The reagent used in Wurtz reaction is

The correct answer is Option (4) → sodium in dry ether

Three named reactions couple halides with sodium in dry ether; they are named by what is joined:

• Fittig reaction: two aryl halides give a diaryl.

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

• Wurtz–Fittig reaction: an alkyl halide and an aryl halide give an alkylarene.

      \( \displaystyle \ce{C6H5-X + R-X + 2Na} \xrightarrow{\text{dry ether}} \ce{C6H5-R + 2NaX} \)

In the Wurtz reaction two alkyl halides are coupled in this way; the ether must be dry, because sodium reacts violently with water. So the reagent used in the Wurtz reaction is sodium in dry ether.

Why the other options are incorrect

• Option (1) alcoholic KOH — Incorrect. It removes HX from a haloalkane (dehydrohalogenation) and gives an alkene.

      \( \displaystyle \ce{CH3-CH2-Br} \xrightarrow{\text{alc. }\ce{KOH}} \ce{CH2=CH2 + HBr} \)

• Option (2) Red P/X₂ — Incorrect. It converts alcohols into alkyl halides and, in the Hell–Volhard–Zelinsky reaction, halogenates carboxylic acids at the α-carbon.

      \( \displaystyle \ce{R-CH2-COOH} \xrightarrow[\text{(ii) }\ce{H2O}]{\text{(i) }\ce{X2}/\text{Red P}} \ce{R-CH(X)-COOH} \)

• Option (3) magnesium in alcohol — Incorrect. Magnesium with a haloalkane forms a Grignard reagent only in dry ether; any alcohol would destroy the Grignard reagent at once.

      \( \displaystyle \ce{R-X + Mg} \xrightarrow{\text{dry ether}} \ce{R-Mg-X} \)

NCERT (2026-27 edition):
Wurtz Reaction

Note: The same/similar question has been asked in different years by NTA as per the details given below: