Question

$C_2{H}_5-CH=C{H}_2\underset{CC{l}_4}{\overset{B{r}_2}{\to }}A\underset{\left(excess\right)}{\overset{alc.KOH}{\to }}B$
A and B in the above reaction sequence are:

• A. $C_2{H}_5-\underset{Br}{\underset{|}{C}}H-\underset{Br}{\underset{|}{C}}{H}_2;C{H}_3-C{H}_2-C\equiv CH$
• B. $C{H}_3-\underset{Br}{\underset{|}{C}}H-\underset{Br}{\underset{|}{C}}{H}_2;C{H}_3-CH=CH-C{H}_3$
• C. $C_2{H}_5-\underset{Br}{\underset{|}{C}}H-\underset{Br}{\underset{|}{C}}{H}_2;C{H}_3-C\equiv C-C{H}_3$
• D. $\underset{Br}{\underset{|}{C}}H-(C{H}_2{)}_2-\underset{Br}{\underset{|}{C}}{H}_2;C{H}_3-C\equiv C-C{H}_3$

Answer 1

The correct option is A $C_2{H}_5-\underset{Br}{\underset{|}{C}}H-\underset{Br}{\underset{|}{C}}{H}_2;C{H}_3-C{H}_2-C\equiv CH$