# Double bond or carbanion for nucleophillic attack [closed]

I came across this question:

According to me the Hydride ion should take hydrogen from left (unsubstituted carbon) because carbanion will be more stable there and this carbanion will then act as a nucleophile and attack $$\ce{CH3I}$$ giving product (b).

However according to the answer, hydride ion takes hydrogen from the right (Substituted carbon) and then enol form is made and then the double bond attacks $$\ce{CH3I}$$. Why so?

• Kindly let me know the reason for the downvote so that I know where to work on?
– asmi
Jul 11 '21 at 17:41
• As the medium is not basic the major enol will be formed on the more substituted side. Jul 11 '21 at 18:08
• @NisargBhavsar Thankyou! What would have happened if the medium was acidic?
– asmi
Jul 11 '21 at 18:11
• It is not necessarily true that "hydride ion takes hydrogen from the right (Substituted carbon) ". All you know is that it is the enolate that undergoes methylation. Under the conditions of the reaction, enolate formation may also occur at the less substituted carbon followed enolate exchange with methylcyclohexanone under equilibrating conditions. Jul 11 '21 at 18:52
• The image should be properly oriented (i.e. not rotated 90°) and its source must be cited (we prefer ACS style). Also, chemical formulas must be properly formatted as such. Please visit this page, this page and this one on how to format your posts better with MathJax and Markdown. Jul 11 '21 at 21:40

The kinetic enolate is the one that forms the quickest. Often, it is promoted by LDA, which is a strong but hindered base, at low temperatures, typically $$\pu{-78^\circ C}$$. Because it is more hindered, it will attack the less hindered and more accessible alpha hydrogen and form the less substituted enolate.
On the other hand, thermodynamic enolates tend to form at higher temperatures with unhindered bases like $$\ce{NaH}$$.
• Why does $\ce{H-}$ not attack $\ce{C=O}$ and form an alcohol? One could argue that acid-base is faster but then why do reagents like $\ce{LiAlH4}$ form alcohols by nucleophilic addition but not $\ce{NaH}$? Jul 13 '21 at 7:32