Question

$H_{3}C-C{H}_2-\stackrel{C{H}_3}{\stackrel{|}{\underset{H}{\underset{|}{C}}}}-C{H}_{2}C{H}_3$

Maximum numbers of isomers(during stereoisomers) that are possible on monochlorination of following compound is:

• A. $2$
• B. $4$
• C. $6$
• D. $8$

Answer 1

The correct option is D $8$

$C{H}_3$

$H_{3}C-C{H}_2-\underset{|}{\stackrel{|}{C}}-C{H}_2-C{H}_3$

$H$

On monochlorination following possibilities are these.

$C{H}_3$

(1) $H_{3}C-C{H}_2-\underset{|}{\stackrel{|}{{}_{\ast }C}}-C{H}_2-C{H}_2-Cl$ $⟶1$ chiral centre gives $2^1$ isomers.

$H$

$C{H}_3$

(2) $H_{3}C-C{H}_2-\underset{|}{\stackrel{|}{{}_{\ast }C}}-\underset{|}{{}^{\ast }CH}-C{H}_3$

$H$ $Cl$

As two carbon become chiral thus it will give $2^2$ isomers.

$C{H}_3$

(3) $H_{3}C-C{H}_2-\underset{|}{\stackrel{|}{C}}-C{H}_2-C{H}_3$

$Cl$

$H$

(4) $H_{3}C-C{H}_2-\underset{|}{\stackrel{|}{C}}-C{H}_2-C{H}_3$

$C{H}_{2}Cl$

$2$ s $4$ isomers$⟶2R,3R,2R,3S$

$2S,3R,2S,3S$

Thus the total number of isomers= $2+4+2=8$ .