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Will Sodium metal donate its electrons to bromine in 1,3-dibromopropane or to the hydroxyl group in ethanol in a polar protic solvent?

This question comes from a problem:

'What is the major product formed when 1,3-dibromopropane is added to Na(sodium)/EtOH(ethanol)? (a)cyclopropane

(b)3-bromoprop-1-ene

(c)propyne

(d)prop-1,2-diene'

The answer was (a)cyclopropane but I was not convinced as then it would have to follow Wurtz reaction- where sodium donates its electron to bromine forming a alkyl radical. But, Wurtz reaction takes place in an ether solvent and here ethanol is used as solvent.

ethanol is shown to readily react with Na to form a sodium ethoxide and liberate hydrogen. The ethoxide ion would react with 1,3-dibromopropane in either Sn1 or E2 mechanism and so it would not form cyclopropane.

If sodium donates electron to bromine however, then it would follow Wurtz reaction mechanism and form (a)cyclopropane.

So, I want to know whether an alkyl halide or alcohol reacts more readily with sodium metal and what your reasoning is to assert your statement.

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  • $\begingroup$ What are your thoughts on the matter? If you include what you think the answer is and your rationale, someone can help steer you in the right direction. $\endgroup$ – jonsca Sep 15 '14 at 10:56

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