Question

$C{H}_3-CH=C{H}_2+NOCl\to P$

Identify the product P?

• A. $C{H}_3-\underset{Cl}{\underset{|}{C}}H-\underset{NO}{\underset{|}{C}}{H}_2$
• B. $C{H}_3-\underset{NO}{\underset{|}{C}H}-\underset{Cl}{\underset{|}{C}}{H}_2$
• C. $C{H}_3-C{H}_2-\stackrel{NO}{\stackrel{|}{\underset{Cl}{\underset{|}{C}}}H}$
• D. $\underset{NO}{\underset{|}{C}}{H}_2-C{H}_2-\underset{Cl}{\underset{|}{C}}{H}_2$

Answer 1

Answer: (A)

$C{H}_3-\underset{Cl}{\underset{|}{C}}H-\underset{NO}{\underset{|}{C}}{H}_2$

$C{H}_3-CH=C{H}_2+NOCl\to C{H}_3-\underset{Cl}{\underset{|}{C}}H-\underset{NO}{\underset{|}{C}}{H}_2$

A molecule of $NOCl$ is added across $C=C$ double bond. $NOCl$ is an asymmetrical reagent and addition follows Markonikov rule. The negative part (chlorine) is attached to more substituted $C$ atom.

Hence, option $A$ is correct.