Question

An alkane (Y) requires 8 moles of oxygen for complete combustion and on chlorination with \(Cl_2/h\nu\), (Y) gives only one monochlorinated product (Z). The total number of primary carbon atoms in (Y) is _____.

Answer 1

Correct Answer: 4

Concept: Combustion of alkane: \[ C_nH_{2n+2} + \frac{3n+1}{2}O_2 \rightarrow nCO_2 + (n+1)H_2O \] Step 1: {Use oxygen requirement}} Given: \[ \frac{3n+1}{2} = 8 \] \[ 3n+1 = 16 \] \[ 3n = 15 \] \[ n = 5 \] Thus the alkane is: \[ C_5H_{12} \] Step 2: {Identify the isomer}} Among pentane isomers:

• \(n\)-pentane
• isopentane
• neopentane

Only neopentane gives a single monochloro product because all hydrogens are equivalent. Structure: \[ C(CH_3)_4 \] Step 3: {Count primary carbons}} Neopentane contains four \(CH_3\) groups. Each is a primary carbon. Thus: \[ \text{Number of primary carbons} = 4 \]