-2
$\begingroup$

What is the major product for the reaction of (1R,2R)-2-methylcyclohexanol[1,2] with hydrogen bromide? What will happen to the stereocenters?

(1R,2R)-2-methylcyclohexanol

I only know the $\ce{HBr}$ will somehow substitute the alcohol, but I'm not sure about the position or stereo chemistry of the final product.

  1. CSID:22437, http://www.chemspider.com/Chemical-Structure.22437.html (accessed 14:03, Oct 12, 2018)
  2. National Center for Biotechnology Information. PubChem Compound Database; CID=24004, https://pubchem.ncbi.nlm.nih.gov/compound/24004 (accessed Oct. 12, 2018).
$\endgroup$

closed as off-topic by Satwik Pasani, bobthechemist, user4076, ManishEarth Jan 14 '14 at 10:28

This question appears to be off-topic. The users who voted to close gave this specific reason:

If this question can be reworded to fit the rules in the help center, please edit the question.

4
$\begingroup$

$\ce{HBr}$ is a fairly strong acid; protonation of (1R,2R)-2-methylcyclohexanol (1) gives 1a. Water is a pretty good leaving group and that's what it does, resulting in the secondary cation 1b.

2-methylcyclohexanol-protonation

Bad news: You have just lost one stereocentre.

You might remember that tertiary cations are more stable than secondary ones. A 1,2 hydride shift converts 1b into the more stable 1c.
More bad news: The rearrangement has flattened the tertiary centre.
The nucleophile $\ce{Br-}$ can add from both sides of cation, yielding 2 as the major product.

hydrogen shift

Depending on the reaction conditions, minor products will arise from the addition of $\ce{Br-}$ to 1b, or from elimination reactions.

$\endgroup$

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