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Reaction of LDA with cyclohexene oxide

The question given above appeared in the 2016 Indian National Chemistry Olympiad (INChO). What I think is that the epoxide ring is being opened by a base-catalysed mechanism ($\mathrm{S_N2}$). I proposed the following few steps for this:

Proposed steps

The diisopropylamide ion, which is a very strong base, would not be a good leaving group. To obtain the desired products, how could the mechanism then proceed? And if not, then what would be the real mechanism? I did not find any reliable information on the Internet regarding ring opening using LDA.

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  • $\begingroup$ For the asymmetric transformation of cyclohexene epoxide to cyclohex-2-en-1-ol, see Tetrahedron, 2002, 58, 4669. DOI:10.1016/S0040-4020(02)00373-3. Also see, J. Am. Chem. Soc., 1970, 92, 2064 for mechanistic studies, doi.org/10.1021/ja00710a045 $\endgroup$
    – user55119
    Dec 31, 2020 at 0:25
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    $\begingroup$ Interestingly enough, I think for cyclohexene oxide you do usually get a simple elimination reaction, but other pathways are possible and with other epoxides can lead to different major products (in particular, one can deprotonate alpha to oxygen, then alpha-elimination to give a carbenoid). The best reference I found from 5 minutes of google is this review thieme-connect.de/products/ejournals/abstract/10.1055/…, but there might be something more recent out there. $\endgroup$
    – orthocresol
    Dec 31, 2020 at 0:46
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    $\begingroup$ Also, I will fix it soon, but you need to be more careful with the number of hydrogens on your nitrogen. If you look back at your scheme, you consistently have too many. Negatively charged nitrogen should only have two bonds, neutral nitrogen should have three. In some places you even have nitrogen with five bonds. $\endgroup$
    – orthocresol
    Dec 31, 2020 at 0:51
  • $\begingroup$ @orthocresol, thanks for pointing out the mistake in my scheme, and for the review, it seems to be very interesting. $\endgroup$
    – Shishir Maharana
    Dec 31, 2020 at 6:58

1 Answer 1

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Because of the steric hindrance of the two iPr groups, LDA is a very poor nucleophile. It is, as you note, a very strong base. The two products observed both arise from deprotonation of cyclohexene oxide.

Deprotonation of one of the methylenes alpha to the epoxide ring opens it to give the allylic alcohol.

Removal of one of the protons on the epoxide ring carbons opens the ring to the enol, which on quench gives cyclohexanone.

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