CBSE Class 12 Chemistry 2016 Outside Delhi Set 1 Solved Paper

Question : 26 of 26

Marks: +1, -0

(a) Write the structures of A and B in the following reactions :
(i)

{\CH}_3 {\COCl} \ {─────▸}↖{ {\H}_2, {\Pd}- {\BaSO}_4} {\A} \ {─────▸}↖{ {\H}_2 {\N}- {\OH}}\B.

(ii)

{\CH}_3 {\MgBr} \ {─────▸}↙{2 ⋅ {\H}_3 {\O}^{+}}↖{1.\CO_2} {\A} \ {─────▸}↖{ {\PCl}_5} {\B}

.
(b) Distinguish between :
(i)

{\C}_6 {\H}_5- {\COCH}_3

and

{\C}_6 {\H}_5- {\CHO}

,
(ii)

{\CH}_3 {\COOH}

and

{\HCOOH}

.
(c) Arrange the following in the increasing order of their boiling points :

{\CH}_3 {\CHO}, {\CH}_3 {\COOH}, {\CH}_3 {\CH}_2 {\OH}

.
OR
(a) Write the chemical reaction involved in WolffKishner reduction.
(b) Arrange the following in the increasing order of their reactivity towards nucleophilic addition reaction :

{\C}_6 {\H}_5 {\COCH}_3, {\CH}_3- {\CHO}, {\CH}_3 {\COCH}_3

{\CH}_3 {\CH}_2 {\CH}= {\CH}- {\CH}_2 {\CN}

\ {───────▸}↙ {\;\text " 2. "\; {\H}_2\O}↖{\;\text " 1. "\;( {\i}- {\Bu})_2 {\AlH} } \;\text " ? "\;

(e) A and B are two functional isomers of compound

{\C}_3 {\H}_6 {\O}

. On heating with

{\NaOH}

and

I_2

isomer

B

forms yellow precipitate of iodoform whereas isomer A does not form any precipitate. Write the formulae of

A

and

B

Solution:

(a) (i)

{\CH}_3 {\COCI}  \ {─────▸}↖{ {\H}_2, {\Pd}- {\BaSO}_4}

{{\CH}_3 {\CHO}}↙{[\A]\ \Acetaldehyde}  \ {─────▸}↖{ {\H}_2 {\N}- {\OH}}

{ {\CH}_3- {\CH}= {\N}- {\OH}}↙{[\B]\ \Acetaldoxime}

(ii)

{\CH}_3 {\MgBr} \ {─────▸}↙{2 ⋅ {\H}_3 {\O}^{+}}↖{1.\CO_2} {\CH_3\COOH}

\ {─────▸}↖{ {\PCl}_5} {\CH}_3 {\COCI}+ {\HCl}+ {\POCl}_3

(b) (i)

{\C}_6 {\H}_5 {\CHO}

being an aldehyde reduces Tollens' reagent to shining silver mirror whereas

{\C}_6 {\H}_5 {\COCH}_3

being a ketone does not.

(ii)

{\HCOOH}

gives silver mirror test with Tollens' reagent whereas ethanoic acid does not.

\; {\HCOOH}+2 [ {\Ag} ( {\NH}_3) _2] ^{+}+2 {\OH}^{-}

→ 2 {\Ag} ↓+2 {\H}_2 {\O}+ {\CO}_2+4 {\NH}_3

\; {\CH}_3 {\COOH} \ {─────▸}↖{\Tollens}↙{\;\text " reagent "\;} \;\text " No silver mirror "\;

OR(a)

(b)

{\C}_6 {\H}_5 {\COCH}_3 < {\CH}_3- {\COCH}_3

< {\CH}_3- {\CHO}

{\COOH}

groups thereby making the carbon atoms less electrophilic.

(d)

{ {\CH}_3 {\CH}_2 {\CH}= {\CH}}↙{\Hex-3-\enenitrile}- {\CH}_2 {\CN}

\ {───────▸}↙ {\;\text " 2. "\; {\H}_2\O}↖{\;\text " 1. "\;( {\i}- {\Bu})_2 {\AlH} } {\CH}_3 {\CH}_2 {\CH}

= {\CH}- {{\CH}_2- {\C}↖{∥}↖{\O}- {\H}}↙{\Hex-3-\enal}

(e)

{{\CH}_3 {\CH}_2 {\CHO}}↙{\Propanal\ [\A]}+ {\NaOH}+ {\I}_2 →

No yellow precipitate

{{\CH}_3- {\C}↖{∥}↖{\O}- {\CH}_3}↙{{\Acetone\ [\B] }}+3 {\NaOH}+4 {\I}_2 \ { ───▸}↖{∆}

{ {\CHI}_3 +3 {\NaI}}↙ {{\Iodoform (\Yellow \precipitate)}}+ {\CH}_3 {\COONa}+3 {\H}_2 {\O}

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