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According to me, the answer should be (A) because that's a Wurtz reaction, but the answer given in the book is (B).

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The answer is (B) because (B) is benzene, which is the prototypical aromatic hydrocarbon. The aromaticity of (B) increases its thermodynamic stability relative to (A), which is called Dewar benzene (mistakenly attributed to James Dewar), which is notoriously difficult to synthesis and isolate. It reverts back to benzene with a half-life of two days.

Generally, if a pathway exists for a reaction to produce an aromatic product, then the reaction likely will do so. In this particular case, the diradical intermediate that might be produced as part of the Wurtz reaction's mechanism, is the triplet excited state of benzene.

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    $\begingroup$ In addition. Even if (A) is immediate product, it quickly converts to (B). It is known. $\endgroup$ – permeakra Jan 15 '13 at 8:03
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From Wikipedia:

Since the reaction involves free radical species, a side reaction occurs to produce an alkene. This side-reaction becomes more significant when the alkyl halides are bulky at the halogen-attached carbon. This is because the activation energy required for the SN2 reaction in the second step becomes significantly high, so the alternate elimination mechanism is favoured.

I can't see the Wurtz reaction happening to make (A) in this case with the two double bonds, there isn't enough flexibility to get the radical to attack across the ring. Also, (B) is making an aromatic ring, which is normally the most stable product when given the option.

So I'll go with (B) for this.

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The answer is (B). (B) is an aromatic Kelukes benzene where as (A) is dewar benzene. Even if (A) is formed it quickly converts into Kelukes benzene ie (B).

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  • $\begingroup$ Actually the conversion to benzene is quite slow and takes several days. $\endgroup$ – bon Feb 4 '16 at 11:34

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