Question

$H_{3}C-C\equiv C-H\stackrel{NaN{H}_2/N{H}_3\left(l\right)}{\to }\left(A\right)\stackrel{H_{3}C-\stackrel{Br}{\stackrel{|}{\underset{C{H}_3}{\underset{|}{C}}}}-C{H}_3}{\to }$

In the reaction : the product (B) is :

• A. $H_{3}C-C\equiv C-\underset{C{H}_3}{\underset{|}{\stackrel{C{H}_3}{\stackrel{|}{C}}}}-C{H}_3$
• B. $H_{3}C-\underset{C{H}_3}{\underset{|}{C}}=C{H}_2$
• C. $H_3-C\equiv C-\underset{C{H}_3}{\underset{|}{\stackrel{C{H}_3}{\stackrel{|}{C}}}}-C{H}_3$ and $H_{3}C-\underset{C{H}_3}{\underset{|}{C}}=C{H}_2$$H_{3}C-\underset{C{H}_3}{\underset{|}{C}}=C{H}_2$ and $H_{3}C-C\equiv C-H$
• D. None of these

Answer 1

The correct option is B $H_{3}C-\underset{C{H}_3}{\underset{|}{C}}=C{H}_2$

(Refer to Image)

In the first reaction Actylenic Hydrogen is abstracted by strong base $NaN{H}_2$. This on reaction with $3^o$- alkyl halide elimination takes place and alkene is formed.