The monochloro isomer that predominates in the mixture resulting from the free radical chlorination of 1- pentene is:

C l 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 H = C H C l

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Solution

The correct option is B $C H_{3} C H_{2} C H C l C H = C H_{2}$
The monochloro isomer that predominates in the mixture resulting from the free radical chlorination of 1-pentene is an allylic chorination product $C H_{3} C H_{2} C H C l C H = C H_{2}$ .

$C H_{3} - C H_{2} - C H_{2} - C H = C H_{2} free radical chlorination - ------------- \to C H_{3} C H_{2} C H C l C H = C H_{2}$

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

An alkane with molecular formula $C_{6} H_{14}$ gives two monochloro alkanes on chlorination. What is the structure of alkane?

C H_{3} - C H (C l) - C H_{2} - C H_{2} - C H_{3}

C H_{3} - C H_{2} - C (C l) (C H_{3}) - C H_{3}

C H C l = C H C l, C H_{2} = C C l_{2}, a n d C C l_{2} = C H C l .

C H C l = C H C l, C H_{2} = C C l_{2}, a n d C C l_{2} = C H C l .

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