9
$\begingroup$

$\hspace{5cm}$enter image description here

I know that the acid protonates the carbonyl oxygen and this opens a pathway to an intramolecular cyclization reaction because the $\ce{-OH}$ group will attack the carbonyl carbon and a six-membered ring is formed.

I know that the carbonyl part is flat, and that the $\ce{-OH}$ can either attack from either the re or the si face. I'm not concerned about that. What happens to the methyl group on the wedge? Does that change depending on which side the $\ce{-OH}$ attacks the carbonyl? I am having a hard time visualizing what might happen to it.

$\endgroup$
12
$\begingroup$

No, the stereochemistry of the methyl group is not dependent on which face the $\ce{-OH}$ attacks from. This is apparent in the products of the cyclization of aldopyranoses; only two isomers (called anomers) are formed, which are classified as either $\alpha$ or $\beta$:

aldopyranose cyclization

Depending on which position we are considering the stereochemistry of, a wedge in the straight chain could appear as either a wedge or a dash in the cyclic form. For the molecule you are considering:

$\hspace{.7cm}$specific case

To visualize this, convert the skeletal structure into its Fischer projection:

$\hspace{4.5cm}$Fischer

$\hspace{2cm}$Fischer cyclization

$\endgroup$
  • 2
    $\begingroup$ I would expect the hemiketal with two equatorial methyl groups to be a tad more preferred in the reaction. $\endgroup$ – Jan May 27 '16 at 14:22

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.