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Problem

enter image description here

Answer

C

Question

Here $\ce{KH}$ obviously acts as a base, and forms the enolate from the ketone, which subsequently act as nucleophiles causing an SN2 reaction on the $\ce{CH3I}$. When this happens twice, we get the compound in option A.
However, after the removal of a third proton from $2,6$-dimethylcyclohexanone, the following enolate is formed -

enter image description here

This enolate will attack through the $\mathrm{C-2}$ atom. This atom is sterically hindered as it is a $3^\circ$ carbon. Hence I reasoned that like tert-butoxide, this enolate wouldn't act as a nucleophile either, so the reaction effectively stops at A.

Since the answer is C, it turns out that the sterically hindered enolate still does react further, and in good yield. Why is my reasoning incorrect? Also, this question is not from a standard textbook, so I would appreciate any reference regarding if and why/why not the enolate will react further.

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    $\begingroup$ T-butoxide anion statically hinders only in elimination reactions. Eg, the Hoffmann product is formed when t-butoxide is used. I have never come across a retardation of the reaction due to steric hindrance of the nucleophile (to an appreciable extent) in substitution reactions like Sn2. Steric hindrance of the substrate generally decreases the rate of Sn2. I may be incorrect though. $\endgroup$ Aug 13, 2021 at 12:39
  • $\begingroup$ @AdityaRoychowdhury - I meant that if we have, say ethyl chloride, then tert-butoxide ion prefers elimination reaction (to give ethene) over substitution, because it's sterically hindered. So the yield of substitution product decreases. I would definitely have understood if the product of option C had been given as one of the products, but with a lower yield. The fact that it's yield is 81% is something I felt doubtful about. $\endgroup$
    – TRC
    Aug 13, 2021 at 13:36
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    $\begingroup$ That's just a methyl substituent, if you were adding t-butyls to the ring then I suspect your reasoning would be true. $\endgroup$
    – Mithoron
    Aug 13, 2021 at 13:44
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    $\begingroup$ @Rishi - it's from a Fiitjee AITS paper - a mock test for JEE Advanced, but then probably you already know what AITS it ;) $\endgroup$
    – TRC
    Aug 13, 2021 at 15:43
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    $\begingroup$ @TRC One could say you kinda tert-butylate MeI - there's no problem with introducing t-butyl for ex. in benzene. $\endgroup$
    – Mithoron
    Aug 13, 2021 at 19:36

3 Answers 3

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One thing that will help you understand how this can happen at a position that otherwise looks hindered is to consider the geometry at that position. Carbon 2 is planar in the enolate anion, and the nucleophilic molecular orbital is more $\mathrm{p}$-like than $\mathrm{sp^3}$-like. The picture below highlights this geometry (note that the orbital shown is a simplification of the conjugated HOMO of the enolate anion):

The geometry of enolate anion

Because the enolate anion is planar at carbon 2, the electrophile approaches from above or below. The methyl group at carbon 2 is not in the way.

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    $\begingroup$ It's important that the ring is not planar, only the O-C1-C2 motive is. You'd probably call this molecular shape half chair, or less specific twisted. But that is hard to draw with a schematic representation. $\endgroup$ Aug 13, 2021 at 22:31
  • $\begingroup$ Agreed. The half-chair / pseudo-twist conformation representation adds cognitive load (and is not necessary to convey this point). $\endgroup$
    – Ben Norris
    Aug 13, 2021 at 22:42
  • $\begingroup$ In my experience, while showing it is difficult and unnecessary, it's important to mention these simplifications. There are not few people who otherwise take the images at face value. This is not a trivial reaction and the other answer references a scientific paper. I also think that the way the question is posed, too much simplification isn't necessary. $\endgroup$ Aug 13, 2021 at 22:50
  • $\begingroup$ Thank you. While I haven't learnt yet about the usage of molecular orbital theory in organic compounds, the structures and planarity at the enolate carbon helps to clear the point. $\endgroup$
    – TRC
    Aug 14, 2021 at 3:50
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The permethylation of cyclohexanone with KH/MeI is described here JOC paper. The yield quoted is 96% and other ketones are also exemplified.

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  • $\begingroup$ Thank you. I visited the paper but could access only the first page. The table at the bottom had a "yield" column with two numbers - like for cyclohexanone, yield says 96(81). Could you please clarify what those two numbers are? $\endgroup$
    – TRC
    Aug 13, 2021 at 15:38
  • $\begingroup$ Also, in the remaining paper (which I could not access), have the authors discussed any effects of steric hindrance on the reaction? $\endgroup$
    – TRC
    Aug 13, 2021 at 15:40
  • $\begingroup$ @TRC you can access the full paper on SciHub (legality is questionable) $\endgroup$ Aug 13, 2021 at 15:49
  • $\begingroup$ @TRC the first yield is GC yield, the second yield is product after distillation. I also can only read the first page. It is, I think significant, that the authors mention that the system needs to be refluxed to get the 4th Me to add in some cases, thoigh not cyclohexanone $\endgroup$
    – Waylander
    Aug 13, 2021 at 15:49
  • $\begingroup$ I must seem like a complete fool now... what is meant by the GC yield? $\endgroup$
    – TRC
    Aug 13, 2021 at 15:55
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Although Ben Norris has given excellent explanation for why the steric hindrence is not playing the role in the reaction of cyclohexanone with excess methyl iodide to give tetramethylated cyclohexanone, there is an enother reason why reaction proceds so smoothly under given condition to give more than 80% yield of 2,2,6,6-tetramethylcyclohexanone. It is the exceptionally high reactivity of $\ce{KH}$ towards ketones compared to that of other strong non-nucleophilic bases such as $\ce{NaNH2}$, and even $\ce{NaH}$ and $\ce{LiH}$ (Ref.1):

Metalation of ketone

As shown in the graph above, methyl tert-butyl ketone (pinacolone) has given 100% metallated enolate within $\pu{5 min}$ in the presence of $\ce{KH}$ at $\pu{20 ^\circ C}$ while it has taken more than $\pu{120 min}$ to give at least 20% metallated enolate with $\ce{NaH}$ under same conditions. For comparition, $\ce{LiH}$ is even less reactive than $\ce{NaH}$ (Ref.1). In a typical reaction ($\pu{25 mmol}$ substrate in $\pu{40 mL}$ of THF in the presence of $\pu{28 mmol}$ of $\ce{KH}$), cyclohexanone is enolated 100% within $\pu{1.5 min}$ and 2-methylcyclohexanone is enolated 100% within $\pu{6 min}$. In the latter case, which can give two enolates, the ratio of isolated enols (as silyl ethers) was found to be $33:67$ less substitured one to more substituted one indicating that reaction is more facile with making more stable enolate even in the presence of steric hindrence.

Thus, it is clear that the question in OP's post is directly related to the reserch done in Ref.2 (which is mentioned by Waylander in his answer. Accordingly, the authors in Ref.2 reported yield of 2,2,6,6-tetramethylcyclohexanone as 96% (GLC analysis) and as 81% (isolated yield by distillation), which is the exact yield given in the question. The procedure was as follows:

To a flask containing $\pu{216 mmol}$ of $\ce{KH}$ in mineral oil $(\pu{40 mL})$, which is in a waterbath maintained at $\pu{25 ^\circ C}$, $\pu{220 mL}$ of THF was added first and then $\pu{5.2 mL}$ of cyclohexanone $(\pu{50 mmol})$ was added dropwise over a $\pu{5 min}$ period. The mixture was stirred for additional $\pu{5 min}$ and $\pu{13.5 mL}$ of $\ce{CH3I}$ $(\pu{216 mmol})$ was added dropwise over a $\pu{15 min}$ period. The reaction mixture was stirred for additional $\pu{15 min}$ before workup with water. The simple distillation of dried organic layer gave $\pu{6.25 g}$ of expected product (81% yield).


References:

  1. Charles Allan Brown, "Saline hydrides and superbases in organic reactions. III. Facile reaction of potassium hydride with ketones. Rapid, quantitative formation of potassium enolates from ketones via kaliation," J. Org. Chem. 1974, 39(9), 1324–1325 (DOI: https://doi.org/10.1021/jo00923a042).
  2. Alan A. Millard and Michael W. Rathke, "Procedure for the permethylation of ketones using potassium hydride and methyl iodide," J. Org. Chem. 1978, 43(9), 1834–1835 (DOI: https://doi.org/10.1021/jo00403a055).
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  • $\begingroup$ Beautiful answer - I'm glad I posted this question on this site :) Is there a way I can mark all the three answers as accepted? $\endgroup$
    – TRC
    Aug 16, 2021 at 9:34
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    $\begingroup$ @TRC: As much as I know, you can't accept all answers. But, it is best if you can decide what answer gives you the best information on your question. It's up to you. $\endgroup$ Aug 16, 2021 at 15:45

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