Question

An isomer (X) of a hydrocarbon of molecular formula $C_6H_{14}$ on monochlorination gives only two structural isomers. How many primary hydrogen atoms are there in a molecule of (X)?

Answer 1

12

Answer 2

Solution:

1. The given information tells us that on monochlorination of the hydrocarbon (X) of formula $C_6H_{14}$ only two structural isomers are formed. This indicates that (X) must be 2,3-dimethylbutane since any other isomer would give more chlorination products.

2. Structure of 2,3-dimethylbutane:

   $$CH_3 - CH(CH_3) - CH(CH_3) - CH_3$$

3. Counting primary hydrogens:

   • Terminal CH$_3$ groups (at both ends) are attached to only one carbon: each contributes $3$ primary hydrogens → $2 \times 3 = 6$.
   • The two methyl groups attached to the inner CH groups are also primary (each attached to just one carbon): each contributes $3$ hydrogens → $2 \times 3 = 6$.

   Hence, total primary hydrogens $= 6 + 6 = 12$.

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Minimal Explanation:
Since (X) is 2,3-dimethylbutane, it contains two terminal CH$_3$ groups (6 hydrogens) and two substituent CH$_3$ groups (another 6 hydrogens). Total $12$ primary hydrogens.