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In the following Grignard reaction, why is aqueous ammonium chloride used to get to the products? I don't see how it participated in or modified the reaction from what is normally expected.

Reaction of (3S)-3-methylcyclohexan-1-one with methylmagnesium bromide

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  • $\begingroup$ This question has been edited to be more obviously on-topic. $\endgroup$ – A.K. Jan 18 at 23:50
  • $\begingroup$ In scientific writing, brevity is highly valued. I do not see the point of padding a post when the question is perfectly clear and answerable. It is a waste of time for the editor as well as for future readers. To make myself abundantly clear, I am not criticising the edit, but rather the situation which necessitates this edit. $\endgroup$ – orthocresol♦ Jan 18 at 23:55
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Ammonium chloride ($\ce{NH4Cl}$) is the work-up reagent that quenches the magnesium alkoxide product of the Grignard addition. It is the reagent of choice as it is a proton source without being acidic; acidic conditions could result in protonation of the tertiary alcohol product and elimination to the alkene. It also ensures that all inorganic salts of Mg will extract into the aqueous phase.

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    $\begingroup$ Oh I actually never knew that! In that case, other weak acids can also be used right? For example, what about weak organic acids like ethanoic acid? $\endgroup$ – Tan Yong Boon Jan 18 at 12:09
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    $\begingroup$ Yes, coud use that or something like sodium dihydrogen phosphate. $\endgroup$ – Waylander Jan 18 at 12:36

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