class 12 chemistry amines formula and structure pdf

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Amines are organic compounds of ammonia in which one or more than one hydrogen atoms are
replaced by other atoms or group of atoms. They are classified as primary (RNH2 or 1°), secondary (R2NH
or 2°) or tertiary (R3N or 3°) depending upon whether one, two or all the three hydrogen atoms of ammonia
are replaced by alkyl groups (R).
Amines are commonly named as alkyl amines. The alkyl groups attached to Natom are named in
the alphabetical order followed by amine. According to IUPAC system of nomenclature they are named as
Alkanamine. The longest carbon atom chain attached to Natom is chosen as the parent compound and in
the name of parent hydrocarbon, the last letter ‘e’ is replaced by
the suffix amine. The substituents are named as prefixes in the alphabetical order. The IUPAC names (in
bold letters) and common names (in parenthesis) of some of the well known amines
are given below.
(a) Primary amines:
CH3CH2CH2NH2 NH2
CH3CHCH2NH2
1Propanamine
(npropylamine)
CH3
2methyl1propanamine
(isobutylamine)
Cyclopropanamine
(cyclopropylamine)
(b) Secondary amines:
CH3CH2NHCH3 (CH3CH2)2NH
NHCH3
Nmethylethanamine
(Ethylmethylamine)
Nmethylcyclohexanamine
(cyclohexylmethylamine)
Nethylethanamine
(Diethylamine)
(c) Tertiary amines:
CH3CH2CH2N(CH3)CH2CH3 (CH3)3N
N(CH3)
NethylNmethyl1propanamine
(Ethylmethylpropylamine)
NmethylNcyclopropylcyclopentanamine
(Methylcyclopentylcyclopropylamine)
N, Ndimethylmethanamine
(Trimethylamine)
(d) Arylamines:
Arylamines have NH2 group directly attached to the benzene ring. They are named as derivatives
of aniline (common name) or benzenamine (IUPAC name).
NOMENCLATURE
1
NITROGEN CONTAINING COMPOUNDS (AMINES)

NH2
Benzenamine
(Aniline)
NHCH(CH3)2
Nisopropylbenzenamine
(Nisopropylaniline)
NH2
3methylbenzenamine
(mtoluidine)
CH3
(e) Heterocyclic amines:
IUPAC names of heterocyclic amines have prefixes aza, diaza or triaza to indicate that one, two or
three nitrogen atoms have replaced carbon atoms in the corresponding hydrocarbons. Numbering of the
ring starts from the hetero atom. The names of some of the well known heterocyclic amines are as follows:
1Azacyclopenta2,4diene
(Pyrrole)
Azabenzene
(Pyridine)
1Azanaphthalene
(Quinoline)
N
N N
H
1
2
3
4
5
1
2
3
4
5
10
6
7
8
9
1Azacyclohexane
(Piperidine)
1,3Diazabenzene
(Pyrimidine)
1,2diazacyclopenta2,4diene
(Pyrazole)
N
N
H
N
N
N
1
2
3
4
5
6
1
2
3
4
5
6
H
2
3
4
5
1
2.1. ALKYLATION OF AMMONIA
Alkyl halildes undergo nucleophilic substitution reaction by SN2 mechanism with NH3 forming
primary amines.
H3N + RX NH3
..
H3NR X

+
H2NR + NH4X
1° amine
The reaction does not stop at this stage. Primary amines being more basic than ammonia further
reacts with alkyl halide forming secondary amine (2°), tertiary amine (3°) and eventually quaternary
ammonium salt, if the alkyl halide is present in excess.
RNH2
RX
R2NH R4N X

+
RX R3N RX
..
HX HX
We can make primary amine as the major product by carrying out the reaction with liquid
ammonia. The reaction between ethyl bromide and ammonia proceeds by the following mechanism:
H3N + CH3CH2Br
NH3
..
CH3CH2NH2Br


H
..
CH3CH2NH2 + NH4Br
METHODS OF PREPARATION OF AMINES
2

CH3CH2NH2 + CH3CH2Br
.. NH3
(CH3CH2)2NHBr


H
..
(CH3CH2)2NH + NH4Br etc.
2.2 ALKYLATION OF AZIDE ION AND REDUCTION
A much better method for preparing a primary amine from an alkyl halide is first to convert the
alkyl halide to an alkyl azide (RN3) by a nucleophilic substitution reaction with sodium azide (NaN3).
NNN:
..
..
2
+
N=N=N:
..
..
+
:
 
Azide ion
:
N=N=N: + RX
..
..
+
:
 
Alkyl azide
N=N=NR + X
..
+
:
 
The alkyl azide is then reduced to a primary amine with Na/C2H5OH or LiAlH4.
RN=N=N:
..
+  Na/C2H5OH
or LiAlH4
RNH2 + N2
2.3 GABRIEL PHTHALIMIDE SYNTHESIS
Another method used for the preparation of primary aliphatic amines only is the
Gabriel Phthalimide synthesis. Phthalimide (pKa = 9) is quite acidic in nature. It can be converted to
potassium phthalimide by its reaction with KOH. The phthalimide anion is a strong nucleophile. It reacts
with alkyl halide, preferably methyl halide and primary alkyl halide only, by an SN2 mechanism to give
Nalkylphthalimide. The secondary and tertiary alkyl halides are not employed because they undergo
elimination reactions also. The Nalkyl phthalimide is hydrolysed with dilute HCl or KOH solution to give
primary aliphatic amine.
NH
O
KOH
O
N: K
O
O
+

RX
NR
O
O
(1° only)
2KOH
COOK
COOK
+ RNH2
KX
Nalkyl phthalimide
Nalkylphthalimide can also be converted to primary amine and phthalazine1, 4dione by
treating it with hydrazine.
NR + NH2NH2
O
O
O
O
CNHNH2
CNHR
.. ..
O
O
NH
NH
+ RNH2
Phthalazine1, 4dione
This method is not suitable for preparing aromatic primary amines as aryl halides are not good
substrates for nucleophilic substitution.
2.4 REDUCTION OF NITRO COMPOUNDS

Aliphatic primary amines can be synthesised by reducing nitroalkanes with metal and acid or with
H2 in presence of nickel as a catalyst.
RNO2 + 3H2 

Ni
RNH2 + 2H2O
RNO2 

 

HCl
/
Sn
3
H
N
R

 
 

Base
RNH2
For example,
O2NCH2CH2OH 

 

Ni
/
H2
H2NCH2CH2OH
2nitro ethanol 2amino ethanol
Primary aromatic amines are also prepared by the reduction of corresponding nitro compounds.
NO2
Sn/HCl
NH3
+
Base
NH2
Other reducing agents such as LiAlH4 and H2/Pt also reduce nitrobenzene to aniline.
Reduction in neutral medium
1. With Fe and steam, a nitro group gets reduced to nitroso group.
C6H5NO2 


 

steam
/
Fe
C6H5NO
2. Zn/NH4Cl/H2O or Zn/CaCl2/H2O reduces nitro group to hydroxylamine.
C6H5NO2
C
50
O
H
/
Cl
NH
/
Zn
o
2
4




 
 C6H5NHOH
Reduction in alkaline medium
Nitro benzene when reduced with Zn/NaOH/C2H5OH gives hydroazobenzene.
2C6H5NO2 




 

OH
H
C
/
NaOH
/
Zn 5
2
C6H5NHNHC6H5
By means of aqueous ethanolic NH4HS, aqueous Na2S or SnCl2 in HCl, nitro groups in a polynitro
compound can be reduced one at a time. For example, mdinitrobenzene can be reduced to mnitroaniline.
NO2
NO2
+ 3NH4HS
NO2
NH2
+ 3NH3 + 3S + 2H2O
It is not always possible to predict that which nitro group will be reduced first. For instance,
2, 4dinitrotoluene when treated with NH4HS, the 4nitro group is reduced whereas treatment with
SnCl2/HCl results in the reduction of 2nitro group.
CH3
NH4HS
NO2
NH2
CH3
NO2
NO2
SnCl2/HCl
CH3
NH2
NO2
2, 4dinitrotoluene
2.5 REDUCTION OF NITRILES, ISONITRILES AND OXIMES
Both aliphatic and aromatic amines can be synthesised by reduction of cyanides
(or nitriles), isocyanides (or isonitriles) and oximes by H2 in the presence of a catalyst or LiAlH4.
For example,
CH3CH2CN
4
2
LiAlH
or
Ni
/
H


 
 CH3CH2CH2NH2

(1° amine)




C
N
H
C 5
6
4
2
LiAlH
or
Ni
/
H


 
 C6H5NHCH3
(2° amine)
C=NOH
CH3
C2H5
Na/C2H5OH
or LiAlH4
CHNH2 + H2O
CH3
C2H5
(oxime) (1°amine)
Cyanides and oximes on reduction gives 1° amines while isocyanide on reduction gives
2° amines.
2.6 REDUCTIVE AMINATION OF CARBONYL COMPOUNDS
Aldehydes and ketones are converted to amines through catalytic or chemical reduction in the
presence of ammonia or amine. Primary, secondary and tertiary amines can be prepared by this method.
C=O
R
R
NH3
RNH2
RCHNH2
[H]
[H]
RNHR
[H]
RCHNHR
R
R
(1° amine)
(2° amine)
RCHN
R
R
R
(3° amine)
(R may be hydrogen or an alkyl group)
The above process appears to proceed through the following mechanism.
C=O + H2NR
R
R
RCNHR
..
R
(1° amine)
OH
H2O
C=NR
R
R (imine)
2[H]
RCHNHR
R
(2° amine)
unstable
2.7 FROM AMIDES
Amides with or without any substituent on the nitrogen can be reduced with LiAlH4 to the
corresponding amines having same number of carbon atoms. For example,
CH3CNH2
LiAlH4
O
CH3CH2NH2 (1° amine)
NHCCH3
O
LiAlH4
NHCH2CH3 (2° amine)
CH3CH2NCCH3
LiAlH4
O
CH3CH2NCH2CH3 (3° amine)
CH3 CH3
2.8 HOFFMANN BROMAMIDE DEGRADATION

Amides with no substituent on the nitrogen when treated with a solution of Br2 or Cl2 in KOH
yield only primary amines having one carbon atom less than the amides.
RCNH2 + Br2 + 4KOH
O
RNH2 + K2CO3 + 2KBr + 2H2O
The reaction mechanism involves basepromoted Nbromination yielding Nbromo amide as an
intermediate. The Nbromo amide then reacts with hydroxide ion to produce Nbromamide anion, which
rearranges with the migration of R group and loss of Br
ion to produce isocyanate.
Base catalysed hydrolysis of isocyanate formed in the reaction mixture produces carbamate ion, which
undergoes spontaneous decarboxylation resulting in the formation of amine.
RCNH + OH
O
..
H

RCN:
O
..
H
 BrBr
RCNBr + Br
O
H

Nbromoamide
..
RCNBr + OH
O
..
H

RCNBr
O
..

RN=C=O + Br

..
Nbromo amide anion
Isocyanate
RN=C=O + OH

RN=CO
OH

RNC=O
OH
IMPE

RNH2 + CO2
O
CO2
H2O
RNHC=O

1° amine
3.1 CARBYLAMINE REACTION:
Both aliphatic and aromatic amines when heated with chloroform and ethanolic KOH form
isocyanide, also called carbylamine, a foul smelling compound.
RNH2 + CHCl3 + 3KOH RN C + 3KCl + 3H2O
or
ArNH2
or
ArN C
This reaction is used as a test for primary amines. In this reaction dichlorocarbene is formed as an
intermediate, which attacks at the Natom of amines resulting in the loss of 2 moles HCl to give
isocyanide.
RNH2 + CCl2
.. ..
RN C:
H
H
+ 
RN C
H
+ 
Cl
Cl
Cl
RN C:

+
HCl HCl
3.2 HOFFMANN MUSTARD OIL REACTION:
When warmed with carbon disulphide, primary amines form dithiocarbamic acid, which is
decomposed by mercuric chloride to the alkyl isothiocyanate.
GENERAL REACTIONS OF AMINES
3

RNH C
H
+
RNH2 + S=C=S
.. S
S
RNH C
S
SH

HgCl2
RN=C=S + HgS + 2HCl
IMPE
3.3 HINSBERG REACTION:
This reaction is used for the separation of amines from a mixture. The mixture containing primary,
secondary and tertiary amines is treated with an aromatic sulphonyl chloride. Originally
benzenesulphonyl chloride, C6H5SO2Cl was used, but has now been replaced by ptoluene
sulphonyl chloride, pCH3C6H4SO2Cl. After treatment with this acid chloride, the solution is
made alkaline with KOH. Primary amines form Nalkylsulphonamides, which dissolve in KOH
forming potassium salt due to the presence of acidic hydrogen attached to Natom.
RNH2
R3N
No reaction
O
R2NH
pCH3C6H4SCl
O
KOH
O
pCH3C6H4SNHR
O
O
pCH3C6H4SNR K+
O
Stabilised by resonance
KOH
O
pCH3C6H4SNR2
O
No reaction
HCl H2O
Secondary amines form N, Ndialkylsulphonamides, which do not dissolve in KOH because there
is no hydrogen atom attached to N. Tertiary amines do not react with ptoluenesulphonyl chloride.
3.4 REACTION WITH CARBOXYLIC ACID DERIVATIVES:
Primary and secondary amines react with acid chlorides, anhydrides or esters to form substituted
acid amides; primary amines forming Nalkyl amides and secondary amines forming N, Ndialkyl
amides.
RNH2 + CH3COCl  CH3CONHR + HCl
R2NH + (CH3CO)2O  CH3CONR2 + CH3COOH
R2NH + CH3COOCH3  CH3CONR2 + CH3OH
3.5 REACTION WITH NITROUS ACID:
Primary amines:
Aliphatic primary amines react with HNO2 with the evolution of N2. Nitrous acid is produced in
the reaction mixture by adding NaNO2 and dilute HCl. The reaction proceeds via diazonium salt,
which is not stable and decomposes to yield a carbocation as one of the intermediates. This
carbocation gives variety of products. For example,
CH3CH2CH2NH2 + NaNO2 + HCl CH3CH2CH2NNCl
+ 
npropyl diazonium salt
CH3CH2CH2NN Cl CH3CH2CH2 + N2 + Cl
+ 
1° carbocation
+ 
CH3CHCH2 CH3CHCH3
+
2° carbocation
+
H
Rearrangement

CH3CH2CH2Cl CH3CH2CH2
+
Cl
H2O
CH3CH2CH2OH + H+
H+
CH3CH=CH2
H+
CH3CHCH3 CH3CHCH3
+
Cl
H2O
CH3CHCH3 + H+
Cl OH
Diazonium salts of aromatic amines are stable at low temperature (05° C) but decompose with the
evolution of N2 on heating. The mechanism involved in the formation of diazonium salt is as
follows:
HNO2 + HNO2 N=O + H2O + NO2
+ 
C6H5NH2 + N=O C6H5NN=O
+
+
..
H
H
..
C6H5NN=OH
+
..
H
IMPE
C6H5NN
+ ..
C6H5N=N
+
H2O
C6H5N=NOH2
+
C6H5N=NOH
..
H
Diazonium cation
+
IMPE ..
Aromatic diazonium chlorides, sulphates, nitrates etc. are reasonably stable in aqueous solution at
room temperature or below but cannot be readily isolated without decomposition. The orbital
system of benzene ring stabilises the diazonium cation by resonance.
+
NN
+
N=N
+
 +
N=N
+
 +
N=N
+

NN
..
N=N
 
The diazonium salts are very important synthetic reagents, being the starting point in the
preparation of various aromatic compounds. Their reactions may be divided into two groups; those
which involve the liberation of N2 gas and the displacement of the diazo group 
 2
N , by another
univalent group and those in which the two Natoms are retained (coupling reactions).
Replacement reactions:

C6H5OH + N2 + H+
H2O, 
C6H6 + N2 + H3PO3 + H+
H3PO2/H2O, 
C6H5Cl + N2 + H+
CuCl/HCl, 
C6H5Br + N2 + H+
CuBr/HCl, 
Sandmeyer reaction
C6H5I + N2 + K+
KI, 
C6H5N2 BF4
HBF4,  +  
C6H5F + N2 + BF3
C6H5CN + N2 + K+
Cu/KCN, 
C6H5NO2 + N2 + Na+
NaNO2/Cu, 
C6H5C6H5 + N2 + H2O + Na+
(Gomberg reaction)
C6H6/NaOH, 
C6H5NN

Baltz Scheimann reaction
Secondary amines
Both the aliphatic and aromatic amines react with HNO2 in which NO+
ion attacks the Natom of
the amine forming Nnitrosamine.
R2NH + HNO2 R2NN=O + H2O
NHR
+ HNO2
RNN=O
+ H2O
Tertiary amines
Aliphatic tertiary amines (3°) do not react with HNO2 whereas aromatic tertiary amines react with
HNO2 forming pnitrosoN,Ndialkylaniline as the major product.
NR2
+ HNO2
NR2
N=O
H2O
pnitrosoN, Ndialkylaniline
3.6 AZO COUPLING
Diazonium salts readily undergo coupling reactions with phenols, naphthols and aromatic amines
to form highly coloured azocompounds. For example, benzenediazonium chloride couples with
phenol in weakly alkaline solution to form phydroxyazobenzene.
C6H5N2Cl
+ C6H5OH
(i)
OH
C6H5N=N OH + Cl

(ii) H
+
The rate of reaction increases as the pH change from 5 to 8. Under mildly alkaline conditions,
phenol behaves as phenoxide ion, which is much more activating than phenol itself.

C6H5NN

C6H5N=N + O C6H5N=N O
H
OH
C6H5N=N

+H+
Coupling with benzene substrates occurs preferentially in the para position to the hydroxyl group.
But if this position is blocked, then the coupling occurs at the ortho position.
For example, pcresol gives oazo compound.
C6H5N=N +
OH
OH
CH3
C6H5N=N
OH
CH3

1and 2naphthols in alkaline solution couple with diazonium salt in the 4and 1position
respectively.

C6H5N2Cl
2naphthol
OH
+OH
1naphthol
+OH
OH
N=NC6H5
N=NC6H5
Aromatic amines are in general somewhat less readily attacked than phenols and coupling is often
carried out in slightly acid solution. Under these conditions not only the concentration of 
2
5
6 N
H
C
is high but also the amine ArNH2 is not significantly converted into the unreactive protonated
cation, ArNH3
+
. The initial diazotisation of aromatic primary amines is carried out in strongly
acidic media to ensure that as yet unreacted amine is converted to the cation and so prevented from
coupling with diazonium salt as it is formed.
With aromatic amines, there is the possibility of attack on either nitrogen or carbon. In the
case of primary amines, the attack of diazonium ion mainly takes place at the nitrogen forming
diazoamino compound (A).
C6H5N=N + H2N
H
H
+
C6H5N=NNH

C6H5N=NNH
(A)

With secondary amines (e.g., Nalkyl anilines), two products are formed; one due to NN
coupling and the other due to NC coupling. For example,
C6H5N=N +

NHCH3 C6H5N=NNCH3
+
NHCH3
N=NC6H5
Tertiary amines (e.g. N, Ndialkylaniline) show only NC coupling.

C6H5N=N +

N(CH3)2 C6H5N=N N(CH3)2
C6H5N=N N(CH3)2
H

H+
Diazonium cation, 
2
5
6 N
H
C is a relatively weak electrophile and reacts with only highly reactive
aromatic compounds such as phenols, aniline and substituted anilines.
It does not react with less reactive compound C6H5OCH3 (anisole), mesitylene etc. Its reactivity
can be increased by introducing strongly electron withdrawing groups such as NO2 at the ortho or
para position. This will enhance positive charge at the diazo group making it a better electrophile.

N
O
O
NN N
O
O
N=N
 
Thus, the 2,4dinitrophenyl diazonium cation will react with C6H5OCH3
and 2,4,6trinitrophenyl diazonium cation will even react with the hydrocarbon 1,3,5trimethyl
benzene (mesitylene).
3.7 OXIDATION REACTIONS
Both the primary and secondary amines undergo oxidation. The oxidation products obtained
depend on the oxidising agent used and on the nature of alkyl group.
Primary amines
(i) With KMnO4 :
[O]
RCH2NH2 RCH=NH
H2O
RCHO + NH3
Aldimine
[O]
R2CHNH2 R2C=NH
H2O
R2C=O + NH3
Ketimine
(ii) With Caro’s acid (H2SO5)/H2O2/Peroxy carboxylic acid :
[O]
RCH2NH2 RCH2NHOH
OH
Nalkyl
hydroxylamine
[O]
H2O
RCH=NOH
[O]
Aldoxime
RC=NOH
Hydroxamic
acid
[O]
R2CHNH2 R2CHNHOH
[O]
H2O
R2C=NOH
Ketoxime
Secondary amines
(i) With KMnO4 :

[O]
2R2NH
Tetraalkyl hydrazine
N N
R
R R
R
H2O
(ii) With Caro’s acid (H2SO5)/H2O2/Peroxy carboxylic acid :
[O]
R2NH
N,Ndialkyl hydroxylamine
N OH
R
R
Tertiary amines
(i) With H2O2/Peroxy carboxylic acid
Tertiary amines are oxidised to amine oxide,



O
N
R3 (a dipolar ion or Zwitterion).
[O]
R3N



O
N
R3
3.8 RING SUBSTITUTION REACTIONS
NH2, NHR and NR2 groups when attached to benzene strongly increases its electron density,
making it highly reactive towards electrophilic aromatic substitution.
(i) Halogenation
When aniline is treated with bromine water, the substitution takes place at all the three places (two
ortho and one para) yielding the solo product 2,4,6tribromoaniline. In order to carry out
monobromination, the NH2 group is first converted to moderately activating acetanilide group
NHCOR and then subjecting it to bromination followed by acid or base catalysed hydrolysis.
NH2
Br2
Br
Br
Br
H2O
NH2
CH3COCl
NHCOCH3
Br2/Fe
NHCOCH3
Br
H
+
/H2O
NH2
Br
HCl
(major product)
(ii) Sulphonation
Aniline reacts with H2SO4 forming a salt, which on heating at 180°C yields sulphamic acid. If the
heating is continued at 180°C for 3 hours, sulphamic acid undergoes rearrangement to give
sulphanilic acid, which exists as a dipolar ion.
NH2
H2SO4
NH3HSO4
180°C
NHSO3H
180°C
NH3
SO3
+ 
H2O
Sulphamic acid
(3 Hrs)


Sulphanilic acid
In contrast, pamino benzoic acid does not exist as a dipolar ion because COOH is a very weak
acid in comparison to SO3H and is unable to transfer proton to the weakly basic NH2 group.
(iii)Nitration
Aniline is very reactive towards nitration and much of it gets oxidised by HNO3. In order to carry
out mononitration of aniline, the aniline is first converted to acetanilide and then acetanilide is
subjected to nitration by nitrating mixture followed by acid or base catalysed hydrolysis.

NH2
CH3COCl
NHCOCH3
HNO3 + H2SO4
NHCOCH3
OH
/H2O
NH2
NO2
HCl
NO2
3.9 REARRANGEMENT REACTIONS
A remarkable property of mono , di , and trialkyl anilinium chlorides (or bromides) is their
ability to undergo rearrangement on strong heating, an alkyl group migrating from the Natom and
entering preferentially the pposition. If this position is occupied, then the alkyl group migrates to
the oposition. For example, when trimethyl anilinium chloride is heated under pressure, the
following rearrangement takes place.
N(CH3)3Cl
300°C
N(CH3)2.HCl NH2.HCl
CH3
CH3

CH3
NH(CH3).HCl
CH3 H3C CH3
 
This reaction is known as the HoffmannMartius rearrangement.
Rearrangements of this kind have been observed to take place with aniline derivatives of the type
C6H5NHZ where Z is R, X, NH2, OH, NO or NO2. For example,
RNNO NHR
+ HCl
NHR
NO
+ NOCl + HCl
This reaction is called FischerHepp rearrangement.
NH2NH2.HCl NH2.HCl
NH2
250°C
Benzidine Rearrangement
Hydroazobenzene, C6H5NHNHC6H5 undergoes rearrangement when heated in the presence of
acid to yield benzidine (4, 4diamino diphenyl).
NH
H
+
NH2
NH
NH2 + NH2
NH2
4,4diaminodiphenyl 2,4diaminodiphenyl
This is known as benzidine rearrangement. The p, pisomer is the major product (70%) and the
rest is o, p and o, oisomer. The reaction probably follows the given mechanism.

NH
2H+
NH NH2 NH2 NH2 NH2
H H
NH2 NH2
H H
NH2
NH2
   
 
2H+
NH2 NH2
NH2

+
+
2H+
NH2

2H+
H
H
NH2
NH2
(o, pisomer)
(p, pisomer)
(o, oisomer)
3.10 EXHAUSTIVE METHYLATION
Primary, amine RCH2CH2NH2 undergoes exhaustive methylation with the MeI to give quaternary
ammonium iodide RCH2CH2NMe3I.

. The quaternary ammonium iodide on treatment with AgOH
gives quaternary ammonium hydroxide RCH2CH2NMe3OH

along with precipitate of AgI. When
quaternary ammonium hydroxide is heated, it undergoes Hoffmann elimination (Hoffmann
degradation) to give mostly less substituted alkene. The alkene formation is governed by the loss
of most acidic H (1° > 2° > 3°). The transition state has more CH
bond breaking than bond
making (double bond formation) because the leaving group is a relatively poor one. Thus, the
acidity of H becomes more important than the stability of alkene that forms. This is called
Hoffmann rule (less substituted alkene is formed in higher amount).
RCH2CH2NH2 


 
 )
excess
(
MeI
3
RCH2CH2NMe3I


 
AgOH
RCH2CH2NMe3OH



RCH=CH2 + H2O + Me3N
But when there are two alkyl groups, each having hydrogens, that alkyl part forms alkene,
whose H is more acidic to give least substituted alkene.
CH3CH2NCH2CH2CH3

CH3
CH3
    
OH CH2=CH2 + Me2NCH2CH2CH3 + H2O
or CH3CH=CH2 + Me2NCH2CH3 + H2O
(major product)
(minor product)
CH3CH2CH2CHNMe3

CH3

OH CH3CH2CH2CH=CH2 + Me3N + H2O
CH3CH2CH=CHCH3
(major product)
(minor product)
+
3.11 REACTION WITH DIETHYL OXALATE
1°, 2° and 3° amines can be distinguished by their reaction with diethyl oxalate. Primary (1°)
amines react with diethyl oxalate forming N, Noxamide, which is a solid.

COOC2H5 H–NH–R CONHR
COOC2H5 H–NH–R
+
CONHR
+ 2C2H5OH
N, Ndialkyl
oxamide (solid)
Secondary (2°) amines react with diethyl oxalate forming oxamic ester, which is a liquid.
COOC2H5 CONHR2
COOC2H5
+ HNR2
COOC2H5
+ C2H5OH
oxamic ester
(liquid)
Tertiary (3°) amines donot react with diethyl oxalate.

SOLVED OBJECTIVE EXAMPLES
Example 1:
Consider the following reaction,
(C)
(B)
(A)
KOH
CHCl
NH
H
C
O
/H
H
Δ
3
2
5
6
2



 






The compounds (B) and (C) are
(a) C6H5COOH and NH3 respectively (b) C6H5NH2 and HCOOH respectively
(c) C6H5NH2 and H2O respectively (d) none of these
Solution:
O
H
3
KCl
3
C
N
H
C
KOH
3
CHCl
NH
H
C 2
)
A
(
5
6
3
2
5
6 





)
C
(
)
B
(
2
5
6
H
2
5
6 HCOOH
NH
H
C
O
H
2
C
N
H
C 





(A)
 (b)
Example 2:
When benzamide is heated with thionyl chloride, the main product of the reaction is
(a) C6H5CN (b) C6H5COCl
(c) C6H5NH2 (d) C6H5CNH2
S
Solution:
SOCl2 here can only act as a dehydrating agent.
C6H5CNH2 + SOCl2  C6H5CN + SO2 + 2HCl
O
 (a)
Example 3:
Which of the following would react with ozone to form an isocyanate?
(a) CH3CN (b) C6H5CN
(c) CH3NC (d) C6H5NH2
Solution:
Isocyanides react with O3 to form isocyanates
CH3N C + O3  CH3N=C=O + O2
 (c)
Example 4:
Phenyl cyanide on reduction with Na/C2H5OH yields
(a) C6H5CH2NH2 (b) C6H5NHCH3
(c) NH2
CH3
(d) C6H5NH2
Solution:
C6H5CN
Na/C2H5OH
C6H5CH2NH2
 (a)

Example 5:
Hydroazobenzene can be obtained by reducing nitrobenzene with
(a) Sn + HCl (b) Zn + NH4Cl
(c) Na3AsO3 + NaOH (d) Zn + NaOH
Solution:
NO2
Zn/NaOH
NHNH
 (d)
Example 6:
Hinsberg reagent is ________ and reacts with ________ amine to form a product soluble in alkali.
(a) SO2Cl, primary
CH3 (b) SO2NH2, secondary
CH3
(c) CH3, tertiary
HO (d) SO2Cl, secondary
CH3
Solution:
Only primary amine reacts with Hinsberg reagent (p-toluene sulphonyl chloride) to form sulphonamide,
which is soluble in alkali because of the presence of acidic hydrogen.
 (a)
Example 7:
The boiling points of the following amines follow the order
(a) C2H5NH2 < (CH3)2NH < (CH3)3N (b) (CH3)2NH < C2H5NH2 < (CH3)3N
(c) (CH3)3N < C2H5NH2 < (CH3)2NH (d) (CH3)3N < (CH3)2NH < C2H5NH2
Solution:
Primary amines and secondary amines can form hydrogen bonds through Hatoms bonded to Natom.
Tertiary amines do not form Hbonds. Boiling point of amines increases as the number of Hbonds formed
by it increases.
 (d)
Example 8:
When aniline is treated with benzene diazonium chloride at low temperature in weakly acidic medium,
the final product obtained is
(a) N=N
NH2
(b) N=N NH2
(c) N=NNH (d) N=N
NH2
Solution:
Diazonium cation reacts with aniline in weakly acidic medium resulting in N, N-coupling rather than N, C-
coupling.
N=N + HN

H
H+
N=NNH
 (c)

Example 9:
CH3CH2NH2 is soluble in
(a) dilute HCl (b) CuSO4 solution
(c) AgNO3 (d) All of these
Solution:
Amines being basic in nature dissolve in dilute HCl. They can also coordinate with Cu2+
and Ag+
ions to
form soluble complexes as they can act as good ligands.
 (d)
Example 10:
The end product (Z) of the following reaction is
N2Cl
–
+ Cu/KCN
(X)
H+
/H2O NaOH
(Y)
CaO, 
(Z)
(a) a cyanide (b) a carboxylic acid
(c) an amine (d) an arene
Solution:
N2Cl
+ Cu/KCN H+
/H2O
COOH
Sodalime,

CN
 (d)

SOLVED SUBJECTIVE EXAMPLES
Example 1:
Complete the following reactions:
(a) NH + CH3CCl (A)
O
LiAlH4
(B)
(b) NN Cl
(C) + (D) (Both organic)
H2O

(c) SCl + C2H5NH2 (E)
O
O
NaOH
(F)
C2H5Br
(G)
H3O+
(H) + (I)
(d) CH3N=C=S + CH3NH2  (J)
Solution:
(a) NH + CH3CCl
(A)
O
LiAlH4
(B)
HCl
NCCH3
O
NCH2CH3
(b) NN Cl
(C)
H2O

N2
OH +
(D)
(c) SCl + C2H5NH2
(E)
O
O
HCl
SNHC2H5
O
O
NaOH
H2O
SNC2H5
(G)
O
O
C2H5Br
SN(C2H5)2
O
O
H3O+
(F)
Br 
(H)
SOH + (C2H5)2NH2
O
O
+
(I)
(d) CH3N=C=S + CH3NH2  CH3NHCNHCH3
S
(J)
Example 2:
What happens when cyclopentanone reacts with
(a) CH3CH2NH2 (1° amine)
(b) (CH3CH2)2NH (2° amine)
Solution:

(a)
O + CH3CH2NH2 N
CH2CH3
Tautomerise
N
H
CH2CH3
–H2O
(b) O + (CH3CH2)2NH
N
CH2CH3
 H+
N
H
CH2CH3
CH2CH3
CH2CH3
OH
Example 3:
Give product of the thermal decomposition of hydroxide of the following quaternary salts.
(a)
N(CH3)3
+
(b)
N(CH3)2CH2CH2CH3
+
(c) CH3CH2NCH2CH2CH3
+
CH3
CH3
(d)
CH3
N(CH3)3
+
Solution:
(a) As there is no  hydrogen atom, alkene is not formed. Instead methanol is formed.
N(CH3)3
+
OH–

CH3OH +
N(CH3)2
(b)
CH3NCH2CHCH3
+
CH3
H  hydrogen
OH–

CH2=CHCH3 +
N(CH3)2
(major)
(minor)
+ N(CH3)2CH2CH2CH3
H  hydrogen
(c) CH2CH2NCH2CH2CH3
+
H CH3
CH3
OH–

CH2=CH2 + N(CH3)2CH2CH2CH3
(d)
CH2H
+
OH–
N(CH3)3

CH2 + N(CH3)3
Example 4:
How will you carry out the following conversions?

(a)
C
NH2
NH
CH3
O
(b) NH2 NH2
Solution:
(a)
C
NH2
NH2
O
Br2/KOH CHCl3
KOH
NC
LiAlH4
NH
CH3
(b) NH2
NaNO2
HCl
OH
H2SO4

KMnO4/H+

BaO
COOH
COOH 
O
NH3

NH
H2/Pt
NH2
Example 5:
An optically active compound (A), C3H7O2N forms a hydrochloride but dissolves in water to give a
neutral solution. On heating with soda lime (A) yields (B) C2H7N. Both (A) and (B) react with NaNO2
and HCl, the former yielding a compound (C) C3H6O, which on heating is converted to (D), C6H8O4
while the latter yields (E), C2H6O. Account for the above reactions and suggest how (A) may be
synthesized.
Solution:
Degree of unsaturation of (A) = 2.
Since (A) forms hydrochloride and dissolves in water to give a neutral solution, it contains both a basic and
an acidic functional group. It is likely to be amino acid as the molecular formula contains one N and 2 O-
atoms. On decarboxylation it forms an amine (B).
Degree of unsaturation of (B) = 0
Therefore, (B) is a saturated amine. (B) reacts with NaNO2 and dilute HCl forming (E) C2H5OH. Thus, (B) is
CH3CH2NH2. (A) also reacts with NaNO2 and dilute HCl forming (C), a hydroxy acid, which forms a
cyclic diester on heating. All the reactions can be given as
CH3CHCOOH
(A)
NH2
NaOH
NaNO2 + HCl
CH3CH2NH2
NaNO2 + HCl
CH3CH2OH
CaO
(B) (E)
CH3CHCOOH
(C)
OH
2CH3CHCOOH
OH

CH3CH
2H2O
O CO
CO O
CHCH3
Preparation of (A):
CH3CH2COOH
P/Br2
CH3CHCOOH
NH3
Br
CH3CHCOOH
NH2
(A)
Example 6:

C5H13N
NaNO2/HCl
(Y) + other products
N2
3° alcohol
(X)
(X) is optically active. Identify the structure of (X) and (Y). Explain the formation of (Y) from (X). Will
(Y) also be optically active? Justify your answer. Draw the structure of important intermediates, if
any.
Solution:
CH3CHCHNH2
CH3CH3
(Optically active)
(X)
CH3CHCHNN
CH3 CH3
NaNO2/HCl  N2
CH3CHCH
CH3CH3

2° carbocation
~H
CH3CCH2CH3
CH3

3° carbocation
H2O/H+
CH3CH2CHCH2NN
CH3
 N2
CH3CH2CHCH2
CH3

1° carbocation
~H
CH3CCH2CH3
CH3
Optically inactive
CH3CH2CHCH2NH2
CH3
NaNO2
HCl
OH
*
*
(Y)
MIND MAP
PREPARATION OF AMINES
R-NO2 R-CN R2C=NOH
R–NH2
NH3
R2NH
R3N
R–C–NH2
O
H2/Ni
LiAlH4
Br2+KOH (Loss of 1C-atom)
R–X
(i) R – X
(ii) KOH
CO
CO
NK+
–
(i) R – X
(ii) LiAlH4
NaN3
R–X
R2CO
NH3
RNH2
R2NH
R–C–NHR
O
R–N =C
R–X
H2/Ni
LiAlH4 R–C–NR2
O
LiAlH4

Ar–NH2
Ar–NO2
Sn/HCl
REACTIONS OF AMINES
CH3CONHR
R–N=C=S
R–OH
(i) CS2
(ii) HgCl2
Ar-SO2NRK+
–
(i) ArSO2Cl
(ii) KOH
CHCl3 + KOH
(Also given by ArNH2)
(CONHR)2
(COOEt)2
CH3COCl or
(CH3CO)2O
R2C=NR
R2CO
R–CHO
or
R2CO
KMnO4
(i) RX (excess)
(ii) Ag2O/H2O, 
Alkene
OH
R–C=N–OH
H2SO5
R–N=C

R–NH2
HNO2
R2N–NR2
CS2
R2N–C=S
SH
HgCl2 No reaction
HNO2
(Also given
by ArNHR
R2N–NO
CONR2
COOEt
ArSO2NR2
KOH
ArSO2Cl
(COOEt)2 KMnO4
R2N–OH
H2SO5
R2NH
No reaction

Multiple Choice Questions
1. In which of the following resonance of – NH2 is possible?
(a) 1-Aminobutane
(b) Ethylamine
(c) Benzyl amine
(c) p-Toluidine [AMU (Med.)]
2. Which one of the following is an example of 30
amide?
(a) O
N
(b) O
N
H
(c) O
N
H
(d) O
NH2 [AMU (Med.)]
3. Which one of the following is a secondary amine?
(a) 2-Butanamine
(b) N-Methylpiperidine
(c) N-Methyl-2-pentanamine
(d) p-Anisidine (J & K CET)
4. Assertion: When acetamide reacts with NaOH & Br2, methyl amine is formed.
Reason: The reaction occurs through intermediate formation of isocyanate.
(a) If both assertion & reason are true & reason is the correct explanation of assertion.
(b) If both assertion & reason are true but reason is not the correct explanation assertion.
(c) If assertion is true but reason is false.

(d) If both assertion & reason are false. (AIIMS)
5. Method by which aniline cannot be prepared is
(a) degradation of benzamide with bromine in alkaline solution.
(b) reduction of nitrobenzene with H2/Pd in ethanol.
(c) potassium salt of phthalimide treated with chlorobenzene followed by hydrolysis with aqueous
NaOH solution.
(d) hydrolysis of phenylisocyanide with acidic solution. (AIPMT)
6. When the following amide is treated with Br2/KOH, it gives
NH2
i) Br2 / ii) KOH
O
(a)
NH2
(b) OH
O
HN
(c) O
(d)
O
N (J & K CET)
7. Which one of the following amines cannot be prepared by Gabriel phthalimide synthesis?
(a) Ethylamine
(b) Iso-propylamine
(c) n-Propylamine
(d) Ethylmethylamine
(e) Allylamine (Kerala PMT)
8. In the given set of reaction:
2-Bromopropane alc. AgCN/Δ X LiAlH4 Y
The IUPAC name of the product Y is
(a) N-isopropylmethanamine

(b) N-methylpropan-2-amine
(c) N-methylpropanamine
(d) but-2-amine (Karnataka CET)
9. The source of nitrogen in Gabriel synthesis of primary amines is
(a) potassium cyanide
(b) potassium phthalimide
(c) sodium azide
(d) sodium nitrite [AMU (Med.)]
10. Based on which method amines are prepared?
(a)Reduction of nitro compounds
(b) Ammonolysis of alkyl halides
(c) Reduction of nitriles & amides
(d) All of the above (J & K CET)
11. Assertion: Acetamide on reaction with KOH & bromine gives acetic acid
Reason: Bromine catalyses hydrolysis of acetamide.
12. Which of the following will not give a primary amine?
(a) CH3CN LiAlH4
(b) CH3NC LiAlH4
(c) CH3CONH2 LiAlH4
(d) CH3CONH2 Br4, NaOH [AMU (Med.)]
13. Which one of the following gives amine on heating with amide?
(a) Br2 in aqueous KOH
(b) Br2 in alcoholic KOH
(c) Cl2 in sodium
(d) Sodium in ether (Karnataka CET)
14. Secondary amines could be prepared by
(a) reduction of nitriles
(b) Hofmann bromide reaction
(c) reduction of amides
(d) reduction of isonitriles
(e) reduction of nitro compounds (Kerala PMT)
15. In a set of reaction m-bromobenzoic acid give a product D. Identify the product D.
COOH
SOCl2 B NH3 C NaOH/Br2 D
A
Br
(a) SO2NH2
Br
(b) COOH

NH2
(c) NH2
Br
(d) CONH2
Br (AIPMT)
16. In the reaction,
C = O R/H+ C = NC2H5 (i) R1 , (ii) H3O+ CH – NHC2H5
Reagent R & R1 are
(a) ethyl amine, ethanol
(b) ethyl amine, sodium borohydride
(c) ethyl amine, hydrogen peroxide
(d) ethyl alcohol, sodium metal [AMU (Med.)]
17. Which of the following reagents can be used to convert acetamide into methanaminr?
(a) P2O5
(b) NaOBr
(c) LiAlH4/H2O
(d) Na(Hg)/C2H5OH (J & K CET)
18. Among the following which one does not act as an intermediate in Hofmann rearrangement?
(a) RNCO
(b) RCON
(c) RCONHBr
(d) RNC (AFMC)
19. A primary amine that can be obtained both by the reduction of cyanides & amides is
(a) methyl amine
(b) benzyl amine
(c) aniline
(d) iso-propyl amine
(e) tertiary butyl amine (Kerala PMT)
20. The compound with foul odour among the following is
(a) NC
(b) CN

(c) NH2
(d) NO2
(J & K CET)
21. Hofmann’s bromamide reaction is to convert
(a) acid to alcohol
(b) alcohol to acid
(c) amide to amine
(d) amine to amide (Karnataka CET)
22. The product formed in this reaction is
Hexanamide + Br2 + KOH
(a) butanamine
(b) pentanamine
(c) hexanamine
(d) pentanamide (OJEE)
23. The correct statement regarding the basicity of arylamines is
(a) arylamines are generally more basic than alkylamines because of aryl group.
(b) arylamines are generally more basic than alkylamines because the nitrogen atom in arylamines is
sp-hybridised.
(c) arylamines are generally less basic than alkylamines because the nitrogen lone-pair electrons are
delocalised by interaction with the aromatic ring 𝜋-electron system.
(d) arylamines are generally more basic than alkylamines because the nitrogen lone-pair electrons are not
delocalised by interaction with the aromatic ring 𝜋-electron system. (NEET Phase – I)
24. Which of the following reagents could be used to distinguish aniline from methanamine?
(i) Bromine water
(ii) CHCl3 & aqueous KOH
(iii) Dilute HCl
(iv) Nitration mixture under heated condition
(a) (i) only
(b) (i) & (ii)
(c) (i) & (iii)
(d) (i) & (iv) (J & K CET)
25. An organic compound A on reduction gives compound B, which on reaction with trichloro methane &
caustic potash forms C. The compound ‘C’ on catalytic reduction gives N-methyl benzenamine,
the compound ‘A’ is
(a) nitrobenzene
(b) nitromethane

(c) methanamine
(d) benzenamine (Karnataka CET)
26. The following reaction
NH2 H
+ Cl NaOH N is known by the name
O
O
(a) Perkin’s reaction
(b) Acetylation reaction
(c) Schotten-Baumann reaction
(d) Friedel-Craft’s reaction (AIPMT)
27. Carbylamine test is given by ____________ amines.
(a) primary
(b) secondary
(c) tertiary
(d) quaternary [AMU (Med.)]
28. Benzene sulphonyl chloride forms a soluble salt in alkali, when it reacts with
(a) primary amine
(b) secondary aamine
(c) tertiary amine
(d) nitrous acid (Kerala PMT)
29. Acid anhydride on reaction with primary amines give
(a) imine
(b) 20
amine
(c) amide
(d) imide [AMU (Med.)]
30. An organic compound (C3H9N) (A), when treated with nitrous acid, gave an alcohol & N2 gas was evolved.
(A) on warming with CHCl3 & caustic potash gave (C) which on reduction gave isopropylmethylamine.
Predict the structure of (A).
(a) CH3
CH – NH2
CH3
(b) CH3CH2 – NH – CH3
(c) CH3 – N – CH3
CH3
(d) CH3CH2CH2 – NH2 [AIPMT (Mains)]
31. An aniline on nitration gives
(a) NH2
NO2

(a) CH3
NO2
NO2
(a) NH2
NO2
(d) both (a) & (c) (AFMC)
32. The weakest base among the following is
(a) phenylmethanamine
(b) N-methylmethanamine
(c) ethanamine
(d) methanamine
(e) benzenamine (Kerala PMT)
33. Nitration of aniline also gives m-nitroaniline in strong acidic medium because
(a) in electrophilic substitution reaction, amino group is meta-directive
(b) inspite of substituents, nitro group always goes to m-position
(c) in strong acidic medium, aniline forms anilinium ion
(d) none of the above. (UP CPMT)
34. NH2 NH2 CH2 – NH2 NH2
OCH3 CH3 Cl
(I) (II) (III) (IV)
The correct decreasing order of 𝑝𝐾 is
(a) I > II > II > IV
(b) III > IV > II > I
(c) II > III > IV > I
(d) IV > II > I > III (AIIMS)
35. Anilinium hydrogensulphate on heating with sulphuric acid at 453-473 K produce
(a) benzene sulphonic acid
(b) anthranilic acid
(c) aniline
(d) m-aminobenzene sulphonic acid
(e) sulphanilic acid (Kerala PMT)
36. Which of the following is most basic in aqueous solution?
(a) CH3NH2
(b) (CH3)2NH

(c) (CH3)3N
(d) NH3 (OJEE)
37. The compound which on reaction with aqueous nitrous acid at low temperature produces an oily
nitrosoamine is
(a) methyl amine
(b) ethyl amine
(c) diethyl amine
(d) triethyl amine (AIIMS)
38. Predict the product.
NHCH3
+ NaNO2 + HCl product
(a) CH3
N – NO2
(b) NHCH3 NHCH3
NO2 +
NO
(c) OH
N – CH3
(d) CH3
N – N = O
(AIPMT)
39. Which of the following is soluble in sodium hydroxide?
(a)
H3C SO2NHCH3
(b)
H3C SO2N(CH3)2
(c)
H3C NH2
(d)
H3C NHCH3
[AMU (Med.)]
40. Carbylamine reaction is given by
(a) N-methyl benzyl amine
(b) 2,4-dimethyl aniline
(c) N,N-dimethyl amine

(d) none of these (WB JEE)
41. Assertion: Anilinium chloride is more acidic than ammonium chloride.
Reason: Anilinium chloride is as resonance stabilised as aniline.
42. During acetylation of amines, what is replaced by acetyl group?
(a) One or more hydrogen atoms attached to nitrogen atom.
(b) Hydrogen atoms attached to either carbon atom or nitrogen atom.
(c) Hydrogen atom attached to nitrogen atom.
(d) One or more hydrogen atom attached to carbon atom. (UP CPMT)
43. Which of the following statement is true?
(a) Trimethyl amines form a soluble compound with Hinsberg reagent & KOH.
(b) Dimethylamines react with KOH & phenol to form an azo dye.
(c) Methylamine reacts with nitrous acid & liberates N2 from aqueous solution.
(d) None of these. (AIIMS)
44. 𝑁 ≡ 𝑁𝐶𝑙 𝑁𝐻
+ H+ (A) [yellow dye]
(a)
– N = N – NH –
(b) NH2
– N = N –
(c) NH2
– N = N –
(d)
– N = N – j – NH2

(AIPMT)
45. Aniline is treated with bromine water to give an organic compound ‘X’ which when treated with NaNO2 &
HCl at 00
C gives a water soluble compound ‘Y’. Compound ‘Y’ on treatment with Cu2Cl2 & HCl gives
compound ‘Z’. Compound ‘Z is
(a) o-bromochlorobenzene
(b) p-bromochlorobenzene
(c) 2,4,6-tribromophenol
(d) 2,4,6-tribromochlorobenzene
(e) 2,4-dibromophenol (Kerala PMT)
46. Aniline in a set of the following reactions yielded a coloured product Y.
NH3
NaNO2/HCl (273-278K) X N,N-dimethylaniline Y
The structure off ‘Y’ would be
(a) CH3
J – N = N – – N
CH3
CH3 CH3
(b)
HN – – NH – – NH
(c)
H3C – K – N = N – J – NH2
CH3 CH3
(c)
HN – K – N = N – J – NH
(AIPMT 2008)
47. Nitrobenzene on reaction with conc. NHO3/H2SO4 at 80 – 1000
C forms which one of the following
products?
(a) 1,4-Dinitrobenzene
(b) 1,2,4-Trinitrobenzene
(c) 1,2-Dinitrobenzene
(d) 1,3-Dinitrobenzene (NEET)
48. The product obtained in the following reaction is
O
CH3CH2CH2 – C – NH2 P2O5
(a) CH3CH2CH2COOH
(b) CH3CH2CH2 – CN
OH
(c) CH3CH2CH2 – C – NH

(d) None of the above [AMU. (Med.) 2011]
49. Nitrobenzene can be prepared from benzene by using a mixture conc. HNO3 & conc. H2SO4.
In the mixture, nitric acid acts as a/an
(a) acid
(b) base
(c) catalyst
(d) reducing agent (AIPMT)
50. In the following sequence of the reactions, what is D?
CH3
[O] A SOCl2 B NaN3 C Heat
(a) Primary amine
(b) An amide
(c) Phenyl isocyanate
(d) A chain lengthed hydrocarbon (AIIMS)
ANSWERS OF MCQ QUESTIONS
1. (d)
2. (a) A tertiary (30
) amide is an amide in the molecule of which the nitrogen atom is bonded to the three
carbon atoms.
3. (c) 2-Butanamine: CH3 – CH2 – CH – CH3
NH2 (10 amine)
N-methylpiperidine:
N (30 amine)
CH3

N-methyl-2-pentanamine: CH3CH2CH2CH –NH – CH3
CH3
OCH3
p-Anisidine:
NH2
N-methyl-2-pentanamine is a secondary amine.
4. (a) CCH3CONH2 + Br2 + 4NaOH CH3NH2 + Na2CO3 + 2NaBr + 2H2O
The reaction occurs through intermediate formation of alkyl isocyanate which on hydrolysis gives
methylamine & sodium carbonate.
O O O O
CH3 – C – NH2 Br2/NaOH (–HBr) CH3 – C – NHBr 𝑂𝐻 (–H2O) CH3 – C – N – Br –Br CH3 – C – N
Rearrangement
CH3 – N – C = O NaOH CH3NH2 + Na2CO3
5. (c) Aniline cannot be prepared by this method because aryl halides do not undergo nucleophilic
substitution reaction with potassium phthalimide under mild conditions.
6. (a) It is an example of Hofmann bromide reaction in which the amine formed has one carbon less than
that of the parent amide.
NH2
i) Br2 / ii) KOH NH2
O
2-Phenylethanamide 1-Phenylmethanamine
7. (d) Gabriel phthalimide synthesis is used for the preparation of 10
aliphatic amines.
Ethylmethylamine is a 20
amine.
8. (b) Br NC NHCH3
CH3 – CH – CH3 alc.AgCN(Δ) CH3 – CH – CH3 LiAlH4 CH3 – CH – CH3
2-Bromopropane (X) N-Methylpropan-2-amine
9.(b) The source of nitrogen in Gabriel synthesis is potassium phthalimide i.e.,
CO
NK
CO
10. (d)
11. (d) O
CH3 – C – NH2 + Br2 + 4KOH CH3NH2 + K2CO3 + 2KBr + 2H2O
Acetamide Methanamine
12. (b) Reduction of isocyanides in presence of LiAlH4 from 20
amines containing methyl as one of the alkyl group.
CH3 – N ≡ C + 4[H] LiAlH4 CH3NHCH3
Methyl isocyanide Dimethylamine
13. (a) Only treatment of amide with Br2 in aqueous solution of Na or KOH will give an amine with lesser no.
of carbon atoms than in the reactant.
R – CONH2 Br2/KOH,Δ R – NH2

Amide Hofmann bromide or degradation Amine
14. (d) 𝑅 − 𝐶 ≡ 𝑁 4[H] R – CH2NH2 (Primary amine)
RCONH2 + Br2 + 4KOH Heat R – NH2 + 2KBr + K2CO3 + 2H2O
(Primary amine)
Hofmann bromamide reaction
RCONH2 4[H] RCH2NH2 + H2O
𝑅 − 𝑁 ≡ 𝐶 4[H] RNH – CH3 (Secondary amine)
R – NO2 6[H] RNH2 + 2H2O
15. (c) COOH O Cl O NH2 NH2
C C
SOCl2 NaOH/Br2
Br NH3 Br
(A) Br Br (D)
(B) (C)
16. (b)
C = O C2H5NH2/H+ C = NC2H5 (i) NaBH4 (ii) H3O+ CH – NHC2H5
So, R is ethylamine & R1 is sodium borohydride.
17. (b) NaOBr is used to convert acetamide to methanamine by Hofmann bromamide reaction.
CH3CONH2 Br2/NaOH CH3NH2 + NaBr
NaOBr serves as the attacking species.
18. (d)
19. (b) Nitriles or cyanides can be reduced by H2/Ni or LiAlH4 to give corresponding 10
amine.
Amides are also reduced by ether H2/Ni, Na/alcohol or LiAlH4 to give corresponding 10
amine.
Out of the given options, only benzyl amine can be obtained by both the methods.
CN CH2NH2 CONH2
H2/Ni LiAlH4
Benzene nitrile Benzyl amine Benzamide
20. (a) Phenyl isocyanide has unpleasant (foul) smell.
21. (c) Hofmann’s bromamide reaction – Amides when heated with bromine & caustic soda or caustic potash
solution, yield primary amine containing one carbon atom less than the amide.
22. (b) This is an example of Hofmann bromamide reaction.
23. (c) In arylamines, lone pair of electrons on nitrogen atom is delocalised over the benzene ring, thus, not
available for donation. So, arylamines are less basic than alkylamines.
24. (d) When bromine water is added to aniline at room temperature decolourisation of the bromine water
occurs & a white ppt. of 2,4,6-tribromoaniline is obtained.
NH2 NH2
Br Br
+ 3Br2 (aq)
(White ppt.)
Br

25. (a) NO2 NH2 𝑁 ≡ 𝐶 NHCH3
Reduction CHCl3/KOH Catalytic reduction
26. (c) Benzoylation of compounds containing an active hydrogen atom such as alcohols, phenols &
amines with benzoyl chloride in the presence of dilute aq. NaOH solution is called
Schotten-Baumann reaction.
27. (a) Carbylamine test is given by only aliphatic & aromatic primary amines. Secondary & tertiary
amines do not give this test.
28. (a)
29. (c) (CH3CO)2O + H2NC2H5 CH3CONHC2H5 + CH3COOH
Ethylamine (10) N-ethyl acetamide
30. (a) As ‘A’ gives alcohol on treatment with nitrous acid thus it should be primary amine. C3H9N has
two possible structure with – NH2 group.
CH3
CH3 – CH2 – CH2 – NH2 or CH3 – CH – NH2
As it gives isopropylmethylamine thus it should be isopropyl amine not n-propyl amine.
CH3 OH
CH3 – CH – NH2 HNO3 CH3 – CH – CH3 + N2↑
(A)
CHCl3/KOH
CH3 CH3
CH3 – CH – NC Reduction CH3 – CH – NH – CH3
(C) Isopropylmethylamine
31. (d) Since –NH2 is o, p-directing group, so aniline on nitration gives both ortho & para-nitro aniline.
32. (e)
33. (c)
34. (d) Substituent with strong +R effect, +I effect & weaker –I effect increases the basicity.
Hence, 𝑝𝐾 decreases. Also, alkylamines are stronger bases than arylamines.
So, the order is IV > II > I > III.
35. (e) 𝑁𝐻 𝐻𝑆𝑂 𝑁𝐻 𝑁𝐻
1800 – 2000C Rearrangement
𝑆𝑂 𝑆𝑂 𝐻
Zwitter ion Sulphanilic acid
36. (b)
37. (c) Secondary (20
) amines (aliphatic as well as aromatic) react with nitrous acid (HNO2) to form
N-nitrosoamines.
(C2H5)2NH + HONO (C2N5)2N – N = O + H2O
(20 amine (Nitrous acid) (N-Nitrosodiethylamine)
38. (d) 20
aliphatic & aromatic amines react with nitrous acid to form N-nitrosoamine.
CH3

NH – CH3 + NaNO2 + HCl N – N = O
N-nitroso-N-methylaniline
39. (a) It is the second step of Hinsberg’s test which is given by 10
amines only.
H
– SO2 – Cl + RNH2 H – SO2 – N – R + HCl
10 amine
Benzenesulphonyl N-alkyl sulphonamide
Chloride (precipitate)
O H
– S – N – R + NaOH J – SO2NNaR
O (soluble)
40. (b) It is a characteristic test given by primary amines only.
R – NH2 + CHCl3 + 3KOH RNC + 3KCl + 3H2O
In options (a) & (c) the – NH2 group is substituted & so the carbylamines reaction fails.
41. (c) Aniline is weaker base than ammonium chloride.
In NH4Cl or aliphatic amines, the non-bonding electron pair of N is localised & is fully available
for coordination with a portion.
On the other hand, in aniline & other aromatic amines, the non-bonding electron pair is
delocalised into benzene ring by resonance.
But anilinium ion is less resonance stabilised than aniline.
42. (c)
43. (c) Aliphatic 10
amines react with cold nitrous acid to give alcohols with quantitative evolution of N2 gas.
CH3NH2 + HONO 273 – 278 K CH3OH + N2 + H2O
Methylamine
This reaction is used as a test for aliphatic primary amines.
44. (d)
45. (d)
46. (a)
47. (d) NO2 NO2
Conc. HNO3 / H2SO4
80 – 1000C
NO2
1,3-Dinitrobenzene
48. (b) O
CH3CH2CH2 – C – NH2 P2O5 CH3CH2CH2 – CN + H2O
Propyl cyanide
49. (b)
50. (c)
IMPORTANT PRACTICE QUESTION SERIES FOR IIT-JEE EXAM – 1

1.
a) b)
c) d)
2. Which of the following cannot react with HNO ?
a) CH CONH b)(CH ) CNO c) (CH CH ) NH d)CH CH NH
3. Primary, secondary, tertiary amines can be separated by the following except:
a) Fractional distillation b)Fractional method using diethyl oxalate
c) Hinsberg's method using C H SO Cl d)Selective crystallisation
4. The correct order of basicities of the following compound is:
a) (2)>(1)>(3)>(4)
b)(1)>(3)>(2)>(4)
c) (3)>(1)>(2)>(4)
d)(1)>(2)>(3)>(4)
5. Stephen's reduction converts nitriles into:
a) Aldehydes b)Ketones c) Amines d)Carboxylic acids
6. Which of the following will yield phenylhydrazine hydrochloride?
a) Benzene and hydrazine b)Hydrazine and HCl
c) Benzenediazonium chloride and SnCl /HCl d)Nitrobenzene and SnCl /HCl
7. Amongst the compound given, the one that would form a brilliant coloured dye on treatment
with 𝑁𝑎𝑁𝑂 in dil. HCI followed by addition to an alkaline solution of β −naphthol is
a) b) c) d)
8. Among the following, the strongest base is:
a) C H NH b)𝑝-NO C H NH c) 𝑚-NO . C H NH d)C H CH NH
9. Mendius reaction involves the:
a) Reduction of aldehydes to give alcohols b)Reduction of nitriles with sodium and
ethanol
c) Oxidation of nitriles d)Hydrolysis of cyanides
10. In order to distinguish between C H NH and C H NH , which of the following reagents is

useful?
a) Hinsberg reagent b)𝑝-Naphthol
c) Benzene diazonium chloride d)None of the above
11. Primary and secondary amines are distinguished by:
a) Br /KOH b)HClO c) HNO d)NH
12. The following reaction constitutes:
RNH + S = C = S ⎯⎯ R − N = C = S + H S
Alkyl isothiocynante
a) Mustard oil reaction b)Test for 3° amine c) Test for 2° amine d)Test for CS
13. Nitrobenzene on electrolytic reduction gives:
a) Azobenzene b)Hydrazobenzene c) Aminophenol d)Aniline
14. When propane is subjected to the treatment with fuming nitric acid at 673 K, which of the
following will not be formed?
a) 1-Nitropropane b)2-Nitropropane c) Nitromethane d)Nitrohexane
15. Which of the following is the weakest base?
a) (CH ) NH b)(CH ) N c) C H NH d)C H NHCH
16. Nitrobenzene is subjected to reduction with zinc dust and ammonium chloride. The main
product formed will be:
a) Benzenamine b)Aniline
c) N-Phenylhydroxylamine d)None of these
17. Nitrogen is likely to be evolved when NaNO in dilute HCl is wanted with:
a) CH NHCH CH b)(C H ) N c) C H NH d)H NCH CH NH
18. The electrolytic reduction of nitrobenzene in strongly acidic medium produces:
a) Phenol b)𝑝-Aminophenol c) Hydroazobenzene d)Azobenzene
19. The conjugate base of (CH ) NH⊕
is:
a) (CH ) N b)(CH ) NH c) (CH ) N⊝ d)(CH ) N⊕
20. A compound (X) has the molecular formula C H NO. On treatment with Br and KOH, (X) gives
an amine (Y); (Y) gives carbylamine test. (Y) upon diazotisation and coupling with phenol gives
an azodye (Z). (X) is:
a) PhCONH b)PhCONHCOCH c) PhNO d)PhCOONH
21. Which of the following amines will form stable diazonium salt at 273-283 K?
a) C H NH b)C H NH c) C H CH NH d)CH NH
22. Diazo coupling is useful to prepare some:
a) Pesticides b)Dyes c) Proteins d)Vitamins
23. Which of the following produces one mononitro and three isomeric dinitro derivatives?
a) 𝑝-Xylene b)Ethyl benzene c) 𝑜-Xylene d)𝑚-Xylene
24. The following reaction is:
a) Benzidine rearrangement b)Pinacol-Pinacolone rearrangement
c) Fries rearrangement d)Benzil-benzilic acid arrangement

25. The compound which on reaction with aqueous nitrous acid at low temperature produces an
oily nitrosoamine is:
a) Methylamine b)Ethylamine c) Diethylamine d)Triethylamine
26. Which of the following are not functional isomers of each other?
a) CH CH NO andCH CH ON = 0 b)C H CHO andCH COCH
c) CH CH NH andCH NHCH d)C H NH and(CH ) CHNH
27. Indicate which nitrogen compound amongst the following would undergo Hofmann reaction
(i.e., reaction with Br and strong KOH) to furnish the primary amine (R − NH )
a) RCONHCH b)RCOONH c) RCONH d)R − CO − NHOH
28. The product of the reaction of alcoholic silver nitrite with ethyl bromide is:
a) Ethane b)Ethene c) Nitroethane d)Ethyl alcohol
29. Hinsberg's reagent is:
a) C H COCl b)CH COCl c) C H CH Cl d)C H SO Cl
30. The product formed by the treatment of ethanol and ethane nitrile in the presence of sulphuric
acid is:
a) Ethyl acetate b)Diethyl ether c) Ethyl methyl ketone d)Butanal
31. Which of the following statements is correct?
a) Methyl amine is slightly acidic b)Methyl amine is less basic than ammonia
c) Methyl amine is less basic than dimethyl
amine
d)Methyl amine is less basic than aniline
32. The compound 1-(N-ethyl-N-methyl)-propanamine forms non-superimoposable mirror images.
But this compound does not show optical activity because of the:
a) Absence of a chiral N atom b)Presence of chiral N atom
c) Presence of lone pair on N atom d)Rapid flipping of one form into the other
33. Tertiary nitro compounds cannot show tautomerism because:
a) They are very stable
b)They isomerise to give secondary nitro compounds
c) They do not have labile hydrogen atom
d)They are highly reactive
34. (A)is subjected to reduction with Zn-Hg/HCl and the product formed is 𝑁-methylmethamine.
(A) can be:
a) Ethane nitrile b)Nitroethane c) Carbylaminoethane d)Carbylaminomethan
e
35. Which of the following reactions does not yield an amine?
a) R − X + NH → b)R − OH = NOH + [H] ⎯⎯⎯⎯⎯⎯
c) R − CN + H O
⊕
d)R − CONH + 4[H] ⎯⎯⎯⎯
36. Which C H N Cl is reduced with Na SnO , the product is:
a) b) c) d)
37. Nitromethane is subjected to treatment with chlorine in the presence of sodium hydroxide, the
main product is:
a) Monochloronitromethane b)Trichloromethane

c) Chloropicrin d)None of the above
38. Which of the following is the weakest base?
a) NH b)C H NH c) C H CH NH d)CH NH
39. Which of the following groups will facilities the electrophilic attack on benzene ring?
a) −NO b)−CHO c) −Cl d)−SO H
40. Pick up the correct statement:
a) The boiling points of alkyl halides are more than those of the corresponding alkenes
b)In water, the solubility of CH OH > C H OH > C H OH
c) C H NH is a weaker base than NH
d)All the above statements are correct
41. Examine the following two structures for the anilinium ion and choose the correct statement
from the ones given below:
a) (II) is not an acceptable canonical structure because carbonium ions are less stable than
ammonium ions
b)(II) is not an acceptable, canonical structure because it is non-aromatic
c) (II) is not an acceptable canonical structure because nitrogen has 10 valence electrons
d)(II) is an acceptable canonical structure
42. Which of the following reagents on treatment with benzenamine in basic medium produces
phenyl isocyanide?
a) CCl b)Trichloromethane c) Methylene dichlorided)Hexachloroethane
43. A compound (X) has the molecular formula C H NO. With Br and KOH, (X) gives (Y). (Y)
responds to mustard oil reaction. (Y) upon treatment with HNO evolves N and gives an alcohol
(Z) which gives iodoform test. (X) is likely to be:
a) C H CONH b)CH CONH c) CH COONH d)C H CNO
44. An amine on treatment with HNO evolved N . The amine on exhaustive methylation with CH I
formed a quaternary salt containing 59.07% iodine. The amine is likely to be:
a) CH NH b)(CH ) NH c) C H NH d)(CH ) N
45. Which of the following nitro compounds will show tautomerism?
a) C H NO b)(CH ) CNO c) CH CH NO d)None of the above
46. The reaction of primary amine with chloroform and ethanoic solution of KOH is called:
a) Hofmann reaction b)Reimer-Tiemann reaction
c) Carbylamine reaction d)Kolbe reaction
47. A primary nitroalkane is treated with nitrous acid, which of the following will be the main
product?
a) Pseudonitriol b)Nitrolic acid c) A primary amine d)Primary alcohol
48. 1°, 2°, and 3° amines can be best distinguished by:
a) HNO treatment b)Exhaustive alkylationc) Mustard oil reaction d)Carbylamine reaction
49. A nitrogenous substance (X) is treated with HNO and the product so formed is further treated
with NaOH solution, which produces blue colouration. Which of the following can (X) be?
a) CH CH NH b)CH CH NO c) CH CH ONO d)(CH ) CHNO

50. Which of the following substances on treatment with P O gives ethanenitrile?
a) Propanamide b)Ethanamide c) Ethanoic acid d)𝑁-Methylethyl amine
51. Methyl cyanide on treatment with methyl magnesium bromide followed by of subsequent
hydrolysis gives:
a) Propanone b)Ethanone c) Ethanal d)Propanal
52. When PhNO is reduced in alkaline medium, the product is:
a) b)Ph − N = N − Ph (Azobenzene)
c) Ph − NH − NH − Ph (Hydrazobenzene) d)All
53. Which of the following represents the poisonous gas which caused the tragedy in Bhopal in
1984?
a) CH C = N = S b)CH − N = C = O c) CH − N = C = S d)CH − O − N = C
54. The major product of the following reaction is
a) b)
c) d)
55. CHCl ⎯⎯ (X)
In the above sequence, (X) is:
a) Nitrochloromethane b)Chloropicrin c) Ethanenitrile d)None of the above
56. The most unlikely representation of resonance structures of 𝑝-nitrophenoxide ion is:
a) b) c) d)
57. Azoxybenzene can be obtained by the treatment of nitrobenzene with:
a) O b)H /Pt c) Na AsO /NaOH d)Zn/NaOH
58. Carbylamine test is performed in alc. KOH by heating a mixture of:
a) Chloroform and silver powder
C
C
N
O CH2
Br
O

b)Trihalogenated methane and a primary amine
c) An alkyl halide and a primary amine
d)An alkyl cyanide and a primary amine
59. Which of the following is not an ambident nucleophile?
a) NO⊝
b) c) CSN⊝ d)CN⊝
60. An organic compound with the formula C H N on hydrolysis forms an acid which reduces
Fehling's solution. The compound can be:
a) Ethanenitrile b)Isocyanoethane c) Ethoxyethane d)Propanenitrile
61. Acetamide is treated separately with the following reagents. Which one of these would give methylamine?
a) PCl b) Soda lime c) NaOH + Br d) Hot conc. H SO
62. CH NH + CHCI + KOH →nitrogen containing compound +KCI + H O. Nitrogen containing compound is
a) CH − C ≡ N b) CH − NH − CH c) d)
63. Gabriel synthesis is used for the preparation of:
a) 1° amine b) 2° amine c) 3° amine d) All can be prepared
64. Which of the following is formed when RNH reacts with RCHO?
a) Hemiacetals b) Acetals c) Ketals d) Imines
65. Ethanamine is treated with nitrous acid at ordinary temperature; the products will be:
a) Ethanol only b) Ethanol, acetic acid, N , and H O
c) Acetic acid, ethane, and H O d) Ethanol, ethene, N , and H O
66. The diamide of carbonic acid is:
a) Acetamide b)Formamide c) Benzamide d)Urea
67. Urea reacts with hydrazine to form:
a) Nitrogen b)Phenyl hydrazine c) Semicarbazide d)Urethane
68. Final product of hydrolysed alkyl cyanide is
a) 𝑅COOH b)𝑅CONH c)
𝑅 − C = NH
|
OH
d)
69. Secondary nitroalkanes can be converted into ketones by using Y. Identify Y from the following
a) Aqueous HCl b)Aqueous NaOH c) KMnO d)CO
70. Among the following compounds, the most basic is
a) Aniline b)Acetanilide c) p-nitroaniline d)Benzyl amine
71. Substitution of one alkyl group by replacing hydrogen of primary amines:
a) Increases the base strength
b)Decreases the base strength
c) Remains the same
d)None of the above
72. Ethylamine undergoes oxidation in the presence of KMnO followed by hydrolysis to form:
a) An acid b)An alcohol c) An aldehyde d)a N-oxide
73. Which of the following compounds will undergo carbylamine reactions?

a) (CH CH ) NH b)(CH ) NH c) C H NH d)(CH ) N
74. Methylethylpropylamine forms non-superimposable mirror images but it does not show optical
activity because:
a) Of rapid flipping
b)Amines are basic in nature
c) Nitrogen has a lone pair of electron
d)Of absences of asymmetric nitrogen
75. Urea on heating with ethanol gives:
a) Urethane b)Urea alcohol c) Ureides d)None of these
76. Aliphatic amines are soluble in water because:
a) They are basic
b)They are amino compounds
c) They are lighter than water
d)Of formation of hydrogen bonds with water
77. A positive carbylamines test is given by
a) N, N-dimethylaniline b)2,4-dimethylaniline
c) N-methy-𝑜 -methylaniline d)N-methylbenzylamine
78. Which of the following amines form maximum hydrogen bonds within themselves?
a) CH NH b)(CH ) NH c) (CH ) N d)None of these
79. Reaction of cyclohexanone with dimethylamine in the presence of catalytic amount of an acid
forms a compound of water during the reaction is continuously removed. The compound formed
is generally known as
a) An amine b)An imine c) An enamine d)A Schiff’s base
80. Benzene diazonium chloride on treatment with hypophosphorous acid and water in presence of
Cu catalyst produce
a) Benzene b)Toluene c) Aniline d)Chlorobenzene
81. Which one of the following does not have 𝑠𝑝 hybridised carbon?
a) Acetone b)Acetic acid c) Acetonitrile d)Acetamide
82. The product of Hofmann elimination of
a) b)
c) d)
83. The best method to synthesise 𝑚-dibromobenzene is by using the reaction
a) Benzene
/ /
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ b)Aniline
,
⎯⎯⎯⎯ [ ]
.
⎯⎯⎯⎯
.
c)
Nitrobenzene
⎯⎯⎯⎯⎯⎯⎯⎯⎯
,∆
[ ]
/
⎯⎯⎯⎯⎯⎯⎯⎯⎯
,
[ ]
.
⎯⎯⎯⎯
.
d)
Bromobenzene
⎯⎯⎯ [ ]
/
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
,
[ ]
.
⎯⎯⎯⎯⎯
.
84. Aniline is reacted with Br water and the resulting product is treated with an aqueous solution
of sodium nitrite in the presence of diluteHCl. The compound so formed is converted into
tetrafluoroborate which is subsequently heated dry. The end product is

a) 𝑝-bromofluorobenzene b)𝑝-bromoaniline
c) 2, 4, 6- tribromofluoro benzene d)1, 3, 5- tribromobenzene
85. Which of the following statements is correct?
a) Aniline is stronger base than ammonia
b)Methylamine is a stronger base than aniline and ammonia
c) Aniline is stronger than ammonia, but weaker base than methylamine
d)Methylamine is stronger than aniline, but weaker base than ammonia
86. 𝑅NH reacts with C H SO Cl in aqueous KOH to give a clear solution. On acidification a
precipitate is obtained which is due to the formation of
a) b)𝑅 − N SO C H K
c) C H SO NH d)𝑅 − NH − SO − C H
87. When NaNO and dilute HCl were added to an amine at 0̊C, a colourless gas was evoloved and an
ionic compound is formed. The amine is:
a) An primary amine
b)An aromatic primary amine
c) Any amine
d)None of the above
88. Choose the incorrect comparision(s)
a) b)
CH CH CH NH > (CH ) N
(basicity in aqueous medium)
c) d)
(basicity in aqueous
medium)
89. Grignard reagent and acetyl chloride does not react with:
a) 𝑅NH b)𝑅 NH c) 𝑅 N d)None of these
90. Which of the following can be used to distinguish acetamide and urea?
a) Fehling’s solution b)Biuret test c) Hofmann’s reaction d)NaOH solution
91. Among the amines (A)C H NH , (𝐵)CH NH , (𝐶)
(CH ) NH, (𝐷)(CH ) N, the order of basicity is
a) A < B < D < C
b)D < C < B < A
c) A < B < C < D
d)B < C < D < A
92. Choose the incorrect statement
a) In the case primary, secondary and tertiary amines, the basic strength depends on the extent
on the extent of hydrogen bonding in the protonated amines

b)The presence of groups like – OCH and – CH increases the basic strength of amines
c) The presence of groups like – NO , −CN and halogens reduces the basic strength of amines
d)The basic strength of amines depends on their concentration
93. An organic amino compound reacts with aqueous nitrous acid at low temperature to produce an
oily nitrosoamine. The compound is
a) CH NH b)CH CH NH c) CH CH NHCH CH d)(CH CH ) N
94. 𝑅Mg𝑋 on reacting with cyanogen chloride gives:
a) 𝑅—NC b)𝑅—Cl c) 𝑅—CN d)None of these
95. Ethyl isocyanide on hydrolysis in acidic medium generated
a) Ethyl amine salt and methanoic acid b)Propanoic acid and ammonium salt
c) Ethanoic acid and ammonium salt d)Methyl amine salt and ethanoic acid
96. Which of the following will give a primary amine on hydrolysis?
a) Nitroparaffin b)Alkyl cyanide c) Oxime d)Alkyl isocyanate
97. Urea when heated a white residue is formed. Its alkaline solution when treated with few drops
of CuSO solution gives:
a) Red colour b)Violet colour c) Green colour d)Yellow colour
98. Which one of the following is most basic?
a) FCH NH b)FCH CH NH c) C H NH d)C H CH NH
99. The basicity of compounds I, II, III and IV
CH NH , (CH ) NH, (CH ) N, C H CH NH
I II III IV
varies in the order
a) I > II > III > IV b)II > I > III > IV c) III >I > II > IV d)IV > I > II > III
100.A gaseous carbon compound is soluable in dilute HCI. The solution on treating with NaNO gives
off nitrogen leaving behind a solution which smells of wood spirit. The carbon compound is
a) HCHO b)CO c) C H NH d)CH NH

1) a 2) b 3) d 4) b
5) a 6) c 7) c 8) d
9) b 10) b 11) c 12) a
13) d 14) d 15) c 16) c
17) d 18) b 19) b 20) a
21) c 22) b 23) a 24) a
25) c 26) d 27) c 28) c
29) d 30) a 31) c 32) d
33) c 34) d 35) c 36) b
37) c 38) b 39) c 40) d
41) c 42) b 43) a 44) c
45) c 46) c 47) b 48) a
49) d 50) b 51) a 52) d
53) b 54) a 55) b 56) c
57) c 58) b 59) b 60) b
61) c 62) b 63) a 64) d
65) d 66) d 67) c 68) a
69) a 70) d 71) a 72) c
73) c 74) a 75) a 76) d
77) b 78) a 79) c 80) a
81) c 82) c 83) c 84) c
85) b 86) d 87) b 88) d
89) c 90) b 91) a 92) d
93) c 94) c 95) a 96) d
97) b 98) d 99) b 100) d
2 (b)
3° nitro compound does not react with HNO
3 (d)
Statement is self explanatory
4 (b)
IMPORTANT PRACTICE QUESTION SERIES FOR IIT-JEE EXAM – 1 (ANSWERS)

(1)>(3)>(2)>(4)
1.
More basic due to the presence of two LP 𝑒's on each N
2. EtNH . Due to +I effect of (Et) group, but +I effect of two Me group is greater than +I
effect of (Et) group. Hence, (3) is more basic than (2)
3. (CH ) NH. Due to (+I) effect of two Me groups
4. (Amides are resonance stabilised; so they are the
weakest bases. Amides behave as amphoteric.)
5 (a)
Stephen''s reduction is partial reduction of RCN to aldehydes
7 (c)
As we know, benzenediazonium salt forms brilliant coloured dye with 𝛽-naphthol, the
compound under consideration must be p-toludine (c) as it is a primary aromatic amine.
Primary aromatic amine, on treatment with NaNO in dil. HCI forms the corresponding
diazonium chloride salt.
8 (d)
(d) is an aliphatic amine; so it is stronger base than aromatic amine. Moreover, EWG
[(−NO )at𝑜, 𝑚, or𝑝] decreases the basic character. The basicity order is:
(d)>(a)>(c)>(b)
9 (b)
R − C ≡ N ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ RCH NH
10 (b)
1° aromatic amine on diazotisation followed by coupling with β-naphthol gives azo dye test
11 (c)
HNO is used to distinguish between 1°, 2°, and 3° amines
13 (d)

PhNO ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ PhNH
15 (c)
Basic order: (CH ) NH > (CH ) N > C H NHCH > C H NH
16 (c)
PhNO ⎯⎯⎯⎯⎯⎯ PhNHOH
𝑁-Phenyl hydroxylamine
17 (d)
Only in 1° aliphatic amines
20 (a)
Reaction of (X) with Br + KOH suggests that (X) is an amide. Carbylamine test coupling
reaction of (Y) suggest that (Y) is 1° aromatic amine
21 (c)
Stable diazonium salts are formed by 1° aromatic amine (PhNH )
22 (b)
Azo dyes
25 (c)
2° amines react with HNO at low temperature to give oily nitrosamine
26 (d)
1°amine to different 1° amines are not functional isomers but (1° and 2°), (1° and 3°),and
(2° and 3°) of same molecular formula are functional isomers.
Therefore, (a), (b), and (c) are functional isomers
27 (c)
Amides give Hofmann bromamide reaction
28 (c)
C H Br
.
⎯⎯⎯⎯⎯ C H NO
Nitroethane
29 (d)
Benzenesulphonyl chloride (PhSO Cl) is called Hinsberg's reagent
30 (a)
Me − C ≡ N is hydrolysed to acid which reacts with alcohol to give ester
31 (c)

Dimethyl amine (Me NH) is more basic than MeNH , due to (+I) effect of two (Me) groups
33 (c)
Nitroalkane containing 𝛼-H atoms, i.e., 1° and 2° nitro alkanes show tautomerism
34 (d)
Reduction with Zn-Hg/HCl is Clemmensen reduction which converts
(R − C ≡ N → RCH NH ), and .Carbylamino methane is
35 (c)
RCN on hydrolysis gives RCOOH and NH
36 (b)
Diazo group can be replaced by (H) on reduction with Na SnO or H PO or warm with
C H OH
37 (c)
H C − NO ⎯⎯⎯⎯⎯⎯⎯ Cl C − NO
Chloropicrin (tear gas)
38 (b)
Aromatic amines are weaker than aliphatic amines and NH
39 (c)
SE reaction is favoured by EDG
(−NO ), (−CHO), and (SO H) are EWG. Cl is 𝑒 donating by (+R) effect and deactivating by
(−I)
40 (d)
All statements are correct and self explanatory
41 (c)
(II) is not acceptable canonical structure because N has 10 valence 𝑒's
43 (a)
Reaction of (X) with Br + KOH suggests that (X) is an amine. Evolution of N and formation
of alcohol suggest that (Y) is a1° aliphatic amine. Iodoform test of (Z) suggests that it is an
alcohol of the type
44 (c)
Evolution of N with HNO suggests that 1° amine [either (a) or (c)]
59.07 gm of (I) is present in 100 gm of compound 127 gm of I (1 mol) in present in
×
.
= 215

Molecular mass = 215
(Molecular mass=215)
Hence, the amine is C H NH (c)
45 (c)
(c) has an 𝛼-H atom and hence shows tautomerism
46 (c)
Carbylamine reaction
49 (d)
The reaction is a test for 2° nitro compound
54 (a)
It is the first step of Gabriel’s phthalimide synthesis. The hydrogen bonded to nitrogen is
sufficiently acidic due to two 𝛼-carbonyls.
The conjugate base forms above act as nucleophile in the subsequent step of reaction. As
shown above, the nucleophile exist in three resonating form, one may think of oxygen being
the donor atom in the nucleophilic attack. However, nitrogen act as donor as it is better
donor than oxygen.

Bromine is not substituted in the above reaction as it is in resonance with benzene ring
giving partial double bo0nd character to C − Br bond, hence difficult to break.
55 (b)
H − C. Cl ⎯⎯ O N − CCl
Chloropicrin
56 (c)
(c) is an unlikely structure because N atoms form five bonds and contain positive charge.
59 (b)
(b) is not an ambident nucleophile, but others are
60 (b)
Isocyanides on hydrolysis give 1° amine and HCOOH. Formic acid reduces Fehling's solution
C H − N ≡ C ⎯
⎯ C H NH + HCOOH
62 (b)
CH NH + CHCI + 3KOH → CH NC + 3KCI + 3H O
CH NC or CH − N ≡ C methyl isocyanide or methyl carbylamine.
This reaction is an example of carbylamine reaction and it is used for the distinction of p-
amines from s- and t-amines or identification of p-amino group.
63 (a)
1° amine or 1° aromatic amine containing EWG at 𝑜-and 𝑝-positions
65 (d)

Under acidic conditions, alcohol is dehydrated to alkene
67 (c)
NH CONH + NH ∙ NH ⟶ NH CONHNH + NH
68 (a)
𝑅 − C ≡ N ⎯⎯⎯⎯⎯⎯ 𝑅 − CONH ⎯⎯⎯⎯⎯⎯ 𝑅COOH
Alkyl cyanide alkyl amide carboxylic acid
69 (a)
Secondary nitroalkanes can be converted into ketones by using aqueous HCI.
70 (d)
Benzyl amine is most basic because positive inductive effect (+I) increases due to presence
of methylene group.
72 (c)
C H NH
[ ]
CH CH=NH ⎯ CH CHO + NH
73 (c)
Carbylamine reaction is given by only primary amines (both aliphatic and aromatic). In this
reation a primary amine reacts with chloroform in basic medium, to form a very bad
smelling compound, called carbylamines
C H NH + CHCl KOH → C H NC + KCl + H O
75 (a)
NH CONH + HOC H
∆
→ H NCOOC H + NH
Urethane
77 (b)
Only primary amines give positive carbylamine test
78 (a)
Primary amines have tendency of forming H-bonds
85 (b)
Due to + ve 𝐼𝐸 in alkyamines and resonance in C H NH .
86 (d)
C H SO Cl + 𝑅NH → 𝑅NHSO C H ⎯⎯⎯⎯ 𝑅 − NKSO C H
Benzene sulphyonyl N-alkyl benzene soluble in KOH
chloride sulphonamide
87 (b)
C H NH ⎯⎯⎯⎯⎯⎯⎯⎯⎯ C H N Cl + H O
89 (c)
Tertiary amines do not have replaceable H-atom.

90 (b)
Urea gives biuret test. Biuret formed gives violet colour with CuSO in alkaline medium.
91 (a)
The order of basicity among the following amines is
(CH ) NH > (CH ) N > 𝐶H NH > C H NH
92 (d)
Concentration does not affect the basis strength of amines
93 (c)
Secondary amine on reaction with aq. HNO at low temperature produces yellow oily
nitrosoamines.CH CH NHCH CH is secondary amine.
95 (a)
Ethyl isocyanide on hydrolysis in acidic medium gives methanoic acid and ethyl amline salt
C H NC + H O ⎯
⎯ HCOOH + C H NH
methanoic acid
C H NH + H → C H NH
Ethylamine salt
96 (d)
R—N=C=O ⎯ 𝑅NH + H CO
97 (b)
Biuret formed gives violet colour with CuSO in alkaline medium.
98 (d)
Benzyl amine (𝐶 𝐻 𝐶𝐻 𝑁𝐻 ) is more basic than aniline (𝐶 𝐻 𝑁𝐻 ) because N-atom of
aniline is delocalized over the benzene ring. However in benzyl amine the lone pair of
electrons on the N-atom is not conjugated with the benzene ring and therefore it is not
delocalized. Hence, the lone pair of electrons on the N-atom in benzyl amine is more readily
available for protonation than that on the N-atom of aniline. Thus, the benzyl amine is a
stronger base than aniline.
99 (b)
Basicity of amines increases with increasing +I effect of alkyl group.3°amine has greater +I
effect than 2° and 1°amines but less basic than these, due to steric hindrance of bulky
groups. Moreover, benzyl amine is a weaker base than aliphatic amines. Hence, the
following compounds has the order of basicity.
CH NH (1°), (CH ) NH(2°), (CH ) N(3°), C H CH NH
I II III IV
II > I > III > V
100 (d)
It is methyl amine which, being basic dissolves in dilute HCI. It with NaNO evolves nitrogen
gas leaving behind methyl alcohol which has smell of wood-spirit.
CH NH ⎯ CH NH . HCI
CH NH + HNO
/
⎯⎯⎯⎯⎯⎯⎯ CH OH + N ↑ +H O
methyl alcohol

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