Question

$ClC{H}_{2}CHC{l}_2\underset{KOH/\Delta }{\overset{alcoholic}{\to }}X\left(majorproduct\right)$
What will be the major product $X$ of the above reaction?

• A. $CHCl=CHCl$
• B. $C{H}_2=CC{l}_2$
• C. $HC\equiv CH$
• D. $ClC{H}_{2}CHO$

Answer 1

The correct option is B $C{H}_2=CC{l}_2$

During dehydrohalogenation of $1,1,2$-trichloroethane in presence of alcoholic $KOH$, the major product is $1,1$-dichloroethene. In $1,1,2$-trichloroethane, the $H$ atom which is attached to $C$ atom having $2Cl$ atoms is more acidic as two electron withdrawing $Cl$ atoms decrease electron density of $C$ to greater extent. Hence, cleavage of $C-H$ bond is easy.