What reaction results in the formation of an amide?

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What reaction results in the formation of an amide?

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Answer 1 · 2016-09-03T14:57:22

$R C (= O) X + 2 H_{2} N R \to R (C = O) N (H) R + H X \cdot H_{2} N R$ $RC(=O)X + 2H_2NR rarr R(C=O)N(H)R + HX*H_2NR$

Explanation:

As shown, this is the amine equivalent to the esterification reaction. An acid halide is treated with 2 equiv of amine, 1 equiv forms the $(O =) C - N$ $(O=)C-N$ bond, the other equiv sops up the liberated $H X$ $HX$ as the ammonium salt. (If the amine were expensive, I could use just the 1 equiv. and use $E t_{3} N$ $Et_3N$ as the base.)

Alternatively (and less likely) you may mean the formation of amido, $^{-} N H_{2}$ $""^(-)NH_2$ anions:

$2 N H_{3} ⇌ N H_{4}^{+} + N H_{2}^{-}$ $2NH_3 rightleftharpoons NH_4^(+) + NH_2^-$ .

This equilibrium operates in liquid ammonia, and is precisely equivalent to the acid/base equilibrium that water undergoes:

$2 H_{2} O (l) ⇌ H_{3} O^{+} + H O^{-}$ $2H_2O(l) rightleftharpoons H_3O^+ + HO^-$

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What reaction results in the formation of an amide?

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