1
$\begingroup$

What will methyl (2-oxocyclohexyl)acetate produce when reacted with sodium borohydride?

methyl (2-oxocyclohexyl)acetate

I feel the answer should be 2-(2-hydroxypropyl)cyclohexan-1-one:

2-(2-hydroxypropyl)cyclohexan-1-one

Because the carbonyl carbon on the aliphatic chain should be more electrophilic than the one on the ring due to the presence of two neighbouring electronegative oxygen atoms. Correct me if I am wrong but I think the reaction begins with the hydride ion which acts as a nucleophile?

On the other hand the answer given in my book is 1-(2-hydroxycyclohexyl)propan-2-one:

1-(2-hydroxycyclohexyl)propan-2-one

Can someone please help me out here?

|improve this question
$\endgroup$
3
$\begingroup$

Both carbonyl carbons (and everything else in the molecule) are in aliphatic environments. The six-membered ring you drew is not aromatic, it is saturated like a cycloalkane.

The carbonyl group in this ring is a ketone function, bonded only to other carbon atoms. The carbonyl group in the chain is bonded to oxygen (did you miss that?) in an ester function, which is a less active electrophile than the ketone function. Yes the additional oxygen atom is electronegative, but with its octet satisfied it can (to a degree) share a nonbonding pair with the carbonyl carbon, and that effect dominates making the ester carbon less electron deficient. The sodium borohydride reacts accordingly.

|improve this answer
$\endgroup$
2
$\begingroup$

It somewhat depends on the conditions used, choice of solvent etc. But sodium borohydride does not routinely react with esters which is what you have at the end of the chain as drawn in pic 1 (not as you have drawn in pics 2 & 3). It does routinely reduce ketones to secondary alcohols and you have an isolated ketone on the cyclohexane ring so this will be reduced.

|improve this answer
$\endgroup$
  • 1
    $\begingroup$ I'd also guess if no solvent was mentioned it will be the one in the ring because the ester is not reactive enough. A similar example you might want to look at if you are already dealing with different carbonyl groups and reductions is the so calle 'Luche-reduction'. This can often differentiate aldeyhdes from ketons and is quite useful to know as well. $\endgroup$ – Justanotherchemist Mar 9 '18 at 11:20
  • $\begingroup$ FYI: The Luche reduction uses a lanthanide chloride to coordinate a carbonyl oxygen accompanied by $\alpha, \beta$ unsatur-ation (hyphenated because device autocorrect is getting weird), activating that carbonyl group for the borohydride attack. $\endgroup$ – Oscar Lanzi Mar 9 '18 at 13:42
  • $\begingroup$ Never had a lot of luck with the selectivity in Luche reduction, tends to reduce both carbonyls in my personal experience $\endgroup$ – Waylander Mar 9 '18 at 13:48
  • $\begingroup$ Reduction of the ketone with NaBH4 will likely effect lactonization, particularly, in the cis isomer and possibly the trans isomer. $\endgroup$ – user55119 Mar 9 '18 at 16:11

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.