The reagent used in dehydrohalogenation is
A.\[{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\]
B.\[{{\text{K}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\]
C. Alcoholic KOH
D. None of the above.

Hint: It occurs in a medium having a strong base. Thus when there is a strong base which favours elimination than substitution, then dehydrohalogenation can occur.

Complete step by step solution:
Let us try to understand Dehydrohalogenation first.
-Dehydrohalogenation is a reaction where a stronger base abstracts \[\text{ }\!\!\beta\!\!\text{ }\] hydrogen followed by eliminating halogens at adjacent carbon.
-Therefore, in this reaction strong base first abstracts the \[\text{ }\!\!\beta\!\!\text{ }\] hydrogen thereby collapsing the carbon \[\text{ }\!\!\beta\!\!\text{ }\]beta hydrogen bonds to form double bonds followed by breaking the carbon halogen bond thereby releasing the halide in that same medium. Therefore it generally proceeds through the E2 mechanism. (where the collapse of Beta hydrogen bond and leaving of halide occurs simultaneously.
Let us understand the concept of dehydrohalogenation through some examples.
-Dehydrohalogenation can be used to obtain alkene from Alkyl halides-

-Dehydrohalogenation can be used to obtain Diene from Alkyl halides-


-Dehydration of Vic-dihalides, Gem-dihalides to form alkynes:


Here, the second step is slower as vinyl halide is resonance stabilized therefore highest temperature and stronger bases are required.
When we use alcoholic KOH is present as a reagent, the negative part of the reagent, that is $\text{O}{{\text{H}}^{-}}$it acts as a base and abstracts the $\text{ }\!\!\beta\!\!\text{ }$ hydrogen from the saturated substrate ( alkyl halide) present and transforms it to an alkene in the product, thereby undergoing elimination reaction.

Therefore, Option C is the correct option.

Additional information: Sodium amide, $NaN{{H}_{2}}$, in liquid ammonia as a solvent was the preferred base for many years. In some cases, $NaN{{H}_{2}}$ is used in mineral oil. Now amide ions such as lithium diisopropylamide, LDA, is routinely used. A hot, alcoholic KOH solution or an alkoxide ion, especially potassium t-butyl alcohol, DMSO, or in THF is also commonly used to effect elimination of HX from vinyl halide. The reaction can be carried out stepwise, via the formation of a vinyl halide, or in one step, generating the alkyne directly. Thus the addition of 1,2-dibromohexane to sodium amide in liquid ammonia followed by evaporation of the solvent and aqueous workup gives 1-hexyne.


Three equivalents of $NaN{{H}_{2}}$ is necessary for the preparation of terminal alkyne because as the alkyne forms, its acidic terminal hydrogen immediately protonates an equivalent amount of base. Because vicinal dihalides are readily available from alkenes through halogen addition and geminal dichlorides are easily available by the treatment of $PC{{l}_{5}}$ on aldehyde or ketone the above double dehydrogenation is a unique technique to obtain alkyne.

Note: Since elimination and substitution can both occur in the same medium therefore for dehydrohalogenation we need to provide better reagent and reaction condition for elimination to undergo dehydrohalogenation quite easily. Therefore we have to use strong bases like alcoholic KOH,$O{{R}^{-}}$,$NH_{2}^{-}$ etc.