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I've studied that when 1,3-dibromopropane (or in general any halogen, not just bromine) is heated with Zn dust, the electrons produced cause the halogens to break away homolytically and two free radicals are formed. These free radicals then combine to form cyclopropane.

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My question is, will this also occur for 1,3-dihalides when the carbon count is increased, say in 1,3-dibromobutane?

Everything I found related to this was either talking about the 3-carbon dihaloalkane case, or seemed to be some complicated free radical cyclization which was way out of my reach.

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    $\begingroup$ Please don't use MathJax for chemical names; they're only supposed to be used for chemical formulas, and then it's better to use the mhchem package as this provides convenient and correct typography, i.e. $\ce{H2O}$ $\ce{H2O}$ not $H2O$ $H2O$. $\endgroup$ Oct 14, 2020 at 10:41
  • $\begingroup$ @orthocresol ahh, thanks. I didn't see any specific details on the Ask Question page so just ended up using MathJax which was the only thing I had used before. $\endgroup$ Oct 14, 2020 at 10:46
  • $\begingroup$ This mechanism there is more like symbolic representation of what may really happen there, or even outright incorrect. $\endgroup$
    – Mithoron
    Oct 14, 2020 at 22:00
  • $\begingroup$ @Mithoron could you point me towards a reference with the correct mechanism? $\endgroup$ Oct 15, 2020 at 5:58

1 Answer 1

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Source: James Grimshaw, Electrochemical Reactions and Mechanisms in Organic Chemistry: Chapter 4

The formation of cyclopropane by reduction of 1,3-dibromopropane was discovered in 1887. Dissolving metals, in particular zinc dust in ethanol, were employed as an electron source. Electrochemical reduction in dimethyl-formamide at a mercury cathode has been found to give good yields of cyclopropane. 1,3-dibromo, 1,3-diiodo and 1-chloro-3-iodopropane all give greater than 90% yield of cyclopropane, the other product being propene.

Since chloroalkanes are not reduced at the cathode potential used, it is concluded that these reactions involve generation of a carbanion by dissociative electron transfer to the most easily reduced carbon-halogen bond followed by SN2 attack of this carbanion on the second carbon-halogen bond. Cyclopropane ring formation under electron transfer conditions shows no stereoselectivity...

...Most of the cyclopropane ring forming reactions can be accommodated in the mechanism that involves a carbanion in SN2 displacement of the second halogen atom. An exception is the reactions of 2,6-dibromobornane, which cannot accommodate the transition state stereochemistry necessary for an intramolecular SN2 displacement. Also, the original preparation of cyclopropane in good yield by dissolving metals in a protic solvent is unlikely to involve carbanions. When the reacting orbitals are parallel, addition of the second electron and carbon-carbon bond formation are likely to be synchronous. If these orbitals are not aligned, reduction leads to a diradical after which the carbon-carbon bond is formed.

The mechanism shown in the question is not wrong, but grossly incomplete.

Cyclopropane ring formation has been achieved in many other compounds, including those with high ring-strain. Although I could not find the specific reaction of 1,3-dibromobutane, it is reasonable to assume that it will undergo the same reaction to give methylcyclopropane.

Cyclobutanes and cyclopentanes are formed in low yields by electrochemical reduction of the appropriate $\alpha,\omega$-dibromoalkane in dimethylformamide. 1,4-Dibromobutane affords 29% cyclobutane along with butane and butene at a vitreous carbon cathode... Similarly, reduction of 1,5-dibromopentane affords 28% cyclopentane... 1,6-Dihalohexanes afford less than 5% of cyclohexane at a vitreous carbon cathode. The cyclization step in these reactions is believed to be an SN2 displacement of the remaining carbon-bromine bond by a carbanion formed in the initial reductive bond cleavage step, Cyclization must be very fast so as to avoid protonation of the carbanion intermediate.

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