Wikipedia mentions that alkynes can be produced from geminal dihalides through the Wurtz reaction. However, I am unsure of the reaction mechanism for this reaction. I have not been able to find any sources which show the mechanism for this reaction. Does anyone have any idea how the reaction is likely to occur?
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asked Apr 8, 2018 at 3:03
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$\begingroup$ This is just a guess but I would guess that if you had 1,2-dichloroethene (idk trans or cis) and added sodium metal, you would get sodium chloride and ethyne. $\endgroup$– tox123Apr 8, 2018 at 3:15
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$\begingroup$ @tox123 But the wikipedia article mentioned "geminal dihalide", not vicinal dihalide. $\endgroup$– Tan Yong BoonApr 8, 2018 at 3:18
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3$\begingroup$ My reading of the Wikipedia article's reference # 2 does not involve "geminal dihalides". The Organic Synthesis (ref #2) article is the Wiberg/Lampman synthesis of bicyclobutane from 1-bromo-3-chlorocyclobutane with sodium. Base treatment of a geminal dihalide would be expected to give an alkyne. Sodium would lead to reductive elimination (carbene) and an alkene. Wikipedia is not a primary resource. $\endgroup$– user55119Apr 8, 2018 at 13:49
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$\begingroup$ I have a better guess now (that takes into account geminal dihalides, that I'll post as an answer just do to it's length. $\endgroup$– tox123Apr 8, 2018 at 21:10
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2 Answers
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@tox123: You have generated two radicals on the same carbon by a two-electron reductive α-elimination (1 --> 2). So far, so good. In other words, you have created a carbene (2). Carbenes can rearrange to alkenes, NOT alkynes. Think of the two orbitals on the carbenoid carbon as one bearing two-electrons (negative) and the other one vacant (positive) (3). When a vicinal hydride migrates to the vacant orbital, one has a negative charge on one carbon and a positive charge on the vicinal carbon (4). Structures 4 and 5 are identical. If dichloride 1 is treated with base, one gets a double elimination to an alkyne. There is no change in oxidation state. Therefore, dichloride 1 cannot give an alkyne via a 2-electron reduction.
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$\begingroup$ In other words, you are saying that the Wurtz reaction cannot be used to produce alkynes directly? $\endgroup$ Apr 8, 2018 at 23:20
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$\begingroup$ what if only one chlorine is removed at a time? also are carbon diradicals necessarily carbenes? I was envisioning the carbon being in a triplet state, rather than the singlet state. $\endgroup$– tox123Apr 9, 2018 at 0:39
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$\begingroup$ A 2-electron reduction is a 2-electron reduction. Singlet or triplet is irrelevant. It is basic oxidation/reduction chemistry. The conversion of a geminal dichloride to an alkyne, no matter the pathway or mechanism, is a net zero-electron change process. Look at the diagram 1 --> 6. As to the whether the chlorine atoms are removed one at a time, they probably are. Doesn't really matter. $\endgroup$ Apr 9, 2018 at 1:16
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$\begingroup$ @user55119 I have learnt that vicinal dihalides can undergo elimination to give alkynes. Can the same be done for geminal dihalide? Could you provide a source? $\endgroup$ Apr 9, 2018 at 11:30
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$\begingroup$ @Tan Yong Boon: This reference may be of help. Bergstrom, F. W. and Fernelius, W. Conard, Chem. Rev., 12, 43-179; 1933 $\endgroup$ Apr 9, 2018 at 14:15
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So, as wikipedia says:
The Wurtz reaction occurs through a free radical mechanism.
So starting off with a geminal dihalide as in the picture below, and then adding sodium metal you get two unpaired electrons on a single carbon.
One of these electrons binds with an already paired electron on the other carbon forming an alkene with free electrons on both sides of the double bond (this should also form a free hydrogen atom, which is not pictured below):
These electrons then pair with one another to form an alkyne (freeing another hydrogen atom, which will react with the first to form hydrogen gas):
As a disclaimer, I don't know if this is the exact mechanism this is just an educated guess on my part, but because the Wurtz reaction has a free radical mechanism, this seems the most obvious way of this occurring.
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1$\begingroup$ I believe after the double bond is made, the molecule remains a radical, and doesn't become a diradical $\endgroup$ Apr 8, 2018 at 22:41
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$\begingroup$ That is what I am saying. You might want to familarize yourself with the concept of oxidation levels. ursula.chem.yale.edu/~chem220/STUDYAIDS/oxidationlevel/… $\endgroup$ Apr 9, 2018 at 0:28