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

Chemistry · Hydrocarbons

JEE Main 2026 — 6 April, Evening Shift — Question 61

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

Answer: 4

Numerical answer — enter this value.

Step-by-step solution

$\quad \mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}+2}+\left(\frac{3 \mathrm{n}+1}{2}\right) \mathrm{O}_{2} \rightarrow \mathrm{nCO}_{2}+(\mathrm{n}+1) \mathrm{H}_{2} \mathrm{O}$ $\frac{3 n+1}{2}=8$ $3 \mathrm{n}+1=8 \times 2$ $\mathrm{n}=5\left(\therefore \mathrm{C}_{5} \mathrm{H}_{12}\right)$ Number of primary carons $=4$

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Exam: JEE Main 2026
Paper: 6 April, Evening Shift
Subject: Chemistry
Chapter: Hydrocarbons
Topic: Preparation of alkanes