Question

How many monochloro structural isomers are expected in free radical monochlorination of $2-$methylbutane?

• A. $2$
• B. $3$
• C. $4$
• D. $5$
• E. $6$

Answer 1

The correct option is E $6$

Number of mono chloro isomers depends upon the types of hydrogen atoms present in the substate alkane. The structure of $2-$methylbutane is
$\stackrel{d}{C}{H}_3-\stackrel{c}{C}{H}_2-\stackrel{b}{H}\underset{\underset{a}{C}{H}_3}{\underset{|}{C}-}\stackrel{a}{C}{H}_3$
Because of the presence of four types of hydrogen, it gives $four+two(d$ and $l$ forms) $=$ six monochloro products on monochlorination as

$C{H}_3-C{H}_2\underset{C{H}_3}{\underset{|}{C}H}-C{H}_3\underset{hv}{\overset{C{l}_2}{\to }}\underset{\begin{array}{c}\text{d and l forms}\\ \text{1-chloro-2-methyl butane}\end{array}}{C{H}_{3}C{H}_2\dot{C}H-C{H}_{2}Cl}+\underset{2-chloro-2-methylbutane}{C{H}_3-C{H}_2-\stackrel{Cl}{\stackrel{|}{\underset{C{H}_3}{\underset{|}{C}}}-}C{H}_3}$

$+\underset{2-chloro-3-methylbutane}{C{H}_3-\underset{Cl}{\underset{|}{\dot{C}}}H-CH(C{H}_3{)}_2}+\underset{1-chloro-3-methylbutane}{C{H}_{2}ClC{H}_2-\underset{C{H}_3}{\underset{|}{C}H}-C{H}_3}$