Dolbier HW Solutions 445

Dolbier HW Solutions 445 - 3 ) 2 O H Cl (CH 3 ) 2 NH 1 C 6...

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In analogy with parts ( a ) and ( b ) of this problem, a proton is lost from the hydroxyl group along with chloride to restore the carbon oxygen double bond. ( d ) The tetrahedral intermediate formed from benzoyl chloride and methylamine has a carbon nitrogen bond. The dissociation of the tetrahedral intermediate may be shown as More realistically, it is a second methylamine molecule that abstracts a proton from oxygen. ( e ) The intermediates in the reaction of benzoyl chloride with dimethylamine are similar to those in part ( d ). The methyl substituents on nitrogen are not directly involved in the reaction. Then 1 C 6 H 5 CN(CH 3 ) 2 O N , N -Dimethylbenzamide Dimethylammonium chloride (CH 3 ) 2 NH 2 Cl 2 1 C 6 H 5 CN(CH
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Unformatted text preview: 3 ) 2 O H Cl (CH 3 ) 2 NH 1 C 6 H 5 CCl O Benzoyl chloride (CH 3 ) 2 NH Dimethylamine Tetrahedral intermediate C 6 H 5 CN(CH 3 ) 2 OH Cl 1 C 6 H 5 CNHCH 3 O N-Methylbenzamide Methylammonium chloride 1 CH 3 NH 3 Cl 2 C 6 H 5 CNHCH 3 O H Cl CH 3 NH 2 1 C 6 H 5 CNHCH 3 O N-Methylbenzamide HCl Hydrogen chloride Tetrahedral intermediate C 6 H 5 CNHCH 3 O H Cl 1 C 6 H 5 CCl O Benzoyl chloride CH 3 NH 2 Methylamine Tetrahedral intermediate C 6 H 5 CNHCH 3 OH Cl 1 C 6 H 5 COCH 2 CH 3 O Ethyl benzoate HCl Hydrogen chloride Tetrahedral intermediate C 6 H 5 COCH 2 CH 3 O H Cl CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION 539...
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