JEE Main 2026 — Hydrocarbons Question with Solution

The general formula of an alkane is $C_n{H}_{2n+2}$.

The balanced chemical equation for the complete combustion of an alkane is:

$C_n{H}_{2n+2}+\left(\frac{3n+1}{2}\right)O_2\to nC{O}_2+(n+1)H_{2}O$

Given that $8$ moles of oxygen are required for complete combustion:

$\frac{3n+1}{2}=8$

$3n+1=16$

$n=5$

The molecular formula of the alkane $(Y)$ is $C_5{H}_{12}$.

Since $(Y)$ yields only one monochlorinated product $(Z)$ upon reaction with $C{l}_2/h\nu$, all the hydrogen atoms in the alkane must be equivalent.

Among the isomers of $C_5{H}_{12}$ (n-pentane, isopentane, and neopentane), only 2,2-dimethylpropane (neopentane) has all equivalent hydrogen atoms.

The structure of neopentane is $C(C{H}_3{)}_4$.

In this structure, the central carbon is quaternary, and the four surrounding methyl carbons are primary carbon atoms.

Therefore, the total number of primary carbon atoms in $(Y)$ is $4$.

Answer: $4$