Question

Question #b9657

Answer 1

The process of this Nucleophilic Acyl Subsctitution is explaied below

Heat helps in dissociation of water

`"RCONR'"_2 + "H"^(+) + "OH"^(-) rarr ("RCO"+"OH"^(-)) + ("R'"_2"N"+"H"^+)`

Mechanism

To begin, we know that in the first step, a strong base is going to be used like `"NaOH"`, where the `color(blue)("OH"^(-) "will act as the")` `color(blue)("nucleophile and attack the electrophile, the partial")` `color(blue)("positive carbon of the carbonyl group")`

Now, when the bond forms between the `"oxygen"` of the `"OH"^-` group and the carbon, the `"pi bond"` of the carbonyl is broken and the electrons are pushed toward oxygen, leaving it with a negative `-1` charge.

`---------------------`

`color(purple)(2.)`

`color(blue)("Oxygen will try to remove this negative charge by")` `color(blue)("reforming a double bond with the carbon.")` `color(blue)("But since we know that the carbon atom")` `color(blue)("cannot form more than 4 bonds, the bond between it ")` `color(blue)("and the amine's nitrogen has to be broken.")` `color(blue)("The electrons will be pushed onto the nitrogen of the")` `color(blue)(" amine and the amine will act as the leaving group (a poor one at that)")`

*The reason the amine is a poor leaving group is because the amine is a strong base, and strong bases are very unstable. Leaving groups are based off their stability in solution, like the halides, tosylate, `"H"_2"O"`

`---------------------`

`color(purple)(3.)`

`color(blue)("Here, the NH"_2^(-) "will act as a base and pluck off the")` `color(blue)(" H from the carboxylic acid -type of acid-base")` `color(blue)("reaction. When this happens, a 'carboxylate anion'")` `color(blue)("forms and NH"_3)`.

Mechanism of acid catalysed amide hydrolysis

Step 1)An acid/base reaction. Since we only have a weak nucleophile and a poor electrophile we need to activate the ester. Protonation of the amide carbonyl makes it more electrophilic

Step 2:
The water O functions as the nucleophile attacking the electrophilic C in the C=O, with the electrons moving towards the oxonium ion, creating the tetrahedral intermediate.

Step 3:
An acid-base reaction. Deprotonate the oxygen that came from the water molecule.

Step 4:
An acid/base reaction. Need to make the -NH2 leave, but need to convert it into a good leaving group first by protonation.

Step 5:
Use the electrons of an adjacent oxygen to help "push out" the leaving group, a neutral ammonia molecule.

Step 6:
An acid/base reaction. Deprotonation of the oxonium ion reveals the carbonyl in the carboxylic acid product and regenerates the acid catalyst.

Answer 2

Coming to the main answer

At first though the formation of a sp3 hybridized carbon from a sp2 carbon leads to a less stable intermediate

https://chemistry.boisestate.edu/richardbanks/organic/amidehydrolysisbasictutorial3.htm

In the second step though the formation of a sp2 hybridized carbon from a sp3 carbon leads to a stabler intermediate,as the amine behaves as the leaving group this reaction is slower because amine is a poor leaving group.

It is unstable so the reaction will not favor it much,

But the next reaction is fast because the unstable leaving group amine is stabilized by a proton from the reagent to form a nice, stable leaving group. So it is energetically favorable.