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Q. Assertion : On free radical monochlorination of $(CH_{3})_{2}CHCH_{2}CH_{3}$ four monochloro structural isomers are possible.
Reason: In $(CH_{3})_{2}CHCH_{2}CH_{3}$ there are four different types of hydrogen atoms.

Haloalkanes and Haloarenes

Solution:

$\left( CH _{3}\right)_{2}- CH - CH _{2}- CH _{3} \xrightarrow{ Cl _{2}}\left( CH _{3}\right)_{2}- CCl - CH _{2}- CH _{3}+$
$\left( CH _{3}\right)_{2}- CH - CH _{2}- CH _{2} Cl +$
$\left( CH _{3}\right)_{2}- CH - CHCl - CH _{3}+$
$\left( CH _{3}\right)- \underset{\overset{|}{CH_2Cl}}{CH} - CH _{2}- CH _{3}$
Due to the different types (four types) of hydrogen atoms are present in it, four monochloro structural isomers are possible.