CBSE Class 12 Chemistry 2019 Outside Delhi Set 2 Solved Paper

Question : 25 of 27

Marks: +1, -0

SECTION - D

(a) Give equations of the following reactions:
(i) Phenol is treated with conc.

{\HNO}_3

.
(ii) Propene is treated with

{\B}_2 {\H}_6

followed by

{\H}_2 {\O}_2 \/ {\OH}^{-}

.
(iii) Sodium t-butoxide is treated with

{\CH}_3 {\Cl}

.
(b) How will you distinguish between butan-1-ol and butan-2-ol?
(c) Arrange the following in increasing order of acidity: Phenol, Ethanol, Water
OR
(a) How can you obtain Phenol from (i) Cumene, (ii) Benzene sulphonic acid, (iii) Benzene diazonium chloride?
(b) Write the structure of the major product obtained from dinitration of 3-methylphenol.
(c) Write the reaction involved in Kolbe's reaction.

Solution:

(a) (i)

(ii)

(iii)

{\Na}^{+} {\O}^{-}- {\C}↖{|}↖{{\CH}_3}↙{|}↙{{\CH}_3}- {\CH}_3+ {\CH}_3 {\Cl} → {\CH}_3 {\O}- {\C}↖{|}↖{{\CH}_3}↙{|}↙{{\CH}_3}- {\CH}_3+ {\NaCl}

Sodium t-butoxide

\;\;\;

Chloromethane

\;\;\;

2-Methoxy 2-Methyl propane

(b) Iodoform test: Butan

-2- {\ol}

(secondary alcohol) reacts with iodine and

{\NaOH}

to give yellow precipitate of

{\CHI}_3

(iodoform) whereas butan -1 -ol does not react with iodine and

{\NaOH}

{\CH}_3 {\CH}_2 {\CH↖{|}↖{\OH}\CH}_3+4 {\I}_2+6 {\NaOH} →

{ {\C}_2 {\H}_5 {\COONa}+ {\CHI}_3+5 {\NaI}+5 {\H}_2 {\O}}↙{(\Chloroform)}

{\OH}- {\CH}_2- {\CH}_2- {\CH}_2- {\CH}_3

+ {\I}_2+ {\NaOH} →

No reaction
OR

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