The monochloro isomer that predominates in the mixture resulting from the free-radical chlorination of $C H_{3} C H_{2} C H_{2} C H = C H_{2}$ is:

C l C H_{2} C H_{2} C H_{2} C H_{2} C H = C H_{2}

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C H_{3} C H C l C H_{2} C H = C H_{2}

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C H_{3} C H_{2} C H C l C H = C H_{2}

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C H_{3} C H_{2} C H_{2} C C l = C H_{2}

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Solution

The correct option is C $C H_{3} C H_{2} C H C l C H = C H_{2}$
Given compound is $C H_{3} C H_{2} C H_{2} C H = C H_{2}$
Primary hydrogens $= 3$
Secondary hydrogens $= 6$
Tertiary hydrogens $= 1$
Mono chloro derivatives relative rates would be
For primary monochloro derivative, the relative rate is $3 \times 1 = 3$
For secondary monochloro derivative, the relative rate is $6 \times 4 = 24$
For tertiary monochloro derivative, the relative rate is $1 \times 5 = 5$
We take selectivity to be $1$ for primary, $4$ for secondary and $5$ for tertiary.
Monochloro isomer that predominates in the mixture resulting from the free-radical chlorination of 1-pentene will be $C H_{3} C H_{2} C H C l C H = C H_{2}$

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