Question

Only one product (including stereoisomers) was obtained by the reaction of $B{r}_2/hv$ with an unknown hydrocarbon. Vapour density of the product is 75.50. The name of the unknown hydrocarbon is: (Molecular weight of Br=80)

• A. Hexamethylcyclopropane
• B. Tetramethylbutane
• C. Neopentane
• D. Ethane

Answer 1

The correct option is C Neopentane

Let, the unknown hydrocarbon be $C_x{H}_y$.

$C_x{H}_{y}B{r}_2/h\nu \to C_x{M}_{y-1}B\gamma +HBr$

We conclude that given unknown hydrocarbon is alkane because

product doesn't show any stereoisomerism.

Thus,

$y=2x+2$

Given, vapour density $=75.50$ (of product)

$\Rightarrow \frac{\text{Molecular weight}}{2}=75.50.$

$\Rightarrow$

weight $=151$

$\text{So,}\begin{array}{c}12x+(2x+1)+80=151\\ \therefore x=5\end{array}$

Thus unknown hydrocarbon is neopentane. [option (C)]