Elimination (E1) with 1,2-alkyl shift

by James

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

Notes: The acids in these reactions is often sulfuric acid (H2SO4) or tosic acid (TsOH) because the conjugate base of these acids are very poor nucleophiles.

Examples:

Notes: In the first and second cases a methyl group shifts over by one carbon (in the second example, silver nitrate  AgNO3 (-) is a Lewis acid that makes halides such as Br better leaving groups).

In the third example the cyclobutane ring expands to give the five-membered ring, which has less ring strain.

In the fourth example the ring expands to form a seven-membered ring from a six-membered ring (recall that primary carbocations are unstable relative to secondary carbocations).

Mechanism: Formation of a carbocation is achieved here through protonation of the alcohol with strong acid (Step 1, arrows A and B) followed by loss of water to give the carbocation (Step 2, arrow C) . Next, migration of a methyl group from the adjacent quaternary carbon through a transition state such as that pictured (Step 3, arrow D) gives a tertiary carbocation which undergoes deprotonation (Step 4, arrows E and F) to form the most substituted alkene (between C-2 and C-3)

Notes:

{ 16 comments… read them below or add one }

gurjit

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.

Reply

Andrew

you are cleared.

Reply

Pankaj Vidh

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?

Reply

james

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.

Reply

Abhinandan Pandey

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

Reply

Mike Harkins

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

Reply

James

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.

Reply

Ekta

In last example. Is stability occure due to formation of tropylium ion formation

Reply

James

No, the tropylium ion has 3 double bonds.

Reply

Alicia

What is there is no leaving group and it is just a carbon chain? will the reaction still occur?

Reply

Karen

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.

Reply

James

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.

Reply

Martin P Boland

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

Reply

James

Oh yes. Sorry about that!

Reply

Krishna Mallem

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?

Reply

James

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.

Reply

Leave a Comment