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I came across this question:

Carbonyl compound reaction

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?

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  • $\begingroup$ Kindly let me know the reason for the downvote so that I know where to work on? $\endgroup$
    – asmi
    Jul 11 at 17:41
  • $\begingroup$ As the medium is not basic the major enol will be formed on the more substituted side. $\endgroup$ Jul 11 at 18:08
  • $\begingroup$ @NisargBhavsar Thankyou! What would have happened if the medium was acidic? $\endgroup$
    – asmi
    Jul 11 at 18:11
  • $\begingroup$ 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. $\endgroup$
    – user55119
    Jul 11 at 18:52
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    $\begingroup$ 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. $\endgroup$
    – andselisk
    Jul 11 at 21:40
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First, there is a little mistake in your reasoning. You said that the carbanion is more stable on the unsubstituted side but actually, forming an enolate on the more substituted side will give a more stable substituted alkene. Remember, the negative charge is not concentrated on carbon. It is concentrated on the oxygen atom as the system is conjugated. That is why electrostatic interactions tend to occur on the oxygen while orbital (HOMO-LUMO) interactions occur on the carbon.

Now, an asymmetric ketone can form two regioisomeric enolates, kinetic and thermodynamic:

enter image description here

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 $\ce{NaH}$.

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    $\begingroup$ 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 $\ce{NaH}$? $\endgroup$ Jul 13 at 7:32
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    $\begingroup$ NaH is not known as a nucleophile in organic chemistry, rather more as a base. The reason for this is likely due to the poor orbital overlap in terms of the HOMO-LUMO interactions. $\endgroup$
    – M.L
    Jul 13 at 18:32

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