Master Organic Chemistry Reaction Guide

Elimination (E1) with 1,2-alkyl shift

Description: When secondary (or primary) carbocations are formed adjacent to a quaternary carbon, 1,2-alkyl shifts can occur.

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Comment section

22 thoughts on “Elimination (E1) with 1,2-alkyl shift

  1. Hi its very good site where I have sort out my many problems.but in 1,2 alkyl shift I just want to confirm is alkyl shift takes place for for stability of carbocation.pls clear me.

  2. About the last example…..why would a six-membered ring expand to form a seven-memebered ring? Wouldn’t a seven-mebered ring have more angle-strain?

    1. There isn’t a strong driving force for the formation of the seven membered ring, but it’s not completely inaccessible. Seven membered rings don’t have significantly prohibitive angle strain.

      1. Instead of becoming a seven membered ring , shoudnt there be a formation of double bond(which is more preferable also)?

  3. How would a benzene ring come into play in sn1 e1 mechanisms? More specifically 1-(1-chloroethly)-3-methly benzene? Is it possible for the ring to expand? Will allylic resonance help with rearranging the carbocation and ultimately produce more products? Thank you

    1. Not at all in terms of ring expansion/contraction. Sometimes the benzene ring can perform what is known as “neighbouring group participation” where it helps to remove the leaving group, but that’s only covered in rare cases. Would need to see the exact molecule you’re talking about.

  4. why during nitration of benzene (any aromantic compound) both of sulphuric acid and nitric acid were used?

    1. The purpose of sulfuric acid is to protonate the nitric acid so that nitric acid loses water, giving NO2+ (the nitronium ion). The nitronium ion is the active electrophile in nitration.

  5. I read that without beta-hydrogens (adjacent protons) to the carbocation of the E1 reaction, elimination cannot occur directly. However, if there is a methide shift, does that override the rule? Example: 3-chloro-2,2,4,4-tetramethylpentane to 2,3-dimethyl-pent-2-ene.

    1. If there is a methide shift, then you are forming a carbocation with hydrogens beta to it, so in that case of course an elimination can occur. In the case you mention it is also worthwhile to point out that it is going from a secondary carbocation to a more stable tertiary carbocation.

  6. Hi James,
    is the third black R-group in the general reaction at the top of the page a typo?

  7. Why is the endocyclic alkene product not formed after doing the methyl shift? Isn’t it possible to eliminate using a beta hydrogen that is part of the ring, thus forming the more stable product?

    1. In this case the exocyclic alkene is tetra substituted and the endocyclic alkene would be trisubstituted, so following Zaitsev’s rule the exocyclic alkene would be favoured.

      In real life, it’s possible that the energy difference between the two products might not be all that significant due to subtle conformational and hyper conjugation effects that favor endocyclic vs. exocyclic alkenes.

  8. Such a great information. But as a student ,I really like this page for more leaning. I think this page should be available for all. The notes and the book are very nice.

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