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I'm tutoring few Class 12 students for their chemistry exams. I'm stuck with few problems. I don't know if the answer key is wrong or I am wrong.

Problem: enter image description here

Answer According to me:

Must be (A) because that's wurtz reaction.

Answer given in the book:


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Could you explain why you think it is (B)? Also, see our homework policy – ManishEarth Jan 14 '13 at 9:55
@Manishearth: I don't think its (B). I think its (A) but the book says answer its (B). No further explanation given. – claws Jan 14 '13 at 10:02
@Manishearth: Kindly, tell me which part of homework policy am I not following? I'm specific. I just didn't ask the community to solve. I first told my point of view, which obviously is not matching with the answer. So, I'm seeking help of community. – claws Jan 14 '13 at 10:04
Nah, I never said you weren't following it -- you come close to not following it though. We expect HW-type problems to show prior effort. There is some prior effort, but it would be better if you expanded on your reasoning (maybe provided a full mechanism). Not necessary though... And sorry for the slip up over there, I meant (A) :) – ManishEarth Jan 14 '13 at 12:13

2 Answers 2

up vote 4 down vote accepted

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

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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