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

Hint: The molecular weight of any organic compound is calculated by the addition of the atomic weights of the individual elements present in the compound. The total weight of the compound is given in the question and we know the atomic weight of bromine too. The molecular weight of the alkane can be now determined by basic mathematical calculations.

Complete answer:
The halogenation of any hydrocarbon includes the replacement of a hydrogen atom of the hydrocarbon with a halogen atom. The reaction can be written as follows:
$R - H + {X_2}(Cl,Br,I) \to R - X + H - X$
In the reaction, we can see that the ‘X’ atom replaces the hydrogen atom from the hydrocarbon. This is a normal substitution method. Now, coming to the mathematical calculation part, we have:
Atomic weight of bromine = $80$
Molecular weight of alkyl bromide ($RBr$) = $151$
The molecular weight of alkyl group ($R - $ ) = $151 - 80 = 71$
Now, we have to consider each of the compounds provided in the options.
(i) In the case of dodecane, the molecular formula is ${C_{12}}{H_{26}}$ .
The molecular weight of dodecane is = $12 \times 12 + 1 \times 26 = 170$
Thus, it is not correct.
(ii) In the case of 2,2-dimethylpropane, the molecular formula is ${C_5}{H_{12}}$ .
The molecular weight of 2,2-dimethylpropane is = $12 \times 5 + 1 \times 12 = 72$
As the alkyl group is formed by the replacement of a hydrogen atom from it, the molecular weight will have a value of 1 reduced from it. Thus, the molecular weight of the 2,2-dimethylpropyl group is equal to $71$ . Hence, this option is correct.
(iii) In the case of 2,2-dimethylhexane, the molecular formula is ${C_8}{H_{18}}$ .
The molecular weight of 2,2-dimethylhexane is = $12 \times 8 + 1 \times 18 = 114$
Thus, it is not correct.

(iv) In the case of 2,2,3-trimethylheptane, the molecular formula is ${C_{10}}{H_{22}}$ .
The molecular weight of 2,2,3-trimethylheptane is = $12 \times 10 + 1 \times 22 = 142$
Thus, it is not correct.

Hence, the correct option is B. 2,2-dimethylpropane.

Note:
Halogenation is a chemical reaction that involves the reaction of a compound with a halogen and results in the halogen atom being added to the compound. The molecular weight of the unknown alkyl group is calculated by reducing the atomic weight of the additional substituents present in the organic compound.