The treatment of alkyl chlorides with aqueous KOH leads to the formation of alcohols but in the presence of alcoholic KOH, alkenes are major product. Explain.

Explanation

KOH when dissolved in water provides `K^(+)` and `OH^(-)` ions.
` KOH + aq hArr K^(+) (aq) +OH^(-) (aq)`
There `OH^(-)` jons act as strong nucleophiles leading to nucleophilic substitution
`(S_(N)^(1) or S_(N)^(2)) R-X +KOH (aq) to R - OH +KX`
KOH when dissolved in alcohol (say ethanol) gives ethoxide ions and `K^(+)` ions.
`underset("n-Butyl bromide")(CH_3 CH_2 CH_2 CH_2 Br ) and underset("Isobutyl bromide ")(CH_3 - overset(CH_3) overset(|)(CH)-CH_2 Br)`
The bulkier nucleophile (alkoxide ion) in this case will prefer to act as a base and abstract a proton rather than approach a tetravalent carbon atom (steric reasons).
However the exact ` - underset(H) underset(|) overset(|)C " " overset(X)overset(|)C +O " " E t to - overset(|)C= overset(|)C= + E`to H nature of products formed also depends upon the nature of alkyl halide used. The ease of elimination follows the order :
` 1^@ lt 2^@ lt 3^@`
Therefore in case of a `3^@` alkyl halide, alkene is the major products even with aq. KOH. However in case of `1^@` alkyl halides almost pure alcohols (with ag. KOH) and almost pure alkenes (with alc. KOH) can be obtained.