Question

$0.0852\,g$ of an organic halide $(A)$ when dissolved in $2.0\,g$ of camphor, the melting point of the mixture was found to be $167^°C$. Compound $(A)$ when heated with sodium gives a gas $(B)$. $280\,mL$ of gas $(B)$ at $STP$ weighs $0.375\,g$. What would be $'A'$ in the whole process? $K_f$ for camphor $= 40$, $m$.$pt$. of camphor $= 179^°C$.

• A. $C_2H_5Br$
• B. $CH_3I$
• C. $(CH_3)_2CHI$
• D. $C_3H_7Br$

Answer 1

Answer: (B)

$\Delta T= 179 -167 = 12$,
$w=0.0852\, g$,
$W= 2\, g$,
$K_{f} = 40$,
molecular weight of $\left(A\right)$
$=\frac{1000\times K_{f}\times w}{\Delta T\times W}$
$=\frac{1000\times40\times0.0852}{12\times2}$
$=142$
$(A)$ undergoes Wurtz reaction to form $(B)$ i.e.
$(A) \xrightarrow{Na}(B) +NaX$
$(B)$ is an alkane say $C_nH_{2n + 2}$
$\because 280\, mL$ of $(B)$ weighs $0.375\, g$ at $NTP$
$\therefore 22400 \,mL$ of $(B)$ weighs $=\frac{0.375\times22400}{280}$
$=30\,g$ at $NTP$
$\therefore $ M.wt. of $(B) = 30$, $12n + 2n + 2 = 30$, $n = 2$
Thus $(B)$ is ethane and therefore $(A)$ is $CH_3X$.
The m.wt. of $CH_3X =142$
At. wt. of $X = 127\, \therefore X$ is iodine
Therefore alkyl halide is $CH_3I$. This reaction is
$\underset{\text{(A)}}{2CH_3I} \xrightarrow[\text{ether}]{Na}\underset{\text{(B)}}{C_2H_6}$