7
$\begingroup$

Under basic conditions a Michael addition reaction can be used to convert compound A into product C. Explain why no Robinson annulation cyclic product is observed?

Compound A
reactant A

Product C
enter image description here

I have worked out that the other compound involved in the Michael addition reaction would be compound B (below), but I don't know why no Robinson annulation reaction could occur after the Michael addition reaction?

Compound B enter image description here

$\endgroup$
4
$\begingroup$

As illustrated in the following figure, the Robinson annulation reaction involves two steps. The first step involves the Michael addition of a ketone enolate (the enolate derived from cyclohexanone in the example pictured below) to an $\alpha,\beta$-unsaturated ketone (methyl vinyl ketone below). In a second step the methyl group adjacent to the carbonyl is deprotonated and undergoes an intramolecular aldol condensation to form the final product. This last step is very favorable since the reaction occurs intramolecularly and involves a favorable 6-membered transition state to form a stable 6-membered ring.

Robinson annulation

Often in Robinson annulations methyl vinyl ketone is used. It is important that the $\alpha,\beta$-unsaturated ketone have an alkyl group attached to the carbonyl so that the enolate required in the second step can be formed.

Explain why no Robinson annulation cyclic product is observed?

In your example, the phenyl vinyl ketone has no alkyl group attached to the carbonyl carbon - only a phenyl group. Hence, the enolizable hydrogens present in your Product C can only react with the other carbonyl through a highly-strained 4-membered transition state preventing any ring formation through an intramolecular aldol from taking place.

$\endgroup$
  • 2
    $\begingroup$ Either of the enolates at either acidic $\alpha$ position in C could form, but the intramolecular aldol would produce an unfavorable four-membered ring (angle strain). $\endgroup$ – Ben Norris Apr 24 '14 at 10:37
1
$\begingroup$

Well, technically it could. There is nothing stopping the Michael addition product from being enolised and when enolised there is little to stop the enolate attacking a different carbonyl.

When I say ‘little’, that was a little lie, though. The main inhibiting factor is the Baeyer strain on three- and four-membered rings. These rings are rather strained (try building them with a molecular model kit!) and thus do not form easily. Especially, a nucleophilic attack onto the carbonyl carbon would be very difficult if not impossible to perform if it generates a four-membered ring. You can also try that with a molecular model kit, and remember that the nucleophile has to attack the carbonyl group in the Bürgi-Dunitz angle of approximately $107^\circ$.

In a regular Robinson annulation, as Ron described, the Michael acceptor system can be enolised on the ‘far’ side of the ketone — the one where your molecule has a phenyl group. It is then this enolate that attacks the other carbonyl group to form a six-membered ring — one that does not have Baeyer strain and is generally the most thermodynamically stable ring system. Your system cannot do this as it is lacking α-hydrogens on the ‘far’ side of the Michael acceptor.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.