Question

Name the reagent used for the dehydrohalogenation of alkyl halides.
a) aqueous $KOH$
b) alcoholic $KOH$
c) Dry $A{g_2}O$
d) $Zn$

Answer 1

Hint: Dehydrohalogenation is an elimination process in which a strong base abstracts a ${\text{β - hydrogen}}$ from an alkyl halide. Recall the difference between reactions of alcoholic $KOH$ and aqueous $KOH$ with an alkyl halide. Alcoholic $KOH$ does elimination while aqueous $KOH$ does nucleophilic substitution reaction with alkyl halides.

Complete step by step solution:
Dehydrohalogenation word itself indicates the reaction process. “De” means removal, hydro refers to hydrogen and halogenations refers to halogens. Thus it is quite clear that dehydrohalogenation is a reaction in which there is the removal of hydrogen and halogen from the alkyl halides, so it is an elimination reaction. We get alkenes as a product after the dehydrohalogenation process.
We know that a base abstracts the hydrogen while a nucleophile attacks a carbon centre. Thus a strong base will act as a reagent for the dehydrohalogenation process.
Aqueous $KOH$ on dissociation in water produces $O{H^ - }$ (hydroxide) ions. Hydroxide ion is a good nucleophile and attacks at carbon. It removes halide ion from an alkyl halide and replaces it with $O{H^ - }$ ion. Hence, the reaction of alkyl halides with aq. $KOH$ is a nucleophilic substitution reaction and we get alcohols as a product here.
$R - Cl + KOH(aq.) \to R - OH + KCl$ (where, R= any alkyl group)
Now, comes to alcoholic KOH. It dissociates in water to give alkoxide ion ($R{O^ - }$) which is a strong base and can abstract ${\text{β - hydrogen}}$ from alkyl halides. Consequently, we get alkenes as a product hereafter elimination of ${\text{β - hydrogen}}$ and halide ion on ${\text{α - carbon}}$. Thus reaction of alkyl halides with alcoholic $KOH$ is an elimination reaction.
$R - C{H_2} - C{H_2} - Cl + KOH(alc.) \to R - CH = C{H_2} + KCl + {H_2}O$ (Where, R= any alkyl group)
Dry $A{g_2}O$ with alkyl halides give ethers ($(R - O - R)$ and $Zn$ do reductive dehalogenation of alkyl halides in the presence of proton donors like acids or water.

Hence, the suitable reagent used for dehydrohalogenation of alkyl halides would be alcoholic $KOH$. The correct option is (B).

Note: Two important points to remember-
-Alkoxide ions are stronger bases than hydroxide ions.
-Aqueous $KOH$ does dehalogenation (removal of halogen) while alcoholic $KOH$ does dehydrohalogenation (removal of hydrogen as well as halogen) of alkyl halides.