Question

$C{H}_3-CH=C{H}_2\underset{\Delta }{\overset{B{r}_2}{\to }}\underset{Dryether}{\overset{Mg}{\to }}\underset{N{H}_{4}Cl}{\overset{C{H}_3-\stackrel{O}{\stackrel{||}{C}}-C{H}_3}{\to }}\underset{\Delta }{\overset{H^{+}}{\to }}\begin{array}{c}⨂\\ \left(Major\right)\end{array}$ End product of above reaction is:

• A. $C{H}_2=CH-C{H}_2-\stackrel{C{H}_2}{\stackrel{||}{C}-}C{H}_3$
• B. $H_{2}C=CH-CH=\underset{C{H}_3}{\underset{|}{C}-}C{H}_3$
• C. $H_{2}C=CH-C{H}_2-\underset{C{H}_3}{\underset{|}{\stackrel{OH}{\stackrel{|}{C}}}-}C{H}_3$
• D. $H_{2}C=CH-C{H}_2-\underset{C{H}_3}{\underset{|}{C}H}-C{H}_2-OH$

Answer 1

The correct option is B $H_{2}C=CH-CH=\underset{C{H}_3}{\underset{|}{C}-}C{H}_3$

$H_{2}C=CH-CH=\underset{C{H}_3}{\underset{|}{C}-}C{H}_3$

First dibromide forms which reacts with Mg and forms grinard reagent, which then reacts with ketone and in presence of acid it looses one molecule of water and forms final product.