Which of the following alkyl halides gives 2,3-dimethylbutane when it is treated with sodium in dry ether?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2-Bromopropane
In the Wurtz reaction, sodium removes the halogen from two molecules of the alkyl halide and joins the two alkyl groups together. The hydrocarbon formed has double the number of carbon atoms of the halide.
2,3-Dimethylbutane, (CH3)2CH–CH(CH3)2, has six carbon atoms, so the halide must have three.
Splitting the molecule at the central C–C bond gives two isopropyl groups, (CH3)2CH–, so the halide is 2-bromopropane:
\( \displaystyle \ce{2(CH3)2CHBr + 2Na} \xrightarrow{\text{dry ether}} \ce{(CH3)2CH-CH(CH3)2 + 2NaBr} \)
Why the other options are incorrect
Option (1) 1-Bromopropane — Incorrect. It has three carbons, but two n-propyl groups joined end to end give the straight chain n-hexane, CH3(CH2)4CH3, not a branched one.
Option (3) 2-Bromobutane — Incorrect. Two sec-butyl groups give 3,4-dimethylhexane, an eight-carbon hydrocarbon.
Option (4) 2-Bromo-2-methylpropane — Incorrect. Two tert-butyl groups give 2,2,3,3-tetramethylbutane, also with eight carbons.
NCERT (2026-27 edition):
6.7.1 Reactions of Haloalkanes – “Wurtz reaction Alkyl halides react with sodium in dry ether to give hydrocarbons containing double the number of carbon atoms present in the halide. This reaction is known as Wurtz reaction.”