Consider the following statements about the reactions of haloalkanes that have β-hydrogen atoms.
(A) In β-elimination, a hydrogen atom is removed from the β-carbon and a halogen atom from the α-carbon atom of the haloalkane.
(B) According to Zaitsev rule, the preferred product in dehydrohalogenation is the alkene that has the smaller number of alkyl groups attached to the doubly bonded carbon atoms.
(C) A bulkier nucleophile prefers to act as a base and abstract a proton rather than approach a tetravalent carbon atom.
(D) When a haloalkane with β-hydrogen atoms reacts with a base or a nucleophile, a primary alkyl halide prefers an SN2 reaction.
Which of the statements given above are correct? Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A), (C) and (D) only
A haloalkane with β-hydrogen atoms meets a base or a nucleophile in two competing routes: substitution (SN1 or SN2) and elimination, which gives an alkene. Which route wins depends on the alkyl halide, the strength and size of the base or nucleophile, and the conditions.
(A) In β-elimination, a hydrogen atom is removed from the β-carbon and a halogen atom from the α-carbon atom of the haloalkane. — True. Heating with alcoholic KOH removes H from the β-carbon and X from the α-carbon, and a C=C bond forms between them.
(B) According to Zaitsev rule, the preferred product in dehydrohalogenation is the alkene that has the smaller number of alkyl groups attached to the doubly bonded carbon atoms. — False. Zaitsev rule favours the alkene with the greater number of alkyl groups on the doubly bonded carbons; so 2-bromopentane gives mainly pent-2-ene.
(C) A bulkier nucleophile prefers to act as a base and abstract a proton rather than approach a tetravalent carbon atom. — True. A crowded carbon is hard to reach, so a bulky reagent takes the easier path and removes a proton, for steric reasons.
(D) When a haloalkane with β-hydrogen atoms reacts with a base or a nucleophile, a primary alkyl halide prefers an SN2 reaction. — True. A primary halide has little crowding at the carbon under attack, so SN2 substitution is preferred.
NCERT (2026-27 edition):
6.7.1 Reactions of Haloalkanes – “When a haloalkane with β-hydrogen atom is heated with alcoholic solution of potassium hydroxide, there is elimination of hydrogen atom from β-carbon and a halogen atom from the α-carbon atom.”
“…Alexander Zaitsev (also pronounced as Saytzeff) who in 1875 formulated a rule which can be summarised as ‘in dehydrohalogenation reactions, the preferred product is that alkene which has the greater number of alkyl groups attached to the doubly bonded carbon atoms.’ Thus, 2-bromopentane gives pent-2-ene as the major product.”
“…Thus, a bulkier nucleophile will prefer to act as a base and abstracts a proton rather than approach a tetravalent carbon atom (steric reasons) and vice versa. Similarly, a primary alkyl halide will prefer a SN2 reaction, a secondary halide- SN2 or elimination depending upon the strength of base/nucleophile and a tertiary halide- SN1 or elimination depending upon the stability of carbocation or the more substituted alkene.”