Question

An alkyl bromide, $RBr$ of molecular weight $151$ is the exclusive product of bromination of which hydrocarbon?

• A. Dodecane
• B. $2,2$ - dimethylpropane
• C. $2,2$ - dimethylhexane
• D. $2,2,3$ - trimethylheptane

Answer 1

Answer: (B)

$2,2$ - dimethylpropane

Molecular weight of $RBr=151$

Molecular weight of $Br=80$

$\therefore$ Molecular weight of $-R=151-80=71$

If we consider $2,2-dimethylpropane$ the molecular formula is $C_5{H}_{12}$ .

After bromination, $C_5{H}_{11}Br$

$\therefore$ Molecular weight of $C_5{H}_{11}=60+11=71$