Question #b9657

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.

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.