Question

Complete the reaction:

Answer 1

To answer this question, you must recall the reactions taking place on use of different reagents. $KOH$ is an ionic compound and in a polar medium like aqueous solution, it breaks up into its constituent ions forming a hydroxide ion.

Complete answer:
When ${\text{KOH}}$ dissociates it forms hydroxide ion. ${\text{O}}{{\text{H}}^ - }$ ion is both a strong base as well as a strong nucleophile. But in an aqueous solution, the hydroxide ions are highly hydrated. This reduces the basic character of ${\text{O}}{{\text{H}}^ - }$ ions and thus it fails to abstract a hydrogen from the $\beta$- carbon of the alkyl chloride to form a double bond. So, it instead acts as a nucleophile which results in a nucleophilic substitution reaction. Both the chlorine molecules are replaced by a hydroxyl group each and a geminal diol is formed.
Oxygen atoms are large in size and have two lone pairs of electrons. Hence, when two hydroxyl groups are attached to a single carbon atom, it is highly unstable due to high lone pair- lone pair repulsion between the oxygen atoms. As a result, geminal diols undergo loss of a water molecule due to the vicinity of two hydroxyl groups to each other and a ketone is formed.
The final product obtained in the reaction is hexan-3-one.

Note:
On the other hand, if we use an alcoholic solution of ${\text{KOH}}$, the attacking species formed is different. In alcoholic solution ${\text{KOH}}$ forms an alkoxide ion, which is a strong base. The alkoxide ion extracts a proton from the $\beta$- carbon atom of the alkyl chloride while the chlorine atom leaves as chloride ion. Hence, if the solution used was alcoholic ${\text{KOH}}$, then the compound would have undergone dehydrohalogenation reaction twice forming an alkyne.