1. Tardigrade
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  4. CH 3 CHO + NH 2 OH → CH 3 CH = N - OH The above reaction is carried out at

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Q. $CH _{3} CHO + NH _{2} OH \to CH _{3} CH = N - OH$ The above reaction is carried out at

Aldehydes Ketones and Carboxylic Acids

Solution:

More Basic condititions: $OH^-$ may attack $CH _{3} CHO$ More Acidic conditions : $H ^{+}$ may attack $NH _{2}- OH$ At optimum pH i.e 4.5 reaction takes place