Q.What is the action of following on ethyl bromide:

Two aryl halide molecules couple over sodium metal in dry ether (Wurtz–Fittig reaction) to give biphenyl.

When chlorobenzene reacts with sodium metal in the presence of dry ether, two aryl radicals couple (Wurtz–Fittig reaction, the aryl-aryl analogue of the Wurtz reaction) to form …

Wurtz coupling doubles the alkyl group; splitting the symmetric product at its central bond gives neopentyl bromide as A.

The Wurtz reaction couples two molecules of an alkyl halide via sodium metal in dry ether, doubling the carbon skeleton: $2R-Br + 2Na \xrightarrow{\text{dry ether}} R-R + 2NaBr$.

The product, 2,2,5,5-tetramethylhexane, is $CH_3-C(CH_3)_2-CH_2-CH_2-C(CH_3)_2-CH_3$ — a symmetric molecule. Splitting it at its central (newly-formed) C–C bond gives two identical fragments, $-CH_2-C(CH_3)_2-CH_3$, i.e. the neopentyl group. So the starting alkyl bromide must be:

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Elimination gives ethene, HBr addition regenerates ethyl bromide, and Wurtz coupling with Na/ether doubles it to n-butane.

Step 1: $CH_3CH_2Br \xrightarrow{\text{alc. KOH}/\Delta}$ B — dehydrohalogenation (elimination) of ethyl bromide with alcoholic KOH gives ethene, $B = CH_2=CH_2$.

Step 2: $B \xrightarrow{HBr}$ C — Markovnikov addition of HBr across the (symmetric) double bond of ethene regenerates ethyl bromide, $C = CH_3CH_2Br$.

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