Question

hi! For part e, I was wondering why the carboxylate wouldn't deprotonate the ammonium and thus the final product be one with a carboxylic acid?

Problem 18.27 Show the product of treating this anhydride with each reagent. MACOOH (b) H2O, NaOH COONa+ (a) HO, HCI heat sweat "COOH heat COO Nat w -d J. CH2OH CH, OH COOCH3 (c) 1. LiAlH4 2. H20 (d) "CH,OH "NICOOH NACONH2 (e) NH3 (2 mol) "ICOONH4+ Problem 18 28 The analarin

Answer 1

In part (e) anhydride when treated with ammonia it form amide.

If 1 mol of NH​​​​​3 is treated it will form one amide and one carboxylic acid but, here 2 mol of NH​​​​​3 is present so two amide will form.

Also carboxylate will not deprotonate amide but on heating dehydration will take place and it form 2 amide.

Solutions x 1 . yao i 2 . --- A unt CONH2 v ICOONHA

The same reaction you can see in  Gabriel phthalimide reaction.

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