, stronger is the base. The pKb values of few amines are given in.
Class_12_Chemistry_Amines_Basicity_Of_Reaction_Of_Amines
  • pKb value of ammonia is 4.75. Aliphatic amines are stronger bases than ammonia due to +I effect of alkyl groups leading to high electron density on the nitrogen atom.
  • Their pKb values lie in the range of 3 to 4.22. On the other hand, aromatic amines are weaker bases than ammonia due to the electron withdrawing nature of the aryl group.

Structure-basicity relationship of amines

  • Basicity of amines to their structure.
  1. Alkanamines versus ammonia
    • Reaction of alkanamines and ammonia in order to compare their basicity.
Class_12_Chemistry_Amines_Structure-Basicity_Relationship_Of_Amines
  •  Due to the electron releasing nature of alkyl group, it (R) pushes electrons towards nitrogen and thus making them available for the unshared electron pair for the proton of the acid.
  • The substituted ammonium ion formed from the amine gets stabilised due to dispersal of the positive charge by the +I effect of the alkyl group.
  • Hence, alkylamines are stronger bases than ammonia. Thus, the basic nature of aliphatic amines should increase with increase in the number of alkyl groups.
  • The order of basicity of amines in the gaseous phase follows the expected order: tertiary amine > secondary amine > primary amine > NH3.
  • But this not same in aqueous state, in that the substituted ammonium cations get stabilised not only by electron releasing effect of the alkyl group (+I) but also by solvation with water molecules.
  • The greater the size of the ion, lesser will be the solvation and the less stabilised is the ion.
  • The order of stability of ions are as follows:
Class_12_Chemistry_Amines_Structure-Basicity_Relationship_Of_Amines1
  • Thus, the order of basicity of aliphatic amines should be: primary > secondary > tertiary, which is opposite to the inductive effect based order.
  • When the alkyl group is small, like –CH3 group, there is no steric hindrance to H-bonding.
  • In case the alkyl group is bigger than CH3 group, there will be steric hindrance to H-bonding.
  • Therefore, the change of nature of the alkyl group, e.g., from –CH3 to –C2H5 results in change of the order of basic strength.
  • Thus, there is a subtle interplay of the inductive effect, solvation effect and steric hindrance of the alkyl group which decides the basic strength of alkyl amines in the aqueous state.
  • The order of basic strength in case of methyl substituted amines and ethyl substituted amines in aqueous solution is as follows:
  • (C2H5)2NH > (C2H5)3N > C2H5NH2 > NH3
  • (CH3)2NH > CH3NH2 > (CH33N > NH3
  1. Arylamines versus ammonia
    • pKb value of aniline is quite high because in aniline or other arylamines, the -NH2  group is attached directly to the benzene ring.
    • As a result the unshared electron pair on nitrogen atom will be less available for protonation as it is in conjugation with the benzene ring.
    • Aniline is resonance hybrid of 5 resonance structures.
Class_12_Chemistry_Amines_Structure-Basicity_Relationship_Of_Amines2
  • On the other hand, anilinium ion obtained by accepting a proton can have only two resonating structures (kekule).
Class_12_Chemistry_Amines_Structure-Basicity_Relationship_Of_Amines3
  • Aniline is more stable than anilinium ion as it has greater number of resonating structure.
  • Therefore, the proton acceptability or the basic nature of aniline or other aromatic amines would be less than that of ammonia.
  • In case of substituted aniline, it is observed that electron releasing groups like –OCH3 , –CH3  increase basic strength whereas electron withdrawing groups like –NO2 , –SO3H, –COOH, –X decrease it.
  • Alkylation
  • Amines undergo alkylation on reaction with alkyl halides.
  • Acylation
  • The reaction of aliphatic and aromatic primary and secondary amines with acid chlorides, anhydrides and esters by nucleophilic substitution reaction is known as Acylation.
  • In this reaction there will be replacement of hydrogen atom of –NH2 or >N–H group by the acyl group.
  • The products obtained by acylation reaction are known as amides. The reaction is carried out in the presence of a base stronger than the amine, like pyridine, which removes HCl so formed and shifts the equilibrium to the right hand side.
Class_12_Chemistry_Amines_Structure-Basicity_Relationship_Of_Amines_Acylation
  • Amines also react with benzoyl chloride (C6H5 COCl). This reaction is known as benzoylation.
Class_12_Chemistry_Amines_Structure-Basicity_Relationship_Of_Amines_Benzoylation
  • They form salts when they react with carboxylic acids at room temperature.
  • Carbylamine reaction
  • Aliphatic and aromatic primary amines on heating with chloroform and ethanolic potassium hydroxide form isocyanides or carbylamines which are foul smelling substances.
  • Secondary and tertiary amines do not show this reaction. This reaction is known as carbylamines reaction or isocyanide test and is used as a test or primary amines.
Class_12_Chemistry_Amines_Structure-Basicity_Relationship_Of_Amines_Carbylamine_Reaction
  • Reaction with nitrous acid
  • Different types of amines react differently with nitrous acid which is prepared in situ from a mineral acid and sodium nitrite.

(i)  Primary Amines:-

  1. Primary aliphatic amines react with nitrous acid to form aliphatic diazonium salts which being unstable, liberate nitrogen gas quantitatively and alcohols.
  2. Quantitative evolution of nitrogen is used in estimation of amino acids and proteins.
Class_12_Chemistry_Amines_Structure-Basicity_Relationship_Of_Primary_Amines_Reaction

   (ii)  Secondary Amines:-

  1.   Aromatic amines react with nitrous acid at low temperatures (273-278 K) to form diazonium salts, a very important class of compounds used for synthesis of a variety of aromatic compounds.
Class_12_Chemistry_Amines_Structure-Basicity_Relationship_Of_Secondary_Amines_Reaction
  • Secondary and tertiary amines react with nitrous acid in a different manner.
  1. Reaction with aryl sulphonyl chloride
  • Reaction with Benzene sulphonyl Chloride: – Benzene sulphonyl chloride (C6H5SO2Cl), which is also known as Hinsberg’s reagent, reacts with primary and secondary amines to form sulphonamides.
  • The reaction of benzenesulphonyl chloride with primary amine yields N-ethylbenzene sulphonyl amide.
Class_12_Chemistry_Amines_Structure-Basicity_Relationship_Of_Secondary_Amines_Reaction_1
  • The hydrogen attached to nitrogen in sulphonamide is strongly acidic due to the presence of strong electron withdrawing sulphonyl group.
  • Hence, it is soluble in alkali.
  • Reaction of secondary Amine benzenesulphonyl chloride: – In this reaction, N, N-diethylbenzenesulphonamide is formed.
  • Since N, N-diethyl benzene sulphonamide does not contain any hydrogen atom attached to nitrogen atom; it is not acidic and hence insoluble in alkali.
Class_12_Chemistry_Amines_Structure-Basicity_Relationship_Of_Secondary_Amines_Reaction_2
  • Tertiary amines do not react with benzenesulphonyl chloride.
  • Note:-
    1. Amine reacting with benzenesulphonyl chloride in a different manner is used for the distinction of primary, secondary and tertiary amines and also for the separation of a mixture of amines.
    2. These days benzenesulphonyl chloride is replaced by p-toluenesulphonyl chloride.
  1. Electrophilic Substitution
  • Bromination :-
  • Aniline reacts with bromine water at room temperature to give a white precipitate of 2, 4, 6-tribromoaniline.
  • Because of high reactivity of aromatic amines problems occur during electrophilic substitution because it occurs at ortho- and para-positions.
Class_12_Chemistry_Amines_Structure-Basicity_Relationship_Of_Bromination
  • In order to prepare monosubstituted aniline derivative activating effect of –NH2 group be controlled done by protecting the -NH2  group by acetylation with acetic anhydride, then carrying out the desired substitution followed by hydrolysis of the substituted amide to the substituted amine.
Class_12_Chemistry_Amines_Structure-Basicity_Relationship_Of_Bromination1
  • The lone pair of electrons on nitrogen of acetanilide interacts with oxygen atom due to resonance as shown below:
Class_12_Chemistry_Amines_Structure-Basicity_Relationship_Of_Bromination2
  • Hence, the lone pair of electrons on nitrogen is less available for donation to benzene ring by resonance.
  • Therefore, activating effect of –NHCOCH3 group is less than that of amino group.
  • Nitration:
  • Direct nitration of aniline yields tarry oxidation products in addition to the nitro derivatives.
  • Moreover, in the strongly acidic medium, aniline is protonated to form the anilinium ion which is meta directing.
  • That is why besides the ortho and para derivatives, significant amount of meta derivative is also formed.
Class_12_Chemistry_Amines_Structure-Basicity_Relationship_Of_Amines_Nitration
  • In order to protect –NH2 group by acetylation reaction with acetic anhydride, the nitration reaction can be controlled and the p-nitro derivative can be obtained as the major product.
Class_12_Chemistry_Amines_Structure-Basicity_Relationship_Of_Amines_Nitration1
  • Sulphonation:
  • Aniline reacts with concentrated sulphuric acid to form anilinium hydrogensulphate which on heating with sulphuric acid at 453-473K produces p-aminobenzene sulphonic acid, commonly known as sulphanilic acid, as the major product.
  • Aniline does not undergo Friedel-Crafts reaction (alkylation and acetylation) due to salt formation with aluminium chloride, the Lewis acid, which is used as a catalyst.
  • Due to this, nitrogen of aniline acquires positive charge and hence acts as a strong deactivating group for further reaction.
Chapter-13-Amines

Class 12 Amines NCERT Solution

NCERT INTEXT QUESTIONS

13.1. Classify the following amines as primary, secondary and tertiary:
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-1
Ans: (i) 1° (ii) -3° (iii) 1° (iv) 2°

13.2. 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?
Ans: Eight isomeric amines are possible
NCERT Solutions for Class 12 Chemistry T2
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.

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:
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-3

13.4. 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, C6H5CH2NH
2
Ans:
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 T4

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 T4A

The Kb values are :

NCERT Solutions for Class 12 Chemistry T4b

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

NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-4

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

NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-5

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

NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-6

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:

NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-7

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

NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-8

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:
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-9

NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-10

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.

NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-11

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

NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-12

(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-13

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 C2H5NHdue 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-14
(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-15
(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-16
(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-17
(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-18
(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-19
(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-20

13.5. 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?

Ans:
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-21
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-22

13.6. 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.
    tiwari academy class 12 chemistry e6
  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 e6a
  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 e6b

(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 e6c
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.
tiwari academy class 12 chemistry e6d
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 e6f
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.

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

(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,
tiwari academy class 12 chemistry e6i
(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 e6j

13.7. 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
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.
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-24
(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-25
(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-26
(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-27
(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-28
(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-29
(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-30

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:
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-31

NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-32
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-33

13.9. Give the structures of A,B and C in the following reaction:
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-34
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-35
Ans:
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-36

13.10. An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’ which on heating with Brand KOH forms a compound ‘C’ of molecular formula C6H7N. Write the structures and IUPAC names of compounds A, B and C.
Ans:
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 e10

13.11. Complete the following reactions:
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-37
Ans:
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-38
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-39

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.
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-40

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
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines-41

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.

Important Questions for Class 12 Chemistry Chapter 13 Amines

Amines: Organic Compounds Containing Nitrogen Class 12 Important Questions Very Short Answer Type

Question 1.
Why is an alkylamine more basic than ammonia? (Delhi 2009)
Answer:
Due to electron releasing inductive effect (+1) of alkyl group, the electron density on the nitrogen atom increases and thus, it can donate the lone pair of electrons more easily than ammonia.

Question 2.
Arrange the following compounds in an increasing order of basic strengths in their aqueous solutions : NH3, CH3NH2, (CH3)2NH, (CH3)3N (All India 2009)
Answer:
Basicity order (due to stability of ammonium cation)
(CH3)2 NH > CH3NH2 > (CH3)3 N > NH3

Question 3.
Give the IUPAC name of H2N — CH2—CH2—CH = CH2. (Delhi 2010)
Answer:
IUPAC name : But-3-ene-1-amine

Question 4.
Arrange the following compounds in an increasing order of their solubility in water : C6H5NH2, (C2H5)2NH, C2HSNH2 (Delhi & All India 2011)
Answer:
C6H5NH2 < (C2H5)2NH < C22H5NH2

Question 5.
Give a chemical test to distinguish between ethylamine and aniline. (All India 2011)
Answer:
Ethylamine and aniline :
By Azo dye test: It involves the reaction of any aromatic primary amine with HNO2(NaNO2 + dil. HCl) at 273-278 K followed by treatment with an alkaline solution of 2-naphthol when a brilliant yellow, orange or red coloured dye is obtained.

Question 6.
Arrange the following in the decreasing order of their basic strength in aqueous solutions: CH3NH2, (CH3)2 NH, (CH3)3N and NH3 (Delhi 2012)
Answer:
(CH3)2 NH > CH3NH2 > (CH3)3 N > NH3

Question 7.
Arrange the following in increasing order of their basic strength in aqueous solution: CH3.NH2, (CH3)3N, (CH3)2NH (Delhi 2013)
Answer:

Question 8.
Write the structure of 2-aminotoluene. (All India 2013)
Answer:

Question 9.
Write the structure of n-methylethanamine. (All India 2013)
Answer:
Structure of n-methylethanamine :
H3C—H2C—NH—CH2

Question 10.
Write the structure of prop-2-en-l-amine. (All India 2013)
Answer:
H2C=CH—H2C—NH2

Question 11.
How may methyl bromide be preferentially converted to methyl isocyanide? (Comptt. Delhi 2013)
Answer:
Bit carbylamine reaction:

Question 12.
Arrange the following compounds in increasing order of solubility in water :
C6H5NH2, (C2H5)2NH, C2H5NH2 (Delhi 2014)
Answer:
C6H5NH2 < (C2H5)2NH < C2H5NH,

Question 13.
Arrange the following in increasing order of basic strength :
C6H5NH2, C6H5NHCH3, C6H5CH2NH2 (Delhi 2014)
Answer:
C6H5NH2 < C6H5NHCH3 < C6H5CH2NH2

Question 14.
Arrange the following in increasing order of basic strength :
C6H5NH2, C6H5NHCH3, C6H5N(CH3)2 (Delhi 2014)
Answer:
C6H5N(CH3)2 > C6H5NHCH3 > C6H5NH2

Question 15.
The conversion of primary aromatic amines into diazonium salts is known as (All India 2014)
Answer:
Diazotization.

Question 16.
Out of CH3—NH2 and (CH3)3N, which one has higher boiling point? (Comptt. Delhi 2014)
Answer:
CH3—NH2 has higher boiling point than (CH3)3N.

Question 17.
Complete the following reaction equation : (Comptt. Delhi 2015)
Answer:

Question 18.
Arrange the following in increasing order of basic strength (Comptt. All India 2015)
Aniline, p-Nitroaniline and p-Toluidine
Answer:

Question 19.
Write the IUPAC name of the given compound: (Delhi 2016)

Answer:
2, 4, 6-Tribromoaniline

Question 20.
Write IUPAC name of the following compound: (CH3CH2)2NCH3 (Delhi 2017)
Answer:
N-Ethyl-N-methylethanamine

Question 21.
Write the IUPAC name of the following compound: (Comptt. All India 2017)
CH3NHCH(CH3)2
Answer:
IUPAC name: N-Methylpropan-2-amine

Question 22.
Write the IUPAC name of the following compound: (Delhi 2017)
(CH3)2N-CH2CH3
Answer:
IUPAC name: N, N Dimethylamine

Question 23.
Write IUPAC name of the following compound : (Comptt. All India 2017)

Answer:
IUPAC name : N, N-Dimethylbutanamine.

Amines: Organic Compounds Containing Nitrogen Class 12 Important Questions Short Answer Type -I [SA – I]

Question 24.
Give the chemical tests to distinguish between the following pairs of compounds :
(i) Ethyl amine and Aniline
(ii) Aniline and Benzylamine (All India 2010)
Answer:
(i) Ethylamine and aniline :
By Azo dye test: It involves the reaction of any aromatic primary amine with HNO2(NaNO2 + dil. HCl) at 273-278 K followed by treatment with an alkaline solution of 2-naphthol when a brilliant yellow, orange or red coloured dye is obtained.

(ii) Distinction between Aniline and Benzylamine :
By Nitrous acid test : Benzylamine reacts with HNO2 to form a diazonium salt which being unstable even at low temperature, decomposes with evolution of N2 gas

Question 25.
Identify A and B in each of the following processes : (All India 20100

Answer:

Question 26.
Give the chemical tests to distinguish between the following pairs of compounds :
(i) Methylamine and Dimethylamine
(ii) Aniline and N-methylaniline (All India 2010)
Answer:
(i) Methylamine and Dimethylamine :
By Carbylamine test: Methylamine being a primary amine gives this test but Dimethylamine being a secondary amine does not.

(ii) Aniline and N-methylaniline
By Carbylamine test : Aniline is a 1° aromatic amine while N-methylaniline is a secondary aromatic amine. Therefore only 1° aromatic amine gives this test.

Question 27.
Describe the following giving the relevant chemical equation in each case :
(i) Carbylamine reaction
(ii) Hofmann’s bromamide reaction (All India 2012)
Answer:
(i) Carbylamine reaction : Aliphatic and aromatic primary amines on heating with chloroform and ethanolic KOH form isocyanides or carbylamines which are foul smelling substances. This reaction is known as carbylamines reaction.

(ii) Hofmann’s bromamide reaction : Primary amines can be prepared by treating an amide with Br2 in an aqueous or alcoholic soln of NaOH.

Question 28.
Complete the following reaction equations : (All India 2012)
(i) C6H5N2Cl + H3PO2 + H2O →
(ii) C6H5NH2 + Br2 (aq) →
Answer:

Question 29.
Give IUPAC names of the following compounds : (Comptt. Delhi 2012)

Answer:
(a) IUPAC name : Methyl prop-2-en-1-amine
(b) IUPAC name : Phenyl acetamide

Question 30.
How are the following conversions carried out :
(a) Aniline to p-hydroxyazobenzene
(b) Ethanoyl chloride to Ethanenitrile. (Comptt. Delhi 2012)
Answer:
(a) Aniline to p-hydroxyazobenzene

Question 31.
How are the following conversions carried out?
(i) CH3CH2Cl to CH3CH2CH2NH2
(ii) Benzene to aniline (Comptt. Delhi 2012)
Answer:

Question 32.
How would you account for the following :
(a) Aniline is a weaker base than cyclohexyl amine.
(b) Methylamine in aqueous medium gives reddish-brown precipitate with FeCl3. (Comptt. All India 2012)
Answer:
(a) In aniline, the lone pair of electrons on the N-atom is delocalised over the benzene ring. As a result, the electron density on the nitrogen decreases.
But in cyclohexylamine, the lone pair of electrons on N-atom is readily available due to absence of reelections. Hence aniline is weaker base than cyclohexylamine.
(b) Methylamine being more basic than H2O, it accepts a proton from water liberating OH ions.

Question 33. ( Class 12 Chemistry Amines )
How would you account for the following :
(a) Electrophilic susbstitution in case of aromatic amines takes place more readily than benzene.
(b) Ethanamide is a weaker base than ethanamine. (Comptt. All India 2012)
Answer:
(a) Aniline exists as a resonance hybrid of the following five structures :

The electron density is maximum at ortho and para positions to the – NH2 group. But in benzene there is no delocalisation of electron at any position and hence electrophilic substitution in case of aromatic amines takes place more readily than benzene.

(b) In ethanamide the lone pair of electron of N-atom is not available due to resonance structure

So it is a weaker base.

Question 34. ( Class 12 Chemistry Amines )
Illustrate the following reactions :
(a) Sandmeyer’s reaction
(b) Coupling reaction (Comptt. All India 2012)
Answer:
(a) Sandmeyer’s reaction : Aniline reacts with NaNO2 in HCl at 273 – 278 K giving diazonium salt which further reacts with cuprous chloride/bromide to give chloro or bromo benzene.

This reaction is Sandmeyer’s reaction.
(b) Coupling reaction : Arene diazonium salts react with highly reactive aromatic compounds such as phenols and amines to form brightly coloured azo compounds.
Ar – N = N – Ar. This reaction is known as coupling reaction.

Question 35. ( Class 12 Chemistry Amines )
Give chemical tests to distinguish between the following pairs of compounds :
(a) Aniline and Ethylamine
(b) Ethylamine and Dimethylamine (Comptt. Delhi 2013)
Answer:
(a) Ethylamine and aniline :
By Azo dye test: It involves the reaction of any aromatic primary amine with HNO2(NaNO2 + dil. HCl) at 273-278 K followed by treatment with an alkaline solution of 2-naphthol when a brilliant yellow, orange or red coloured dye is obtained.

(b) Ethylamine and dimethylamine can be distinguished by the carbylamine test.
Carbylamine test: Aliphatic and aromatic amines on heating with chloroform and ethanolic potassium hydroxide form foul smelling isocyanides or carbylamines. Ethylamine (being an aliphatic primary amine) gives a positive carbylamine test, but dimethylamine does not.

Question 36. ( Class 12 Chemistry Amines )
Explain the following reactions :
(a) Gabriel Phthalimide reaction
(b) Coupling reaction (Comptt. Delhi, Comptt. All India 2013)
Answer:
(a) Gabriel phthalimide synthesis : It is used to prepare 1° amine (Primary amine). The starting compound is a phthalimide. But aromatic primary amines cannot be prepared by this method because aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide.
Example :

(b) Coupling reaction : Arene diazonium salts react with highly reactive aromatic compounds such as phenols and amines to form brightly coloured azo compounds.
Ar – N = N – Ar. This reaction is known as coupling reaction.

Question 37. ( Class 12 Chemistry Amines )
Give reasons :
(a) Aniline is a weaker base than cyclohexyl amine.
(b) It is difficult to prepare pure amines by ammonolysis of alkyl halides. (Comptt. All India 2013)
Answer:
(a) In aniline, the lone pair of electrons on the N-atom is delocalised over the benzene ring. As a result, the
electron density on the nitrogen decreases.
But in cyclohexylamine, the lone pair of electrons on N-atom is readily available due to absence of π-electrons. Hence aniline is weaker base than cyclohexylamine.
(b) Because the primary amine formed by ammonolysis itself acts as a nucleophile and produces further 2° and 3° alkyl amine.

Question 38. ( Class 12 Chemistry Amines )
Give reasons :
(i) Electrophilic substitution in aromatic amines takes place more readily than benzene.
(ii) CH3CONH2 is weaker base than CH3CH2NH2. (Comptt. All India 2013)
Answer:
(i) Due to the strong activating effect of the NH2 group, aromatic amines undergo electrophilic substitution reactions readily than benzene.
(ii) In case of acetamide due to resonance, the lone pair of electrons on the nitrogen atom is delocalized over keto group which decreases electron density hence less basic while in ethylamine due to +1 effect of ethyl group electron density increases on N-atom and hence basic character increases.

Question 39. ( Class 12 Chemistry Amines )
(i) Arrange the following compounds in an increasing order of basic strength :
C6H5NH2, C6H5N(CH3)2, (C2H5)2NH and CH3NH2
(ii) Arrange the following compounds in a decreasing order of pKb values :
C2H5NH2, C6H5NHCH3, (C2H5)2NH and C6H5NH (Comptt. Delhi 2014)
Answer:
(i) Increasing order of basic strength

(ii) Decreasing order of pKb values
C6H5NH2 < C6H5NHCH3 < C2H5NH2 < (C2H5)2NH

Question 40. ( Class 12 Chemistry Amines )
Give a chemical test to distinguish between each of the following pairs of compounds :
(i) Ethylamine and Aniline
(ii) Aniline and Benzylamine (Comptt. All India 2014)
Answer:
(i) Ethylamine and aniline :
By Azo dye test: It involves the reaction of any aromatic primary amine with HNO2(NaNO2 + dil. HCl) at 273-278 K followed by treatment with an alkaline solution of 2-naphthol when a brilliant yellow, orange or red coloured dye is obtained.

(ii) Distinction between Aniline and Benzylamine :
By Nitrous acid test : Benzylamine reacts with HNO2 to form a diazonium salt which being unstable even at low temperature, decomposes with evolution of N2 gas

Question 41. ( Class 12 Chemistry Amines )
Write the chemical equations involved in the following reactions: (All India 2016)
(i) Hoffmann-bromamide degradation reaction
(ii) Carbylamine reaction
Answer:
(i) Hoffmann’s bromamide reaction : In this reaction, migration of an alkyl or aryl group takes place from carbonyl carbon of the amide to the nitrogen atom. Therefore the amine so formed has one carbon atom less than that of amide.

(ii) Carbylamine reaction. This reaction is used to distinguish primary amines from 2° and 3° amines as it is only given by 1° amines with the production of a very bad smelling organic compound.
For example :

Amines: Organic Compounds Containing Nitrogen Class 12 Important Questions Short Answer Type -II [SA – II]

Question 42. ( Class 12 Chemistry Amines )
Giving an example for each describe the following reactions :
(i) Hofmann’s bromamide reaction
(ii) Gatterman reaction
(iii) A coupling reaction (Delhi 2009)
Answer:
(i) Hofmann’s bromamide reaction : When amide is treated with bromide in alkaline solution, an amide yields an amine containing one carbon less than the starting amide.

(ii) Gatterman reaction: When benzene or its derivative is treated with carbon monoxide and hydrogen chloride in the presence of anhydrous aluminium chloride or cuprous chloride, it gives benzaldehyde or substituted benzaldehyde.

Gattermann Koch reaction : Diazonium salt reacts with hydrogen halide in presence of copper powder giving haloarene.

(iii) A coupling reaction : Arene diazonium salts react with highly reactive aromatic compounds such as phenols and amines to form brightly coloured azo compounds.
Ar – N = N – Ar. This reaction is known as coupling reaction.

Question 43. ( Class 12 Chemistry Amines )
Complete the following reaction equations: (All India 2009)

Answer:
Reduction reaction

Question 44. ( Class 12 Chemistry Amines )
Complete the following reaction equations :
(i) C6H5Cl + CH3COCl →
(ii) C2H5NH2 + C6H5SO2Cl →
(iii) C2H5NH2 + HNO2 →
Answer:

Question 45. ( Class 12 Chemistry Amines )
In the following cases rearrange the compounds as directed : (Delhi 2010)
(i) In an increasing order of basic strength :
C6H5NH2, C6H5 N(CH3)2, (C2H5)2NH and CH3NH2
(ii) In a decreasing order of basic strength :
Aniline, p-nitroaniline and p-toluidine
(iii) In an increasing order of pKb values :
C2H5NH2, C6H5 NHCH3, (C2H5)2NH and C6H5NH2
Answer:
(i) Order of basic strength :

Since a stronger base has a lower pKb value therefore basic strength order.
(C2H5)2NH > C2H5NH2 > C6H5NHCH3 > C6H5NH2

Question 46. ( Class 12 Chemistry Amines )
Complete the following chemical equations : (Delhi)

Answer:

Question 47. ( Class 12 Chemistry Amines )
(a) Explain why an alkylamine is more basic than ammonia?
(b) How would you convert
(i) Aniline to nitrobenzene (ii) Aniline to iodobenzene (Delhi 2011)
Answer:
(a) Due to electron releasing inductive effect (+1) of alkyl group, the electron density on the nitrogen atom increases and thus, it can donate the lone pair of electrons more easily than ammonia.

(b) (i) Aniline to nitrobenzene

Question 48. ( Class 12 Chemistry Amines )
Complete the following chemical equations: (Delhi 2011)

Answer:

Question 49. ( Class 12 Chemistry Amines )
State reasons for the following :
(i) pKb value for aniline is more than that for methylamine.
(ii) Ethylamine is soluble in water whereas aniline is not soluble in water.
(iii) Primary amines have higher boiling points than tertiary amines. (All India 2011)
Answer:
(i) Higher the pKb value, lower will be the basicity therefore aniline is less basic than methylamine because the
lone pair of electrons on nitrogen atom gets delocalized over the benzene ring are unavailable for protonation due to resonance in aniline which is absent in case of alkylamine.
(ii) Ethylamine is soluble in water due to its capability to form H-bonds with water while aniline is insoluble in water due to larger hydrocarbon part which tends to retard the formation of H-bonds.
(iii) Due to 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 a H-atom on the N-atom, do not undergo H- bonding. As a result, primary amines have higher boiling points than 3° amines.

Question 50. ( Class 12 Chemistry Amines )
Write chemical equations for the following conversions :
(i) Nitrobenzene to benzoic acid.
(ii) Benzyl chloride to 2-phenylethanamine
(iii) Aniline to benzyl alcohol. (Delhi 2012)
Answer:
(i) Nirtobenzene to benzoic acid :

(ii) Benzyl chloride to 2-phenylethanamine

(iii) Aniline to benzyl alcohol

Question 51. ( Class 12 Chemistry Amines )
Give the structure of A, B and C in the following reactions: (Delhi 2013)

Answer:

Question 52. ( Class 12 Chemistry Amines )
Give the structure of A, B and C in the following reactions : (Delhi 2013)

Answer:

Question 53. ( Class 12 Chemistry Amines )
Complete the following reactions : (All India 2013)

Answer:

Question 54. ( Class 12 Chemistry Amines )
Write the main products of following reactions: (All India 2013)

Answer:

Question 55. ( Class 12 Chemistry Amines )
Give the structures of A, B and C in the following reactions : (Delhi 2014)

Answer:

Question 56. ( Class 12 Chemistry Amines )
How will you convert the following :
(i) Nitrobenzene into aniline
(ii) Ethanoic acid into methanamine
(iii) Aniline into N-phenylethanamide
(Write the chemical equations involved) (Delhi 2014)
Answer:
(i) Nitrobenzene into aniline

Question 57. ( Class 12 Chemistry Amines )
Account for the following :
(i) Primary amines (R-NH2) have higher boiling point than tertiary amines (R3N).
(ii) Aniline does not undergo Friedel – Crafts reaction.
(iii) (CH3)2NH is more basic than (CH3)3N in an aqueous solution. (All India 2014)
Answer:
(i) Due to 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 a H-atom on the N-atom, do not undergo H- bonding. As a result, primary amines have higher boiling points than 3° amines.
(ii) Aniline being a Lewis base reacts with Lewis acid AlCl3 to form a salt.

As a result, N of aniline acquires positive charge and hence it acts as a strong deactivating group for electrophilic substitution reaction. Consequently, aniline does not undergo Freidel Craft reaction.
(iii) Due to more steric hindrance in (CH3)3N it is less basic than (CH3)2NH.

Question 58. ( Class 12 Chemistry Amines )
Account for the following:
(i) Aniline does not give Friedel-Crafts reaction.
(ii) Ethylamine is soluble in water whereas aniline is not.
(iii) pKb of methylamine is less than that of aniline. (Comptt. Delhi 2014)
Answer:
(i) Aniline being a Lewis base reacts with Lewis acid AlCl3 to form a salt.

As a result, N of aniline acquires positive charge and hence it acts as a strong deactivating group for electrophilic substitution reaction. Consequently, aniline does not undergo Freidel Craft reaction.
(ii) Ethylamine is soluble in water due to its capability to form Fl-bonds with water while aniline is insoluble in water due to larger hydrocarbon part which tends to retard the formation of H-bonds.
(iii) In aniline due to resonance lone pair of electron of nitrogen atom is delocalised due to which it is weaker base than methyl amine.
Hence Aniline has high pKb molecule i.e., methylamine has less pKb molecule.

Question 59. ( Class 12 Chemistry Amines )
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. (Comptt. Delhi 2015)
Answer:
The data shows that C6H7N may be C6H5NH2 i.e. Aniline. Since it is obtained by heating with Br2 and KOH (Hoffmann bromamide reaction), then the compound ‘B’ is Benzamide C6H5CONH2 which is in turn obtained by reaction with aqueous ammonia then the compound ‘A’ can be Benzoic acid i.e. C6H5COOH

Question 60. ( Class 12 Chemistry Amines )
Write the structures A, B and C in the following : (Delhi 2016)

Answer:

Question 61. ( Class 12 Chemistry Amines )
Give reasons for the following:
(i) Aniline does not undergo Friedal-Crafts reaction.
(ii) (CH3)2 NH is more basic than (CH3)3 N in an aqueous solution.
(iii) Primary amines have higher boiling point than tertiary amines. (All India 2016)
Answer:
(i) Aniline being a Lewis base reacts with Lewis acid AlCl3 to form a salt.

As a result, N of aniline acquires positive charge and hence it acts as a strong deactivating group for electrophilic substitution reaction. Consequently, aniline does not undergo Freidel Crafts reaction.
(ii) In (CH3)N there is maximum steric hindrance and least solvation but in (CH3)2NH the solvation is more and the steric hindrance is less than in (CH3)3NH; although + I effect is less, since there are two methyl groups; di-methyl amine is still a stronger base than tri-methyl.
(iii) Due to 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 a H-atom on the N-atom, do not undergo H- bonding. As a result, primary amines have higher boiling points than 3° amines.

Question 62. ( Class 12 Chemistry Amines )
Write major product(s) in the following reactions : (Comptt. Delhi 2016)

Answer:

Question 63. ( Class 12 Chemistry Amines )
Describe a method for the identification of primary, secondary and tertiary amines. Also write chemical equations of the reaction involved. (Comptt. Delhi 2016)
Answer:
Benzenesulphonyl chloride (C6H5SO2Cl), which is also known as Hinsberg’s reagent, reacts with primary and secondary amines to form sulphonamides and tertiary amine does not react.

It does not react with tertiary amines.

Question 64. ( Class 12 Chemistry Amines )
Write the products A and B in the following : (Comptt. All India 2016)

Answer:

Question 65. ( Class 12 Chemistry Amines )
Give reasons:
(i) Acetylation of aniline reduces its activation effect,
(ii) CH3NH2 is more basic than C6H5NH2.
(iii) Although —NH2 is o/p directing group, yet aniline on nitration gives a significant amount of m-nitroaniline. (Delhi 2017)
Answer:
(i) Acetylation of aniline reduces its activation effect because acetyl group being electron withdrawing group attracts the lone pair of electrons of the N-atom towards carboxyl group and the lone pair of electrons on N is less available for donation to benzene ring by resonance.
(ii) CH3NH2 is more basic than aniline due to availability of lone pair of electrons for donation while in aniline lone pair of electrons on the nitrogen atom is delocalised over benzene ring and thus unavailable for donation.
(iii) Because of nitration in an acidic medium, aniline gets protonated to give anilinium ion which is indirecting.

Question 66. ( Class 12 Chemistry Amines )
Give reasons for the following:
(a) Acetylation of aniline reduces its activation effect.
(b) CH3NH2 is more basic than C6H5NH2.
(c) Although —NH2 is o/p directing group, yet aniline on nitration gives a significant amount of m-nitroaniline. (All India 2017)
Answer:
(i) Acetylation of aniline reduces its activation effect because acetyl group being electron withdrawing group attracts the lone pair of electrons of the N-atom towards carboxyl group and the lone pair of . electrons on N is less available for donation to benzene ring by resonance.
(ii) CH3NH2 is more basic than aniline due to availability of lone pair of electrons for donation while in aniline lone pair of electrons on the nitrogen atom is delocalised over benzene ring and thus unavailable for donation.
(iii) Because of nitration in an acidic medium, aniline gets protonated to give anilinium ion which is indirecting.

Question 67. ( Class 12 Chemistry Amines )
Write the structures of compounds A, B and C in each of the following reactions: (All India 2017)

Answer:

Question 68. ( Class 12 Chemistry Amines )
Illustrate Sandmeyer’s reaction with the help of a suitable example. (Comptt. Delhi 2017)
Answer:
Sandmeyer’s reaction: The substitution of diazo group of benzene diazonium chloride by Chloro, Bromo and Cyano group with the help of solution of CuCl dissolved in HCl, CuBr/HBr and CuCN/KCN respectively is known as Sandmeyer’s reaction.

Question 69. (Class 12 Chemistry Amines )
Identify A, B and C in the following reactions : (Comptt. Delhi 2017)

Answer:

Question 70. ( Class 12 Chemistry Amines )
Identify A, B and C in the following reactions : (Comptt. All India 2017)

Answer:

Amines: Organic Compounds Containing Nitrogen Class 12 Important Questions Long Answer Type [LA]

Question 71. ( Class 12 Chemistry Amines )
An aromatic compound JA’ of molecular formula C7H27ON undergoes a series of reactions as shown below. Write the structures of A, B, C, D and E in the following reactions : (Delhi 2015)

Answer:

Question 72. ( Class 12 Chemistry Amines )
(a) Write the structures of main products when aniline reacts with the following reagents :
(i) Br2 water (ii) HCI (iii) (CH3CO)2O/pyridine
(b) Arrange the following in the increasing order of their boiling point :
C2H5NH2, C2H5OH, (CH3)3N
(c) Give a simple chemical test to distinguish between the following pair of compounds : (CH3)2NH and (CH3)3N (Delhi 2015)
Answer:
(a) (i) Br2 water

(b) Increasing order of boiling point :
(CH3)3 < C2H5NH2 < C2H2OH

(c) By Hinsberg test, secondary amines or (CH3)3NH shows precipitate formation which is insoluble in KOH. Tertiary amines or (CH3)3N do not react with Hinsberg’s reagent (benzene sulphonyl chloride).

Question 73. ( Class 12 Chemistry Amines )
An aromatic compound ‘A’ of molecular formula C7H6O3 undergoes a series of reactions as shown below. Write the structures of A, B, C, D and E in the following reactions : (All India 2015)

Answer:

Question 74. ( Class 12 Chemistry Amines )
(a) Write the structures of main products when benzene diazonium chloride reacts with the following reagents :
(i) H3PO2 + H2O (ii) CuCN/KCN (iii) H2O
(b) Arrange the following in the increasing order of their basic character in an aqueous solution :
C2H5NH2, (C2H5)2NH, (C3H5)3N
(c) Give a simple chemical test to distinguish between the following pair of compounds :
C6H5—NH2 and C6H5—NH—CH3 (All India 2015)
Answer:
(a) The structure of main products when aniline (benzene diazonium chloride) reacts with the following reagents :

(b) C2H5NH2 < (C3H5)3N < (C2H5)2NH
(c) Aniline and Benzylamine can be distinguished by the Nitrous acid test. Benzylamine reacts with HNO2 to form a diazonium salt which being unstable even at low temperature, decomposes with evolution of N2 gas.

Related Links

Class 9th

Class 10th

Class 11th

Class 12th

Leave a Reply

Your email address will not be published. Required fields are marked *