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Question: Name the reagent used for the dehydrohalogenation of alkyl halides. a) aqueous \(KOH\) b) alcoho...

Name the reagent used for the dehydrohalogenation of alkyl halides.
a) aqueous KOHKOH
b) alcoholic KOHKOH
c) Dry Ag2OA{g_2}O
d) ZnZn

Explanation

Solution

Dehydrohalogenation is an elimination process in which a strong base abstracts a β - hydrogen{\text{β - hydrogen}} from an alkyl halide. Recall the difference between reactions of alcoholic KOHKOH and aqueous KOHKOH with an alkyl halide. Alcoholic KOHKOH does elimination while aqueous KOHKOH 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 KOHKOH on dissociation in water produces OHO{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 OHO{H^ - } ion. Hence, the reaction of alkyl halides with aq. KOHKOH is a nucleophilic substitution reaction and we get alcohols as a product here.
RCl+KOH(aq.)ROH+KClR - 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 (ROR{O^ - }) which is a strong base and can abstract β - hydrogen{\text{β - hydrogen}} from alkyl halides. Consequently, we get alkenes as a product hereafter elimination of β - hydrogen{\text{β - hydrogen}} and halide ion on α - carbon{\text{α - carbon}}. Thus reaction of alkyl halides with alcoholic KOHKOH is an elimination reaction.
RCH2CH2Cl+KOH(alc.)RCH=CH2+KCl+H2OR - 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 Ag2OA{g_2}O with alkyl halides give ethers ((ROR)(R - O - R) and ZnZn 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 KOHKOH. The correct option is (B).

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