Amine IIT JEE Organic Chemistry

8/13/2019 Amine IIT JEE Organic Chemistry

1/37

TOPIC PAGE NO.

1. INTRODUCTION 2

2. PREPARATION OF AMINES 2

3. PHYSICAL PROPERTIES 6

4. CHEMICAL PROPERTIES 7

5. DIAZONIUM SALTS 10

CONTENTS

S.NO. TOPIC PAGE NO

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Amine

1. INTRODUCTION

(i)

(ii)

Amines are called alkyl derivative of NH3.

If a hydrogen atom of NH3is replaced by an alkyl group then it is called primary amineand possess NH2(amino) group.

If two hydrogen atoms of NH3are replaced then it is called secondary amine and it

posses > NH (mino) group.

If all hydrogen atoms of NH3are replaced then it is called tert. amine and has a nitrilo Ngroup.

N is in sp3 hybridisation and tetrahedral geometry.

Bond angle increases from ammonia to 3 amines.

NH3(107) < RNH2< R2NH < R3N

(iii)

(iv)

(v)

(vi)

2. PREPARATION OF AMINES(i) Reductionof nitro compounds

(ii)Ammonolysis of alkyl halides

Thefreeamine can beobtained from theammonium salt bytreatment with a strongbase:

R NH3 XNaOHR NH2H2O Na Xprimary amine isobtainedas a major product by taking largeexcessof ammonia.The order of reactivity of halideswith amines isRI> RBr >RCl.

(iii) Reduction of nitriles

2 2R CN H 2Ni R CH NH

Na(Hg )C3H4OH

(iv) Reduction of amides

AMINE

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Amine

(v) Gabriel phthalimide synthesis

(vi) CurtiusReaction:

RCON RNCOH2OR NH3

Acyl azide

2N CO

2Alkyl isocyanate

2

Alkyl amide

Mechanism :

RCOCl NaN3 RCON3NaCl

OO O

N NR N NR C NR C N C N

R N N2 R N COalkyl isocyanate

C

OH2O

C OHR N R N

OH OHO

R NH C OH CO2 R NH2 CO2

N2

H2O

C O

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Amine

(vii) Schmidt Reaction :Carboxylic acid reacts with hydrozoic acid in presence of concentrated

H2SO4 to give isocyanates.

R COOH + NH H2SO4R N H + C O + H O3 2 2 2

Mechanism :

O

C H

OH OH

HN3R

H

R C OHR C

+OH N N NOH

H2O

OR C O H

R NCON2

isocyanate

R C +N N N

+

H 2O N N

R NH2 CO2

(viii) Lossen Reaction : Hydroxylamine on treatment with acid chloride gives acyl derivatives of hy-

droxyl amine. the acyl derivativesexist intwo tautomeric formketo form called hydroxamic form and

enol formcalled hydroximic acid. The hydroxamic form.

O

ROH

RNH2CO2NH

HO

Mechanism :

O

R H2N OH

Cl

OHO

RR

NNH

HO

(enol form)

HOHydroxamic acid

(keto form)

O

R

NH

+ HCl

HO

H2

N

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Amine

Thehydroxamic form(keto form) formso-acyl derivatives of hydroxamic form which on heating with

bases forms isocyanates and finallyamines uponhydrolysis

H O H O

RR C N OH R C Cl C N O C R

O O

O

H2ORNH2R NCO R C N O C R CO2

O

(ix) Hoffmann bromamide degradation reaction : The reaction of amid with bromin in present ofbase to formprimaryamine.

Mechanism:

2NaOH Br2NaOBrNaBrH2O

OO

BrOH OBr R C NR C NH2

HN-bromoacetone

O O

OHR BrH2OR C N Br C NH 2O

H Rearrangement

H2OR NH2 CO2 R NCO (Isocyanate)

Ex.1 Compound (A) has molecular formula C9H14NCl. (A) gives an immediate precipitate withAgNO3. It isvery 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, itis resistant to oxidation, heat, nitration etc. it must be a quanternary ammoinium salt. Therefore, possiblestructure of (A) may be :

Sol.

+CH3

C6H5N CH

Cl, quaternary-salt

CH3

(A)(C9H14NCl)

N,N,N-Trimethylaniliniumchloride

3

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Amine

Ex.2 Identify (A)through (E) in the following sequenceH3O

HCl NaOH EtBrPhSO2Cl EtNH2 (A)(B)(C)(D) (E)

PhSO NHEt PhSO NEt Ph SO NEtSol. 2 2(B)

2 2(A) (C)

H3OPhSO OHEt NH2(D )

2 2(E )

How theethylamine ispreparedfrom :Ex.3(a)(e)(a)

Propanamide (b) Ethanamide (c) Nitroethane (d)AcetaldehydeEthanol (onlarge scale) ?

Whenpropanamide is treated with Br2 and KOH, ethanamine isobtained.CH3CH2CONH2 + Br2+ 4KOHCH3CH2NH2 + K2CO3 + 2KBr + 2H2O

Ethanamine

Whenethanamide isreduced with lithiumaluminium hydride, ethanamine is

obtained. CH3CONH2 + 4[H] CH3CH2NH2+ H2OEthanamide

Sol.

(b)

(c) Nitroethane can beconverted into ethanamine bythereduction with Sn/HCl.

CH3CH2NO2 + 6[H]CH3CH2NH2 + 2H2OSn/HClNitroethane

Ethanamine

(d) Whenacetaldehyde is treated withammonia, imine isformed whichonreductiongivesethanamine.

H|

CH3 CO

H|

CH3 CNH H 2/ Ni+ NH3 CH CH NHH2O 3 2 22[H]

Ethanal Acetaldimine Ethanamine

(e) Whena mixture of ethanol and ammonia ispassed over alumina at 723 K and high pressure,than amine isformed.

Al2O3CHCH3CH2OH + NH3 CH NH H O3 2 2 2723KEthanamine

3. PHYSICAL PROPERTIES :

(i) Unlike other organic compounds, amines are much more soluble in water. BecauseAllamines form a stronger H- bond with water.

Like ammonia, amines are polar compounds and except 3 aminescan formintermoleculer H-

bonds thats why theyhave 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 1 and 2 than that in alcohols andcarboxylic 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 1, 2 and 3 amines follow the order.

1 > 2 > 3 amine.

Solubility in water follow the order.

1 > 2 > 3 amine.

This is all due to H- Bonding.

(ii)

(iii)

(iv)

(v)

LiAlH4

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Amine

4. CHEMICAL PROPERTIES :

Alkylation :Amines undergo alkylation on reaction with alkyl halides

R1|

R N RR1X R2X R3X

RNH2 HX RNHR HX R - N - R HX X21 2|R3(quartenary amm. salts.)

Acylation: 1 and2 amines react withacetyl chloride or acetic anhydride to form acetyl derivatives.

R-NH2+ CH3COClRNHCOCH3+ HClR -NH2+ (CH3CO)2O

(CH3)2NH + CH3COCl

RNHCOCH3 + CH3COOH

(CH3)2N-COCH3 + HCl

Note :(a) Tertiary amines donot undergo this reaction because of absence of replacable H- atom.

(b) When Benzoyl chloride is

Schotten - Baumann reaction.

used in place of acetyl chloride reaction is called

Benzoylation : WhenAmines react with benzoyl chloride (C H COCl).6 5

CH3NH2 C6H5COClCH3NHCOC6H5HCl

Methanamine Benzoyl chloride N

Methylbenzamide

HeatCarbylamine reaction :R -NH 2 CHCl 2 3KOHR -NC 3KCl 3H 2O

Cl

R NH2 CCl2 R HN CMechanism :

ClH

-HCl

+R N C R N C Cl

2

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Amine

Reactionwithnitrousacid :

-NaOH HClR -NH 2HNO 22[R - N 2 Cl ] ROHN 2HCl

-NaNO 2HClC6H5- NH 22 C6H5- N 2ClNaCl 2H 2O

Ani ln e BenzenediamoniumChloride

Reactionwitharylsulphonylchloride : Benzenesulphonyl chloride (C H SO Cl),which isalso known6 5 2

as Hinsbergs reagent, reacts with primaryand secondary amines to form sulphonamides.

(i)Thereaction ofbenzenesulphonyl chloridewithprimaryamineyields N-ethylbenzenesulphonyl amide.

The hydrogen attached to nitrogen insulphonamide is strongly acidic due to the presence of strong

electron withdrawing sulphonyl group. Hence, it issoluble inalkali.

(ii) In the reaction with secondary amine, N,N-diethylbenzenesulphonamide is formed. It is in soluble

alkali.

(iii) ter.Amine does not react withHinsbergs reagent it ispresent above solution.

Electrophilic substitution : Due to +Meffect of -NH genrate electron dencity at ortho and para2

position hence, aniline active toward Electrophilic substitution

(i) Bromination: Aniline reacts with bromine water at roomtemperature to give a white precipitate of

2,4,6-tribromoaniline.

If we have to prepare monosubstituted aniline derivative, This can be done byprotecting the -NH2

groupbyacetylation with acetic anhydride, thencarrying out the desired substitution followed by hy-drolysis ofthe substituted amide to thesubstituted amine.

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ss

Amine

The lone pair of electrons on nitrogen of acetanilide interacts with oxygen atom due to resonanceasshownbelow:

Hence,the lone pair of electrons on nitrogen is less available for donation to benzene ring byresonance. Therefore, activating effect of -NHCOCH group is less thanthat of amino group.3

(ii) Nitration:Direct nitration ofaniline isnot possiblebecause inthe stronglyacidicmedium, aniline isprotonated to formtheanilinium ionwhich ismeta directing.

However, by protecting the -NH group byacetylation reaction with acetic anhydride, the nitration2

reaction can becontrolled and the p-nitro derivative can be obtained asthe major product.

(iii) Sulphonation:

Consider the following reaction,Ex.4

H

C H NH CHCl KOH (A)/2(B) (C) H O6 5 2 3

Find the compounds (B) and (C).

C6 Hl5NH2 CHCl3 3KOHC6 H5NC 3Kl 3H2O(A)

Sol.

C H NC2H OHC H NH HCOOH6 5

(A )2 6 5 2

(B) (C )

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Amine

Ex.5

Sol.

Sulphanilic acid is insoluble in water and acid but soluble in caustic alkali. Comment.Sulphanilicacidexistas a Zwitter ion andexhibits strongdipole-dipole interactions. Therefore, it is insolublein water. on adding acid, SO 3fails to accept H ion, thus sulphanilic acid is insoluble in acid. However,

when alkali is added, strongly basic hydroxyl ion can abstract a proton fromNH3to formsoluble salt.

- +

+

+NH3 NH2

+ H2O+ NaOH

SO3Na

Sodiump-aminobenzenesulphonateSO3

-

Zwitter ion structureof sulphanilic acid

How willyou carryout thefollowing conversions?

O

Ex.6

C NHNH2 CH3(a)

NH2 NH2(b)

O

C NH2

KOH

NC

NH2 CHCl3(a) Br2/KOHSol.

NHCH3

LIAlH4

NaNO2 OHH2SO4

-

KMnO / HNH 42 HCl

(b)

COOHCOOH

BaOH2/ PtNH3 NHO NH 2

5. DIAZONIUM SALTS : - -

Thediazonium saltshavethe generalformula R N2 X where R stands for anaryl group andXionmay

be Cl-, Br-, HSO-, BF -, etc.4 3

Resonance of benzenediazonium ionis

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Amine

Methodof Preparation of Diazonium Salts

-273-278KC6H5NH2NaNO2 2HClC6H5N2 ClNaCl 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 retentionof diazo group.(i) Reactions involvingdisplacement of nitrogen

Diazonium group being a very good leaving group, is substituted by other groups such as

Cl-, Br- , I-, CN- and OH- whichdisplace nitrogen from the aromatic ring.

(a)Replacement byhalide or cyanide ion: TheCl-, Br- and CN- nucleophiles caneasilybeintroduced atthe benzene ring in the presenceof Cu(I) ion. Thisreaction iscalled Sandmeyer reaction.

Alternatively, chlorine or brominecanalso be introducedbyGatterman reaction.

-(b)

(c)

Replacement byiodide ion:

Replacement byfluoride ion:Ar N 2 Cl KIArI KClN 2

- -Ar N2 Cl HBF4 Ar - N 2BF4 Ar -F BF2N 2

Replacement by H:ArN2 ClH3PO2ArHN2H3PO3 HCl -(d)

-

ArN2 ClCH3CH 2OHArHN2CH3CHOHCl -

(e)

(f)

Replacement byhydroxyl group:

Replacement by NO group:

ArN 2 Cl H 2OArOHN2HCl

2

(ii) Reactions involving retentionof diazogroup couplingreactions

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Amine

Ex.7 Provide structures for the products of the reaction of PhN2with

(a) PhNMe (b) 2-naphthol (c) PhCH .2 3

ArN2 is a weak electrophile that undergoes diazo couplignonly with rings activated by OH, NH2,

Sol.

NHR or NR2 .

Ph N N NMe2

(a) N,N-dimethylaminoazobenzene(butter yellow

N N Ph

OH

(b)

1-Phenylazo-2-naphthol

(c) No reaction.The substrate ring is insufficientlyactivated.

Ex.8 A weakly acidic medium isComment.

provided for coupling of benzene diazonium chlor ide with aniline.

+ -N2Cl

pH 4-5

273-278 K

-HCl

N = N+H NH2Sol. NH2

Benzene diazonium chloride p-AminoazobenzeneAniline

H+If the high conc. of ions are used during these reactions, then protonation of aniline takes

place.

+H3N

H+NH2

Anilinium ionAniline

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.

Theend product (Z) of thefollowing reaction is ?Ex.9

N2Cl Cu / KCN

H / H2O NaOH

Cu / KCNN2Cl CN

Sol.

(X) (Y) (Z)

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Amine

Ex.1 Account for the fact that 2-aminoethanoic acid (glycine) exists as a dipolar ion, as does p-aminobenzenesulfonic acid(sulfanilic acid) but p-aminobenzoic acid does not.

The aliphatic NH2 issufficiently basic to accept anHfromCOOH. the COOH isnot strongenoughSol.

to donate to the weakly basic ArNH2 , but SO3H is a sufficiently strong acid to do so.H

H N CH COO-

p H N C H COOH p H N C H SO-3 2 2 6 4

pA min obenzoic acid3 6 4

Sulfanilic acid3

Glycine

Ex.2 Mixture oftwo aromatic compounds (A) and (B)wasseparated bydissolving inchloroformfollowed byextraction with aqueous KOH solution. The organic layer containing compound (A), when heated withalcoholic solution of KOH produced a compound (C) (C7H5N) associated withanunpleasant odour.

The alkaline aqueous on the other hand, when heated with chloroform and then acidified give a mixture oftwo isomeric compounds (D) and (E) of molecular formulaC7H6O2. Identify thecompounds (A),

(B), (C), (D), (E) and writetheir structures.

(a) One of the compound A and B is soluble in aq.KOH and thus it must be acidic in nature whereasanother compound is soluble in chloroformis either basic or neutral in nature.(b) The compound (A) on heating with alcoholic KOH solution (chloroform already present) producescompound(C) (C7H5N)havingunpleasant odour. The compound(C) isPhenylisocyanide and, therefore,

compound (A)must be aniline which issoluble inchloroform but insoluble inaq.KOH.

(c) The compound (B) must be phenol as it is soluble in aqueous KOH and produces isomers o-hydroxybenzaldehyde (D) and p-hydroxylbenzaldehyde (E) on heating, with chloroform followed byacidification.Thereactions areas follows :

Sol.

NH2 N C

Warm+ CHCl3 + 3KOH + 3KCl + 3H3O (carbylamine reaction)

Aniline (A)(Soluble in chloroform)

Phenylisocyanide

(C)

OHOH OHCHO

CHCl3/KOH

H+ +

Phenol(B)

(Soluble in KOH)

Salicylaldehyde

(D) CHOp-Hydroxybenzaldehyde

(E)

Ex.3 Aneutral compound (A) C H ON on treatment with sodium hypobromite forms an acidsoluble sub-8 9

stance (B), C H N.Onaddition of aqueoussodiumnitrite to a solutionof (B) in dil. HCl at 0 -5C, an7 9

ionic compound (C). C H N Cl is obtained. (C) givesa red dye with alk -naphthol solution. When7 7 2treated with potassium cuprocyanide, (C) yields a neutral substance(D), C H N. ON hydrolysis (D)8 7gives (E), C H O . (E) liberates CO from aqueous sodium bircarbonate. (E) on permanganate

8 8 2 2

oxidation furnishes (F), C H O , (F)on nitration yields two isomeric mononitroderivatives (G) and (H)8 6 4

having molecular formula, C H NO . Write the reaction involved indifferent steps.8 5 6

SOL VED EXAMPLE

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Amine

CNCONH2 NH2 N2Cl

(C) :(A) : (D) :(B) :

CH3

COOH

CH3

COOH

CH3

COOH

CH3

COOHSol.

(F) : (H) :

O2N

(G) :(E) :

CH3 COOH COOH COOH

NO2

Ex.4 Anopticallyactive amine (A) issubjected to exhaustive methylation andHoffman elimination to yield onalkene (B). (B) on ozonolysis gives anequimolar mixture of formaldehyde andbutanol. Deduce thestructuresof(A) and (B). Is three any structural isomer of (A), if yesdrawits structure?

(A) : H3C CH CH2CH2CH3 (B) :H2C CHCH2CH2CH3Sol.

NH2

NH2CH2CH2CH2CH2CH3ispositional isomer of (A).

Amixtureof two aromatic compounds (A)and (B)wasseparatedbydissolving inchloroform followedby extraction with aqueous KOHsolution.Theorganic layer containing compound (A), when heatedwith alcoholic solution of KOHproduced a compound (C) (C H N) associated with an unpleasant

Ex.5

7 5

odour. Thealkaline aqueous layer on the other hand,whenheated with chloroform andthen acidifiedgive a mixture of two isomeric compounds (D) and (E) of molecular formula C H O . Identify the

7 6 2

compounds (A), (B), (C), (D) and (E).

NH2 OH N = C OH OHCHO

(E) :Sol. (A) : (B) :

CHO

Ex.6 Compound(A), C H N O whenheated with dilute sulphuricacid gives ammonium sulphate, com-14 10 2

pound(B), C H O andcompound(C), C H N (as its sulphate). Compound (B) on heating gives (D),8 6 4 6 7

C H O and onheatingwith sodalime gives benzene. Compound (C)with dilute H SO /NaNO in the8 4 3 2 4 2

cold followed by heatingwith water gives (E), C H O. Compound (E) onheating withZndust again6 6gives benzene. Identify(A), (B), (C)and (e) giving proper reactions?

CNCO2H

C6H5NH2(NH4)2SO4 dil.H2SO4

C NH (C)

H2SO4

CO2H(B)

O(A)Sol. O

(C6H5NH3 )2SO4CO2H C

O-H 2O

CO2H C(B) (D)

O

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Amine

Ex.7

Sol.

PCl5 Na+

C2H5OH

KCN -NH3

ROH(A)

RCl(B)

ER CH NHR -CN(C)

2 2 Nitrobenzene

(D)

Start with propane-1, 3-diol

CH2OH CH Cl CH2CN

CH CN

PCl5 2CH2 CH 2KCN Na + C H OHCH22 2 5CH2OH

(A)

CH2Cl

(B)2

(C)

CH2CH2NH2

CH2CH2NH2

(D)

CH2

-NH3

Nitrobenzene

-6HN

HPiperidine

(Hexahydropyridine)(E)

Ex.8

Sol.

Glycine existsas a Zwitter ionbut anthranilic acid doesnot. Comment.-COOH group of glycine releases H+ ion which is accepted by -NH2 group. Thus glycine exist of aZwitter ion.

HH

H2N -C - COOH

H

Glycine

+

H3N - C - COO-

H

Zwitter ion of glycine

NH2COOH

In anthranilic acid,Anthranilic acid

(2- Aminobenzoic acid)

Electron withdrawing -COOH and phenyl group reduces electron density of N of NH2 group,

therefore,-NH2fails to accept a proton. Thus anthranilic acid can not form Zwitter ion.

Write reactions of thefinal alkylation product of aniline with excessof methyl iodide inthe presence of

sodium carbonate Sol.

Ex.9

+

[C6H5NHCH3]IN-Methylanilinium iodide

-H6H5NH2 + CH3ISol.Aniline Methyliodide

2[C6H5NHCH3]I- + Na2CO3 2C6H5NHCH3 CO2 2NaI

NMethylaniline

An organic compound (A) composed of C, H and O gives a characteristic colour with cericammoniumnitrate. Treatmentof (A)with PCl5 gives (B) whichreacts with KCNto form(C). The reductionof(C)with Na and C2H5OH produces (D) which on heating gives (E) with evolution of NH3. Pyridine isobtained on treatment of (E) with nitrobenzene. Give thestructures of (A) to (E) with proper reasoning.(A) isanalcohol which gives characteristic redcolour with cericammonium nitrate.

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Amine

2CH I2C6H5NH CH3 3 2C6H5N(CH3)2 + CO2+ 2NaI

Na 2CO3

+ -C6H5NHCH3 + CH3I [C6H5NH(CH3)] IN.N.N-Trimethyl anilinium ioidide

[2C6H5N (CH3)3] I- + Na2CO3

2-[C6H5N(CH3 )3 ]2 CO3 2NaIN, N, NTrimethylanilinium Carbonate

Ex.10 What happenswhen(a)(b)(c)(d)(e)(a)

Aniline is treated with sodium nitrateand dil. HCl at 273K,Aniline is treated with Benzaldehyde,Ethylamine reacts withAgCl,Amixture ofalcohol and ammonia ispassedover heated aluminiumoxide as catalyst.Sodium nitrite is added to a solution of ethylamine indil. HCl at 273 K ?Benzenediazonium chloride is formed.

N NCl

Sol.

NH2NaNO2 + HCl

273 K

Aniline

Benzalaniline isformed.

H

Benzenediazonium chloride

(b)

H

NH2 O C N CH+

Benzaldehyde BenzalanilineSchiff's base

(c) Soluble complex-Bis-(ethylamine) silver (I) chloride is formedAgCl(s) + 2C2H5NH2(aq) [Ag(C2H5NH2)2]Cl(aq)

Bis-(ethylamine) sliver (I) chloride (soluble)

(d) Primary amine isformed. If ammonia is taken in excess, a mixture of 1, 2, 3amines andquaternaryammoniumsalt is formed.

Al O

ROH + NH3 2

3

R NH + H O2 2723K1 alcohol 1 alcohol

(e) Ethyl alcoholis formedNaNO2+ HClNaCl + HNO2

Nitrous acid

C2H5NH2 + HNO2C2H5OH + N2+ H2OEx.11 Giveonechemical test to distinguishbetweenthe following pairs ofcompounds :

(i)(iii)(v)

Methylamine and dimethylamineEthylamine and aniline

Aniline andN-methyl aniline

(ii) Secondaryand tertiaryamines(iv)Anilineand benzylamine

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Amine

Sol. (i) Methyl amine, beinga primary amine will give carbylamine reaction (offensive smell of isocyanidewhen treated with CHCl3+ KOH) while dimethyl amine will not show any reaction with

CHCl3 + KOH (alc.)

CH3NH2 + CHCl3 + 3KOH (alc)CH3NC 3KCl 3H2OMethyl isocyanide(offensive smell)

(ii)Whensecondaryamine istreated with HNO2(NaNO2 + HCl), if formsyellowcolouredoilycompoundN-nitrosoamine, which on warmingwith a crystal of phenol andH2SO4 gives a green solution and onfurther addition ofaqKOH redsolution(Libermannnitroso test).This test isnot given bytertiaryanines.

(CH3)2NH + HONSecondary amine Nitricacid

O (CH3)2NN O + H2ON-Nitrosodimethylamine

Phenol + H2SO4

KOH +WaterGreen colourRed colour

(iii) Whenaniline is treated byNaNO2 + HClat 273K,benzenediazonium chloride isformed which on

treatment with b-naphtholgives a bright orange dye, 1-phenylazo-2-naphthol.

NaNO2/HClNH2 N N Cl273K

OH

N N

(iv) Benzyl amine on treatment withNaNO2 and HCl forms benzyl alcohol with the evolution of N2 gas

whileaniline froms benzene drazonium chloride whichgivesorgane dye with alkaline b-naphthol.

(v)Aniline, being a primaryamine, gives carbylamine reaction while N-methyl aniline (sec. amine) willnot give this rest.

NH2 + CHCl3 + 3KOH(alc) NC + 3KCl + 3H2O

Phenyl isocyanide(offensive smell)

OH

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Amine

Q.1 Arrange thefollowing :

(i) Indecreasing order of the pK values:b

C H NH , C H NHCH , (C H ) NH and C H NH2 5 2 6 5 3 2 5 2 6 5 2

(ii) In increasing order ofbasic strength:

C H NH , C H N(CH ) , (C H ) NH and CH NH6 5 2 6 5 3 2 2 5 2 3 2

(iii) In increasing orderofbasic strength:

(a)Aniline, p-nitroaniline andp-toluidine

(b) C H NH , C H NHCH , C H CH NH .6 5 2 6 5 3 6 5 2 2

(iv) In decreasing order of basic strengthin gas phase:

C H NH , (C H ) NH, (C H ) N and NH2 5 2 2 5 2 2 5 3 3

(v) In increasing orderofboiling point:

C H OH, (CH ) NH, C H NH2 5 3 2 2 5 2(vi) In increasing orderof solubility in water:

C H NH , (C H ) NH, C H NH .6 5 2 2 5 2 2 5 2

Q.2 Describe a method for the identification ofprimary, secondaryand tertiaryamines.Alsowrite chemical

equationsof the reactions involved.

ExplainHofmannBromanide reaction withMechanism?

Whycannot aromatic primary amines be preparedbyGabriel phthalimidesynthesis?

Write the reactions of (i)aromatic and (ii) aliphatic primary amines with nitrous acid.

Write onechemical reaction each to illustrate the following

(i) HofmannBromanide reaction.

(ii) Garbriel Phthalimide reaction.

Assigna reasonfor the following statements

(a)Alkylamines are stronger bases than arylamines.

(b) How would you convert methylamine into ethylamine ?

Illustrate thefollowing with anexample of reaction ineach case :

Q.3

Q.4

Q.5

Q.6

Q.7

Q.8

(i) Sandmeyer reaction (ii) Coupling reaction

Q.9 Write thechemical reaction equationsfor oneexample each of thefollowing

(a)Acoupling reaction

(b) Hofmann'sbromamide reaction

(c)Acetylation

Account for thefollowing :

(i)Aniline isweaker base thanmethylamine.

(ii)Aryl cyanides cannot be formed bythe reaction ofaryl halides andsodium cyanide.

Describe tests to distinguish between : Secondary amine andtertiary amine.

Q.10

Q.11

EXERCISE-I

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Amine

Q.12 Account for thefollowing observations :

(i) pK for aniline ismore thanthat for methylamine.b(ii) Methylamine solution in water reacts with ferric chloride solution to give a precipitate of ferric

hydroxide.

(iii)Aniline does not undergo FriedelCrafts reaction.

State the reactionsand reaction conditions for the following conversions

(i)Benzene diazoniumchloride to nitrobenzene.

(ii)Aniline to benzene diazoniumchloride.

(iii)Ethyl amide to methylamine.

Write the physicalpropertyof aniline ?

Write themethodof formation of benzene diazonium chloride ?

Account for the following:

(i) pKb ofaniline is more thanthat of methylamine.

(ii) Ethylamine issoluble inwater whereas aniline isnot.

(iii) Methylamine inwater reacts with ferric chloride to precipitate hydrated ferric oxide.

(iv)Althoughamino group is o- andp- directing inaromaticelectrophilic substitutionreactions,anilineon nitrationgivesa substantialamount ofm-nitroaniline.

(v)Aniline doesnot undergoFriedel-Crafts reaction.

(vi) Diazonium saltsof aromatic amines aremore stable than those of aliphatic amines.(vii) Gabrielphthalimide synthesis ispreferred for synthesising primary amines.

How willyou convert:

Q.13

Q.14

Q.15

Q.16

Q.17

(i)Ethanoic acid into methanamine

(iii) Methanol to ethanoic acid(v) Ethanoic acid into propanoic acid

(vii) Nitromethaneintodimethylamine

Write short noteson thefollowing:(i) Carbylamine reaction(iii) Hofmanns bromamide reaction

(v)Ammonolysis

(vii) Gabrielphthalimide synthesis.

Accomplish the following conversions

(i) Nitrobenzene to benzoic acid

(iii) Benzoicacid to aniline(v) Benzyl chloride to 2-phenylethanamine

(vii)Anilineto p-bromoaniline

(ix)Aniline to benzyl alcohol.

(ii) Hexanenitrile into 1-aminopentane

(iv) Ethanamine intomethanamine(vi) Methanamine into ethanamine

(viii) Propanoic acid into ethanoic acid ?

Q.18

(ii) Diazotisation(iv) Coupling reaction

(vi)Acetylation

Q.19

(ii) Benzene to m-bromophenol

(iv)Aniline to 2,4,6-tribromofluorobenzene(vi)Chlorobenzene to p-chloroaniline

(viii) Benzamideto toluene

Q.20

Q.21

Write theequation ofcurtius reaction with mechanism ?

Complete the following reactions:(ii) C H N Cl H PO H O(i) C H NH CHCl alc.KOH

6 5 2 3 6 5 2 3 2 2

(iii) C H NH H SO (conc.) (iv) C H N Cl C H OH6 5 2 2 4

(v) C H NH + Br (aq)6 5 2 2 5

(vi) C H NH + (CH CO) O6 5 2 2

(i) HBF4

6 5 2 3 2

(vii) C H N Cl 6 5 2 (ii) NaNO2/ Cu.

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Amine

Q.22 Give plausibleexplanation foreachofthefollowing:

(i)Whyare aminesless acidicthanalcohols of comparable molecular masses?

(ii) Whydo primaryamines havehigherboiling point thantertiaryamines?

(iii) Whyare aliphatic amines strongerbases thanaromatic amine?

Write the reaction andconditions for the following conversionsQ.23

(i)Aniline to benzene(iii) Propanenitrile to ethylamine

(v) Nitrobenzene to 2, 4, 6-tribromoaniline.

Write the method of formation ofzwitter ion ?

Explainnitration ofaniline ?

Whyaniline do not give fridelcraft reaction ?

(ii) Methylamine to methyl cyanide(iv) m-Bromoaniline to m-bromophenol

Q.24

Q.25

Q.26

Q.27 What isGabrielphthalimide synthesis ? For what purpose isit used ? Giveequation onlyto explainyour

answer.

Write theequation of carbyl amine reaction withmechanism ?

How willyou convert 4-nitrotoluene to 2-bromobenzoic acid ?

Draw thestructure of trimethylamine andtell theshapeof themolecule. Show theangle between two

methyl groups.

Q.28

Q.29

Q.30

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Amine

Q.1 Aspartame, anartificial sweetener, isa peptide and has thefollowing structures :

NH2 CH2C6H5

HOOC - CH2CH - CONH - CH -COOCH3

(a)(b)(c)(d)

Identifythefour functionalgroups.Write thezwitterionic structure

Write the structures of the amino acids obtained from thehydrolysis of aspartame.Which of thetwo amino acids ismore hydrophobic ?

Q.2 Compound ofA(molecular formula C9H11NO) gives a positiveTollens test and is soluble in dilute HCl.It gives no reaction with benzene sulphonyl chlorideor with NaNO2 and HClat 0C. (A), upon oxidationwith KMnO4 gives an acid (B). When(B) is heated with soda-lime, compound (C) is formed whichreacts with NaNO2 and HCl at 0 - 5C. What is (A) ?

Q.3 An organic compoundA, when treated withnitrous acid yields an alcohol B, C4H10O with the evolutionof N2. B on careful oxidationyields a substance C of vapour density36 which formsoxime; B can reactwith NaHSO3but does not reduce Fehling solution. Identify compoundAand write the structural formulaeof 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, momentis zero. Deduce the structure of (A).

Q.5 Explainthefollowing observations :(1)

(2)(3)(4)

Aniline dissolves inaqueousHCl.

The amino group in ethylamine isbasic whereas that inacetamide it is not basic.Dimethylamine isa stronger base thantrimethylamine.Sulphanilic acid although has acidic as well as basic group, it is soluble inalkali but insoluble inmineralacids.

Q.6 Explain,why?Glycine exists as H3N

+CH2COO- while anthranilic acid, p-NH2.C6H4.COOHdoesnot exist

asdipolar ion.Benzenesulphonic acid isa stronger acid thanbenzoic acid.Aweaklybasic solution favours coupling with phenol.

It is difficult to prepare pure amines byammonolysis of alkyl halides.

(1)

(2)(3)

(4)

Q.7 Explainwith reason ?(1) Although trimethylamine and n-propylamine havesamemolecular weight, theformer boilsat a

lower temperature (3C) than the latter (49C).

Dimethylamine isa stronger base thanmethylamine but trimethylamine isa weaker basethan both dimethylamine and methylamine.Although trimethylamine and n-propylamine have same molkecular weight, the former boilsata lower temperature (3C) than the latter (49C). Explain.Silverchloride dissolves inaqueous solutionof methylamine. Explain.

(2)

(3)

(4)

EXERCISE-II

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Amine

Q.8 Explain it ?(1)(2)(3)

Anaqueous solution ofethylamine gives a red precipitate withferric chloride. Explain.Tertiary amines do not undergo acetylation. Comment

2, 6-Dimethyl -N N-dimethylaniline, although hasa free p-position, doesnot undergo couplingwithbenezenediazoniumchloride.Comment.In the followingcompounds :(4)

O

N

H(I)

N

H(IV)

N

=(III)

N

(II)

The order of basicity is I > III > II > IV. Explain.

Q.9 Explain it with reason ?(1)

(2)(3)(4)(5)

tert-Butylamine cannot be prepared by the action of NH3 on tert-butyl bromide.

Isocyanides are hydrolysed bydilute acids but not byalkalies to formamine and formic acid.Howwill you explain theacidic nature of 1 and 2 nitroalkanes ?Aniline doesnot undergo Friedal Crafts reaction ?Althoughborontrifluoride adds on trimethylamine, it doesnot addon triphenylamine.Comment.

Q.10 Complete thefollowing reactions :

PCl5 NH3 P2O5 H2/Hi(1) C6H5COOH [C] [D] C6H5CN [E]

H+P2O5(2) CONH2 F G

KOH(3) EtNH2+ KCN + Br2 KBr + H

N(CH3)2+ HNO2(4) I

(i) NaNO2/HCl, 5C

(ii) anisole(5) 2, 4-Dinitroaniline (J)

NaOH

heatNaOHOleum(6) C6H6 (K) (L) (M)

SO3H OH

1

(7)

fuming

H SO

(i) NaOH fuse

(ii)H+ SO3H(8) O P

2 4

HCN, HCl

AlCl3

Et2SO4NaOH

PhNH.NH2(9) Phenol (Q) (R) (S)

Br2, Fe(10)

(11)

CH3CONHC6H5 T + U

(A)

HClC6H5N2Cl (V) Gattermann reaction

CHCl3/NaOH N

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Amine

Q.11 Give structures for thecompounds (A) to (I) :NaNO2/HCl KCNC8H11N

(A)

B C0-5C CuCN

Hot H2SO4

Hot

aq. KMnO4

Heat tom.p.

CH3OHat 30C

I G D

Cl2, 2 moles u.v.

E

H

F

Q.12 When2.25g of an unknown amine was treated with nitrous acid, the evolved nitrogen, corrected toS.T.P. measured 560ml.Thealcoholisolated fromthereactionmixturegave a positive iodoformreaction. What isthe structural formula of the unknown amine ?

The aqueoussolution of a nitrogen and chlorine containing organic compound (A) is acidic towards litmus.(A) ontreatment with aqueousNaOHgives acompound (B), containing nitrogen, butnotchlorine.

Compound (B) on treatment with C6H5SO2Cl in the presence of NaOH gives an insoluble product

(C), C13H13NO2S. give the structuresof compounds (A) and (B).

Anorganic compound (A) composed ofC, H and O gives characteristic colour with ceric ammonium

nitrate. Treatment of(A) with PCl5gives (B), which reacts with KCNto form(C). the reduction of (C)

with warmNa/C2H5OH produces (D), which on heatinggives (E) with evolutionof ammonia Pyridine

is obtained on treatment of (E) with nitrobenzene. Give structure of compounds (A) to (E) with properreasoning.

One moleofbromoderivative (A) and moleof NH3react to give one mole ofanorganic compound(B).

(B) reacts with CH3I to give (C). Both (B) and (C)react with HNO2 to give compounds, (D)and (E)

respectively. (D) on oxidation and subsequent decarboxylation gives 2-methoxy-2-methyl propane.Give structuresof compounds (A) to (E) with proper reasoning.

What happens with cyclopentanoen reacts with

Q.13

Q.14

Q.15.

Q.16

(a)

(b)

CH3CH2NH2 (1 amine)

(CH3CH2)2NH (2amine)

Q.17

Q.18

Cyclohexylamineis a stronger base thananiline.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 astronger base thanmethylamine but trimethylamine isaweaker base thanbothdimethyl amineand methylamine. Why?

From analysis and molecular weight determination, the molecular formula of (A) is C3H7NO . the

compound gave following reactions.

(i) Onhydrolysis it gives anamine (B) and a carboxylic acid (C)(ii)Amine (B) reacts with benzene sulphonjyl chloride and gives a productwhich isinsoluble inaqueous

sodiumhydroxide solution.

(iii)Acid (C) on reaction withTollens reagent gives a silver mirror when areA, B and C. Explain thereactions.

Q.19

Q.20

Q.21

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Amine

Q.22 Anopticallyactive amine (A)issubjected to exhaustive methylation and Hofmannelimination to yield analkene (B). (B) on ozonolysis gives an equimolar mixture of formaldehyde and butanal. Deduce thestructures of(A) and (B). Is there any structural isomer to (A), ifyes draw its structure.

An aromatic compound (a) having molecular formula C7 H7NO2dissolves in NaHCO3to evolve CO2 and

when reacted with NaNO2/ HCl forms (b), C7H6O3 . (B) dissolves in NaHCO3and gives colour reactionwith FeCl3and can be prepared by the action of CCl4and NaOH on phenol. When (B) is reacted with

excess HNO3 , it forms (C), C6H3N3O7 . (C) undergoes acetylation and decomposes NaHCO3to evolve

CO2 . On reaction with PCl5 . (C) is converted to (D), C6H5N3O6Cl which when reacted with water gives

back (C). Identify compounds (A) to (D).

Compound (A)havingM.F. C8H8O on treatment with NH2OH.HCl gives (B) and (C). (B) and (C)

rearrange to give (D)and (E), respectivelyon treatment with acid. Compounds (B), (C),(D) and (E)are

allisomersofmolecular formula C8H9NO. When(D)isboiled with alcoholic KOH, anoil(F) C6H7Nseparated out. (F) reacts rapidly with CH3COCl togive back (D). On the other hand, (E) on boiling

withalkali followed by acidification gives a white solid (G), C7H6O2 . Identify the compounds (A) to

(G).

Anaromatic compound (A), having M.F C7H5NO2Cl2onreduction with Sn/HClgives (B), which on

reactionwith NaNO2/ HCl gives(C).Compound(B) isunableto forma dye with -naphthol. However,(C) gives red colour with ceric ammonium nitrate and on oxidation gives an acid (D), having equivalentweight 191. Decarboxylation of (D) gives (e) which forms a single mononitro derivative (F), on nitration. Give

thestructures of (A) to (F) with proper reasoning.

An organiccompound (A) of molecular weight 135, on boiling with NaOH evolves a gas which gives whitedenso fumes on bringing a rod dipped in Hcl near it. The alkaline solution thus obtained on acidificationgives theprecipitate of acompound (B) havingmolecular weight 136.Treatment of(A)with

HNO2also yields (B), whereas it treatment with Br2/ KOH gives (C). Compound (C) reacts with

cold HNO2to gives (D), whichgive red colourwith ceric ammoniumnitrate. Ontheother hand, (E)an

isomer of (A) on boiling with dilute HCl gives an acid (F), having molecular weight 136. On oxidationfollowed by heating, (F) gives an anhydride (G), which condenses with benzene in the presence of

anhydrous AlCl3to give anthraquinone.Give the structures of (A) to (G)with proper reasoning.

Anorganic compound(A) havingM.F C7H9N ontreatment with NaNO2and HClat roomtemperature

forms another compound (B), C7H8O . 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 C8H7N. Hydrolysis of(C) followed byreaction with NaNO2and HClat low temperature and

subsequent reaction withHCN in the presence of Cu(D), which is isomeric to (C). (D) on hydrolysisfollowed by oxidation gives a dibasic acid, which on halogenation forms only one monohalo derivative.Identify the compounds (A) to (E).

Q.23

Q.24

Q.25

Q.26

Q.27

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Amine

Q.28 An optically active compound (A), C3H7O2N forms a hydrochloride but dissolves in water to give a

neutral solution.Onheating with sodalime (A) yields (B) C2H7N . Both (A) react with NaNO2and

HCl, theformer yielding a compound (C) C3H6O, whichon heating is converted to (D), C6H8O4while

the latter yields (E), C2H6O .Account for the above reactions and suggest how (A) maybesynthesized.

An optically inactive acid (A), C5H8O5 , on being heated lost CO2to give an acid (B), C4H8O3

capable ofbeing resolved. Onaction ofsulphuricacid, B gave anacid C whose ethyl ester gave(D)on

theaction of hydrogen and platinum. (D)with conc. NH3gave E, C4H9OH which with Br2and KOH

solution gave (F), C3H9N . Fwith HNO2gaveG. (G) onmild oxidationgave H.BothAand H gave the

iodoform reaction. Elucideate the reaction mechanismand suggest a synthesis of (C).

Aneutral compound (A) C8H9OH ontreatment withNaOBr forms anacidsolublesubstance C7H9N.

On addition of aqueous NaNO2to a solution of B in dilute HCl at 0-5C an ionic compound (C)

Q.29

Q.30

C7H7N2Clis obtained. (C) gives a red dye with alkaline -napththol solution. When treated with

potassium cuprocyanide (C) yields a neutral substance (D) C8H7N . ON hydrolysis (D) gives E

(C8H8O2 ) . E liberates CO2 from aqueous NaHCO3 . (E) on permanganate oxidation furnishes (E)

C8H6O4 . (F) onnitration yields two isomericmononitroderivatives (G andH)havingmolecular formula

C8H5NO6 . Write the reactions involved indifferent steps.

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Amine

Q.1 When aniline is treated with fumingsulphuric acid at 475K, it gives(a) Sulphanilic acid

(b)Aniline sulphate

(c)o-aminobenzenesulphonic acid(d)m-aminobenzenesulphonic acid.

When nitrobenzene is treated with Br2 in presence of FeBr3, the major product formed is m-bromonitrobenzene. 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 leastdestablilzed

(c) Loss of aromaticity, whenBr+ attacks at theortho and para positions, and not at meta position(d) Easier loss of H+ to regain aromaticity fromthemeta position than fromthe ortho and para positions.

Examine thefollowing two structures for theanilinium ion and choose the correct statement from the ones

given below.

Q.2

Q.3

(a) II isnot anacceptable canonicalstructure, because carboniumions are less stable than ammoniumions

(b) II isnot an acceptable canonical structure, because it is non aromatic(c) II isnot anacceptable canonicalstructure, because the nitrogen has10 valence electrons(d) II is an acceptable canonical structure.

The correct order of basic strengthof in CCl4Q.4(1) NH3 (2) RNH2 (3) R2NH (4) R3NWhere R is CH3 group is

(a) 3 > 2 > 1 > 4 (b) 2 > 3 > 4 > 1 (c) 3 > 2 > 4 > 1 (d) None of these

Q.5 Place the following inthe decreasing order of basicity.(1) Ethylamine(a) 1 > 3 > 2

(2) 2-aminoethanol(b) 1 > 2 > 3

(3) 3-aminopropan-1-ol(c) 2 > 1 > 3 (d) Noneof these

Q.6 Which of the following willgive a positive carbylamine test ?1. H3CNH2 2. H3C-NH-CH3 3. (CH3)3N 4. C6H5NH2Select thecorrect answer using the codes given below.(a) 1 and 3 (b) 2 and 4 (c) 3 and 4 (d) 1 and 4.

Q.7 Match thecompounds in list I with the appropriate test that will beanswered byeachoneof them inlistII from the combinations shown.

Selects thecorrect answer using thecodesgiven below the list .List IPropyne

Ethyl benzoateAcetaldehydeAniline 4.

List II1.

2.

3.

(A)(B)(C)(D)

ReducesFehlings solutionFormsa precipitate withAgNO3 inethanol

Insoluble inwater, but dissolves inaqueous NaOH uponheatingDissolves in diluteadditionof alkali

HCl in the cold and is reprecipitated by the

(a)(c)

A-3, B-2, C-1, D-4A-2, B-3, C-4, D-1

(b)(d)

A-2, B-3, C-1, D-4A-1, B-3, C-2, D-4

EXERCISE-II

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Amine

Q.8 In the following 2- reaction sequenceR-CH = CH2+ H2SO4 R-CH-CH3 the end product would be|SO4H

NaOH

R-CH-CH3(Where R = C14H29) |

SO4Na+

usefulas(a)Afertilizer (b)An explosive (c)Adetergent (d) None of these

Q.9 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 (d) Noneof these

Q.10 Which of the following will have highestKbvalue.

NH2NH2

(a) (b) (c) (d)N

HN ClMe

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 aminoacid is

(a) H2N CH COOH (b) H3N CH COO

R

CH

R

(d) Noneof these.(c)H3N COOR

Q.13 -NH2 group shows acidic nature while reacts with reagent.

(c) Br2 NaOH(a) Na (b) CS2 (d) Water

Q.14 Which of the following does nto giveethylamine on reduction(a) methyl cyanide

In thereaction,

(b) Ethyl nitrile (c) Nitro ethane (d)Acetamide

Q.15

excess CH3Cl

CH3NH2 (X) (Y)(Z) the final product (Z) is - -(a) (CH3 )3N (b) (CH3 )4N Cl (c) (CH3 )4N OH (d) (CH3 )2NH

Q.16 The product not obtained in the following reaction, CH3 NO2Cl2NaOHis(a) ClCH2NO2 (b) Cl2CHNO2 (c) Cl3CNO2 (d) CH3NH2

Q.17 Asequential reaction may beperformed as represented below:

R CH2CO2H(1)R CH2COCl( 2)R CH2CONH2

R CH NHR CH OH R CO H(3) 2 (4) 2 (5) 2

Theappropriate reagent for step (3) is

SO2 Cl2 NH3

(a) NaBr (b)Bromine and alkali (c) HBr (d) P2C5

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Amine

Q.18 Which of the following amine form N-nitroso derivative when treated with NaNO2and HCl?

(a)H3C NH2 NH2(b)

N(CH3)2 NH(CH3)(c) (d)

Q.19 Thestrongest baseamong thefollowing is

H2N H2N H2N H2N

C NH C NH2(a) (b) (c) C O (d) CH OH

H2N H2N H2N H2N

Q.20 Identify compound (A) inthe following oxidation reaction.

(A) 2 2 7 2 4

OH OHNH2

(a) (b) (c) (d)Allof these.

NH2 OHNH2

Q.21 Aniline isa weaker base than ethyl amine because(a) Phenyl gp inaniline isa +Rgp

(b)Ethyl gp inethyl amine decreases the electron density on nitrogen atom

(c) The lone pair ofelectron on nitrogenatomin aniline isdelocalized over aniline

(d)Aniline is less soluble inwater than ethylamine

Diazonium coupling reaction with aniline should becarried out inQ.22(a) Weaklybasic medium(c)Strongly basic medium

For CH3CHO, CH3NO2, CH3COOH(a)Allhave same chemical property(c)All are basic

Bromine in CS2 reacts withaniline to give

(b) Weaklyacidic medium(d) Stronglyacidic medium

Q.23

(b)All haveone common chemical behaviour(d) Noneof these

Q.24

NH2 NH2NH2 BrBr Br(a) (b) (c) (d) Both (a) and (b)

BrBr

Q.25 RNC cannot undergo(a)Acidic hydrolysis

(c) Basehydrolysis(b)Electrophillic, nucleophillic, addition on carbon(d) Both(b) & (c)

K

Cr

O

, H S

O

O O

28


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Amine

NH2

Q.26 + phosgene X. Here X is

ClO N = C = O

CH -C - HNH -C -Cl(a) (b) (c) (d) None of these

Cl

Q.27 Ethylamine undergoes oxidation in the presence of KMnO4 to give(a) CH3COOH (b) CH3CH2OH (c) CH3CHO (d) N-oxide

Q.28 Baker Mullikens test isused to detect the presence(a) -COOH gp (b) -NO2 (c) -OH (d) -NH2

Q.29 t-amines with different alkyl gphas a chiral nitrogen atomstillit isopticallyinactive because(a) ChiralN-atoms cannot rotates planepolarized light

(b) The lone pair prevents the rotation ofplane polarized light

(c) Both of these(d) Noneof these

In CH3NO2 we can observeQ.30(a) H-bonding(c) Tautomerism

Thereaction:

O

(b)-halogenation reaction(d)All of these

Q.31

(a) Carbylamine reaction(c) Gabrielphthalimide synthesis

The conjugate acid of HO(CH2)3NH2 is

(b)Hofmannreaction(d) Cope reaction .

Q.32+ + +-

(a)H2O(CH3)3NH2 (b)HO(CH2)3NH3 (c)O(CH2)3NH2 (d)HO(CH2)3NH

Q.33 Consider thefollowing compounds :1. H2C = CHCH2NH2 2. CH3CH2CH2NH2 3. HCCCH2NH2The increasing order of basicity is(a) 3 < 1 < 2 (b) 3 < 2 < 1 (c) 2 < 1 < 3 (d) None of these

NH2Q.34 During the conversion of with HNO2 at high temperatures the following substances or

entermediates are formed.

-N2Cl OH

1. N2O3

(a) 1 only

2. 3. 4. C6H5NH -N = O

(b) 1, 2, 4 only (c) 2, 4only (d) Allof these

O||C

||C +n-BuBrKOH :N:KNH

C||O

C||O

O||C 1. aq.NaOH n-BuNH2N Bu-n + iscalled2. H O+

3C||O

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Amine

Q.35 Match list I (condition of reaction of nitrobenzene) with list II (products formed ) andselect thecorrectanswerusing the codesgiven below.

List ISnand HCl

Zn and NH4ClMethanolic NaOMeZn and KOH

List II1.

2.

3.

4.

(A)(B)(C)(D)

HydrazobezeneAzoxybenzene

Phenyl hydroxylamine

Aniline(a) A-2, B-1, C-3, D-4(c) A-1, B-4, C-2, D-3

(b)A-4, B-3, C-2, D-1(d)A-1, B-3, C-2, D-4

Q.36 The increasing order of basicity of RCN, RCN = NR and RHN2is(a) RCN < RCH = NR < RH2N

(c) RCH = NR < RNH2 < RCN

(b) RNH2 < RCN < RCH = NR

(d) RH2N < RCH = NR < RCN.

(i) NH3 (A). What is (A) ?CH3 CO COOH(ii) H /Pd 2

(a) CH3CONH2

(c) CH3CH2CONH2

Q.37

(b) CH3 CO CONH2

(d) CH3CH(NH2 )COOH.

Q.38 How many isomeric amines with that formula C7H9N contain a benzene ring?

(a) two (b) three (c) four (d) five.Q.39 Isopropylamine can be obtainedby

LiAlH4 H2/ Ni(a) (CH3 )2 CHO NH2OH ?

H3C

(b)(CH3 )2 CHONH3 ?

CHOH NH3(c) (d)Allof these.H3C

Q.40 Reaction of RCONH2with a mixture of Br2 and KOH gives RNH2 as the main product. The interme-

diate involved inthe reaction is

OO

Br

(a)R C NHBr (b) RNHBr (c)R C N (d) RC = N = O.Br

Q.41 Amines arehighlysoluble in:(a)Alcohol (b) Diethylether (c) benzene (d) Water.

Q.42 Which of the following reagents canconvert benzene diazonium chloride into benzene?(a) Water(c) Hypophosphorous acid

Thebrominationof aniline produces

(b)Acid(d) HCl.

Q.43

(a)2-bromoaniline (b) 4-bromoaniline (c) 2,4,6-tribromoaniline (d) 2, 6-dibromoaniline.

Q.44 Thecompound, which on reaction with aqueous nitrousacid at lowtemperature produces anoilyni-trosoamine is

(a) Methyl amine (b) Ethylamine (c) Diethylamine (d)Triethylamine

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Amine

Q.45 Carbylamine test is performed inalcoholic KOH byheating a mixtureof(a) Chloroformandsilver powder

(b) Trihalogenated methaneanda primaryamine

(c)An alkyl halie anda primary amine

(d)An alkyl cyanide and a primary amine.

Comprehension :Arenediazonium saltsaremore stable thanalkanediazonium saltsdue to dispersalof thepositive charge

on the benzene ring. Obviously electron donating groups favour diazotisation by retarding the

decomposition ofdiazoniumsalts to phenyl cation.Thehighreactivelyof arenediazoniumsalts isdue to

theexcellent leaving abilityof thediazogroup asN gas.Therefore, diazonium saltsundergo a number2ofsubstitution reactions inwhich the diazo group is replaced bya monovalent atom/group suchasH(by

H PO in presence of Cu+ ions, CH CH OH, NaBH etc.), OH (by boiling in presence of mineral3 2 3 2 4acids), OCH (byheating with CH OH) Cl (by CuCl/HCl or Cu/HCl), Br (byCuBr/HBr or Cu/HBr)

3 3

I (by KI in presence of Cu+ ions), F (byfirst converting into N F followed byheating), CN (byfirst2 4

neutralizing with Na CO andthenreacting with KCN/CuCN),NO (byfirst neutralizing with Na CO2 3 2 2 3and then treating with NaNO ) phenyl or substituted phenyl (bytreating withbenzene or substituted

2

benzene inpresence of NaOH) etc.

Diazoniumsalts also couple with phenols and aromatic amines to formcoloured azo dyes. The reactivity

of diazoniumsalts towards coupling reactions isfavoured bypresence ofelectron withdrawing groups;

the reactivityof 2, 4, 6-trinitrobenzenediazonium chloride is so high that it even couples with reactive

hydrocarbons suchas mesitylene.

Consider the following ions:Q.46

I. Me N N+ N II. O N N+ N2 2

III. CH O N+ N IV. CH N+ N3 3

Thereactivity of these ions towardsazo 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 diazoniumsaltswhenboiledwith dil.H SO gives the corresponding phenolmost2 4

readily?

OMe

+

(a) N N (b) MeO N+ N

N+ N N+ N(c) Me (d)

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Q.48 Which of the following arylamines undergoes diazotisation most readily?

(a) NO NH (b) Cl NH2 2 2

(c) CH O NH (d) CH NH3 2 3 2

Q.49 Theproduct formed when bromobenzene reactswith benzenediazonium chloride inpresence of NaOH

is

(a) Diphenyl

(c)p, p-Dibromodiphenyl

(b)p-Bromodiphenyl

(d)p-Bromoazobenzene

Q.50 Benzendiazonium chloride onreaction withphenol inweaklybasic medium gives:

(a) Diphenyl ether

(c) ChlorobenzeneAssertion and Reason :

(b)p-Hydroxyazobenzene

(d) Benzene

Each of the questionsgiven belowconsists of twostatements, an assertion (A) andreason (R).

Select the numbercorresponding to the appropriate alternative as follows

(a) Ifboth A andRare true and Ris the correct explanation ofA, then mark(a)

(b) If both AandRare true but Risnot the correct explanation ofA, then mark(b)

(c) IfAis truebut Ris false, then mark(c)

(d) If both AandRare false, then mark(d)Q.51 A.

R.

A.R.

A:

R:

A:

R:

A:

R:

Benzyl amine ismore basic than aniline.Positive inductive effect of phenyl group creates high electron densityaroundN atom.

White precipitate of silver chloride get dissolved inNH4OH soln.NH3 reacts withAgCl to form a solution complex with formula [Ag(NH3)2]Cl.

o-nitrophenol ismore acidic thanp-nitrophenol.Nitro grouphas+M and -I effect.

3 amine isproved to be less basic inaq. solution.

Conjugate acid of3amine is poorlysolvated inaq. solution.

In order to convert R-Cl to pure R-NH , Gabriel-phthalimide synthesis canbe used.

Q.52

Q.53

Q.54

Q.552

With properchoice of alkyl halides,phthalmide synthesis canbeusedto prepare 1,2and 3amines.

4-Nitrochlorobenzene undergoes nucleophilic substitution more readily than chlorobenzene.Chlorobenzene undergoes nucleophilic substitution byelimination-addition mechanism while4-nitrochlorobenzene undergoes nucleophilic substitution byaddition-elimination mechanism.

1 amides react with Br + NaOHto give 1amines with one carbon atom less than the parent

Q.56 A:R:

Q.57 A:2

amide.R:

A:

Thereactionoccurs through intermediate formationof acylnitrene.

Acetamide reacts with Br in presence of methanoic CH ONa to form methylQ.582 3

N-methylcarbonate.R: Methyl isocyanate isformed as anintermediate which reacts with methanol to form methyl N-

methylcarbamate.

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Amine

EXERCISE - II

The hydrolysed products are aspartic acid andphenylalanine.Q.1

CHO COOH

N(CH3)2

Q.2 A = B = C =

N(CH3)2 N(CH3)2

Q.3 CH3C HC2H3 , The other isomers should be p-amines only|

NH2

NO2

Q.4

Q.10 C = C6H5COCl D = C6H5CONH2 E = C6H5CHNH2F = C6H5CN G = C6H5COOH

N(CH3)2

(3) H = EtNHBr (4) I =

NO

NO2-N = N -(5) J =O2N Me

(6) M = C6H5OH

OHCHO

(7) I = NaOH N =

SO3H OH

(8) O = P =SO

3

H OH

OEt

CH = N- NH- Ph(9) Q = Ph -O - Et R = S =EtO

CHO

(10) T =CH3- C - NH + para isomer

O Br

Cl

(11) A = Cu Powder V = + N2

ANSWER KEY

(1)(2)

K = C6H5SO3H L = C6H5SO3Na

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Amine

NH2Cl

Et

NH2

Et

CN COOH

Q.11 A B C DEt E

COOH COOHCOOH COOH

(E) (F) (G) (H)C -CH C - OMeC -CH COOH3 3Cl O OCl

O

CO(I)

C

O

Q.12 C2H5NH2

CH3 N H2Cl CH3-NH-

Q.13

(A) (B)

CH2 CH2CH2OH CH2Cl CH2CN CH2CH2NH2

Q.14 H2C H2C H2C H2C H2C NH

CH2OH

(A)

CH2Cl

(B)

CH2CN

(C)

CH2CH2NH2

(D)

CH2 CH2

(E)

OCH3 OCH3 OCH3

Cl C CH2CH2Br H3C C CH2CH2NH2H3C C CH2CH2NH-CH3

CH3(A)

CH3

(B)

CH3

(C)

OCH3 CH3OCH3

Q.15 CH2CH2OHH3C C CH2CH2N CH3H3C C

CH3

(E)

CH3

(D)

CH2CH3CH2CH3

TautomeriseOCH3CH2NH2 NQ.16 a N

H

CH2CH3CH2CH3

-H+

O N Nb

CH CH CH CH2 3H 2 3H

(CH3CH2 )2NH

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Amine

Q.17

Q.18

Aniline isa weak base thancyclohexylamine because of resonance.Use of excess ammonia reduces chances of reaction of 1 amine withalkyl halide to form2 and 3

amines.

Amino group, being activating group, activates bromination of aniline andformstribromoaniline.Dimethyl amine is stronger base because of inductive effect. Trimethyl amine isa weaker base because

positive charge onnitrogen couldbestabilized and due to crowdingbyalkyl groupsaround thenitrogen atom protonationcannot take place.

Q.19

Q.20

CH3

(N, N dim ethyl formamide)(A) NOCH

CH3

CH3

Q.21

HN(B)

CH3

C HCOOH

NH2

H3C CH3(A)

Q.22H2C(B)

CH3

ClOHCOOHCOOHO2N NO2O2N NO2OHNH

2Q.23

(A)Anthranilic acid

(B)salicylic acid NO2

(D)Picryl chloride

NO2(C)

picric acid

C6H5

C

H3C

(A)

HC OH H3C

O C

C6H5(B)

N

CC6H5

NOH

(C)Q.24 Acetophenone

O

E-Acetophenone oxime

O

Z-Acetophenone oxime

C6H5 NH2 C6H5 COOHC6H5 NH C CH3 C6H5 C NH CH3(G)

Benzoic acid

(D)Acetanilide

(F)

Aniline

(E)

N-methyl benzamide

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Amine

Cl Cl Cl Cl Cl Cl

CH2NO2 CH2NH2 CH2OH COOH NO2

Q.25

Cl

(A)

Cl

(B)

Cl

(C)

Cl

(D)

Cl

(E)

Cl

(F)

CH2CONH2CH2COOH CH2NH2 CH2OH O

CONH2 COOH C

Q.26 OCH3(C) CH3(A) (B) C(D) (E) (F) (G)

O

NH2 OH NC CN COOH

Q.27

COOHCH3

(A)

CH3

(B)

CH3

(C)

CH3

(D) (E)

Q.28 Degree of unsaturation ofA= 2.Since Aforms hydrochloride and dissolves inwater to givea neutral solution, it containsbotha basicandan acidic functional group. It is likely to be amino acid as the molecular formula contains one N and 2O- atoms. Ondecarboxylation it forms anamineB. Degree of unsaturationof B = 0.

Therefore, B isa saturatedamine. B reacts with NaNO2and dilute HCl forming (E) C2H5OH. Thus, B

is CH3CH2 NH2 .Aalso reacts with NaNO2and dilute HCl formingC, a hydroxyl acid, which

forms a cyclic diester on heating.All the reactionscan begiven as

H3C CH COOHNaOHCH CH -NH NaNO2HClCH CH -OH3 2 2(B)

3 2(E )

CaO

NH2(A)

NaNO2 HCl

H3C CH COOH

OH(C)

CH COOHCHHCH3C2 3-2H2 O CHCH3

O CO

O OOH

COOH H3C2 CH COOHCH3CH2 COOHH3C2 CH NH3P/Br 2

NH2Br

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Amine

CH2CH2OH CH2OH

H3C C CO2HH3C C CO2H C =Q.29 H3C C CO2HA = B =

HCO2H

D (CH3 )2 CHCO2C2H5

HCH3COCH3E (CH3 )2 CHCONH2 F (CH3 )2 CHNH2 G(CH3 )2 CHOH

CONH2 NH2 N2Cl

A = B = C =

CH3

CN

CH3

CO2H

CH3

COOH

E =Q.30 D = F =

CH3 COOHCH3

CO2H CO2H

G, H are

CO2H O2N CO2H

NO2

EXERCISE - III

Q.1Q.8

Q.15Q.22

Q.29

Q.36

Q.43

Q.50

Q.57

ac

cbdac

ba

Q.2Q.9

Q.16Q.23

Q.30

Q.37

Q.44

Q.51

Q.58

(a, b)d

dbdccca

Q.3Q.10

Q.17Q.24

Q.31

Q.38

Q.45

Q.52

ca

bcc

bba

Q.4Q.11

Q.18Q.25

Q.32

Q.39

Q.46

Q.53

dd

dcba

bd

Q.5Q.12

Q.19Q.26

Q.33

Q.40

Q.47

Q.54

bc

abaaaa

Q.6Q.13

Q.20Q.27

Q.34

Q.41

Q.48

Q.55

dc

ddddcc

Q.7Q.14

Q.21Q.28

Q.35

Q.42

Q.49

Q.56

bb

cbbc

bb

Documents

Amine IIT JEE Organic Chemistry