CHAPTER

4What are the missing compounds/

reagents ?

Q1. Complete the following :

(A) Hydrocarbons and halogen derivatives.

(i) CH3CH2CH = CH2 + HBr (AI 2009, Foreign 2011) (ii) CH3CH2CH = CH2 + HBr

Peroxide (Delhi 2008)

(iii) CH2 = CH2 + Br2 CCl

4 (Delhi 2008, AI 2008) (iv) C6H5N2Cl + KI

(v) C = C

H

H

H+ HBr

(Foreign 2011)

(vi) (CH3)3CBr + H2O Heat

(vii) CH3CH2Cl + SbF3 Heat

(viii) CH3CH2C ≡ CNa + CH2 = CHCH2Br Liq. NH

3

(ix)

CH

CH3

+ Br2

heat

CH3

(x) OH

red P

I2

1

2 (xi) CH3CH2Cl + KNO2 Two products

(xii) CH3

Br C H2 5

ONa

C H2 5

OH

(xiii) + Br2

heat

light

(Delhi 2012C)

(xiv)

OH

CH2OH

+ HCl

(xv)

CH3

+ HI (Delhi 2012C, AI 2009, Foreign 2011)

(xvi) CH3CH = CHCH2Cl + CN– (A) + (B) (2012C)(B) Alcohols, phenols and ethers.

(i) OH + SOCl2

(Delhi 2009)

(ii) CH OH2

HO

+ HCl

(Delhi 2009)

(iii) OC H2 5

+ HBr (Foreign 2012)

(iv) (CH3)3C – O – C2H5 + HI (Foreign 2012) (v) CH3CH2CH2OCH3 + HBr (Foreign 2012)

(vi) + HI

(C) Aldehydes, ketones and carboxylic acids.

(i) CH3C≡CH Hg2+

H SO2 4

(Delhi 2012)

(ii) O3

H O2

2 O..... (Delhi 2008, AI 2011)

(iii) CH2

CHO (Delhi 2008, AI 2011)

3

(iv)

CH3

NO2

(i) CrO Cl2 2

(ii) H O3

+ (Delhi 2012)

(v) + .....anhy. AlCl

3

C HOC3

(Delhi 2009)

(vi) + .....

C

O

..... (Delhi 2009)

(vii) + C H COCl2 5

anhy. AlCl3

CS2

(Delhi 2011C, 2012, AI 2012C)

(viii) CH

3

O

+ C H NH3 2

H C2

H+

(AI 2011C)

(ix) C6H5CHO H NCONHNH

2 2 (AI 2011) (x)

O

C HConc. KOH (Delhi 2013)

(xi) C6H5CHO + C6H5COCH3 293 K

OH–

(Delhi 2011C)

(xii)

O

C CH3

Conc. HCl

Zn Hg (AI 2013)

(xiii) C6H5–COCl H

2

Pd BaSO4

(Delhi 2009A, 2011C, AI 2013)

(xiv) CH

3

O

H CrO2 4 (AI 2011C)

(xv)

CHO

HNO /H SO3 2 4

273-283 K

(Delhi 2013)

5

(xvi) CH

2CH

3KMnO

4

KOH, heat

......

(Delhi 2008, 2011, Foreign 2011)

(xvii) KMnO4

H SO , heat2 4

(AI 2011C)

(xviii) COOH

SOCl2

heat

COOH

(Delhi 2011, AI 2011, Foreign, 2011)

(xix) CH3COOH Br P

2/

(Delhi 2013)

(xx)

COOH

Br /FeBr2 3 (AI 2013)

(xxi)

O

R NH2

LiAlH4

H O2

(AI 2009)

(xxii) C6H5CONH2 H O

3

+

heat (AI 2009)

(xxiii)

O

CH3

NH2

Br + NaOH2 (Delhi 2011, Foreign 2011)

(xxiv) O

CH2

OCH3

O

+ NaBH4

(D) Amines.

(i) C2H5NH2 + C6H5SO2Cl (AI 2009) (ii) C6H5NH2 + CH3COCl (AI 2009) (iii) C2H5NH2 + HNO2 (AI 2009) (iv) RNH2 + CHCl3 + KOH (Delhi 2010, AI 2013) (v) C6H5NH2 + Br2 (aq.) (AI 2009, 2012, 2013, Foreign 2012) (vi) C6H5NH2 + HCl (aq.) (AI 2013) (vii) C6H5N2Cl + KI ........ (Delhi 2008) (viii) Diazonium group + –I (Delhi 2008, AI 2008)

6

(ix) C6H5N2Cl + H3PO2 + H2O (AI 2009, 2012, Foreign 2012, AI 2013) (x) C6H5N2Cl + CH3CH2OH (Delhi 2010) (xi) C6H5N2Cl + C6H5NH2

OH (Delhi 2010)

(xii) C6H5N2+Cl– H O

2

room temp. (AI 2013)

Q2. Identify the missing compounds (reagents) in the following:

(i) C2H5Cl NaCN A Reduction

Ni/H2

B (AI 2010)

(ii) CH3CH2Br KCN A LiAlH

4 B HNO

2

0°C C (AI 2013)

(iii) C6H5NH2 NaNO

2

HCl A

C H NH6 5 2

OH

B (AI 2010)

(iv) C6H5N2+Cl– CuCN A

H O/H2

+

B NH

3

heat C (Delhi 2013)

(v) C6H5NO2 Sn + HCl A NaNO + HCl

2

273 K B

H O/H2

+

heat C

(Delhi 2013)

(vi) NH3

∆NaOH + Br

2 CHCl3

alc. KOHCH COOH

3A B C (AI 2013)

(vii) CH CHO3

(i) C H MgCl2 5

(ii) H O2

AConc. H SO

2 4

BHBr

peroxideC

(Delhi 2011C)

(viii)

(AI 2008C) (ix)

(AI 2009C)

(x)

COOH

+ NH3

COOH

HeatA B C

strong

heating

(AI 2011C)

7

(xi)

COOH

Conc. HNO3

Conc. H SO2 4

ASOCl

2

B

(i) NaBH4

(ii) H O3

+

SOCl2

DH , Pd, BaSO

2 4

S or quinolineE

C

(xii) 273-278 K

H O , heat3

+

KMnO , KOH,4

heat

D

A

CH3

+ CrO3+ (CH CO) O

3 2

H3

+

O

E

B

Conc. NaOH

COONa

C +

(AI 2010C)

8

SOLUTION

(A) Hydrocarbons and halogen derivatives.

1. (i) CH3CH2CH = CH2 + HBr Mark. addition CH3CH2CHCH3

| Br

(ii) CH3CH2CH = CH2 + HBr Peroxide

anti-markow.

addition

CH3CH2CH2CH2Br

(iii) H2C = CH2 + Br2 CCl

4 BrCH2 – CH2Br (iv) C6H5N2Cl + KI C6H5I + N2 + KCl

(v) C

H

H

H

Br H

(vi) (CH3)3COH (vii) 3CH3CH2Cl + SbF3

Heat 3CH3CH2F + SbCl3 (viii) CH3CH2C ≡ CCH2CH = CH2

(ix) ( )

(x) OH

red P

I

I2

(xi) CH CH Cl + KNO3 2 2

N�O

O

+ CH3CH

2O N=O

Nitroethane Ethyl nitrite

(xii) CH3

BrC H ONa/C H OH

2 5 2 5

( HBr)

CH3

1-Methylcyclohexene

(xiii)

Br

3-Bromocyclohexene

9 In presence of heat and light, allylic bromination takes place, while

in dark bromine adds on double bond to form dibromoderivative.

(xiv)

OH

CH2Cl

HCl reacts with alcohols (–CH OH) and not

with phenols.

2

(xv) CH

3

I HI adds according to Markownikof’s rule.

(xvi) The reactant is allylic compound, so it can form two products. CH3–CH = CH–CH2Cl CH3CH = CH–CH2 + Cl–

CH3CH= CH CH

2

+

CN–

CH3CH =CH CH

2

CN–

CH3CH= CH CH

2CN CH

3CH =CH CH

2

CN

+

(B) Alcohols, phenols and ethers.

(i) Cl + SO + HCl2

(ii) CH OH2

HO

+ HCl

CH2Cl

HO

+ H O2

(Phenolic –OH is not easily replaced, hence does not react with HCl)

(iii) OC H2 5

+ HBr

OH

+ C H Br2 5

C6H5–O bond has some double bond character (see below), thus it is shorter and stronger, and hence difficult to be cleaved.

C H2 5

C H2 5

+

C H2 5

+

10 Moreover, C2H5

+ is more stable than C6H5+.

C6H5OC2H5 H

+

C6H5O+ C2H5

C6H5+ + HOC2H5 C6H5O

+C2H5 C6H5OH + C H2 5

(more stable)

Br

C H2 5

Br

+

(less stable)

(iv) (CH3)3C – O – C2H5 + HBr (CH3)3CBr + C2H5OH (CH3)3C

+ (a 3° carbocation) is more stable than the C2 H5

+, hence it is easily formed.

(v) CH3CH2CH2OCH3 + HBr CH3CH2CH2Br + CH3OH + CH3CH2CH2 is more stable than CH3

+ due to inductive effect (CH3CH2 →– CH2

+) as well as hyperconjugation.

(vi) + HI CH I2

+ HO

CH2

+

(benzyl carbocation) is more stable due to resonance than

+

(C) Aldehydes, ketones and carboxylic acids.

(i) Hg , H SO

2+

2 4

(hydration)CH CH

3≡C CH

3= CH

2

OH

CH3

CH3

O

(ii) OO

3

Zn-H O2

2

In ozonolysis, C=C is always cleaved to C=O + O=C .

(iii) CH2

CH3B H /THF

2 6

3

BH

2 2O

NaOH

11

CH2OH3

PCCCHO3

Cyclohexane carbaldehyde

(iv)

CH3

NO2

(i) CrO Cl2 2

(ii) H O3

+

(Etard oxidation)

CHO

NO2

(v) + CH COCl3

anhy. AlCl3

(Friedel-Craft reaction)

COCH3

Acetophenone

(vi) C

O

+

COClanhy. AlCl

3

(Friedel-Craft

reaction)

Benzolphenone

(vii) + C H COCl2 5

anhy. AlCl , CS3 2

(Friedel-Craft

reaction)

Propiophenone

COC H2 5

(viii) H+

CH3

O + H NCH CH2 2 3

CH3

Schiff’s base

NCH CH2 3

(ix) C6H5CHO + H2NNHCONH2 H

+

C6H5CH = NNHCONH2 + H2O Benzaldehyde semicarbazone

Note that the –NH2 end of –CONH2 does not react because its

N C N H2

O

H

N C = N H2

O

H +

nucleophilic character is decreased due to resonance.

(x) 2H HConc. KOH

(Cannizzaro reaction)

O

CH OH + HCOONa3

Methanol Sod. methanoate

12 (xi) OH , 293K

–

crossed aldol

condensation

C H CHO + H CCOC H6 5 3 6 5

Benzaldehyde Acetophenone

C H CH = CHCOC H6 5 6 5

Benzal acetophenone

(xii) CH3COCH3 Zn–Hg/HCl

Clemmensen reduction CH3CH2CH3

(xiii) H

2Pd BaSO

4

(Rosenmund reduction)C H COCl

6 5C H CHO

6 5

Benzaldehyde

(xiv) CH

3

O

H CrO2 4

(oxidation)

COOH

(xv) CHO

HNO /H SO3 2 4

273 283 K

m-Nitrobenzaldehyde

CHO

NO2

(xvi) Alkyl group present on benzene ring is always oxidised to –COOH group, irrespective to its length, when oxidised by a strong oxidis-ing agent.

CH CH2 3

KMnO4

KOH, heat

COOK

(xvii) KMnO , H SO4 2 4

(Oxidation)

COOH.......

COOH

Cyclohexene Hexan-1, 6-dioic acid

(Adipic acid)

(xviii) COOH

COOH

SOCl2

heat

COCl

COCl

Phthaloyl chloride

(xix) Br /P2

HVZ reactionCH COOH

3BrCH COOH

2

Acetic acid α-Bromoacetic acid

(xx)

COOH

Br /FeBr2 3

COOH

Br

13

(xxi)

O

R NH2

LiAlH , H O4 2

(reduction)R – CH – NH

2 2

Alkanamide Alkanamine

(xxii) H O , heat

3

+

(hydrolysis)C H CONH

6 5 2C H COOH + NH

6 5 3

Benzoic acidBenzamide

(xxiii)

O

Hofmann

bromamide reactionCH – C – NH

3 2+ Br + 4NaOH

2

2 3 2+ 2NaBr + Na CO + 2H OCH NH

3 2

(xxiv) Sodium borohydride (NaBH4) is a weak reducing agent, hence it reduces only aldehydes/ketones, but not esters.

O

CH2

OCH3

ONaBH

4 O

CH2

OCH3

OH

(D) Amines.

1. (i) C2H5NH2 + C6H5SO2Cl C2H5NHSO2C6H5 N-Ethyl benzenesulphonamide

(ii) C6H5NH2 + CH3COCl C6H5NHCOCH3 Acetanilide

(iii) C2H5NH2 + HNO2 C2H5OH + N2 + H2O

(iv) Carbylamine

reactionRNH + CHCl + 3KOH

2 3N == C + 3KCl + 3H O

2

�

Alkyl isocyanide

(v)

NH2

Aniline

NH2

+ Br2(aq.)

Br Br

Br

2, 4, 6-Tribromoaniline

(–NH2 group is highly activating and aqueous Br2 solution is highly ionising to give Br+ ions, hence tribromo product is formed)

(vi) C6H5NH2 + HCl (aq.) C6H5NH3+Cl–

(vii) C6H5N2Cl + KI C6H5I + N2 + KCl

14

(viii)

N2Cl

+ KI

I

+ KCl + I2

Benzene diazonium

chloride

Iodobenzene

+

(ix) C6H5N2Cl + H3PO2 + H2O Reduction

C6H6 + N2 + H3PO3 + HCl

(x) C6H5N2Cl + CH3CH2OH Reduction

C6H6 + CH3CHO + N2 + HCl Benzene

(xi) N+ ≡ + NH

2

OH

Coupling reaction

p-Aminoazobenzene

(azo dye)

N = N NH2

(xii) C6H5N2+ Cl–

H O2

room temp. C6H5OH + N2 + HCl

2. (i) C H

2 5CH NH

2 2

(B)

C H Cl2 5

NaCN Reduction

Ni/H2

C H CN2 5

(A)

(ii) CH CH3 2

CH NH2 2

(B)

CH CH Br3 2

KCNCH CH

3 2CN

(A)

LiAlH4

HNO2

0°CCH CH

3 2CH OH

2

(C)

(iii) C H NH6 5 2

NaNO2

HClC H N

6 5 2Cl

C H NH6 5 2

OH

(A)

C H N = N

6 5NH

2

(B)

(iv) CuCN H O/H2

+

C H CN6 5

C H COOH6 5

NH3

Benzonitrile (A) (B)

C H N6 5 2

Cl+

C H COONH6 5 4

heatC H CONH

6 5 2

Benzamide (C)

15

(v) C H NO6 5 2

Sn + HCl NaNO + HCl2

273 KC H NH

6 5 2C H N Cl

6 5 2

Aniline (A) (B)

H O/H2

+

heatC H OH

6 5

Phenol (C)

(vi) CH COOH3

NH3 ∆

(–H2O)

CH CONH3 2

CH COONH3 4

(A)

CH NH3 2

NaOH + Br2

Hofmann

bromamide reaction

CHCl + alc. KOH3

(Carbylamine

reaction)Methyl amine

CH N3

==�

C

Methyl isonitrile

(C)(B)

(vii) CH CHO3

(i) C H MgCl2 5

(ii) H O2

Conc. H SO2 4

CH C H52

OH

(A)

HBr + peroxide

anti-Mark. additionCH CH = CHCH

3 3C H

2H C CH CH

3 3

Br

But-2-ene (B)

(Major)

2-Bromobutane (C)

(viii)

(ix)

16

(x)

(xi)

SOCl2

COOH

COCl

(D)

H2, Pd, BaSO + S or quinoline

4

(Rosenmund reduction)

CHO

(E)

Conc. HNO3

COOH

NO2

SOCl2

(A)

(i) NaBH

(ii) H O

4

3

+

COCl

NO2

(B)

NO2

(C)

CH2OH

NaBH does not

reduce –NO gp.

4

2

Conc. H SO42

17

(xii)