JEE Main 2026 Apr 6 Shift 2, Chemistry Q22: Naming of Alkanes

JEE Main2026Apr 6, Shift 2Chemistry

An alkane $(Y)$ requires 8 moles of oxygen for complete combustion and on chlorination with Cl $_{2}$ / $h ν$ , $(Y)$ gives only one monochlorinated product $(Z)$ . The total number of primary carbon atoms in $(Y)$ is _____.

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Solution

Combustion: $C_{n} H_{2 n + 2} + \frac{3 n + 1}{2} O_{2} \to n CO_{2} + (n + 1) H_{2} O$ .

Given $\frac{3 n + 1}{2} = 8$ $\Rightarrow$ $n = 5$ . So $Y$ is a C $_{5}$ H $_{12}$ isomer.

For only one monochlorinated product, every hydrogen must be equivalent. The C $_{5}$ H $_{12}$ isomer with this property is neopentane, (CH $_{3}$ ) $_{4}$ C: all 12 hydrogens are on equivalent primary carbons.

Number of primary carbons in neopentane = 4 (the four methyl groups attached to the central quaternary carbon).

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JEE Main 2026 Apr 6 Shift 2, Chemistry Q22: Naming of Alkanes

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