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In cannizzaro reaction, why doesn't the intermediate formed in second step show nucleophilic attack on another aldehyde by virtue of the negative charge on the oxygen. Why departure of hydride is favoured more?

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I am a high school student so please, answer keeping this in mind.

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  • $\begingroup$ Because it doesn't do anything? Things like that happen all the time, but there's no point in wasting time on drawing them, when they don't result any new compound. $\endgroup$
    – Mithoron
    Mar 16 at 15:41
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    $\begingroup$ It does, but as @Mithoron says there is no forward pathway for the molecule that is formed. The reaction is completely reversible so any material formed goes back to the aldehydes which are slowly consumed by the reaction that does have a forward pathway i.e. the formation of alcohol and acid $\endgroup$
    – Waylander
    Mar 16 at 15:54

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