Which of the following, on treatment with sodium in dry ether, gives…

The correct answer is Option (2) → Bromobenzene and bromoethane

Sodium in dry ether removes the halogen atoms and joins the two carbon groups that carried them. A mixture of an alkyl halide and an aryl halide gives an alkylarene; this is the Wurtz-Fittig reaction. Ethylbenzene, C₆H₅–CH₂CH₃, is a phenyl group joined to an ethyl group, so it needs bromobenzene and bromoethane:

    \( \displaystyle \ce{C6H5Br + CH3CH2Br + 2Na} \xrightarrow{\text{dry ether}} \ce{C6H5CH2CH3 + 2NaBr} \)

Why the other options are incorrect

Option (1) Bromobenzene and bromomethane — Incorrect. This is also a Wurtz-Fittig mixture, but it joins a phenyl group to a methyl group and gives toluene, C₆H₅CH₃, not ethylbenzene.

Option (3) Bromobenzene alone — Incorrect. An aryl halide alone joins two aryl groups (Fittig reaction), giving biphenyl:

    \( \displaystyle \ce{2C6H5Br + 2Na} \xrightarrow{\text{dry ether}} \ce{C6H5-C6H5 + 2NaBr} \)

Option (4) Bromoethane alone — Incorrect. An alkyl halide alone gives a hydrocarbon with double the number of carbon atoms (Wurtz reaction), here butane, CH₃CH₂CH₂CH₃, with no benzene ring.

NCERT (2026-27 edition):
6.7.2 Reactions of Haloarenes – “Wurtz-Fittig reaction … A mixture of an alkyl halide and aryl halide gives an alkylarene when treated with sodium in dry ether and is called Wurtz-Fittig reaction.”