I am looking over mechanism of the oxidation of alcohol by chromic acid and I am confused on why $\ce{H2O}$ abstracts hydrogen from alpha carbon and form double bond between $\ce{C}$ and $\ce{O}$ atom rather than beta hydrogen and form alkene like in usual elimination reaction we learned. Like to start from beginning i am confused why in chromate ester, which is an intermediate, entire group of $\ce{O-CrO3H}$ does not leave as learning group. Why does only $\ce{CrO3H}$ leave, leaving $\ce{O}$ atom behind which thus makes $\ce{H2O}$ abstact hydrogen from alpha carbon instead of abstracting beta hydrogen and creating alkene.
$\begingroup$
$\endgroup$
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
Technically, it would be possible for an entire chromate group $\ce{CrO4^2-}$ to be eliminated along with a β-hydrogen to give a double bond. However, that elimination would have to proceed via either an $\mathrm{E2}$ or an $\mathrm{E1_{cb}}$ mechanism as the chromium(VI) centre by itself is much more happy being reduced to chromate(IV) (the intermediate chromium species) in the electrocyclic breakdown of the complex you have been taught.
The entire reaction proceeds in acidic conditions, so any conjugate base mechanism is basically impossible. And likewise, there is no sufficiently strong base present that would allow an $\mathrm{E2}$ mechanism to occur, either.
