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For this problem, Can I use weinreb ketone synthesis? I found the product but not sure it is the right one

If the amide react with either R1-Mgbr or R2-Li, then it produces the ketone?

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    $\begingroup$ Why does this addition happen to the amide and not the ketone? $\endgroup$
    – Zhe
    Sep 22, 2020 at 16:40
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    $\begingroup$ The Weinreb ketone synthesis i know does not involve this amide but rather an O-methyl-N-methyacetamide. en.wikipedia.org/wiki/Weinreb_ketone_synthesis $\endgroup$
    – user55119
    Sep 22, 2020 at 17:21
  • $\begingroup$ What is the source of the reaction shown? $\endgroup$
    – user55119
    Sep 22, 2020 at 22:13

1 Answer 1

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All the reaction doesn't occur just in one pot. First, BuLi removes the tributylstannyl substituent from the compound 1 in nearly 0.5 h. Then, by adding the amide 3 the reaction pot, the anion 2 attacks to the carbonyl group only once to provide the ketone 4 without any tertiary alcohol, as expected by the Weinreb–Nahm reaction.

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Isono, .N., & Mori, .M. (1996). A NOVEL CARBON-CARBON BOND FORMING REACTION USING BIS(TRIBUTYLSTANNYL)PROPANOL. Main Group Metal Chemistry, 19(5),

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    $\begingroup$ Is the selectivity of the amide over the ketone explained? $\endgroup$
    – Waylander
    Sep 22, 2020 at 19:57
  • $\begingroup$ I think this is associated with the chelation of Li and the oxygen of the ketone. This looks some weird but if the above path is confirmed as a Weinreb–Nahm reaction, the ketone must play the same role as the hydroxylamine. Also, if the Weinreb–Nahm reaction could turn ketones to tertiary alcohols, it would primarily do this with amide groups. See the mechanism section of en.wikipedia.org/wiki/Weinreb_ketone_synthesis $\endgroup$
    – Reihani
    Sep 22, 2020 at 20:50
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    $\begingroup$ Here is the reference cited above: DOI:10.1515/MGMC.1996.19.5.277 It has little to do with the reaction shown by the OP and Chem. Absts. doesn't know about the reaction. $\endgroup$
    – user55119
    Sep 22, 2020 at 21:14
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    $\begingroup$ @Reihani I am very familiar with the Weinreb-Nahm and this does not look like it. I do not find this explanation convincing $\endgroup$
    – Waylander
    Sep 22, 2020 at 21:20
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    $\begingroup$ @Waylander: Neither do I. $\endgroup$
    – user55119
    Sep 22, 2020 at 22:15

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