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In the following reaction we have to find A and B.

reaction scheme: propionyl amide reacts with A to give ethyl amine; this reacts with B to give ethanol

I thought that A could be $\ce{Br_2}$/$\ce{KOH}$ resulting in a Hoffmann bromide reaction.

For B, I thought I would need a base such as $\ce{KOH,NaOH}$. But the answer is $\ce{HNO2}$. Why?

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    $\begingroup$ Think about what forms when a primary amine reacts with HONO... $\endgroup$ – Eashaan Godbole Dec 8 '16 at 9:28
  • $\begingroup$ @EashaanGodbole I don't know $\endgroup$ – user123733 Dec 8 '16 at 9:39
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    $\begingroup$ Well, your choice of reagents is definitely not going to do anything. The amide is a terrible leaving group and will not undergo substitution. You should look at diazonium... $\endgroup$ – Zhe Dec 8 '16 at 19:48
  • $\begingroup$ @Zhe does it is good substituting compound in almost every cases $\endgroup$ – user123733 Dec 9 '16 at 2:19
  • $\begingroup$ I think that actually highlights your misunderstanding of the substitution reaction. You must have a good leaving group. Having a good nucleophile doesn't hurt, but if the leaving group is bad, no amount of nucleophilicity is going to force that substitution. $\endgroup$ – Zhe Dec 9 '16 at 3:49
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Hydroxide may be a somewhat good nucleophile but the amide anion is one of the most lousy leaving groups you can imagine. If you want to remove an amino group in an $\mathrm{S_N2}$ reaction, it can only be eliminated as an ammonium cation — but protonating it means we can no longer use hydroxide as a base. Under basic conditions an amino group will never be a leaving group.

Instead, you must turn the amino group into something that is actually a good leaving group. If you include nitrogen and let your thoughts wander, you may arrive at the diazonium group $\ce{-N2+}$. Upon nucleophilic attack, this will be released as nitrogen gas which is great both thermodynamically and kinetically. This leaving group is so good that even relatively bad nucleophiles such as water will react with it.

However, you need to produce it from the amine. This is where nitrous acid comes in. Under acidic conditions, $\ce{HONO}$ will react with primary amines to give diazonium salts according to the following general equation:

$$\ce{R-NH2 + HONO + H+ -> R-N2+ + 2 H2O}\tag{1}$$

Since nitrous acid is produced by acidifying aquaeous solutions of nitrite salts, you have all three components in hand if you ‘just’ add $\ce{HONO}$ or $\ce{HNO2}$.

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