Question

$CH_3 - CH = CH_2 \ce{->[Br_2/hv][\text{(low conc.)}]} (A) $;
Product $(A)$ of the reaction is :

• A. $CH_3 - \underset{\overset{|}{B}r}{C}H - CH_2 - Br$
• B. $H_2C = CH - CH_2 - Br$
• C. $CH_3 - \underset{\overset{|}{B}r}{C} - CH_2$
• D. $Br - CH_2 - CH_2-CH_2-Br$

Answer 1

Answer: (B)

$CH_3 - CH = CH_2 \ce{->[Br_2/hv][\underset{(FRSR)}{\text{(low conc.)}}]} \underset{\overset{|}{B}r}{C}H_2 - CH = CH_2$