Question

$C{H}_3-CH-C{H}_3+C{l}_2\stackrel{hv}{\to }P$ (Major product)

$|$ (on monochlorination)

$C{H}_3$

The product $P$ is:

• A. $C{H}_3-\stackrel{Cl}{\stackrel{|}{\underset{C{H}_3}{\underset{|}{C}}}}-C{H}_3$
• B. $C{H}_3-\stackrel{Cl}{\stackrel{|}{\underset{H}{\underset{|}{C}}}}-C{H}_2-C{H}_3$
• C. $H_{3}C-CH-C{H}_2-Cl$ $|$ $C{H}_3$
• D. $Cl$ $|$$C{H}_2-CH-C{H}_2$ $|$ $|$$Cl$ $Cl$

Answer 1

Answer: (C)

$H_{3}C-CH-C{H}_2-Cl$

$|$

$C{H}_3$

We know that in presence of light radical formation takes place and generally the most stable free radical led to the formation of respective alkyl halide. But there are exceptions as well. If the ratio of $1^{o}H$ to $2^{o}H$ or $3^{o}C$ is quite large then the major product will be $1^o$ alkyl halide.

Here, the primary H atom is abstracted in preference to tertiary H atom.

There are nine primary H atoms and only one tertiary H atom.

Ratio of $1^{o}H$ to $3^{o}H$ is $9:1$

$C{H}_3-CH-C{H}_3+C{l}_2\stackrel{hv}{\to }{H}_{3}C-CH-C{H}_2-Cl$

$|$ $|$

$C{H}_3$ $C{H}_3$

P (Major product)

(on monochlorination)

The major product of the monochlorination of isobutane is represented by option (C).