CBSE Class 12 Chemistry 2017 Outside Delhi Set 1 Solved Paper

Question : 6 of 26

Marks: +1, -0

(a) Arrange the following compounds in the increasing order of their acid strength: p-cresol, p-nitrophenol, phenol.
(b) Write the mechanism (using curved arrow notation) of the following reaction:

{\CH}_2= {\CH}_2 { {─────▸}}↖{ {\H}_3 {\O}^{+}} {\CH}_3- { {\C}}↖{+} {\H}_2+ {\H}_2 {\O}

OR
Write the structures of the products when Butan-2ol reacts with the following :
(a)

{\CrO}_3

(b)

{\SOCI}_2

Solution:

(a) Increasing order of acidic strength:

p-nitrophenol in most acidic due to presence of

- {\NO}_2

group i.e., electron with drawing group (-I group) which stabilises phenoxide ion so increases acidity while p-serve is least acidic due to presence of

- {\CH}_ 3

group i.e., electron releasing group (+ I group) which destablises phenoxide ion so decreases acidity.
(b) Mechanism:
(i) Electrophilic attack of

{\H}_3 {\O}^{+}

{\H}_2 {\O}+ {\H}^{+} \;\text "(From acid) "\; → {\H}_3 {\O}^{+}

(ii) Nucleophilic attack of water on carbocation.

(iii) Deprotonation:

OR
(a) When Butan-2-ol reacts with

{\CrO}_3

, it forms butan-2-one.

{\CH}_3- {\CH}↙{|}↙{\OH}- {\CH}_2- {\CH}_3 {  {─────▸}}↖{ {\CrO}_3}

{ {\CH}_3 - {\C}↙{∥}↙{\O}- {\CH}_2- {\CH}_3}↙{\Butan-2-\one}+ {\H}_2 {\O}

(b) Butan-2-ol reacts with

{/SOCl}_2

to form 2-chlorobutane.

{\H}- {\C}↖{|}↖{\H}↙{|}↙{\H}- {\C}↖{|}↖{\H}↙{|}↙{\OH}- {\C}↖{|}↖{\H}↙{|}↙{\H}- {\C}↖{|}↖{\H}↙{|}↙{\H}- {\H}+ {\SOCl}_2 →

{{\H}- {\C}↖{|}↖{\H}↙{|}↙{\H}- {\C}↖{|}↖{\H}↙{|}↙{\Cl}- {\C}↖{|}↖{\H}↙{|}↙{\H}- {\C}↖{|}↖{\H}↙{|}↙{\H}- {\H}}↙{2-\Chlorobutane}+ {\SO}_2+ {\HCl}

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