CONTENTS S.NO.
TOPIC
TOPIC
PAGE NO
PAGE NO.
1.
INTRODUCTION
2
2.
PREPARATION OF AMINES
2
3.
PHYSICAL PROPERTIES
6
4.
CHEMICAL PROPERTIES
7
5.
DIAZONIUM SALTS
10
1
Amine
AMINE 1. INTRODUCTION (i) (ii) (iii) (iv) (v) (vi)
Amines are called alkyl derivative of NH3. If a hydrogen atom of NH3 is replaced by an alkyl group then it is called primary amine and possess –NH2(amino) group. If two hydrogen atoms of NH3 are replaced then it is called secondary amine and it posses > NH (mino) group. If all hydrogen atoms of NH3 are replaced then it is called tert. amine and has a nitrilo N group. N is in sp3 hybridisation and tetrahedral geometry. Bond angle increases from ammonia to 3º amines. NH3 (107º) < RNH2 < R2NH < R3N
2. PREPARATION OF AMINES (i) Reduction of nitro compounds
(ii)Ammonolysis of alkyl halides
The free amine can be obtained from the ammonium salt by treatment with a strong base:
–
–
R – NH3 X NaOH R – NH2 H2O Na X primary amine is obtained as a major product by taking large excess of ammonia. The order of reactivity of halides with amines is RI > RBr >RCl. (iii) Reduction of nitriles R –C N
H 2 Ni Na (Hg )C3H 4OH
R – CH2 – NH 2
(iv) Reduction of amides
2
Amine
(v) Gabriel phthalimide synthesis
(vi) Curtius Reaction : H2 O R N C O R NH 2 RCON N
3 Acyl azide
CO
2
2
Alkyl isocyanate
Alkyl amide
Mechanism : RCOCl NaN3 RCON3 NaCl
R
C
O
O
O N
N
C
N R
R
N C
N2
N
N
N R
N
N2
R—NCO alkyl isocyanate
O
R
C
H2 O
N
C OH
R
H2 O
OH R
C
O
OH
O
NH C
N
R — NH 2 CO2 OH – CO 2
3
Amine
(vii) Schmidt Reaction : Carboxylic acid reacts with hydrozoic acid in presence of concentrated H 2SO 4 to give isocyanates. R – COOH + N 3H H2SO4 R – N2H + C2 O + H 2O
Mechanism :
R
O
OH
C H R
C
OH
OH
OH H N3
R
C
OH
H
N
N
+
N
–H2O
O C H2
R — N C O R N2
R
isocyanate
N
–H 2O
+
N
C
O
N
N
+
H N
N
R — NH2 CO2 (viii) Lossen Reaction : Hydroxylamine on treatment with acid chloride gives acyl derivatives of hydroxyl amine. the acyl derivatives exist in two tautomeric form keto form called hydroxamic form and enol form called hydroximic acid. The hydroxamic form. O R
OH
NH
R NH 2 CO 2
HO
Mechanism : O
O R
R
OH
H2N
+ HCl
Cl
NH HO
O
OH R
R NH HO Hydroxamic acid (keto form)
N HO (enol form)
4
Amine
The hydroxamic form (keto form) forms o-acyl derivatives of hydroxamic form which on heating with bases forms isocyanates and finally amines upon hydrolysis H R
C
O
N
OH R
C
H R Cl
O
C
N
O O
C
R
O O
H2 O R NH2 R — N C O R CO2
C
N
O
C
R
O (ix) Hoffmann bromamide degradation reaction : The reaction of amid with bromin in present of base to form primary amine.
Mechanism: 2NaOH Br2 NaOBr NaBr H 2O O
O R
C
– R NH2 OBr
C
N
Br OH –
H N-bromoacetone
O R
C
O N
R Br OH – –H 2 O
H
C
N – Br H2O Rearrangement
H O R — NH2 CO2 2 R — N C O (Isocyanate)
Ex.1
Sol.
Compound (A) has molecular formula C9H14NCl. (A) gives an immediate precipitate with AgNO3. It is very resistance to bromination in either acid or alkaline solution. It is also resistance to heat, nitration and oxidation by KMnO4. Suggest structure for (A). Since (A) (C9H14NCl) gives immediate precipitate with AgNO3, it must be an ionic compound. Further, it is resistant to oxidation, heat, nitration etc. it must be a quanternary ammoinium salt. Therefore, possible structure of (A) may be : +
CH3 C6H5 – N – CH
3
–
Cl, quaternary-salt
CH3 (A) (C9H14NCl) N,N,N-Trimethylaniliniumchloride
5
Amine
Ex.2
Identify (A) through (E) in the following sequence H3 O
– HCl NaOH EtBr PhSO2Cl EtNH2 (A) (B) (C) (D) (E)
Sol.
PhSO2 N – Et
PhSO 2 NHEt
Ph SO2 NEt2
(B)
(A )
(C)
H 3O
PhSO2OH Et2 NH2 (D)
Ex.3
Sol.
(E)
How the ethylamine is prepared from : (a) Propanamide (b) Ethanamide (c) Nitroethane (d)Acetaldehyde Ethanol (on large scale) ? (e) (a) When propanamide is treated with Br2 and KOH, ethanamine is obtained. CH3CH2CONH2 + Br2 + 4KOH CH3CH2NH2 + K2CO3 + 2KBr + 2H2O Ethanamine
(b)
When ethanamide is reduced with lithium aluminium hydride, ethanamine is LiAlH4
obtained. CH3CONH2 + 4[H]
CH3CH2—NH2+ H2O
Ethanamide
(c)
Nitroethane can be converted into ethanamine by the reduction with Sn/HCl. Sn / HCl CH3CH2—NO2 + 6[H] CH3CH2NH2 + 2H2O Nitroethane
(d)
Ethanamine
When acetaldehyde is treated with ammonia, imine is formed whichonreduction gives ethanamine. H H | | H 2 / Ni CH 3 — C O + NH3 CH — C NH CH 3 — CH 2NH 2 3 H2O 2[H] Ethanal
(e)
Acetaldimine
Ethanamine
When a mixture of ethanol and ammonia is passed over alumina at 723 K and high pressure, than amine is formed. O3 l2 CH3 CH2 NH2 H2 O CH3CH2OH + NH3 A723K Ethanamine
3. PHYSICAL PROPERTIES : (i) (ii) (iii)
(iv) (v)
Unlike other organic compounds, amines are much more soluble in water. Because All amines form a stronger H- bond with water. Like ammonia, amines are polar compounds and except 3º amines can form intermoleculer Hbonds that’s why they have higher boiling points. Boiling points of amines are lesser than alcohols and acids of comparable mol. weight. Because H- bonding in amines is less pronounced in 1º and 2º than that in alcohols and carboxylic acids. Because nitrogen is less electronegative than oxygen. Thus every question regarding boiling point can be answered on the basis of H - bonding. Boiling points of 1º, 2º and 3º amines follow the order. 1º > 2º > 3º amine. Solubility in water follow the order. 1º > 2º > 3º amine. This is all due to H- Bonding.
6
Amine
4. CHEMICAL PROPERTIES : Alkylation : Amines undergo alkylation on reaction with alkyl halides
RNH2
R1 X HX
RNHR1
R2 X HX
R3 X
R -N -R
2
HX
R2
R1 | N R2 | R3
X¯
(quartenary amm. salts.) Acylation : 1º and 2º amines react with acetyl chloride or acetic anhydride to form acetyl derivatives. R-NH2 + CH3COCl RNHCOCH3 + HCl R -NH2 + (CH3CO)2O RNHCOCH3 + CH3COOH (CH3)2N-COCH3 + HCl (CH3)2NH + CH3COCl Note : (a) Tertiary amines donot undergo this reaction because of absence of replacable H- atom. (b) When Benzoyl chloride is used in place of acet yl chloride react ion is called ‘Schotten - Baumann’ reaction.
Benzoylation : WhenAmines react with benzoyl chloride (C6H5COCl). CH 3 NH 2 C6 H5COCl CH 3 NHCOC6 H5 HCl Methanamine
NMethylbenzamide
Benzoyl chloride
Heat Carbylamine reaction : R - NH 2 CHCl 2 3KOH R - NC 3KCl 3H 2 O
Cl R R — NH2 CCl2
Mechanism :
HN C Cl
H -HCl R — N C R
N
+
C
Cl
7
Amine
Reaction with nitrous acid : -
NaOH2 HCl R - NH 2 HNO 2 [R - N 2 Cl ]
ROH N 2 HCl
-
NaNO2 2HCl C6 H5 - NH 2 C6 H5 - N 2 Cl NaCl 2H 2O Ani ln e
Benzenedia monium Chloride
Reaction with arylsulphonyl chloride : Benzenesulphonyl chloride (C6 H5 SO2 Cl), which is also known as Hinsberg’s reagent, reacts with primary and secondary amines to form sulphonamides. (i) The reaction of benzenesulphonyl chloride with primary amine yields N-ethylbenzenesulphonyl amide.
The hydrogen attached to nitrogen in sulphonamide is strongly acidic due to the presence of strong electron withdrawing sulphonyl group. Hence, it is soluble in alkali. (ii) In the reaction with secondary amine, N,N-diethylbenzenesulphonamide is formed. It is in soluble alkali.
(iii) ter.Amine does not react with Hinsberg’s reagent it is present above solution. Electrophilic substitution : Due to +M effect of -NH 2 genrate electron dencity at ortho and para position hence, aniline active toward Electrophilic substitution (i) Bromination: Aniline reacts with bromine water at room temperature to give a white precipitate of 2,4,6-tribromoaniline.
If we have to prepare monosubstituted aniline derivative, This can be done by protecting the -NH 2 group by acetylation with acetic anhydride, then carrying out the desired substitution followed by hydrolysis of the substituted amide to the substituted amine.
8
Amine
The lone pair of electrons on nitrogen of acetanilide interacts with oxygen atom due to resonance as shown below:
Hence, the lone pair of electrons on nitrogen is less available for donation to benzene ring by resonance. Therefore, activating effect of -NHCOCH 3 group is less than that of amino group. (ii) Nitration: Direct nitration of aniline is not possible because in the strongly acidic medium, aniline is protonated to form the anilinium ion which is meta directing.
However, by protecting the -NH 2 group by acetylation reaction with acetic anhydride, the nitration reaction can be controlled and the p-nitro derivative can be obtained as the major product.
(iii) Sulphonation:
s Ex.4
Consider the following reaction,
H / H2O C 6 H 5 NH 2 CHCl 3 KOH ( A) (B) (C )
Find the compounds (B) and (C). Sol.
C6 Hl5NH 2 CHCl3 3KOH C6 H5 N C 3Kl 3H 2 O (A)
C6 H5 N C 2H 2 O H C6 H5 NH2 HCOOH (A)
(B)
(C )
9
Amine
Ex.5 Sol.
Sulphanilic acid is insoluble in water and acid but soluble in caustic alkali. Comment. Sulphanilic acid exist as a Zwitter ion and exhibits strong dipole-dipole interactions. Therefore, it is insoluble in water. on adding acid, SO 3- fails to accept H + ion, thus sulphanilic acid is insoluble in acid. However, + when alkali is added, strongly basic hydroxyl ion can abstract a proton from —NH 3 to form soluble salt.
NH2
+
NH3
+ H2 O
+ NaOH SO3Na
SO3-
Sodium p-aminobenzene sulphonate
Zwitter ion structure of sulphanilic acid
Ex.6
How will you carry out the following conversions? O
(a)
C
NH2
NH
NH2
NH2
(b)
CH3
O C Sol.
(a)
NC
NH2 CHCl3 KOH
KOH NH2 Br2 /
NH LIAlH4
CH3
-
NH NaNO 2
(b)
2
H 2SO
OH
HCl
BaO COOH COOH
K Mn 4O / H
4
NH O
t NH H2 / P
3
NH 2
5. DIAZONIUM SALTS :
-
-
The diazonium salts have the general formula R N 2 X where R stands for an aryl group and X ion may be Cl -, Br-, HSO 4-, BF -3, etc. Resonance of benzene diazonium ion is
10
Amine
Method of Preparation of Diazonium Salts
-
273-278K C6 H5 NH 2 NaNO2 2HCl C6 H5 N 2 Cl NaCl 2H 2O
Chemical Reactions The reactions of diazonium salts can be broadly divided into two categories, namely (A) reactions involving displacement of nitrogen and (B) reactions involving retention of diazo group. (i) Reactions involving displacement of nitrogen Diazonium group being a very good leaving group, is substituted by other groups such as Cl-, Br- , I-, CN- and OH- which displace nitrogen from the aromatic ring. (a) Replacement by halide or cyanide ion: The Cl-, Br- and CN- nucleophiles can easily be introduced at the benzene ring in the presence of Cu(I) ion. This reaction is called Sandmeyer reaction.
Alternatively, chlorine or bromine can also be introduced byGatterman reaction.
(b) (c)
Replacement by iodide ion: Replacement byfluoride ion:
-
-
Ar N 2 Cl KI ArI KCl N 2
-
Ar N 2 Cl HBF4 Ar - N 2 BF4 Ar - F BF2 N 2
(d)
-
Replacement by H: Ar N 2 Cl H 3PO2 ArH N 2 H3PO3 HCl
-
Ar N 2 Cl CH 3CH 2OH ArH N 2 CH 3CHO HCl (e)
Replacement by hydroxyl group:
(f)
Replacement by –NO 2 group:
-
Ar N 2 Cl H 2O ArOH N 2 HCl
(ii) Reactions involving retention of diazo group coupling reactions
11
Amine
Ex.7
Provide structures for the products of the reaction of PhN 2with (a) PhNMe2 (b) 2-naphthol (c) PhCH3.
Sol.
ArN 2 is a weak electrophile that undergoes diazo couplignonly with rings activated by OH, NH2 , NHR or NR 2 . Ph
(a)
N
N
NMe2
N,N-dimethylaminoazobenzene (butter yellow
N
N
Ph OH
(b) 1-Phenylazo-2-naphthol
(c) Ex.8
No reaction. The substrate ring is insufficiently activated.
A weakly acidic medium is provided for coupling of benzene diazonium chloride with aniline. Comment. +
-
N2Cl
Sol.
+H
NH2
Benzene diazonium chloride
pH 4-5 273-278 K
N=N
-HCl
Aniline
NH2
p-Aminoazobenzene
If the high conc. of H+ ions are used during these reactions, then protonation of aniline takes place. H+
NH2
+
H3 N Anilinium ion
Aniline
The positive charge on protonated amine exerts -I effect, thus coupling of amine with diazonium salt is not favoured, at low pH (or high acid strength). Also, we know high pH is not desirable for coupling reactions. Therefore, optimum pH for coupling reactions with amines is 4-5. Ex.9
The end product (Z) of the following reaction is ? N2Cl
Sol.
Cu / KCN
N2Cl Cu / KCN
(X)
H / H 2O
(Y)
NaOH
(Z)
CN
12
Amine
SOL VED EXAMPLE Ex.1
Account for the fact that 2-aminoethanoic acid (glycine) exists as a dipolar ion, as does paminobenzenesulfonic acid (sulfanilic acid) but p-aminobenzoic acid does not.
Sol.
The aliphatic NH 2 is sufficiently basic to accept an H from COOH. the COOH is not strong enough to donate H to the weakly basic ArNH2 , but SO3H is a sufficiently strong acid to do so. H3 N — CH 2COO
-
p — H 2N — C 6H 4 — COOH
p — H 3N — C6 H4 — SO-3
pA min obenzoic acid
Sulfanilic acid
Glycine
Ex.2
Sol.
Mixture oftwo aromatic compounds (A) and (B) was separated by dissolving in chloroform followed by extraction with aqueous KOH solution. The organic layer containing compound (A), when heated with alcoholic solution of KOH produced a compound (C) (C7H5N) associated with an unpleasant odour. The alkaline aqueous on the other hand, when heated with chloroform and then acidified give a mixture of two isomeric compounds (D) and (E) of molecular formula C7H6O2. Identify the compounds (A), (B), (C), (D), (E) and write their structures. (a) One of the compound A and B is soluble in aq.KOH and thus it must be acidic in nature whereas another compound is soluble in chloroform is either basic or neutral in nature. (b) The compound (A) on heating with alcoholic KOH solution (chloroform already present) produces compound (C) (C7H5N) having unpleasant odour. The compound (C) is Phenylisocyanide and, therefore, compound (A) must be aniline which is soluble in chloroform but insoluble in aq.KOH. (c) The compound (B) must be phenol as it is soluble in aqueous KOH and produces isomers ohydroxybenzaldehyde (D) and p-hydroxylbenzaldehyde (E) on heating, with chloroform followed by acidification. The reactions are as follows : NH2
N C
+ CHCl3 + 3KOH
Warm
+ 3KCl + 3H3O (carbylamine reaction)
Aniline (A) (Soluble in chloroform)
Phenyl isocyanide
(C)
OH
OH
OH CHO
CHCl3/KOH
+
H+ Phenol (B) (Soluble in KOH)
Salicylaldehyde
(D)
CHO p-Hydroxybenzaldehyde
(E)
Ex.3
Aneutral compound (A) C 8H 9ON on treatment with sodium hypobromite forms an acid soluble substance (B), C7H9N. On addition of aqueous sodium nitrite to a solution of (B) in dil. HCl at 0 -5°C, an ionic compound (C). C7H7N2Cl is obtained. (C) gives a red dye with alk -naphthol solution. When treated with potassium cuprocyanide, (C) yields a neutral substance (D), C8H7N. ON hydrolysis (D) gives (E), C8H8O2. (E) liberates CO2 from aqueous sodium bircarbonate. (E) on permanganate oxidation furnishes (F), C8H6O4 , (F) on nitration yields two isomeric mononitro derivatives (G) and (H) having molecular formula, C 8H NO . Write the reaction involved in different steps. 5 6
13
Amine CONH2 (A) :
(C) :
(B) : CH3
Sol.
COOH
(D) :
CH3
CH3
COOH
COOH
(F) :
(E) :
CH3 COOH (H) :
(G) :
COOH O2N
COOH
CH3
CN
N2Cl
NH2
COOH
NO2
Ex.4
An optically active amine (A) is subjected to exhaustive methylation and Hoffman elimination to yield on alkene (B). (B) on ozonolysis gives an equimolar mixture of formaldehyde and butanol. Deduce the structures of (A) and (B). Is three any structural isomer of (A), if yes draw its structure? (A) : H3C
CH CH2CH2CH3
(B) : H2C
Sol.
CHCH2CH2CH3
NH2 NH 2 CH 2 CH 2 CH 2 CH 2 CH 3 is positional isomer of (A). Ex.5
A mixture of two aromatic compounds (A) and (B) was separated by dissolving in chloroform followed by extraction with aqueous KOH solution. Theorganic layer containing compound (A), when heated with alcoholic solution of KOH produced a compound (C) (C 7H 5N) associated with an unpleasant odour. The alkaline aqueous layer on the other hand, when heated with chloroform and then acidified give a mixture of two isomeric compounds (D) and (E) of molecular formula C7H 6O 2. Identify the compounds (A), (B), (C), (D) and (E). NH2
OH
N=C
OH
OH CHO
Sol.
(A) :
(E) :
(B) :
CHO Ex.6
Compound (A), C14H10N2O whenheated with dilute sulphuric acid gives ammonium sulphate, compound (B), C8H6O 4 and compound (C), C 6H 7N (as its sulphate). Compound (B) on heating gives (D), C8H4O3 and on heating with sodalime gives benzene. Compound (C) with dilute H2SO 4/NaNO 2 in the cold followed by heating with water gives (E), C 6H6O. Compound (E) on heating with Zn dust again gives benzene. Identify (A), (B), (C) and (e) giving proper reactions? CN
CO2H C6 H5 NH 2
dil. H2SO4 (NH 4 ) 2 SO 4
C
Sol.
(B)
NH (B)
O (A)
O
CO2H
C
- H 2O CO2H
C
CO2H
(C) H 2SO4
(C6 H5 NH3 ) 2 SO 4 O
(D)
O
14
Amine
Ex.7
Sol.
An organic compound (A) composed of C, H and O gives a characteristic colour with ceric ammonium nitrate. Treatment of (A) with PCl5 gives (B) which reacts with KCN to form (C). The reduction of (C) with Na and C2H5OH produces (D) which on heating gives (E) with evolution of NH3. Pyridine is obtained on treatment of (E) with nitrobenzene. Give the structures of (A) to (E) with proper reasoning. (A) is an alcohol which gives characteristic red colour with ceric ammonium nitrate. PCl5
ROH (A)
RCl KCN (B)
Na+ C2H5OH
R -C N (C)
R CH2NH2 (D)
-NH3
E
Nitrobenzene
Start with propane-1, 3-diol CH2OH CH2
PCl5
CH2OH (A)
CH2
CH2Cl
CH2
2KCN
CH2Cl (B)
CH2CH2NH2
CH2CH2NH2
-NH3
CH2CN
CH2
Na + C2H5OH
CH2CN (C)
Nitrobenzene -6H
N
(D)
H Piperidine (Hexahydropyridine) (E)
Ex.8 Sol.
Glycine exists as a Zwitter ion but anthranilic acid does not. Comment. -COOH group of glycine releases H+ ion which is accepted by -NH2 group. Thus glycine exist of a Zwitter ion. H
H +
H2N - C - COOH
H3N - C - COOH
H Glycine
Zwitter ion of glycine
NH2 COOH
In anthranilic acid, Anthranilic acid (2- Aminobenzoic acid)
Ex.9
Electron withdrawing -COOH and phenyl group reduces electron density of N of –NH 2 group, therefore, -NH2 fails to accept a proton. Thus anthranilic acid can not form Zwitter ion. Write reactions of the final alkylation product of aniline with excess of methyl iodide in the presence of sodium carbonate Sol.
Sol.
H6H5NH2 + CH3I Aniline
Methyliodide
+
[C6H5NHCH3]IN-Methylanilinium iodide
2[C6H5 NH CH3]I- + Na2CO3 2C6 H 5 NHCH3 CO 2 2NaI NMethylaniline
15
Amine
2CH I3 2C6H5 NH CH3 2C6H5N(CH3)2 + CO2 + 2NaI Na 2CO 3
[C6H5NH(CH3)] +I -
C6H5NHCH3 + CH3I
N.N.N-Trimethyl anilinium ioidide
2[2C6H5 N (CH3)3] I- + Na2CO3 [C6 H5 N(CH3 )3 ]2 CO3 2NaI N, N, NTrimethylanilinium Carbonate
Ex.10 What happens when (a) Aniline is treated with sodium nitrate and dil. HCl at 273 K, (b) Aniline is treated with Benzaldehyde, (c) Ethylamine reacts withAgCl, (d) Amixture of alcohol and ammonia is passed over heated aluminium oxide as catalyst. (e) Sodium nitrite is added to a solution of ethylamine in dil. HCl at 273 K ? Sol. (a) Benzenediazonium chloride is formed. N NCl NH2 NaNO2 + HCl 273 K Aniline
(b)
Benzenediazonium chloride
Benzal aniline is formed. H NH2
O
H
C
H+
Benzaldehyde
(c)
N
C
Benzalaniline Schiff's base
Soluble complex-Bis-(ethylamine) silver (I) chloride is formed AgCl(s) + 2C2H5NH2(aq) [Ag(C2H5NH2)2]Cl(aq) Bis-(ethylamine) sliver (I) chloride (soluble)
(d)
Primary amine is formed. If ammonia is taken in excess, a mixture of 1°, 2°, 3° amines and quaternary ammonium salt is formed. Al O
2 3 R — NH2 + H2 O ROH + NH3 723K
1° alcohol
(e)
1° alcohol
Ethyl alcohol is formed NaNO2 + HCl NaCl + HNO2 Nitrous acid
C2H5NH2 + HNO2 C2H5OH + N2 + H2O Ex.11 Give one chemical test to distinguish between the following pairs of compounds : (i) Methylamine and dimethylamine (ii) Secondary and tertiary amines Ethylamine and aniline (iv)Aniline and benzylamine (iii) Aniline and N-methyl aniline (v)
16
Amine
Sol.
(i) Methyl amine, being a primary amine will give carbylamine reaction (offensive smell of isocyanide when treated with CHCl3 + KOH) while dimethyl amine will not show any reaction with CHCl3 + KOH (alc.) CH3NH2 + CHCl3 + 3KOH (alc) CH 3 NC 3KCl 3H 2 O Methyl isocyanide (offensive smell)
(ii) When secondary amine is treated with HNO2(NaNO2 + HCl), if forms yellow coloured oily compound N-nitrosoamine, which on warming with a crystal of phenol and H2SO4 gives a green solution and on further addition of aq KOH red solution(Libermann nitroso test). This test is not given by tertiaryanines. (CH3)2 NH + HO—N
(CH3)2 N—N
O
Secondary amine Nitric acid
O + H2O
N-Nitrosodimethylamine Phenol + H2SO4 KOH + Water
Green colour Red colour (iii) When aniline is treated by NaNO2 + HCl at 273 K, benzenediazonium chloride is formed which on treatment with b-naphthol gives a bright orange dye, 1-phenylazo-2-naphthol.
NH2
NaNO2/HCl 273 K
N
N
Cl OH
OH
N
N
(iv) Benzyl amine on treatment with NaNO2 and HCl forms benzyl alcohol with the evolution of N2 gas while aniline froms benzene drazonium chloride which gives organe dye with alkaline b-naphthol.
(v)Aniline, being a primary amine, gives carbylamine reaction while N-methyl aniline (sec. amine) will not give this rest. NH2 + CHCl3 + 3KOH(alc)
NC + 3KCl + 3H2O Phenyl isocyanide (offensive smell)
17
Amine
EXERCISE-I Q.1
Arrange the following : (i) In decreasing order of the pK values: b C2H5NH2, C6H5NHCH3 , (C2 H5 )2 NH and C6 H5 NH2 (ii) In increasing order of basic strength: C6H5NH2, C6H5N(CH3)2, (C2H5)2NH and CH 3NH 2 (iii) In increasing order of basic strength: (a)Aniline, p-nitroaniline and p-toluidine (b) C6H5NH2, C 6H 5NHCH 3 , C6H5CH2NH2. (iv) In decreasing order of basic strength in gas phase: C2H5NH2, (C2H5) 2NH, (C 2H 5) 3N and NH 3 (v) In increasing order of boiling point: C2H5OH, (CH3) 2NH, C 2H 5NH 2 (vi) In increasing order of solubility in water: C6H5NH2, (C2H5) 2NH, C 2H 5NH 2.
Q.2
Describe a method for the identification of primary, secondary and tertiary amines.Also write chemical equations of the reactions involved.
Q.3
Explain Hofmann Bromanide reaction with Mechanism ?
Q.4
Why cannot aromatic primary amines be prepared by Gabriel phthalimidesynthesis?
Q.5
Write the reactions of (i) aromatic and (ii) aliphatic primary amines with nitrous acid.
Q.6
Write one chemical reaction each to illustrate the following (i) Hofmann Bromanide reaction. (ii) Garbriel Phthalimide reaction.
Q.7
Assign a reason for the following statements (a)Alkylamines are stronger bases than arylamines. (b) How would you convert methylamine into ethylamine ?
Q.8
Illustrate the following with an example of reaction in each case : (i) Sandmeyer reaction (ii) Coupling reaction
Q.9
Write the chemical reaction equations for one example each of the following (a)Acoupling reaction (b) Hofmann's bromamide reaction (c)Acetylation
Q.10
Account for the following : (i)Aniline is weaker base than methylamine. (ii)Aryl cyanides cannot be formed by the reaction of aryl halides and sodium cyanide.
Q.11
Describe tests to distinguish between : Secondary amine and tertiary amine.
18
Amine
Q.12
Account for the following observations : (i) pKb for aniline is more than that for methylamine. (ii) Methylamine solution in water reacts with ferric chloride solution to give a precipitate of ferric hydroxide. (iii)Aniline does not undergo Friedel Crafts reaction.
Q.13
State the reactions and reaction conditions for the following conversions (i) Benzene diazonium chloride to nitrobenzene. (ii)Aniline to benzene diazonium chloride. (iii) Ethyl amide to methylamine.
Q.14
Write the physical property of aniline ?
Q.15
Write the method of formation of benzene diazonium chloride ?
Q.16
Account for the following: (i) pKb of aniline is more than that of methylamine. (ii) Ethylamine is soluble in water whereas aniline is not. (iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide. (iv)Although amino group is o- and p- directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline. (v)Aniline does not undergo Friedel-Crafts reaction. (vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines. (vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines.
Q.17
How willyou convert: (i) Ethanoic acid into methanamine (iii) Methanol to ethanoic acid (v) Ethanoic acid into propanoic acid (vii) Nitromethane into dimethylamine
Q.18
Q.19
Q.20 Q.21
Write short notes on the following: (i) Carbylamine reaction (iii) Hofmann’s bromamide reaction (v)Ammonolysis (vii) Gabriel phthalimide synthesis.
(ii) Hexanenitrile into 1-aminopentane (iv) Ethanamine into methanamine (vi) Methanamine into ethanamine (viii) Propanoic acid into ethanoic acid ? (ii) Diazotisation (iv) Coupling reaction (vi)Acetylation
Accomplish the following conversions (ii) Benzene to m-bromophenol (i) Nitrobenzene to benzoic acid (iv)Aniline to 2,4,6-tribromofluorobenzene (iii) Benzoic acid to aniline (vi) Chlorobenzene to p-chloroaniline (v) Benzyl chloride to 2-phenylethanamine (viii) Benzamide to toluene (vii)Aniline to p-bromoaniline (ix)Aniline to benzyl alcohol. Write the equation of curtius reaction with mechanism ? Complete the following reactions: (i) C6H5NH2 CHCl3 alc.KOH (iii) C 6H5NH2 H2SO4 (conc.) (v) C6 H5NH2 + Br2(aq)
(ii) C6H5N2Cl H3PO2 H 2O (iv) C6H5N2Cl C2H5OH (vi) C6H5NH2 + (CH3CO) 2O
( i)H B F4 (vii) C6H5N2 Cl (ii)NaNO / Cu. 2
19
Amine
Q.22
Give plausible explanation for each ofthe following: (i) Whyare amines less acidic than alcohols of comparable molecular masses? (ii) Whydo primaryamines have higher boiling point than tertiaryamines? (iii) Whyare aliphatic amines stronger bases than aromatic amine?
Q.23
Write the reaction and conditions for the following conversions (i)Aniline to benzene (ii) Methylamine to methyl cyanide (iii) Propanenitrile to ethylamine (iv) m-Bromoaniline to m- bromophenol (v) Nitrobenzene to 2, 4, 6-tribromoaniline.
Q.24
Write the method of formation of zwitter ion ?
Q.25
Explain nitration of aniline ?
Q.26
Why aniline do not give fridel craft reaction ?
Q.27
What is Gabriel phthalimide synthesis ? For what purpose is it used ? Give equation only to explainyour answer.
Q.28
Write the equation of carbyl amine reaction with mechanism ?
Q.29
How will you convert 4-nitrotoluene to 2-bromobenzoic acid ?
Q.30
Draw the structure of trimethylamine and tell the shape of the molecule. Show the angle between two methyl groups.
20
Amine
EXERCISE-II Q.1
Aspartame, an artificial sweetener, is a peptide and has the following structures : NH2
CH2C6H5
HOOC - CH2CH - CONH - CH - COOCH3
(a) (b) (c) (d)
Identify the four functional groups. Write the zwitterionic structure Write the structures of the amino acids obtained from the hydrolysis of aspartame. Which of the two amino acids is more hydrophobic ?
Q.2
Compound ofA(molecular formula C9H11NO) gives a positive Tollen’s test and is soluble in dilute HCl. It gives no reaction with benzene sulphonyl chloride or with NaNO2 and HCl at 0ºC. (A), upon oxidation with KMnO4 gives an acid (B). When (B) is heated with soda-lime, compound (C) is formed which reacts with NaNO2 and HCl at 0 - 5ºC. What is (A) ?
Q.3
An organic compoundA, when treated with nitrous acid yields an alcohol B, C4H10O with the evolution of N2. B on careful oxidation yields a substance C of vapour density 36 which forms oxime; B can react with NaHSO3 but does not reduce Fehling solution. Identify compoundAand write the structural formulae of the isomeric compounds that behave with HNO2 in the same manner.
Q.4
An organic compound (A), C6H4N2O4, is insoluble in both dilute acid and base and its dipole, moment is zero. Deduce the structure of (A).
Q.5
Explain the following observations : (1) Aniline dissolves in aqueous HCl. (2) The amino group in ethylamine is basic whereas that in acetamide it is not basic. (3) Dimethylamine is a stronger base than trimethylamine. (4) Sulphanilic acid although has acidic as well as basic group, it is soluble in alkali but insoluble in mineral acids.
Q.6
Explain, why ? Glycine exists as H3N+CH2COO- while anthranilic acid, p-NH2.C6H4.COOH does not exist (1) as dipolar ion. Benzenesulphonic acid is a stronger acid than benzoic acid. (2) Aweakly basic solution favours coupling with phenol. (3) It is difficult to prepare pure amines by ammonolysis of alkyl halides. (4)
Q.7
Explain with reason ? (1) Although trimethylamine and n-propylamine have same molecular weight, the former boils at a lower temperature (3ºC) than the latter (49ºC). (2) Dimethylamine is a stronger base than methylamine but trimethylamine is a weaker base than both dimethylamine and methylamine. (3) Although trimethylamine and n-propylamine have same molkecular weight, the former boils at a lower temperature (3ºC) than the latter (49ºC). Explain. Silver chloride dissolves in aqueous solution of methylamine. Explain. (4)
21
Amine
Q.8
Explain it ? (1) An aqueous solution of ethylamine gives a red precipitate with ferric chloride. Explain. Tertiary amines do not undergo acetylation. Comment (2) (3) 2, 6-Dimethyl -N N-dimethylaniline, although has a free p-position, does not undergo coupling with benezenediazonium chloride. Comment. In the following compounds : (4) O N
N
= (III)
H (IV)
N N (II)
H (I)
The order of basicity is I > III > II > IV. Explain. Q.9
Q.10
Explain it with reason ? tert-Butylamine cannot be prepared by the action of NH3 on tert-butyl bromide. (1) Isocyanides are hydrolysed by dilute acids but not by alkalies to form amine and formic acid. (2) How will you explain the acidic nature of 1º and 2º nitroalkanes ? (3) Aniline does not undergo Friedal Craft’s reaction ? (4) Although borontrifluoride adds on trimethylamine, it does not add on triphenylamine. (5) Comment. Complete the following reactions : (1) (2) (3)
PCl5
C6H5COOH
P2O5
CONH2
KOH
2, 4-Dinitroaniline
(6)
C6H6
Oleum
H2/Hi
[E]
KBr + H I
(ii) anisole
SO3H
C6H5CN
G
(i) NaNO2/HCl, 5ºC
NaOH
(K)
P2O5
[D]
H+
— N(CH3)2 + HNO2
(5)
(L)
(J) NaOH heat
(M)
OH 1
(7)
CHCl3/NaOH
— SO3H fuming
O
H SO 2
Et2SO4 NaOH
(9)
Phenol
(10)
CH3CONHC6H5
(11)
F
EtNH2 + KCN + Br2
(4)
(8)
NH3
[C]
C6H5N2Cl
(A) HCl
4
(Q)
(i) NaOH fuse (ii) H+
HCN, HCl AlCl3
Br2, Fe
N
(R)
P
PhNH.NH2
(S)
T+U
(V) Gattermann reaction
22
Amine
Q.11
Give structures for the compounds (A) to (I) : C8H11N (A)
NaNO2/HCl 0-5ºC
Heat to m.p.
I
KCN CuCN
B
C Hot H2SO4
Hot aq. KMnO4
G
CH3OH at 30ºC
D Cl2, 2 moles u.v.
E H F
Q.12
When 2.25 g of an unknown amine was treated with nitrous acid, the evolved nitrogen, corrected to S.T.P. measured 560 ml. The alcohol isolated from the reaction mixture gave a positive iodoform reaction. What is the structural formula of the unknown amine ?
Q.13
The aqueous solution of a nitrogen and chlorine containing organic compound (A) is acidic towards litmus. (A) on treatment with aqueous NaOH gives a compound (B), containing nitrogen, but not chlorine. Compound (B) on treatment with C6 H 5SO 2Cl in the presence of NaOH gives an insoluble product (C), C13H13 NO 2S. give the structures of compounds (A) and (B).
Q.14
An organic compound (A) composed of C, H and O gives characteristic colour with ceric ammonium nitrate. Treatment of (A) with PCl5 gives (B), which reacts with KCN to form (C). the reduction of (C) with warm Na/ C 2 H5OH produces (D), which on heating gives (E) with evolution of ammonia Pyridine is obtained on treatment of (E) with nitrobenzene. Give structure of compounds (A) to (E) with proper reasoning.
Q.15. One mole of bromoderivative (A) and mole of NH 3 react to give one mole of an organic compound (B). (B) reacts with CH 3I to give (C). Both (B) and (C) react with HNO 2 to give compounds, (D) and (E) respectively. (D) on oxidation and subsequent decarboxylation gives 2-methoxy-2-methyl propane. Give structures of compounds (A) to (E) with proper reasoning. Q.16
Q.17 Q.18 Q.19 Q.20 Q.21
What happens with cyclopentanoen reacts with (a)
CH3CH2 NH2 (1° amine)
(b)
(CH3CH 2 )2 NH (2° amine)
Cyclohexylamineis a stronger base than aniline. Why? How does the formation of 2° and 3° amines can be avoided during the preparation of 1° amines by alkylation? It is necessary to acetylate aniline first for preparing bromoaniline. Why? Dimethylamienis a stronger base than methylamine but trimethylamine is a weaker base than both dimethyl amine and methylamine. Why? From analysis and molecular weight determination, the molecular formula of (A) is C3H 7 NO . the compound gave following reactions. (i) On hydrolysis it gives an amine (B) and a carboxylic acid (C) (ii)Amine (B) reacts with benzene sulphonjyl chloride and gives a product which is insoluble in aqueous sodium hydroxide solution. (iii)Acid (C) on reaction with Tollen’s reagent gives a silver mirror when areA, B and C. Explain the reactions.
23
Amine
Q.22
An optically active amine (A) is subjected to exhaustive methylation and Hofmann elimination to yield an alkene (B). (B) on ozonolysis gives an equimolar mixture of formaldehyde and butanal. Deduce the structures of (A) and (B). Is there any structural isomer to (A), if yes draw its structure.
Q.23
An aromatic compound (a) having molecular formula C7 H 7 NO 2 dissolves in NaHCO3 to evolve CO2 and when reacted with NaNO 2 / HCl forms (b), C7 H 6O3 . (B) dissolves in NaHCO3 and gives colour reaction with FeCl3 and can be prepared by the action of CCl 4 and NaOH on phenol. When (B) is reacted with excess HNO3 , it forms (C), C6 H 3 N 3O 7 . (C) undergoes acetylation and decomposes NaHCO3 to evolve CO 2 . On reaction with PCl5 . (C) is converted to (D), C6 H 5 N 3O6 Cl which when reacted with water gives back (C). Identify compounds (A) to (D).
Q.24 Compound (A) having M.F. C8 H8O on treatment with NH 2OH.HCl gives (B) and (C). (B) and (C) rearrange to give (D) and (E), respectively on treatment with acid. Compounds (B), (C), (D) and (E)are all isomers of molecular formula C8 H 9 NO . When (D) is boiled with alcoholic KOH, an oil (F) C6 H 7 N separated out. (F) reacts rapidly with CH 3COCl togive back (D). On the other hand, (E) on boiling with alkali followed by acidification gives a white solid (G), C7 H 6O 2 . Identify the compounds (A) to (G). Q.25 An aromatic compound (A), having M.F C7 H5 NO 2Cl2 on reduction with Sn/HCl gives (B), which on reaction with NaNO 2 / HCl gives (C). Compound (B) is unable to forma dye with -naphthol. However, (C) gives red colour with ceric ammonium nitrate and on oxidation gives an acid (D), having equivalent weight 191. Decarboxylation of (D) gives (e) which forms a single mononitro derivative (F), on nitration. Give the structures of (A) to (F) with proper reasoning. Q.26 An organic compound (A) of molecular weight 135, on boiling with NaOH evolves a gas which gives white denso fumes on bringing a rod dipped in Hcl near it. The alkaline solution thus obtained on acidification gives the precipitate of a compound (B) having molecular weight 136. Treatment of(A)with HNO 2 also yields (B), whereas it treatment with Br2 / KOH gives (C). Compound (C) reacts with cold HNO 2 to gives (D), which give red colour with ceric ammonium nitrate. On the other hand, (E) an isomer of (A) on boiling with dilute HCl gives an acid (F), having molecular weight 136. On oxidation followed by heating, (F) gives an anhydride (G), which condenses with benzene in the presence of anhydrous AlCl3 to give anthraquinone. Give the structures of (A) to (G) with proper reasoning. Q.27 An organic compound (A) having M.F C7 H 9 N on treatment with NaNO 2 and HClat room temperature forms another compound (B), C7 H 8O . When (A) or (B) is treated with bromine water, they form dibromo derivatives, When (A) is reacted with chloroform and alkali, it forms (C) having the molecular formula C8 H 7 N . Hydrolysis of (C) followed by reaction with NaNO2 and HCl at low temperature and subsequent reaction with HCN in the presence of Cu(D), which is isomeric to (C). (D) on hydrolysis followed by oxidation gives a dibasic acid, which on halogenation forms only one monohalo derivative. Identify the compounds (A) to (E).
24
Amine
Q.28
An optically active compound (A), C3H7O2 N forms a hydrochloride but dissolves in water to give a neutral solution. On heating with soda lime (A) yields (B) C 2 H 7 N . Both (A) react with NaNO2 and HCl, the former yielding a compound (C) C3H 6O , which on heating is converted to (D), C6 H 8O 4 while the latter yields (E), C2 H 6O .Account for the above reactions and suggest how (A) may be synthesized.
Q.29
An optically inactive acid (A), C5H 8O5 , on being heated lost CO 2 to give an acid (B), C 4 H 8O 3 capable of being resolved. On action of sulphuric acid, B gave an acid C whose ethyl ester gave (D) on the action of hydrogen and platinum. (D) with conc. NH 3 gave E, C 4 H9OH which with Br2 and KOH solution gave (F), C3H9 N . F with HNO 2 gave G. (G) on mild oxidation gave H. BothAand H gave the iodoform reaction. Elucideate the reaction mechanism and suggest a synthesis of (C).
Q.30
Aneutral compound (A) C8 H 9OH on treatment with NaOBr forms an acid soluble substance C7 H 9 N. On addition of aqueous NaNO2 to a solution of B in dilute HCl at 0-5°C an ionic compound (C) C7 H7 N 2Cl is obtained. (C) gives a red dye with alkaline -napththol solution. When treated with potassium cuprocyanide (C) yields a neutral substance (D) C8 H7 N . ON hydrolysis (D) gives E (C8 H8O 2 ) . E liberates CO 2 from aqueous NaHCO3 . (E) on permanganate oxidation furnishes (E) C8 H 6O 4 . (F) onnitration yields two isomeric mononitro derivatives (G and H) having molecular formula C8 H5 NO 6 . Write the reactions involved in different steps.
25
Amine
EXERCISE-II Q.1
When aniline is treated with fuming sulphuric acid at 475K, it gives (a) Sulphanilic acid (b)Aniline sulphate (c) o-aminobenzenesulphonic acid (d) m-aminobenzenesulphonic acid.
Q.2
When nitrobenzene is treated with Br2 in presence of FeBr3, the major product formed is mbromonitrobenzene. Statements which are related to obtain m-isomer are: (a) The electron-density on meta carbon is more than that on ortho and para positions (b) The intermediate carbonium ion formed after initial attack of Br+ at the meta position is least destablilzed (c) Loss of aromaticity, when Br+ attacks at the ortho and para positions, and not at meta position (d) Easier loss of H+ to regain aromaticity from the meta position than from the ortho and para positions.
Q.3
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 the nitrogen has 10 valence electrons (d) II is an acceptable canonical structure. Q.4
The correct order of basic strength of in CCl4 (3) R2NH (1) NH3 (2) RNH2 Where R is CH3 group is (a) 3 > 2 > 1 > 4 (b) 2 > 3 > 4 > 1 (c) 3 > 2 > 4 > 1
(4) R3N (d) None of these
Q.5
Place the following in the decreasing order of basicity. (1) Ethylamine (2) 2-aminoethanol (3) 3-aminopropan-1-ol (a) 1 > 3 > 2 (b) 1 > 2 > 3 (c) 2 > 1 > 3 (d) None of these
Q.6
Which of the following will give a positive carbylamine test ? 3. (CH3)3N 1. H3CNH2 2. H3C-NH-CH3 Select the correct answer using the codes given below. (a) 1 and 3 (b) 2 and 4 (c) 3 and 4
Q.7
4. C6H5NH2 (d) 1 and 4.
Match the compounds in list I with the appropriate test that will be answered by each one of them inlist II from the combinations shown. Selects the correct answer using the codes given below the list . List II List I 1. Reduces Fehling’s solution (A) Propyne 2. Forms a precipitate withAgNO3 in ethanol (B) Ethyl benzoate 3. Insoluble in water, but dissolves in aqueous NaOH upon heating (C) Acetaldehyde (D) Aniline 4. Dissolves in dilute HCl in the cold and is reprecipitated by the addition of alkali (a) A-3, B-2, C-1, D-4 (b) A-2, B-3, C-1, D-4 (c) A-2, B-3, C-4, D-1 (d) A-1, B-3, C-2, D-4
26
Amine
Q.8
In the following 2- reaction sequence R-CH = CH2 + H2SO4
R-CH-CH3 | SO4H
the end product would be
NaOH (Where R = C14 H29)
Q.9 Q.10
R-CH-CH3 | SO4– Na+
useful as (a)Afertilizer (b)An explosive (c)Adetergent The basic strength of amines (ethyl) and ammonia in H2O is (a) NH3 > p > s > t (b) p > s > t > NH3 (c) s > p > t > NH3 Which of the following will have highest Kb value. (a)
(d) None of these NH2
NH2
(b)
(c)
(d)
N
N
(d) None of these
Cl
Me
H
Q.11
Activation of benzene ring by — NH2 in aniline can be reduced by treating with (a) Dilute HCl (b) Ethyl alcohol (c)Acetic acid (d)Acetyl chloride
Q.12
Dipolar ion structure for amino acid is (a) H2N
(b) H3N
CH COOH R
(c) H3N
CH COO R
CH COO
(d) None of these.
R Q.13
- NH2 group shows acidic nature while reacts with reagent.
Q.14 Q.15
(c) Br2 NaOH
(b) CS2
(a) Na
Which of the following does nto give ethylamine on reduction (a) methyl cyanide (b) Ethyl nitrile (c) Nitro ethane
(d) Water (d)Acetamide
In the reaction, excess CH Cl
3 CH3 NH2 (X)
(a) (CH3 )3 N
(Y) (Z) the final product (Z) is
(b) (CH3 )4 N Cl -
(c) (CH3 )4 N OH -
(d) (CH3 )2 NH
Q.16
The product not obtained in the following reaction, CH3 — NO2 Cl 2 NaOH is (a) ClCH 2 NO2 (b) Cl 2CHNO2 (c) Cl3CNO2 (d) CH3 NH 2
Q.17
Asequential reaction may be performed as represented below: SO 2 Cl 2 NH3 R — CH 2 CO 2 H R — CH 2 COCl R — CH 2 CONH 2 (1) ( 2)
R — CH2 NH R — CO2 H R — CH2 OH (3) ( 4) (5) The appropriate reagent for step (3) is (a) NaBr (b) Bromine and alkali (c) HBr
(d) P2 C5
27
Amine
Q.18
Which of the following amine form N-nitroso derivative when treated with NaNO2 and HCl? (a) H3C
NH2 N(CH3)2
(c) Q.19
NH2
(d)
NH(CH3)
The strongest base among the following is H2N
H2N C
(a)
NH
H2N C
(b)
H2N Q.20
(b)
NH2
H2N C
(c)
O
(d)
H2N
H2N
CH OH H2N
Identify compound (A) in the following oxidation reaction. K Cr O , H S O 2 2 7 2 4 O (A)
O OH
NH2 (a)
(b)
OH
(c)
(d)All of these.
NH2
NH2
OH
Q.21
Aniline is a weaker base than ethyl amine because (a) Phenyl gp in aniline is a +R gp (b) Ethyl gp in ethyl amine decreases the electron density on nitrogen atom (c) The lone pair of electron on nitrogen atom in aniline is delocalized over aniline (d)Aniline is less soluble in water than ethylamine
Q.22
Diazonium coupling reaction with aniline should be carried out in (a) Weakly basic medium (b) Weakly acidic medium (c) Strongly basic medium (d) Strongly acidic medium
Q.23
For CH3CHO, CH3NO2, CH3COOH (a)All have same chemical property (c) All are basic Bromine in CS2 reacts with aniline to give
Q.24
NH2
(a)
Br
NH2
(b)
RNC cannot undergo (a)Acidic hydrolysis (c) Base hydrolysis
Br
NH2 Br
(c) Br
Q.25
(b)All have one common chemical behaviour (d) None of these
(d) Both (a) and (b)
Br
(b) Electrophillic, nucleophillic, addition on carbon (d) Both (b) & (c)
28
Amine NH2
+ phosgene
Q.26
X. Here X is
O NH - C - Cl
(a) Q.27 Q.28 Q.29
Q.30
Q.31
(b)
CH - C - H
(c)
Cl
(d) None of these
Ethylamine undergoes oxidation in the presence of KMnO4 to give (a) CH3COOH (b) CH3CH2OH (c) CH3CHO
(d) N-oxide
Baker Mulliken’s test is used to detect the presence (a) -COOH gp (b) -NO2 (c) -OH
(d) -NH2
t-amines with different alkyl gp has a chiral nitrogen atom still it is optically inactive because (a) Chiral N-atoms cannot rotates plane polarized light (b) The lone pair prevents the rotation of plane polarized light (c) Both of these (d) None of these In CH3NO2 we can observe (a) H-bonding (b) -halogenation reaction (c) Tautomerism (d)All of these The reaction: O || C
NH
O || C :N:K + n-BuBr C || O
KOH
C || O
O || C
N Bu-n
1. aq.NaOH 2. H O+
n-BuNH2
3
C || O
Q.32
Cl
N=C=O
(a) Carbylamine reaction (c) Gabriel phthalimide synthesis The conjugate acid of HO(CH2)3 NH2 is +
(a) H2O(CH3)3NH2
is called
+
+
(b) HO(CH2)3NH3
(b) Hofmann reaction (d) Cope reaction . -
(c) O(CH2)3NH2
+
(d) HO(CH2)3NH
Q.33
Consider the following compounds : 2. CH3CH2CH2NH2 3. HC CCH2NH2 1. H2C = CHCH2NH2 The increasing order of basicity is (a) 3 < 1 < 2 (b) 3 < 2 < 1 (c) 2 < 1 < 3 (d) None of these
Q.34
During the conversion of
NH2
with HNO2 at high temperatures the following substances or
entermediates are formed.
1. N2O3 (a) 1 only
2.
N2 Cl
-
(b) 1, 2, 4 only
3.
OH
4. C6H5NH - N = O
(c) 2, 4 only
(d)All of these
29
Amine
Q.35
Match list I (condition of reaction of nitrobenzene) with list II (products formed ) and select thecorrect answer using the codes given below. List I List II Sn and HCl 1. (A) Hydrazobezene Azoxybenzene 2. (B) Zn and NH4Cl Methanolic NaOMe Phenyl hydroxylamine (C) 3. Zn and KOH (D) Aniline 4. (a) A-2, B-1, C-3, D-4 (b) A-4, B-3, C-2, D-1 (c) A-1, B-4, C-2, D-3 (d) A-1, B-3, C-2, D-4
Q.36
The increasing order of basicity of RCN, RCN = NR and RHN2 is (a) RCN < RCH = NR < RH 2 N (b) RNH 2 < RCN < RCH = NR (c) RCH = NR < RNH 2 < RCN
Q.37
(i ) N H
3 (A). What is (A) ? CH3 — CO — COOH ( ii ) H / Pd
2
(a) CH 3CONH 2 Q.38 Q.39
(d) RH 2 N < RCH = NR < RCN.
(b) CH3 — CO — CONH2 (d) CH 3CH(NH 2 )COOH .
(c) CH 3CH 2CONH 2 How many isomeric amines with that formula C7 H9 N contain a benzene ring? (a) two (b) three (c) four (d) five. Isopropylamine can be obtained by LiAlH4 (a) (CH3 )2 CHO NH 2OH ?
H 2 / Ni (b) (CH3 )2 CHO NH3 ?
H3C CHOH NH3
(c)
(d)All of these.
H3C Q.40
Reaction of RCONH 2 with a mixture of Br2 and KOH gives RNH2 as the main product. The intermediate involved in the reaction is O (a) R
C
O
NHBr
(b) R—NHBr
(c) R
C
Br
(d) R—C = N = O.
N Br
Q.41
Amines are highly soluble in: (a)Alcohol (b) Diethyl ether
(c) benzene
(d) Water.
Q.42
Which of the following reagents can convert benzene diazonium chloride into benzene? (a) Water (b)Acid (c) Hypophosphorous acid (d) HCl.
Q.43
The bromination of aniline produces (a) 2-bromoaniline (b) 4-bromoaniline
Q.44
(c) 2,4,6-tribromoaniline (d) 2, 6-dibromoaniline.
The compound, which on reaction with aqueous nitrous acid at low temperature produces an oily nitrosoamine is (a) Methyl amine (b) Ethylamine (c) Diethylamine (d) Triethylamine
30
Amine
Q.45 Carbylamine test is performed in alcoholic KOH by heating a mixture of (a) Chloroform and silver powder (b) Trihalogenated methane and a primary amine (c)An alkyl halie and a primary amine (d)An alkyl cyanide and a primary amine. Comprehension : Arenediazonium salts are more stable than alkanediazonium salts due to dispersal of the positive charge on the benzene ring. Obviously electron donating groups favour diazotisation by retarding the decomposition of diazonium salts to phenyl cation. The high reactively of arenediazonium salts isdue to the excellent leaving ability of the diazo group as N2 gas. Therefore, diazonium salts undergo a number of substitution reactions in which the diazo group is replaced by a monovalent atom/group such as H(by H3 PO2 in presence of Cu+ ions, CH3 CH2OH, NaBH4 etc.), OH (by boiling in presence of mineral acids), OCH3 (by heating with CH3OH) Cl (by CuCl/HCl or Cu/HCl), Br (by CuBr/HBr or Cu/HBr) I (by KI in presence of Cu+ ions), F (by first converting into N 2F 4 followed by heating), CN (by first neutralizing with Na2 CO3 and then reacting with KCN/CuCN), NO2 (by first neutralizing with Na2CO3 and then treating with NaNO2) phenyl or substituted phenyl (by treating with benzene or substituted
Q.46
benzene in presence of NaOH) etc. Diazonium salts also couple with phenols and aromatic amines to form coloured azo dyes. The reactivity of diazonium salts towards coupling reactions is favoured by presence of electron withdrawing groups; the reactivity of 2, 4, 6-trinitrobenzenediazonium chloride is so high that it even couples with reactive hydrocarbons such as mesitylene. Consider the following ions: I.
Me2N
N+ N
II.
O2N
N+ N
III.
CH3O
N+ N
IV.
CH3
N+ N
The reactivity of these ions towards azo coupling reactions under similar conditions is (a) I < IV < II < III (b) I < III < IV < II (c) III < I < II < IV (d) III < I < IV < II Q.47
Which of the following diazonium salts when boiled with dil. H2 SO4 gives the corresponding phenol most readily? OMe
(a)
(c) Me
N
+
N
(b) MeO
N+ N
(d)
N+ N
N+ N
31
Amine
Q.48
Which of the following arylamines undergoes diazotisation most readily? (a) NO2
(c) CH3O Q.49
NH2
NH2
(b) Cl
NH2
(d) CH3
NH2
The product formed when bromobenzene reacts with benzenediazonium chloride in presence of NaOH is (a) Diphenyl (b) p-Bromodiphenyl (c) p, p´-Dibromodiphenyl (d) p-Bromoazobenzene
Benzendiazonium chloride on reaction with phenol in weakly basic medium gives: (a) Diphenyl ether (b) p-Hydroxyazobenzene (c) Chlorobenzene (d) Benzene Assertion and Reason : Each of the questions given below consists of two statements, an assertion (A) and reason (R). Select the number corresponding to the appropriate alternative as follows (a) If both A and R are true and R is the correct explanation of A, then mark (a) (b) If both Aand R are true but R is not the correct explanation of A, then mark (b) (c) If Ais true but R is false, then mark (c) (d) If both Aand R are false, then mark (d) Q.51 A. Benzyl amine is more basic than aniline. R. Positive inductive effect of phenyl group creates high electron density around N atom. Q.50
Q.52
A. R.
White precipitate of silver chloride get dissolved in NH4OH soln. NH3 reacts withAgCl to form a solution complex with formula [Ag(NH3)2]Cl.
Q.53
A: R:
o-nitrophenol is more acidic than p-nitrophenol. Nitro group has +M and -I effect.
Q.54
A: R:
3° amine is proved to be less basic in aq. solution. Conjugate acid of 3° amine is poorly solvated in aq. solution.
Q.55
A: R:
In order to convert R-Cl to pure R-NH , Gabriel-phthalimide synthesis can be used. 2 With proper choice of alkyl halides, phthalmide synthesis can be used to prepare 1°, 2° and 3° amines.
Q.56
A: R:
4-Nitrochlorobenzene undergoes nucleophilic substitution more readily than chlorobenzene. Chlorobenzene undergoes nucleophilic substitution by elimination-addition mechanism while 4-nitrochlorobenzene undergoes nucleophilic substitution byaddition-elimination mechanism.
Q.57
A:
1° amides react with Br2 + NaOH to give 1° amines with one carbon atom less than the parent amide. The reaction occurs through intermediate formation of acylnitrene.
R: Q.58
A: R:
Acetamide reacts with Br 2 in presence of methanoic CH 3ONa to form methyl N-methylcarbonate. Methyl isocyanate is formed as an intermediate which reacts with methanol to form methyl Nmethylcarbamate.
32
Amine
ANSWER KEY EXERCISE - II Q.1
The hydrolysed products are aspartic acid and phenylalanine. CHO
COOH N(CH3)2
Q.2
A=
B=
C=
N(CH3)2
Q.3
N(CH3)2
CH3 C H C 2 H3 , The other isomers should be p-amines only |
NH 2 NO2
Q.4 Q.10
(1) (2)
C = C6H5COCl D = C6H5CONH2 F = C6H5CN G = C6H5COOH
E = C6H5CHNH2
N(CH3)2
(3)
H = EtNHBr
(4)
I= NO
NO2 -N = N -
(5)
J = O2 N
(6)
K = C6H5SO3H
Me
L = C6H5SO3Na
M = C6H5OH
OH
(7)
I = NaOH
CHO
N=
SO3H
(8)
OH
O=
P= SO3 H
OH OEt
(9)
Q = Ph - O - Et
S = EtO
R=
CH = N- NH- Ph
CHO
(10)
T = CH3 - C - NH O
+ para isomer Br Cl
(11)
A = Cu Powder
V=
+ N2
33
Amine NH2 Cl
NH2
Q.11
A
B
Et
COOH
C - CH 3 Cl Cl
(F)
COOH
D
Et
Et COOH
(E)
CN
C
E COOH
COOH
(G)
C - CH 3 O
(H)
COOH
C - OMe O
O C O
(I)
C O
Q.12
C2H5NH2
CH3-NH
CH3 — N H 2 Cl -
Q.13
Q.14
(A)
(B)
CH2 OH
CH2Cl
H2 C
H2 C
H2 C
CH2 OH
CH2Cl
(A)
(B)
Cl
C
CH2CH2Br
H3 C
C
CH2 CH2NH2
C
CH2CH2OH H3C
OCH3 C
CH2CH2NH-CH3
CH3
CH3
(B)
(C)
CH3
CH3
(D)
(E)
CH2
(E)
CH2CH2NH2H3C
C
NH CH2
(D)
OCH3
OCH3 H3C
CH2 CN
CH2
H2 C
OCH3
CH3 (A)
Q.15
H2 C (C)
OCH3
CH2
CH2 CH2NH2
CH2 CN
CH3
CH2 CH2 N
CH3
CH2CH3
CH2CH3 Q.16
a
O CH3 CH2 NH 2
N Tauto mer ise
N H
b
+
O (CH3 CH 2 )2 NH
N H
H
CH2CH3
CH2CH3 H
-
CH2CH3
N CH2CH3
24
Amine
Q.17 Q.18 Q.19 Q.20
Aniline is a weak base than cyclohexylamine because of resonance. Use of excess ammonia reduces chances of reaction of 1° amine with alkyl halide to form 2° and 3° amines. Amino group, being activating group, activates bromination of aniline and forms tribromoaniline. Dimethyl amine is stronger base because of inductive effect. Trimethyl amine is a weaker base because positive charge on nitrogen could be stabilized and due to crowding by alkyl groups around the nitrogen atom protonation cannot take place. CH3 (N, N dim ethyl formamide)
(A) NOCH CH3
Q.21
CH3 (B) HN CH3
C HCOOH NH2 (A) H3C
CH3
Q.22 (B) H2C
CH3
Cl
OH
COOH
COOH
OH O2N
NH2
NO2
O2 N
NO2
Q.23 (A)
(B)
Anthranilic acid
salicylic acid
C 6 H5
HC
C
O
Q.24
NO2 (D)
picric acid
Picryl chloride
OH C
H3C (A)
NO2 (C)
N
C
C6H 5 (B)
Acetophenone
Acetanilide
N
C6H 5
OH (C)
E-Acetophenone oxime
Z-Acetophenone oxime
O
O C6H5 NH C (D)
H3C
CH3
C 6 H5
C
NH CH3
(E) N-methyl benzamide
C6H5 NH2 (F) Aniline
C6H5 COOH (G) Benzoic acid
35
Amine
Cl
Cl CH2NO2
Cl
Cl
Cl
COOH
CH2OH
CH2NH2
Cl
NO2
Q.25 Cl (A)
Cl (B)
Cl (C)
Cl (D)
Cl (E)
Cl (F)
CH2CONH2 CH2COOH CH2NH2 CH2OH
O CONH2
COOH
C
Q.26
O (A)
(C)
(B)
(D)
CH3
(E)
(F)
CH3
(G)
C O
NH2
OH
NC
CN
COOH
CH3
CH3
CH3
CH3
COOH
(A)
(B)
(C)
(D)
Q.27
Q.28
(E)
Degree of unsaturation ofA= 2. Since Aforms 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) C 2 H5OH . Thus, B is CH 3CH 2 — NH 2 . Aalso reacts with NaNO 2 and dilute HCl forming C, a hydroxyl acid, which forms a cyclic diester on heating.All the reactions can be given as H3C
aOH aNO 2 HC l CH COOHN CH3 CH2 - NH2 N CH3 CH2 -OH CaO (B)
( E)
NH2 (A) NaNO2 HCl H3C
CH COOH OH (C)
H3 C2
O
CO
-2 H2 O
CH COOH CH 3HC
CHCH O
OH P / Br
2 CH3CH2 — COOH H3C2
3
O
NH3 H3C2 CH COOH
CH COOH
Br
NH2
36
Amine
Q.29
A = H3C
CO2H
C
CH2
CH2OH
CH2OH B = H3C
CO2H
CO2H C = H3C
C
C
CO2H
H
D (CH 3 )2 CHCO2 C2 H 5 E (CH3 )2 CHCONH 2 F (CH3 )2 CHNH 2 G (CH3 )2 CHOH H CH 3COCH 3 CONH2 A=
Q.30
NH2 B=
N2 Cl C=
CH3
CH3
CH3
CN
CO2H
COOH
E=
D=
F= CH3
CH3 CO2H
COOH CO2H
G, H are CO2H
O2 N
CO2H
NO2
EXERCISE - III Q.1 Q.8 Q.15 Q.22 Q.29 Q.36 Q.43 Q.50 Q.57
a c c b d a c b a
Q.2 Q.9 Q.16 Q.23 Q.30 Q.37 Q.44 Q.51 Q.58
(a, b) d d b d c c c a
Q.3 Q.10 Q.17 Q.24 Q.31 Q.38 Q.45 Q.52
c a b c c b b a
Q.4 Q.11 Q.18 Q.25 Q.32 Q.39 Q.46 Q.53
d d d c b a b d
Q.5 Q.12 Q.19 Q.26 Q.33 Q.40 Q.47 Q.54
b c a b a a a a
Q.6 Q.13 Q.20 Q.27 Q.34 Q.41 Q.48 Q.55
d c d d d d c c
Q.7 Q.14 Q.21 Q.28 Q.35 Q.42 Q.49 Q.56
b b c b b c b b
37
