Question

(a) When alkyl chlorides are treated with aqueous KOH, alcohol is formed, whereas when alkyl chlorides are treated with alcoholic KOH, alkenes are formed. Explain.

(b) Complete the given equation :

\( \mathrm{C_6H_5Cl} + \mathrm{CH_3COCl} \xrightarrow{\text{Anhyd. } \mathrm{AlCl_3}} \)

Hint:

Aqueous KOH acts as a nucleophile and gives substitution. Alcoholic KOH has alkoxide, a strong base, and gives elimination.

Answer 1

Step 1: Part (a), look at the species present in aqueous KOH.
In aqueous KOH, the KOH is ionised and the hydroxide ion, \( \mathrm{OH^-} \), is heavily hydrated.
Hydrated \( \mathrm{OH^-} \) is a good nucleophile but a poor base.
So it attacks the carbon bearing Cl and replaces Cl. This is nucleophilic substitution.
\[ \mathrm{R-Cl + KOH(aq) \rightarrow R-OH + KCl} \]

Step 2: Look at the species present in alcoholic KOH.
In alcoholic KOH, the solution holds alkoxide ions, \( \mathrm{RO^-} \), formed from KOH and the alcohol.
An alkoxide ion is a much stronger base than \( \mathrm{OH^-} \).
It pulls off a beta hydrogen, and HCl is lost from the alkyl halide. This is dehydrohalogenation (elimination).
\[ \mathrm{CH_3CH_2Cl + KOH(alc) \rightarrow CH_2=CH_2 + KCl + H_2O} \]

Step 3: Part (b), do the Friedel-Crafts acylation.
Anhydrous \( \mathrm{AlCl_3} \) turns acetyl chloride into the acylium ion \( \mathrm{CH_3CO^+} \).
This electrophile attacks the ring of chlorobenzene.
Chlorine deactivates the ring but directs the new group to the ortho and para positions. The para product is the major one because it has less crowding.
\[ \mathrm{C_6H_5Cl + CH_3COCl \xrightarrow{Anhyd.\ AlCl_3} p-ClC_6H_4COCH_3 + HCl} \]
The main product is 4-chloroacetophenone, with a little 2-chloroacetophenone.

Final Answer:
(a) Aqueous KOH gives \( \mathrm{OH^-}\), a nucleophile, so substitution gives alcohol. Alcoholic KOH gives \( \mathrm{RO^-}\), a strong base, so elimination gives alkene.
(b) 4-Chloroacetophenone (major) and 2-chloroacetophenone (minor).