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I recently came across two contrasting products for the same set of reagents - carboxylic acid and alkyllithium, and wish to understand why exactly any difference exists between the two.

Reaction 1

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Reaction 2

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I understand that in steps A, B, and E, the alkylide (hydride in E) is acting as a base and abstracting the acidic proton. I'm fine with that.

However, compare step C and F. Both involve the same reagents (carboxylate and an alkylide). In step C, nucleophilic addition of the ethylide has taken place on the carbonyl group, which is a reasonable step according to me. However, in step F, an ester has been formed instead. I am unable to grasp how an ester can even form in this condition (considering both ethylide and carboxylate are negatively charged), and even if it can form, why wasn't it formed in C as well?


Source: MS Chouhan; Advanced Problems In Organic Chemistry; 11th ed; Q64 and 27 in Alcohol, Ethers and Epoxides

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    $\begingroup$ There has to be a mistake somewhere. Based on charge conservation, the $\ce{Li}$ ends up associated with the counterion, $\ce{Li^+}$, i.e. a $\ce{Li2}$ molecule gets formed, which I believe has only been observed in the gas phase. $\endgroup$ – ralk912 Mar 15 '18 at 8:37
  • $\begingroup$ @ralk912 Yes, that's reasonable. So, you're suggesting that reaction 1 is correct, and reaction 2 is incorrect? $\endgroup$ – Gaurang Tandon Mar 15 '18 at 8:40
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    $\begingroup$ Yes, I don't see anyway reaction 2 would be right. $\endgroup$ – ralk912 Mar 15 '18 at 8:42
  • $\begingroup$ Step F has to be wrong. I suspect it is a typo and should be MeI $\endgroup$ – Waylander Mar 15 '18 at 9:35
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    $\begingroup$ @GaurangTandon I mean that the reagent should be MeI instead of MeLi as currently written $\endgroup$ – Waylander Mar 15 '18 at 11:31
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Reaction 2 is wrong. Reaction 1 is correct.

Later in the book (q87), another question is asked with the same set of reagents (carboxylic acid and methyllithium). Reaction 1 has been used in that as well.

Reaction 2 is wrong because, as ralk912 noted above, charge conservation has to be followed. If methyl lithium reacts with a negatively charged species, the product has to be negatively charged as well. If the ester is neutral, then lithium must be negatively charged, and would associate with the $\ce{Li+}$ formed in step E to form $\ce{Li2}$, which is only observed in gas phase.

Not to forget that there simply isn't any mechanism wherein a carboxylate would react with an alkylide to form an ester. It does not simply occur. Reaction 2 is wrong.


Waylander has suggested a good reason for a misprint i.e. the reagent $\ce{CH3Li}$ might have been intended to be $\ce{CH3I}$ instead. Then, it would make sense for the carboxylate anion to attack an alkyl halide, and displace an iodide ion (which could easily attach with the $\ce{Li+}$ ion to form lithium iodide) forming the methyl ester.


user55119 has confirmed that the product of the step F is cyclohexylmethyl ketone, and that it is obtained by reaction 1 only. The book containing this is "Margret Jorgensen in Organic Reactions."

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  • $\begingroup$ Maybe add that cyclohexyl methyl ketone would form after aqueous workup, as in Reaction 1. $\endgroup$ – ralk912 Mar 15 '18 at 16:13
  • $\begingroup$ @ralk912: The geminal bis-lithium alkoxide is implicit from reaction 1. Its stability is the reason the methyl ketone is not formed, and over reacts with methyllithium, prior to addition of water. $\endgroup$ – user55119 Mar 15 '18 at 19:35
  • $\begingroup$ @user55119 I'm afraid I don't follow? $\endgroup$ – ralk912 Mar 15 '18 at 19:37
  • $\begingroup$ If the product of C were not stable in solution, it would collapse to Li2O and a ketone. The ketone would react with methyllithium to form a tertiary alcohol. The same is true of F. Read "Reaction of Organolithium Reagents with Carboxylic Acids", Margaret J. Jorgenson ,Vol.18, in Organic Reactions. $\endgroup$ – user55119 Mar 15 '18 at 19:55
  • $\begingroup$ But I never said it is not stable. I actually say that to get the ketone back from the geminal dialkoxide, an acidic workup is needed. $\endgroup$ – ralk912 Mar 15 '18 at 20:10

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