Question

An alcohol $(A)$ gives Lucas test within $5 \,\min$. $7.4\, g$ of alcohol when treated with sodium metal liberates $1120\, mL$ of $H_2$ at $STP$. What will be alcohol $(A)$?

• A. $CH_3(CH_2)_3OH$
• B. $CH_3CH(OH)CH_2CH_3$
• C. $(CH_3)_3COH$
• D. $CH_3CH(OH)CH_2CH_2CH_3$

Answer 1

Answer: (B)

$R O H+N a \rightarrow K O N a+1 / 2 H_{2} \uparrow$
We have to get molecular mass of alcohol corresponding to half mole of $H_{2}$ only.
No. of moles of $H _{2}= No$. of moles of alcohol.
$\frac{1120}{112000}=\frac{7.4}{M M} $
$\Rightarrow M M=74 $
$C_{n} H_{2 n+1} O H=74 \rightarrow C_{n} H_{2 n+1}=74-17=57 $
$\Rightarrow C_{n} H_{2 n}=57-1=56$
i.e. $ 12 n+2 n=14 m=56 $
$\Rightarrow n=56 / 14=4$
Thus molecular formula of $(A)$ is $C_{4} H_{9} O H .$
As $(A)$ gives Lucas test within $5 \min$., thus $2^{\circ}$ alcohol corresponding to molecular formula $C _{4} H _{9} OH$ is $CH _{3} CH ( OH ) CH _{2} CH _{3}$ (butan-$2$-ol).