Question

An organic compound with molecular formula $\mathrm { C } _ { 4 } \mathrm { H } _ { 10 } \mathrm { O }$ does not react with sodium. With excess of HI it gives only one type of alkyl halide. The compound is

• A. $\mathrm { C } _ { 2 } \mathrm { H } _ { 5 } \mathrm { OC } _ { 2 } \mathrm { H } _ { 5 }$
• B.
• C. $\mathrm { CH } _ { 3 } \mathrm { CH } _ { 2 } \mathrm { CH } _ { 2 } \mathrm { OCH } _ { 3 }$
• D. $\mathrm { CH } _ { 3 } \mathrm { CH } _ { 2 } \mathrm { CH } _ { 2 } \mathrm { CH } _ { 2 } \mathrm { OH }$

Answer 1

Answer: (A)

$\mathrm { C } _ { 2 } \mathrm { H } _ { 5 } \mathrm { OC } _ { 2 } \mathrm { H } _ { 5 }$

Answer 2

$\mathrm { C } _ { 4 } \mathrm { H } _ { 10 } \mathrm { O }$ can have two structures:

$\mathrm { CH } _ { 3 } \mathrm { CH } _ { 2 } \mathrm { CH } _ { 2 } \mathrm { CH } _ { 2 } \mathrm { OH }$ and $\mathrm { C } _ { 2 } \mathrm { H } _ { 5 } \mathrm { OC } _ { 2 } \mathrm { H } _ { 5 }$

Since it does not react with Na metal, it cannot be an alcohol.

$\mathrm { C } _ { 2 } \mathrm { H } _ { 5 } \mathrm { OC } _ { 2 } \mathrm { H } _ { 5 } + \underset { \text { excess } } { \mathrm { HI } } \longrightarrow 2 \mathrm { C } _ { 2 } \mathrm { H } _ { 5 } \mathrm { I } + \mathrm { H } _ { 2 } \mathrm { O }$