Download organic compound containing nitrogen

CON – 1
ORGANIC COMPOUND
CONTAINING NITROGEN
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CON – 2
CONCEPTS
C1A Structure : of the organic compounds that show appreciable basicity (e.g. those strong enough to
turn litmus blue), by for the most important are the amines. An amine has the general formulae
RNH2, R2NH or R3N where R is an alkyl or aryl group. For e.g.
Nomenclature : Aliphatic amines are named by naming the alkyl group or groups attached to
nitrogen and following these by the word-amine e.g.
Salts of amine are generally named by replacing amine by ammonium (or aniline by anilinium), and
adding the name of the anion (chloride, nitrate, sulfate etc.) e.g.
C6H5NH3+Cl–
Anilininum
Chloride
C1B
(C2H5NH3+)2SO42–
Ethylammonium
Sulfate
Physical Properties of amines : Like ammonia, amines are polar compounds and except for tertiary
amines, can form intermolecular hydrogen bonds. Amines have higher boiling points than
non-polar compounds of same molecular weight, but lower boiling points that alcohols or carboxylic
acids.
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CON – 3
Amines of all three classes are capable of forming hydrogen bonds with water. As a result smaller
amines are quite soluble in water, with borderline solubility is reached with six carbon atoms.
Amines are soluble in less polar solvents like ether, alcohol, benzene etc.
C1C Stereochemistry of Nitrogen : Consider quaternary ammonium salts, compounds in which four
alkyl groups are attached to nitrogen. Here all four sp3 orbitals are used to form bonds and quatenary
nitrogen is tetrahedral. Thus quaternary ammonium salts in which nitrogen holds four different
groups have been found to exist as configurational enantiomers, capable of showing optical activity.
C2
Methods of Preparation :
1.
Reduction of Nitro Compounds :
2.
Reaction of halides with ammonia or amines :
R
|
RX
RX
R
|
RX
R
|

RX
NH 3  R  NH 2  R 0 N  H  R  N  R  N  R X 
|
R
0
3 Amine
2 Amine
10 Amine
|
R
Quaternary
ammonium
salts
( 40 )
RX  must be alkyl or aryl with electron withdrawing substituents.
The presence of large excess of ammonia lessens the importance of these last reactions and increases
the yield of primary amine.
3.
Reductive Amination :
H , Ni
C = O + NH3 2 
or NaBH 3 CN
H , Ni
C = O + RNH2 2 
CH – NH2
10 Amine
CH – NHR
20 Amine
CH – NR2
30 Amine
NaBH 3 CN
H , Ni
C = O + R2NH 2 
NaBH 3 CN
4.
Reduction of nitriles (Higher carbon number is obtained)
2 H , Catalyst
R  C  N 2  R  CH 2 NH 2
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CON – 4
NaCN
H , Ni
ClCH 2 CH 2 CH 2 CH 2 Cl  NC(CH 2 ) 4 CN 2  H 2 NCH 2 (CH 2 )4 CH 2 NH 2
Adiponitrile
5.
Hexamethylene diamine
(10 )
Hoffman degradation of amides :
OBr 
RCONH 2 or ArCONH 2  R  NH 2 or ArNH 2  CO 23 
KOBr
CH 3 (CH 2 ) 4 CONH 2  CH 3 (CH 2 ) 4 NH 2
Capromide
(Hexanamide)
n  Pentylamine
Discussion : From the above reaction it is clear that in this reaction, the rearrangement occurs, since
the group joined to carbonyl carbon in the amide is found joined to nitrogen in the product.
The reaction is believed to proceed by the following steps :
1.
2.
5.
H O
2
R  N  C  O  2OH  

 RNH 2  CO 23
Steps (3) and (4) are generally takes place simultaneously. The attachment of R to nitrogen helps to
pushout halide ion.
(3,4)
Step (5) is the hydrolysis of an isocyanate (R – N = C = O) to form amine and carbonate ion.
If the Hoffman degradation is carried in absence of water an isocyanate is actually isolated.
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*
When the migrating group is aryl the rate of degradation is increased by the presence of electron
releasing substituents in the aromatic ring. Thus substituted benzamide show the following order of
reactivity :
G : –OCH3 > –CH3 > –H > –Cl > – NO2
6.
Gabriel Phthalimide Synthesis :
Practice Problems :
1.
Boiling of C2H5NCO + NaOH leads to the formation of
(a)
C2H5COOH + NH3
(b)
C2H5NH2 + Na2CO3
(c)
CH3NH2 + CH3COONa
(d)
None
2.
Intermediates of this reaction are except :
(a)
R–N=C=O
(b)
(c)
3.
(d)
a, c
Which of the following would you predict as incorrect
(a)
(b)
can be hydrolysed to C2H5COOH
(CN)2 can be hydrolysed to
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CON – 6
can be hydrolysed to C2H5NH2 and NO3—
(c)
(d)
4.
can be hydrolysed to C2H5NH2 and CH3COOH
Which of the following reactions does not yield an amine

(a)
H
R  C  N  H 2O 
......
(c)
R  CH  NOH  [H] C

 ......(d)
2 H 5OH
(b)
Na
R  X  NH 3 
 ......
RCONH 2  4[H] LiAlH
4  ......
[Answers : (1) b (2) b (3) c (4) a]
C3
Chemical properties of Amines : The tendency of nitrogen to share the unpaired electrons
underlines the entire chemical behaviour of amines : their basicity, their action as nucleophiles - in
both aliphatic and acyl substitution - and the usually high reactivity of aromatic rings bearing amino
or substituted amino groups.
1.
Basicity of Amines : Salt formation :
R – NH2 + H+
R+NH3
R2NH + H+
R2NH2+
R3N + H+
R3NH+
Example :
Structure and Basicity :
Let us see how basicities are related to the structure.
We shall compare the stabilities of amines with the stabilities of their ions; the more stable the ion
relative to the amine from which it is formed, the more basic the amine.
First of all, amines are more basic than alcohols, ethers, esters etc. for the same reason that
ammonia is more basic than water, Nitrogen is less electronegative than oxygen and can better
accomodate the positive charge of the ion.
An aliphatic amine is more basic than ammonia : because the electron-releasing alkyl groups tend to
disperse the positive charge of the substituted ammonium ion;
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How can be account for the fact that aromatic amines are weaker bases than ammonia ?
Let us compare the structure of aniline with anilinium ion with the structures of ammonia and the
ammonium ion.
Aniline i.e. Aromatic amines are less basic due to the fact that amine is stablized by resonance to the
greater extent than its ion.
From another point of view we can say that its electron pair is partly shared by ring and is less
available for sharing with a hydrogen ion.
Effect of substituents on basicity of aromatic amines :
Electron releasing substituents like CH3, increases the basicity of aniline, and electron withdrawing
substituents like –X, –NO2 decreases the basicity.

The electron releasing substituents tends to disperse the positive charge of the anilinium ion, and
thus stablizes the ion relative to amine. The electron withdrawing tends to intensify the positive
charge of the anilinium ion, and thus destablizes the ion relative to the amine.

We notice that base strengthening substituents are the ones that activate an aromatic ring towards
electrophilic substitution; the base-weakining substituents are the ones that deactivate an aromatic
ring towards electrophilic substitution.
2.
Alkylation :
RX
RX
RX
RNH 2  R 2 NH  R 3 N  R 4 N  X 
RX
RX
RX
ArNH 2  ArNHR  ArNR 2  ArNR 3 X 
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Example :
CH 3 I
H
|
CH 3 I
CH 3
|
CH I
3
n  C 3 H 7 NH 2  n  C 3 H 7  N  CH 3  n  C 3 H 7  N  CH 3 
 n  C 3 H 7 N(CH 3 ) 3 I 
Trimethyl  n  propyl
ammonium iodide
( 40 )
3.
Hoffmann elimination from quatenary Ammonium Salts :
Example :
This reaction is called Hoffmann elimination, is quite analogous to the dehydrohalogenation of an
alkyl halide. Most commonly reaction is E2 :
Hydroxide ion abstracts a proton from carbon; a molecule of tertiary amine is expelled.
4.
The Cope Elimination :
Tertiary amine oxide are prepared easily by treating tertiary amines with H2O2.
5.
Reactions of Amines with Nitrous acid :
Nitrous acid is a weak acid & is unstable also. It is prepared by treating sodium nitrite (NaNO2) with
an aqueous solution of a strong acid :
HCl (aq) + NaNO2(aq)  HONO(aq) + NaCl(aq)
H2SO4(aq) + NaNO2(aq)  HONO(aq) + Na2SO4(aq)
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Nitrous acid reacts with all kinds of amines. The products that we obtain from these reactions
depends on whether the amine is primary, secondary or tertiary and whether the amine is aliphatic
or aromatic.
Reactions of primary aliphatic amines with nitrous acid :

Primary aliphatic amines react with nitrous acid through diazotisation reaction giving high yield of
unstable diazonium salts.

Even at low temperature they decompose to form nitrogen (N2) and carbocation (R+)

R+ reacts with H2O to form ROH, or alkene or R – X. It means mixture of products produced.
Reactions of Primary Arylamines with Nitrous acid :
Primary arylamine react with HONO acid to give arenediazonium salts. These salts are although
unstable but are more stable than the diazonium salt of aliphatic primary amine.
ArNH 2
Primary arylamine
HONO, H O

2
  
 Ar N  N : X 
0
Temp. 0  5 C
Reaction of Secondary amine with Nitrous acid :
Secondary amines both aliphatic and aromic react with nitrous acid to yeild N-nitroamines usually
separate from reaction mixture as oily yellow liquid. Specific Examples :
..
HONO
(CH 3 ) 2 N H  HCl  NaNO 2  
 (CH 3 ) 2 N  NO
H 2O
N  Nitrosodimethyl amine
(a yellow oil)
Reaction of Tertiary amine with nitrous acid :
When tertiary aliphatic amine is mixed with nitrous acid an equilibrium is established among the
tertiary amine, its salt and an N-nitrosoammonium ion compound

R 3 N :  HX  NaNO 2

..
R 3 N HX   R 3 N  N  OX 
Amine salt
Tertiary aryl amine react with nitrous acid to form p-nitroso aromatic compound.
Nitrosation exclusivery takes place at para position.
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Replacement Reactions of Arene Diazonium Salts :
[H3PO2 is known as hypophosphorous acid]
6.
Ring Substitution in Aromatic Amines :
Example :
(a)
(b)
(c)
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
Nitric acid not only nitrates but also oxidizes the highly reactive ring as well.

In the strongly acidic condition aniline is converted into anilinium ion (–NH3+) because of its positive
charge it directs substitution to the meta position.

To overcome this difficulty, we protect the amino group : we acetylate the amine, then carry out the
substitution and finally hydrolyze the amide to the desired substituted amine
For Example :
(a)
(b)
7.
Sulfonation of Aromatic Amines :
Sulphanilamide. The Sulfa drug :
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8.
Analysis of Amines : Hinsberg Test :
H
KOH
R  NH 2  C 6 H 5SO 2 Cl 
 C 6 H 5SO 2 NHR  C 6 H 5 SO 2 NR  K   C 6 H 5 SO 2 NHR
10
Clear Solution
R
|
KOH
R 2 NH  C 6 H 5 SO 2 Cl 
 C 6 H 5 SO 2  N  R  no rxn.
2
0
Insoluble in KOH solution
R 3 N  C 6 H 5 SO 2 Cl 
 No reaction
30
9.
Carbyl amine test : It is given by 10 alkyl amine and aryl amine.

RNH 2  CHCl 3  3KOH  RNC  3KCl  H 2 O
unpleasent smell of alkyl isocyanide is obtained. 20 amine and 30 amine does not give this test.
Practice Problems :
1.
Which compound is obtained at the end of the following reaction,
Ethylamine HNO
2 (A) PCl

5 (B) NH

3 (C)
(a)
2.
3.
4.
ethyl cyanide
(b)
ethyl amine
(c)
methyl amine
(d)
acetamide
When ethyl amine is heated with chloroform and alcoholic KOH, a bad odour compound is formed.
The compound is
(a)
a secondary amine
(b)
an acid
(c)
a cyanide
(d)
an isocyanide
CH3NH2 + CHCl3 + 3KOH  X + Y + 3H2O; compounds X and Y are
(a)
CH3CN + 3KCl
(b)
CH3NC + 2KCl
(c)
CH3CONH2 + 3KCl
(d)
CH3NC + K2CO3
In the following series of reaction, A is
( A ) Re
duction
( B) HNO
2  C 2 H 5 OH
(a)
5.
7.
8.
(b)
CH3NC
(c)
C2H5CN
(d)
CH3NO2
(c)
(C2H5)2NH
(d)
C6H5NH2
Which one of the following is least basic
(a)
6.
CH3CN
C2H5NH2
(b)
NH3
Which of the following is least basic
(a)
aniline
(b)
p-methylaniline
(c)
diphenylamine
(d)
triphenylamine
A positive carbylamine test is given by
(a)
N, N-dimethyl aniline
(b)
2, 4-dimethyl aniline
(c)
N-methyl-o-methylaniline
(d)
p-methyl benzylamine
A nitrogenous substance X is treated with HNO2 and the product so formed is further treated with
NaOH solution, which produces blue colouration. X can be
(a)
CH3CH2NH2
(b)
CH3CH2NO2
(c)
CH3CH2ONO
(d)
(CH3)2CHNO2
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CON – 13
9.
X is :
(a)
(b)
(c)
10.
(CH3)3N + CH3CH2CH2OH
(CH3)3N + CH3 – CH = CH2 + H2O
(d)
C4 H11N HNO2  C 4 H10O (30 alcohol) hence X will give :
(x)
11.
(a)
carbyl amine reaction
(b)
Hofmann mustard oil reaction
(c)
diazonium salt (as the intermediate) with HNO2
(d)
all are correct
In the following compounds,
the order of basicity is
(a)
IV > I > III > II
(b)
III > I > IV > II
(c)
II > I > III > IV
(d)
I > III > II > IV
POCl
2 CH MgBr
3
 A
NH2 3   

12.
H 3O
(a)
(b)
(c)
(d)
[Answers : (1) b (2) d (3) b (4) a (5) d (6) d (7) b (8) d (9) b (10) d (11) d (12) a]
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INITIAL STEP EXERCISE
1.
The product (D) in the following sequence of
reaction is
7.
CH 3COOH NH

3 ( A ) heat
  ( B)
P2 O 5
Na  C 2 H 5 OH

(C)   
( D)
2.
3.
5.
6.
(a)
homocyclic but not aromatic
(b)
aromatic but not homocyclic
(a)
ester
(b)
amine
(c)
homocyclic and aromatic
(c)
acid
(d)
alcohol
(d)
heterocyclic
Ethyl amine on oxidation with acidified KMnO4
gives
8.
By the action of bromine and alkali on benzamide
it gives
(a)
an acid
(a)
benzene
(b)
an alcohol
(c)
bromo benzene (d)
(c)
an aldehyde
(d)
a nitro compound
9.
The correct increasing order of basic strength in,
CH3CH2CN, CH3CH2NH2, CH3N = CHCH3 is
4.
An organic compound X having molecular formula
C6H7O2N has 6 carbon atoms in a ring system, two
double bonds and also a nitro group as substitution, X is
(a)
CH3N = CHCH3, CH3CH2NH2,
CH3CH2CN
(b)
CH3CH2NH2, CH3N = CHCH3,
CH3CH2CN
(c)
CH3CH2CN, CH3N = CHCH3,
CH3CH2NH2
(d)
CH3CH2CN, CH3CH2NH2, CH3N =
CHCH3
10.
HCl
(b)
Cu2Cl2
(c)
Chlorine in presence of anhydrous AlCl3
(d)
nitrous acid followed by heating with
Cu2Cl2
aniline
acetanilide
Nitroso amines (R2N – N = O)
are
water
soluble. On heating with conc. HCl, they give
secondary amines. The reaction is called
(a)
Perkin reaction
(b)
Fries reaction
(c)
Liebermann nitroso reaction
(d)
Etard reaction
Which nitro compound will show tautomerism
(a)
C6H5NO2
(b)
(CH3)3CNO2
(c)
CH3CH2NO2
(d)
o-nitrotoluene
POCl 3
11.
Chlorobenzene can be prepared by reacting aniline
with
(a)
(b)
12.
  ? is
HCONHR Pyridine
(a)
RCN
(b)
RNC
(c)
RCNO
(d)
RNCO
Identify X in the series,
HNO
3 
H 2SO 4
Nitrobenzene on reduction with Zn and NH4Cl
forms
intermediate
2O
H

X
(a)
aniline
(b)
nitrobenzene
(c)
hydrazobenzene
(d)
phenyl hydroxylamine

In the reaction,
(a)
(b)
(c)
(d)
2  HCl
C 6 H 5 NH 2 NaNO


( A ) CuCN
( B)
0
KCN
0 5 C

2O
H/ H

(C)
the product (C) is
(a)
C6H5CH2NH2
(b)
C6H5COOH
(c)
C6H5OH
(d)
none of these
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13.
The reactions,
(c)
—CH3, 30
HO—
RI

(d)
—SO2Cl, 20
15.
H 2O
2
 RNH 2 
Which is correct alternate ?
illustrate :
14.
(a)
Gabriel’s phthalimide reaction
(b)
A good method to prepare pure second
ary amine
(c)
A reaction which can be extended to
preparation of -amino acid
(d)
None
(a)
Hinsberg reagent is.................and reacts
with....................amine to form a product soluble in
alkali :
(a)
(b)
— SO2Cl, 10
(c)
(b)
—SO2NH2, 2
0
(d)
none is correct
FINAL STEP EXERCISE
1.
2.
3.
A compound A when reacted with PCl5 and then
with ammonia gave B, B when treated with
bromine and cuastic potash produced C. C on
treatment with NaNO2 and HCl at 00C and then
boiling produced orthocresol. Compound A is
(a)
o-toluic acid
(b)
o-chlorotoluene
(c)
o-bromotoluene
(d)
m-toluic acid
An aromatic amine (A) was treated with alcoholic
potash and another compound (Y) when a foul
smelling gas was formed with formula C6H5NC. (Y)
was formed by reacting a compound (Z) with Cl2
in presence of slaked lime. Compound (Z) is
(a)
C6H5NH2
(b)
CH3OH
(c)
CH3COCH3
(d)
CHCl3
Deamination of n-BuNH2 with NaNO3/HCl gives
two butanols, two butenes and two butyl chlorides.
The possible explanation of these products is
Einstein Classes,
Intermediate formed is RN2+ which is
very stable
(b)
As intermediate is carbocation
(c)
Both (a) and (b)
(d)
None
What products would you expect to get by the
application of the Hofmann exhaustive methylation
to Me2CHCH2CH2NH2
(a)
Me2C = CH2
(b)
Me2HCCH = CH2
(c)
CH2 = CH2
(d)
All
An organic compound (A) on reduction gave a
compound (B). Upon treatment with HNO2, (B)
gave ethyl alcohol and on warming with CHCl3 and
alcoholic KOH, (A) gave
offensive smell. The
compound (A) is
(a)
CH3CN
(b)
C2H 5CN
(c)
CH3NH2
(d)
CH3NC
(a)
4.
5.
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6.
Examine the following two structures for the anilinium
ion and choose the correct
statement from the ones
give below :
(d)
11.
End product of following sequence of reaction :
3 / H 2O
— Br NH

3  A O

 B BaO

 C

=O
(b)
—OH
(a)
7.
8.
9.
10.
II is not an acceptable canonical
structure because carbonium ions are less
stable than ammonium ions
(b)
II is not an accepted canonical structure
because it is non-aromatic
(c)
II is not an acceptable canonical
structure because the nitrogen has 10
valence electrons
(d)
II is an acceptable canonical structure
What major products would you expect to get by
the application of the Hofmann exhaustive
methylation respectively
Me2CHCH2NHCH2CH2Me;
EtNHCH2CH2Cl
(A)
(B)
(a)
Me2C = CH2, CH2 = CH2
(b)
MeCH = CH2, CH2 = CH2
(c)
MeCH = CH2, CH2 = CHCl
(d)
Me2C = CH2, CH2 = CHCl
A compound X has the molecular formula C7H7NO.
On treatment with Br2 and KOH, X gives an amine
Y. The latter gives carbylamine test. Y upon
diazotisation and coupling with phenol gives an azo
dye. Thus X is
(a)
C6H5CONH2
(b)
C6H5NO2
(c)
C6H5COONH4 (d)
None
Methyl
ethyl
propylamine
forms
non-super-imposable 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 electrons
(d)
Of absence of asymmetric nitrogen
Which is maximum basic in nature ?
(a)
(a)
(b)
(c)
=O
(d)
ANSWERS (INITIAL STEP
EXERCISE)
1.
2.
3.
4.
5.
6.
7.
8.
b
c
c
d
d
b
a
b
9.
10.
11.
12.
13.
14.
15.
c
c
b
b
a
a
a
ANSWERS (FINAL STEP EXERCISE)
1.
2.
3.
4.
5.
6.
a
c
b
b
a
c
7.
8.
9.
10.
11.
c
a
a
b
c
(c)
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
TEST YOURSELF
CON – 17
7.
Resonance hybrid of nitrate ion is
(a)
1.
2.
3.
Reaction CH 3CONH 2 NaOBr

 gives
(a)
CH3Br
(b)
CH4
(c)
CH3OBr
(d)
CH3NH3
Hoffmann bromamide reaction is used to prepare.
Which degree of amine from amides
(a)
10
(b)
20
0
(c)
3
(d)
all
Primary amines give
(a)
Iodoform test
(b)
Victor mayer test
(c)
Lucas test
(d)
Carbylamine test
(b)
(c)
(d)
8.
4.
5.
6.
is a
(a)
primary amine
(b)
secondary amine
(c)
tertiary amine
(d)
quaternary salt
A positive carbylamine test is given by
(a)
N, N dimethylaniline
(b)
2, 4-dimethylaniline
(c)
N-methyl-o-methylaniline
(d)
P-methylbenzylamine
How many primary amines are possible for the
formula C4H11N
(a)
1
(b)
2
(c)
3
(d)
4
9.
10.
An organic compound having molecular mass 60 is
found to contain C = 20%, H = 6.67 % and
N = 46.67 % while rest is oxygen. On heating it gives
NH3 alongwith a solid residue. The solid residue
gives violet colour with alkaline copper sulphate
solution. The compound is
(a)
CH3NCO
(b)
CH3CONH2
(c)
(NH2)2CO
(d)
CH3CH2CONH2
The correct order of basicities of the following
compounds is
(a)
2>1>3>4
(b)
1>3>2>4
(c)
3>1>2>4
(d)
1>2>3>4
Which of the following carbocations is expected to
be most stable ?
(a)
(b)
(c)
(d)
ANSWERS
Einstein Classes,
1.
d 3.
d 5.
d 7.
c 9.
b
2.
a 4.
a 6.
d 8.
c 10.
d
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111