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Anees_5379
#1 Posted : Monday, June 15, 2020 6:16:22 PM
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It is a question from the ICE 2, pg. 269, Q37.
Would you be able to explain why the answer to this question is D, that is NaH is required in the methylation to form a carbanion nucleophile? I read the answer key and it only confused me more.

I was thinking that since the C donates an H , it makes it as a carbocation (positive charged).
and hence a nucleophile.
Thank you in advance!
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INSTR_Katerina_102
#2 Posted : Wednesday, June 17, 2020 10:23:40 PM
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Hello,

https://ibb.co/3pBhKxs

If you look at the image in the link I have provided, it shows that NaH is a base that deprotonates the acidic proton by the ester, forming a carbanion.

A carbanion has a lone pair (which I've omitted accidentallly but it is represented by the negative formal charge), and recall species with an excess of electrons are Negative and therefore Nucleophiles.Therefore the correct answer is C because you've formed a carbanion nucleophile (pink text in photo).

If it is unclear what the distinction between a carbanion is and a carbocation is, recall that cations are PAWsitive (positive).

I have omitted the full structure in the example in order to not post the full question, but the other parts of the structure can be seen in Reaction 1, they're just not significant in answering the question.

Please let me know if this is still unclear, I hope this helps!
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