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 ${167}^{\circ }$C. Compound (A) when heated with sodium gives a gas (B). 280 mL of gas (B) at STP weighs 0.375g. What would be 'A' in the whole process? $K_f$ for camphor = 40, m.pt. of camphor = ${179}^{\circ }$C.

• A. $C_2{H}_{5}Br$
• B. $C{H}_{3}I$
• C. $(C{H}_3{)}_{2}CHI$
• D. $C_3{H}_{7}Br$

Answer 1

The correct option is B $C{H}_{3}I$

$\delta$=179-167=12, w=0.0852g, W=2g,Kf= 40
Moleculer weight of (A)
=$\frac{1000\times Kf\times w}{\delta T\times W}$
=$\frac{1000\times 40\times 0.0852}{\delta 12\times 2}$
=142
(A) undergoes wortz reaction to form (B) i.e
$\left(A\right)\to Na\left(B\right)+NaX$
(B) is an alkane say $C_n{H}_{2}n+2$
$\therefore$ 280 mL of (B) weighs 0.375 g at NTP
$\therefore$ 22400 mL of (B) weighs
=$\frac{0.375\times 22400}{280}$
=30 g at NTP
M.wt of (B)=30, 12n+2n+2=30,n=2
Thus (B) is ethane and therefore (A) is $C{H}_{3}X$
The m.wt of $C{H}_{3}X$=142
At.wt of X=127
$\therefore$ is iodine
Therefore alkyl halide is $C{H}_{3}I$.
This reaction is
2$C{H}_{3}I\underset{Na}{\overset{ether}{\to }}{C}_2{h}_6$
(A) (B)