What will be the product formed the if the reaction of Mg in ether with 1,4-dibromobut-2-yne occurs? At first, I thought the answer would be cyclobutyne, but the correct answer is buta-1,2,3-triene.

Can anyone please explain how?

  • $\begingroup$ Cyclobutyne cannot exist. $\endgroup$ – orthocresol May 18 '19 at 17:46

Cyclobutyne cannot exist. There is no way to enforce 90° bends at both ends of a triple bond, but the carbanion formed by the Grignard reagent synthesis finds an easy way around this problem.

Let's say the magnesium reacts at position 1. Then the negative charge would couple with the conjugated triple bond to spread to position 3:

$\ce{Br\overset{+}{Mg}\overset{-}{C}H2-C#C-CH2Br <-> Br\overset{+}{Mg}CH2=C=\overset{-}{C}-CH2Br}$

Then the carbon atom at position 3 can act as a nucleophile, displacing the remaining bromine as a bromide ion to give the product named in the book.

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    $\begingroup$ Sorry, I've noticed you've been struggling with formatting for a while and IMO it's visually better to put charges above with \overset{<above>}{<below>} command so that the skeletal structure is preserved – it's then arguably easier to visually compare both structures, like so (you can right-click on the formula and select Show Math As → TeX Commands): $$\ce{Br\overset{\large +}{Mg}-\overset{\large -}{C}H2-C#C-CH2Br <-> Br\overset{\large +}{Mg}-CH2=C=\overset{\large -}{C}-CH2Br}$$ Just a suggestion:) $\endgroup$ – andselisk May 18 '19 at 19:23

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