Free radical chlorination of butane gives 2-chlorobutane and 1-chlorobutane. 2-chlorobutane formed is

$(\pm)$ isomer

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Meso

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(+)

isomer

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(-)

isomer

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Solution

The correct option is A
$(\pm)$ isomer

The free radical $C H_{3} ∙ C H C H_{2} C H_{3}$ during formation of 2-chlorobutane is planar. Its collision with $C l_{2}$ gives ( $\pm$ )2 chlorobutane as $∙ C l$ may attack on either side forming two molecules with opposite configuration.

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Similar questions

Which is the correct increasing order of boiling points of the following compounds?

1-iodobutane , 1-bromobutane, 1-chlorobutane, Butane

(a) Butane < 1-chlorobutane < 1-bromobutane < 1-iodobutane

(b) 1-iodobutane < 1-bromobutane < 1-chlorobutane < Butane

(d) Butane < 1-chlorobutane < 1-iodobutane < 1-bromobutane

n

72 %

2

28 %

1

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