With reference to the elimination reaction of alkyl halides, the correct statements are;
(A) Zaitsev rule predicts the major product in dehydrohalogenation reaction.
(B) Grignard's reagent is prepared under perfectly anhydrous conditions.
(C) Alkyl halides on reaction with aqueous KOH forms an eliminated product but with alcoholic KOH forms a substituted product.
(D) Alkyl halides on reaction with aqueous KOH forms a substituted product but with alcoholic KOH forms an eliminated product.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A), (B) and (D) only
An alkyl halide with a β-hydrogen can undergo either substitution (the halogen is replaced by –OH) or elimination (H and X are lost and an alkene forms). The solvent decides the path: in water, \( \ce{OH^{-}} \) acts as a nucleophile; in alcohol, the alkoxide ion formed acts as a strong base and removes a β-hydrogen.
(A) Zaitsev rule predicts the major product in dehydrohalogenation reaction. — Correct. The preferred alkene is the one with more alkyl groups on the doubly bonded carbons: \( \ce{CH3CH2CH(Br)CH3} \xrightarrow{\text{alc. KOH}} \ce{CH3CH=CHCH3} \) (major) + \( \ce{CH3CH2CH=CH2} \) (minor).
(B) Grignard’s reagent is prepared under perfectly anhydrous conditions. — Correct. Grignard reagents react with even traces of water and are destroyed: \( \ce{RMgX + H2O -> RH + Mg(OH)X} \). So they are made in dry ether.
(C) Alkyl halides on reaction with aqueous KOH forms an eliminated product but with alcoholic KOH forms a substituted product. — Incorrect. It swaps the two reagents; aqueous KOH gives the alcohol and alcoholic KOH gives the alkene.
(D) Alkyl halides on reaction with aqueous KOH forms a substituted product but with alcoholic KOH forms an eliminated product. — Correct. \( \ce{C2H5Br + KOH(aq) -> C2H5OH + KBr} \), while \( \ce{C2H5Br + KOH(alc) -> CH2=CH2 + KBr + H2O} \).