Benzaldehyde and p-anisidine are dissolved in dichloromethane and can react to form (E)-N-(4-methoxyphenyl)-1-phenylmethanimine. Without using an...
Benzaldehyde and p-anisidine are dissolved in dichloromethane and can react to form (E)-N-(4-methoxyphenyl)-1-phenylmethanimine. Without using an acid catalyst or under any acidic reaction conditions, draw a step-by-step mechanism to show this reaction. Magnesium sulfate was also used in the reaction.
Homework Answers
It is the reaction of -NH2 group with -CHO.
Nucleophilic attack from -NH2 of anisidine due to l.P.present on Nitrogen to the carbonyl group of benzaldehyde.
Trigonal carbonyl convert to Tetrahedral C (sp2 to sp3) and oxygen of CO will be negatively charged, it means oxyanion will be formed and -NH2 will be carried positive charged.
The proton exchange reaction takes place from -NH2 (+) to CO(minus) and neutral condense adduct will be formed.
But it is unstable and remove water as -OH as negative and H+ from -NH then system will be conjugated, more stable due to delocalisation of double bonds (extend of conjugation).
And removal of water is done by anhydrous magnesium sulphate therefore reaction will not undergo backward direction
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Benzaldehyde and p-anisidine are dissolved in dichloromethane
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