Question

a. The hydrogen atom of chloroform is acidic. Explain.
b. Why is dehydrohalogenation reaction in haloalkanes terms as Beta-elimination reaction?

Answer 1

Chloroform is a substance that has a strong non- irritating smell. It was used a s an anaesthetic in the old times but was discontinued due to deaths it caused. Chloroform acts as a Lewis acid due to the presence of the acidic hydrogen. The hydrogen is acidic due to electronegativity differences between $Cl$ atoms and it.
Dehydrohalogenation of haloalkanes is a β-elimination reaction as the hydrogen eliminated along with halogen is at the $\beta$- position.

Complete step by step solution:
(a) The hydrogen in chloroform is acidic and dehydrohalogenation reaction in haloalkanes is termed beta-elimination reaction. We have to give a reason why this occurs.
Hydrogen atoms of chloroform are acidic due to three electronegative chlorine atoms present on carbon. The latter requires a partial positive charge due to the $-{\text{ }}I$ effect of chlorine with the result it tends to attract electrons to the $C - H$ bonds towards itself. So the removal of the hydrogen atom as a proton becomes easy.
The structure of Chloroform is:

Since the cl attract the shared pair of electrons of the $C - Cl$ bond, there is a partial $+ ve$ charge on carbon, and as you know $C$ is a little more electronegative than $H$ , it in turn tries to pull the electrons from H. The presence of $3 - Cl$ groups aggravates the pull. Also CL accepts the lone pair left behind by $H$ and stabilizes the $-ve$ charge on $C$. thus chloroform behaves as an acid.

(b) In elimination reactions some molecules leave the reactant or are eliminated to form a double or triple bond. In a carbon chain, the carbon which carries the functional group is termed as $\alpha -$carbon while the carbon next to it is termed as $\beta$- carbon. The hydrogen attached to the $\beta$- carbon is termed as $\beta$- hydrogen.
In dehydrohalogenation of haloalkanes, alcohol $KOH$ is used. When alkyl halides have $\beta$- hydrogen are heated with alc. $KOH$, the halogen is eliminated from $\alpha -$position and hydrogen from $\beta$-position leading to formation of alkenes.

This reaction is called $\beta$-elimination because the relative position of eliminated substituents is 1,2 (and hydrogen is eliminated from $\beta$-position).

Note: Students must go through the steps to understand the concept. Also learning about the $-{\text{ }}I$ effect, $+ I$ effect, $+ R/ - R$ effect, etc is helpful in understanding the chemistry behind it. Also one must not confuse between elimination, addition, substitution and rearrangement reactions.