12th Standard Chemistry English Medium Organic Nitrogen Compounds Reduced Syllabus Important Questions With Answer Key 2021

Download Tamil Nadu 12th Standard Chemistry question papers, model tests, one-mark questions, important questions, and public exam papers in PDF format. Free study materials and answer keys for TN State Board students.

Questions:

Multiple Choice Question

15 x 1 = 15

Sandmeyer's reaction occurs in the presence of _________ and __________

(a)

cuprous halide and halo acid

(b)

cupric halide and halogen

(c)

copper - zinc couple and H₂

(d)

none of these

p-amino phenol is the product of reducing nitrobenzene in _________

(a)

acid medium

(b)

basic medium

(c)

electrolytic reduction

(d)

neutral medium

Amine that cannot be prepared by Gabriel Phthalimide synthesis is ___________

(a)

benzyl amine

(b)

methyl amine

(c)

ethyl amine

(d)

phenyl amine

An organic compound 'A' with molecular formula C₃H₉N neither give nitrogen nor yellow oily liquid with HNO₂ Then A is

(a)

CH₃CH_2-CH₂-NH₂

(b)

CH₃CH₂-CH₂-NH₂

(c)

\({ CH }_{ 3 }-\underset { \overset { | }{ { CH }_{ 3 } } }{ N } -{ CH }_{ 3 }\)

(d)

All of these

Aniline differs from ethylamine by the reaction with

(a)

metallic sodium

(b)

an alkyl halide

(c)

chloroform and caustic potash

(d)

nitrous acid

The reduction of CH₃ - CH2 - C ☰ N with sodium and alcohol results in the formation of

(a)

\({ CH }_{ 3 }-\underset { \overset { | }{ { NH }_{ 2 } } }{ CH } -{ CH }_{ 3 }\)

(b)

CH₃- CH₂- CH₂- OH + N₂

(c)

CH₃ - CH₂ - CH₂ - NH₂

(d)

CH₃ - CH₂ - NH₂

In nitro alkanes -NO₂ group is converted to - NH₂ group by the reaction with

(a)

Sn/HCI

(b)

Zn dust

(c)

Zn/NH₄CI

(d)

Zn/NaOH

Identify the amide which will not undergo Hofmann bromamide reaction

(a)

CH₃CONHCH₃

(b)

CH₃CH₂CONH₂

(c)

CH₃CONH₂

(d)

both CH₃CH₂CONH₂and CH₃CONH₂

The gas leaked from storage tank of the union Carbide plant in Bhopal gas tragedy was

(a)

Phosgene

(b)

methyl isocyanide

(c)

methyl isocyanate

(d)

methyl isothiocyanide

10.

(CH₃)₂ - CH - NO when boiled with HCI/H₂O gives compound 'X'. Identify 'X'.

(a)

CH₃COOH

(b)

(CH₃)₂C = 0

(c)

CH₃CHO

(d)

CH₃NO₂

11.

Among the following, the reaction that proceeds through an electrophilic substitution is : _______.

(a)

(b)

(c)

(d)

12.

When is reduced with Sn / HCl the pair of compounds formed are _______.

(a)

Ethanol, hydrozylamine hydrochloride

(b)

Ethanol, ammonium hydroxide

(c)

Ethanol, NH₂OH

(d)

C₃H₅NH₂, H₂O

13.

_______.

(a)

H₃PO₂ and H₂O

(b)

H⁺/H₂O

(c)

HgSO₄/H₂SO₄

(d)

Cu₂Cl₂

14.

When aniline reacts with acetic anhydride the product formed is_______.

(a)

o – aminoacetophenone

(b)

m-aminoacetophenone

(c)

p – aminoacetophenone

(d)

acetanilide

15.

Aniline + benzoylchloride \(\overset { NaOH }{ \longrightarrow } \)C₆H₅ - NH - COC₆ H₅ this reaction is known as ______.

(a)

Friedel – crafts reaction

(b)

HVZ reaction

(c)

Schotten – Baumann reaction

(d)

none of these

Generic 5

9 x 5 = 45

16.

Outline the mechanism of
(a) Nitration of aniline,
(b) Acetylation of aniline.

17.

Write a note on
(i) Carbylamine reaction
(ii) Mustard oil reaction
(iii) Acetylation of benzylamine
(iv) Formation of Schiff's base
(v) Diazotisation reaction

18.

Write the differences between nitro methane and nitro benzene.

19.

How can the following coversion be effected?
i) Nitrobenzene ⇾ Nitrosobenzene
ii) Nitrobenzene ⇾ Azoxybenzene
iii) Nitrobenzene ⇾ Hydrazobenzene

20.

Account for the following:
(i) Nitroethane is soluble NaOH
(ii) Nitroethane reacts with nitrous acid
(iii) 2-methyl-2-nitro propane has neither of the properties.

21.

What happens when ethylamine is treated with
(i) CHCl₃/NaOH
(ii) CS₂
(iii) C₆H₅CHO.

22.

Account for the following:
i) (CH₃)₂ NH is stronger base than NH₃
ii) CH₃CH₂NH₂ is more basic than CH₃CONH₂
iii) Aniline is less basic than Ethyl amine
iv) On sulphonation of aniline, p-amino benzene sulphonic acid is formed.

23.

Arrange the following
i. In increasing order of solubility in water, C₆H₅ NH₂, (C₂H₅)₂NH, C₂H₅NH₂
ii. In increasing order of basic strength
a) aniline, p- toludine and p – nitroaniline
b) C₆H₅ NH₂, C₆H₅ NHCH₃, C₆H₅NH₂, p-Cl-C₆- H₄-NH₂
iii. In decreasing order of basic strength in gas phase
(C₂H₅)NH₂, (C₂H₅)NH, (C₂H₅)₅N and NH₃
iv. In increasing order of boiling point
C₆H₅OH, (CH₃)₂NH, C₂H₅NH₂
v. In decreasing order of the pK_b values
C₂H₅NH₂, C₆H₅NHCH₃.(C₂H₅)₂ NH and CH₃NH₂
vi. Increasing order of basic strength
C₂H₅NH₂,C₆H₅N(CH₃)₂, (C₂H₅)₂ NH and CH₃NH₂
vii. In decreasing order of basic strength

24.

How will you distinguish between primary secondary and tertiary alphatic amines.

Generic 3

8 x 3 = 24

25.

An organic compound 'A' on reduction gives compound 'B' which on reaction with trichloromethane and caustic potash forms 'C'. Compound 'C' on catalytic reduction given N-methyl benzenamine. Identify A, B and C and write the reactions involved.

26.

Write a note on structure of amine.

27.

Aniline reacts with Br₂/H₂O to give a tribromo derivative. How would you convert aniline to get a monobromo derivate?

28.

Explain Levine and hauser acetylation.

29.

Write a note on the basicity of amines.

30.

How will convert nitrobenzene to benzoic acid?

31.

Identify A,B,and C
CH₃- NO₂ \(\overset { { L }_{ 1 }{AlH }_{ 4 } }{ \underset { {} }{ \longrightarrow } }\) A \(\overset { { 2CH_3 }{Ch_2Br } }{ \underset { {} }{ \longrightarrow } }\) B \(\overset { {H}_{ 2 }{SO}_{ 4 } }{ \underset { {} }{ \longrightarrow } } \) C

32.

How will you prepare propan – 1- amine from
i) butane nitrile
ii) propanamide
ii) 1- nitropropane

Generic 2

10 x 2 = 20

33.

Bring about the conversion of Chlorobenzene to aniline

34.

\({ C }_{ 2 }{ H }_{ 3 }N\xrightarrow [ Ether ]{ LiAl/{ H }_{ 4 } } B\underrightarrow { HN{ O }_{ 2 } } \) Identify A, B and C.

35.

\({ C }_{ 6 }{ H }_{ 5 }CON{ H }_{ 2 }\xrightarrow [ KOH ]{ { Br }_{ 2 } } X\xrightarrow [ HCl ]{ { HNO }_{ 2 } } Y\xrightarrow [ { Cu }^{ + } ]{ { NaNO }_{ 2 } } Z\)
Identify X, Y and Z.

36.

Identify (A), (B) and (C)
\({ CH }_{ 4 }\xrightarrow [ { 400 }^{ o }C ]{ { HNO }_{ 3 } } A \stackrel{\mathrm{Sn} / \mathrm{HCl}}{\longrightarrow} \mathrm{B}\xrightarrow [ { HgCl }_{ 2 } ]{ { CS }_{ 3 } } C\)

37.

When benzamide is treated with bromine and alkali gives compound A. Also when benzamide is reduced by LiAlH₄, compound B is formed. Find A and B. Write the equations.

38.

Explain why aniline is less basic than methylamine?

39.

What happens when aniline is treated with bromine?

40.

How will you convert benzene to m-dinitro benzene?

41.

Write about the functional isomerism of nitro methane?

42.

Nitrobenzene does not undergo Friedel Crafts reaction - Give reason.

Visit QB365-SLMS - https://www.qb365.in/slms/

Answers Key:

Multiple Choice Question

15 x 1 = 15

(a)
cuprous halide and halo acid

(d)
phenyl amine

(d)
nitrous acid

(a)
Sn/HCI

(a)
CH₃CONHCH₃

10.

(b)
(CH₃)₂C = 0

11.

12.

(a)
Ethanol, hydrozylamine hydrochloride

13.

(a)
H₃PO₂ and H₂O

14.

(d)
acetanilide

15.

Generic 5

9 x 5 = 45

16.

(i) Nitration of aniline is accompanied by oxidation. But a mixture of cone. HNO₃ and cone. H₂SO₄ gives m-nitroaniline also. Nitric acid is a strong acid. It protonates aniline forming anilinium ion C₆H₅NH₃⁺ because of positive charge on nitrogen it is meta directing.

p-nitro aniline is prepared in the following three stages

(ii) Aniline reacts with acetyl chloride and acetic anhydride to form corresponding amides called anilides.
\({ C }_{ 6 }H_{ 5 }{ NH }_{ 2 }+\underset { Acetylchloride }{ ClCO{ CH }_{ 3 } } \rightarrow \underset { Acetanilide }{ { C }_{ 6 }{ H }_{ 5 }NHCO{ CH }_{ 3 } } +HCl\)
\({ C }_{ 6 }H_{ 5 }{ NH }_{ 2 }+\underset { Aceticanhydride }{ { CH }_{ 3 }COOCO{ CH }_{ 3 } } \rightarrow \underset { Acetanilide }{ { C }_{ 6 }{ H }_{ 5 }NHCO{ CH }_{ 3 } } +{ CH }_{ 3 }COOH\)

17.

(i) Carbylamine reaction: Primary amines
on heating with chloroform and alcoholic potash forms a foul smelling substance called carbylamine or alkyl isocyanide
\(\underset { Chloroform }{ CH_{ 3 }{ NH }_{ 2 }+{ CHCl }_{ 3 } } +3KOH\rightarrow \underset { Methyl \ isocyanide }{ { CH }_{ 3 }NC } +3KCl+{ 3H }_{ 2 }O\)
This reaction is characteristic of primary amines. Secondary and tertiary amines do not undergo this reaction.
(ii) Mustard oil reaction: This is also a reaction characteristic of primary amine. When primary amines are warmed with carbon disulphide and mercuric chloride, alkyl isothiocyanate, having a pungent Mustard like odour is obtained
\(CH_{ 3 }{ NH }_{ 2 }+\underset { Carbondisulphide }{ S=C=S } \rightarrow \underset { Methyl \ isothicyanate }{ { CH }_{ 3 }-N=C=S } +{ H }_{ 2 }S\)
(iii) Aniline reacts with acetyl chloride and acetic anhydride to form corresponding amides called anilide.
\({ C }_{ 6 }H_{ 5 }{ NH }_{ 2 }+\underset { Acetyl \ chloride }{ ClCO{ CH }_{ 3 } } \rightarrow \underset { N-phenyl \ acetamide }{ { C }_{ 6 }{ H }_{ 5 }NHCO{ CH }_{ 3 } } +HCl\)
\({ C }_{ 6 }H_{ 5 }{ NH }_{ 2 }+\underset { Acetic \ anhydride }{ { CH }_{ 3 }COOCO{ CH }_{ 3 } } \rightarrow \underset { Acetanilide }{ { C }_{ 6 }{ H }_{ 5 }NHCO{ CH }_{ 3 } } +{ CH }_{ 3 }COOH\)
(iv) Formation of Schiff's base: Primary amines condense with aromatic aldehydes forming Schiff's base
\({ C }_{ 6 }H_{ 5 }CHO+{ H }_{ 2 }{ NCH }_{ 3 }\rightarrow \underset { Schiff's \ base\\ (benzal-N-methyl \ amine) }{ { C }_{ 6 }{ H }_{ 5 }CH=NC{ H }_{ 3 } } \)
(v) Reaction with nitrous acid: A cold solution of sodium nitrite reacts with aniline dissolved in hydrochloric acid, a clear solution is obtained. This solution contains 'benzene diazonium chloride'. This reaction is known as 'diazotisation'.
\({ C }_{ 6 }{ H }_{ 5 }N\boxed { { H }_{ 2 }+O } =N-OH\underrightarrow { HCl } \underset { Bensene\quad diazonium\quad chloride }{ { C }_{ 6 }{ H }_{ 5 }-N=N-Cl } \)

18.

S. No.	Nitromethane	Nitrobenzene
i.	It is colourless	It is yellow coloured.
ii.	It is soluble in water	Insoluble in water.
iii.	Has pleasant smell	Has smell of bitter almonds.
iv.	Exhibits tautomerism	Doesnot exhibit tautomerism
v.	Used as good solvent	Used to prepare explosives Eg. TNT
vi.	Undergoes reduction in acidic and neutral medium.	Undergoes reduction in acidic, basic and neutral medium

19.

i) Nitrobenzene ⇾ Nitrosobenzene
When nitrobenzene is treated with glucose and NaOH (alkaline medium) Nitrosobenzene is formed.
\({ C }_{ 6 }{ H }_{ 5 }N{ O }_{ 2 }\rightarrow \underset { Nitrosobenzene }{ { C }_{ 6 }{ H }_{ 5 }NO+{ H }_{ 2 }O } \)
ii) Nitrobenzene ⇾ Azoxybenzene
When nitrobenzene is subjected to reduction with glucose and NaOH forms the intermediate products nitrosobenzene and phenyl hydroxyl amine. These undergo bimolecular condensation reaction to give azoxy benzene.
\({ C }_{ 6 }{ H }_{ 5 }N{ O }_{ 2 }\underrightarrow { Glucose+NaOH } \underset { Nitrosobenzene }{ { C }_{ 6 }{ H }_{ 5 }NO+{ H }_{ 2 }O } \xrightarrow [ { H }_{ 2 }O ]{ \triangle } \underset { Azoxy\quad benzene }{ { C }_{ 6 }{ H }_{ 5 }-N=N{ C }_{ 6 }{ H }_{ 5 } } \)
iii) Nitrobenzene ⇾ Hydrazobenzene
When nitrobenzene is subjected to alkaline reduction in the presence of Zn+NaOH, hydrozo benzene is formed.
\({ C }_{ 6 }{ H }_{ 5 }N{ O }_{ 2 }\underrightarrow { Zn/NaOH } \underset { Nitrosobenzene }{ { C }_{ 6 }{ H }_{ 5 }NH+{ NHC }_{ 6 }{ H }_{ 5 }} \)

20.

(i) Nitroethane exhibits tautomerism of nitro form and aci form. Aciform of nitroethane contains (α - Hydrogen) replaceable hydrogen atom. Hence it dissolves in sodium hydroxide solution forming salt like compounds.

(ii) CH₃CH₂NO₂ has two a-hydrogen and that reacts with HNO₂
(iii) 2-methyl - 2 - nitro propane neither soluble in alkali nor reacts with nitrous acid. Because it contains a tertiary nitro group where aciform is not possible and . it does not contain α - H atoms. So it has neither of the properties.

21.

(i) When ethylamines is treated with chloroform and an alkali, carbylamine reaction takes place and a foul smelling substance called carbylamlne (or) alkyl isocyanide is formed.
C₂H₅NH₂ + CHCl₃ + 3NaOH ➝ \(\underset { Ethyl \ isocyanide }{ { C }_{ 2 }{ H }_{ 5 }NC+3NaCl+3{ H }_{ 2 }O } \)
(ii) When ethylamine is warmed with CS₂ and mercuric chloride, alkyl isothiocyanate having a pungent mustard like odour is obtained. This reaction is called mustard oil reaction.
C₂H₂NH₂ + S = C-S \(\underrightarrow { Hg{ Cl }_{ 2 } } \) C₂H₅ - N = C = S + H₂S
(iii) When ethylamine condense with aromatic aldehyde (benzaldehyde) Schiff's base is formed.
C₆H₅CHO + H₂NC₂H₅ ➝ \(\underset { Schiffi's \ base \ Benzal-ethylamine }{ { C }_{ 6 }{ H }_{ 5 }CH=N{ C }_{ 2 }{ H }_{ 5 }+{ H }_{ 2 }O } \)

22.

i) (CH₃)₂ NH is stronger base than NH₃:
(a) (CH₃)₂NH (Secondary amine) is a stronger base than NH₃(ammonia).
(b) In dimethylamine, the methyl group with +1 effects tends to Increase the electron density on the nitrogen atom.
(c) As a result, the electron releasing tendency or basic strength of amine increases.
ii) CH₃CH₂NH₂ is more basic than CH₃CONH₂:
(a) Resonance and inductive effect possible in acetamide involving the non bonding pair of electrons on the nitrogen atom.
(b) Another important factor is that amides containing a powerful electron is not readily available for donation and it is less basic.
(c) Thus ethylamine is more basic than acetamide.
iii) Aniline is less basic than Ethyl amine:
(a) The lone pair of electrons on the nitrogen atom of aniline is involved in resonance and is not easily available for donation to protons.
(b) The positive charge on nitrogen makes protonation difficult. Thus aniline is less basic than ethylamine.
vi) When aniline is heated with fuming sulphuric acid, p-amino benzene sulphonic acid is formed.

Aniline does not give o-amino benzene sulphonic acid because it is sterically less favoured.

23.

i) In increasing order of solubility in water:
\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2},\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH}, \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2} \)
\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}>\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}>\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH}\)
ii) In increasing order of basic strength:
a. Aniline, p - toluidine and p - nitro aniline
p - toluidine > aniline >p - nitro aniline

b. \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}, \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NHCH}_{3}, \mathrm{p}-\mathrm{Cl}-\mathrm{C}_{6} \mathrm{H}_{4}-\mathrm{NH}_{2} \)
\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NHCH}_{3}>\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}>\mathrm{p}-\mathrm{Cl}-\mathrm{C}_{6} \mathrm{H}_{4}-\mathrm{NH}_{2}\\ 2^{o} \text { amine } \quad \quad \quad \quad e^{\ominus} \text { with drawing (group) }\)
(iii) In decreasing order of basic strength in gas phase:
\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2},\left(\mathrm{C}_{2} \mathrm{H}_{5}\right) \mathrm{NH},\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{3} \mathrm{~N} \text { and } \mathrm{NH}_{3} \)
\(\mathrm{NH}_{3}<\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}<\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{~N}<\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{3} \mathrm{NH}\\ \quad \quad \quad 1^{0} \text { amine } \quad 2^{0} \text { amine } \quad \quad \quad 3^{0} \text { amine }\)
(iv) In increasing order of boiling point:
\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH},\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}, \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2} \)
\(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}>\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}>\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH} \\ 2^{0} \text { amine } \quad \quad1^{0} \text { amine }\)
Generally amines have lower boiling point than alcohol. Due to comparable molecular mass and weaker H-bonds in Amines.
(v) In decreasing order of the pK_b values:
\( \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}, \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NHCH}_{3},\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH} \text { and } \mathrm{CH}_{3} \mathrm{NH}_{2} \)
\(\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH}<\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}<\mathrm{CH}_{3} \mathrm{NH}_{2}<\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NHCH}_{3} \)
\(\quad \quad \quad 2^{0} \text { amine } \quad 1^{0} \text { amine } 1^{0} \text { amine } \quad 2^{0} \text { amine } \)
\(\mathrm{PK}_{b}: \quad 3.00\quad < \quad 3.29 \quad < \quad 3.38 \quad<\quad 9.30\)
pK_b ,Due to + 1 effect of C₂H₅ group. Higher the value of pK_b lower is the basicity
(vi) Increasing Order of basic strength:
\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}, \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~N}\left(\mathrm{CH}_{3}\right)_{2},\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH} \text { and } \mathrm{CH}_{3} \mathrm{NH}_{2} \)
\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}>\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~N}\left(\mathrm{CH}_{3}\right)_{2}>\mathrm{CH}_{3} \mathrm{NH}_{2}>\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH} \\ \left(\mathrm{pK}_{6}: 9.38 \quad > \quad \quad 8.92 \quad \quad > \quad 3.38 \quad > \quad \quad 3.00\right)\)
Due to + 1 effect of C₂H₅ group.
In decreasing order of basic strength:

24.

S.No	Reagents or Reaction	Primary amine RNH₂	Secondary amine R₂NH	Tertiary amine R₃N
1.	Carbylamine reaction or with CHCl₃/KOH	Carbylamine is formed (unpleasant smell)	-	-
2.	Mustard oil reaction or CS₂/HgCl₂ (Hoffmann's mustard oil test)	Alkyl isothiocyanate is formed (Mustard oil odour)	-	-
3.	HNO₂ (or) NaNO₂/ HCl	Alcohol is formed +H₂	Yellow oily nitrosoamine is formed, insoluble in water. (Liberman's Test)	Forms nitrite in cold, soluble in water.
4.	CH₃COCl	N-acetyl derivative is formed	N,N- diacetyl derivative is formed	-
5.	Diethyl oxalate Hoffmann's method	Solid oxamide is formed	Liquid oxamic ester is formed	-
6.	Benzene sulphonyl chloride in presence of excess. KOH (Hinsberg's reaction)	N- alkyl benzene sulphonamide is formed (soluble)	N, N - dialkyl benzene sulphonamide is formed (Insoluble).	-
7.	With RX	1 mol → 2^o amine 2 mol → 3^o amine 3 mol → Quarternary salt	1 mol → 3^o amine 2 mol → Quarternary salt	1 mol → Quarternary salt

Generic 3

8 x 3 = 24

25.

(i) A - C₆H₅NO₂ - Nitro benzene
(ii) B - C₆H₅NH₂ - Aniline
(iii) C - C₆H₅NC - Phenyl Carbylamine

26.

Structure of amines:
(i) Like, ammonia, nitrogen atom of amines is trivalent and carries a lone pair of electron and sp³ hybridised, out of the four sp³ hybridised orbitals of nitrogen, three sp³orbitals overlap with orbitals of hydrogen (or) alkyl groups of carbon, the fourth sp³ orbital contains a lone pair of electron.
(ii) Hence, amines possess pyramidal geometry. Due to presence of lone pair of electron C-N-H (or) C-N-C bond angle is less than the normal tetrahedral bond angle 109.5°.
For example, the C-N-C bond angle of trimethylamine is 108° which is lower than tetrahedral angle and higher than the H-N-H bond angle of 107°.
(iii) This increase is due to the repulsion between the bulky methyl groups.

27.

To get mono bromo compounds, -NH₂ is first acylated to reduce its activity.

28.

(i) The nitriles containing a-hydrogen also undergo condensation with esters in the presence of sodamide in ether to form ketonitriles.
(ii) This reaction is known as "Levine and hauser" acetylation.
(iii) This reaction involves replacement of ethoxy (OC₂H₅) group by methylnitrile (-CH₂CN) group and is called as cyanomethylation reaction.

29.

(i) The nitrogen in amines possess an unshared pair of electrons (Lone Pair).
(ii) The lone pair of electrons is available for the formation of a new bond with a proton or lewis acid.
(iii) Thus amines are basic in nature and they react with acids to form salts.
(iv) The greater is the number of electron releasing alkyl. groups, the greater the availability of nitrogen's lone pair and stronger the base.
NH₃< CH₃ - NH₂< CH₃ - NH - CH₃

30.

31.

32.

+ 2H₂O

Generic 2

10 x 2 = 20

33.

When chlorobenzene is heated with alcoholic ammonia, aniline is obtained

34.

35.

X = C₆H₅NH₂; Y = C₆H₅N₂Cl; Z = C₆H₅NO₂.

36.

\({ CH }_{ 4 }\xrightarrow [ { 400 }^{ o }C ]{ { HNO }_{ 3 } } \underset { Nitro \ methane }{ { CH }_{ 3 }{ NO }_{ 2 } } \stackrel{\mathrm{Sn} / \mathrm{HCl}}{\longrightarrow} \underset { Methylamine }{ { CH }_{ 3 }{ NH }_{ 2 } } \xrightarrow [ { HgCl }_{ 2 } ]{ { CS }_{ 3 } } \underset { Methyl \ isothiocyanate }{ { CH }_{ 3 }-N=C=S } \)

37.

(i) When benzamide react with bromine and alkali, Hoffmann reaction take place and aniline C₆H₅NH₂ is formed as (A).
\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CONH}_{2} \stackrel{\mathrm{Br}_{2} / \mathrm{KOH}}{\longrightarrow} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}+\mathrm{CO}_{2}\\ Benzamide \quad \quad \quad \quad Aniline(A)\)
(Ii) When Benzamide is reduced by LiAIH₄, Benzylamine C₆H₅CH₂NH₂ is formed as (B).

38.

(i) The lone pair of electrons on the nitrogen atom of aniline is involved in resonance and not available for donation to protons.
(ii) Protonation becomes difficult.
(iii) So aniline is a weaker base than methylamine.

39.

Aniline decolourises bromine water with the formation of white precipitate which is 2,4,6-tribromo aniline

40.

Benzene on reaction with a mixture of cone. HNO₃ and cone. H₂SO₄ produces nitro benzene as product which on further nitration will result in the formation of m-dinitrobenzene.