# Which mechanism is not seen during hydrolysis of esters? [closed]

Which mechanism is not seen during the hydrolysis of esters?

1. AAC2
2. BAC2
3. AAL2
4. BAL2
5. AAC1

I know that hydrolysis of ester is pseudo-first order reaction but I found this question where there are completely unknown reaction mechanisms.

## closed as off-topic by Karl, Pritt Balagopal, aventurin, Mithoron, Todd MinehardtApr 15 '18 at 16:23

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The reaction mechanism said actually corresponds to the following mechanisms,

1. $A_{AC^2}$ which means acid catalysed acyl carbon oxygen bond breaking bimolecular mechanism.
2. $A_{AL^2}$ which means acid catalysed alkyl carbon oxygen bond breaking bimolecular mechanism.
3. $B_{AC^2}$ which means base catalysed acyl carbon oxygen bond breaking bimolecular mechanism.
4. $B_{ALC^2}$ which means base catalysed alkyl carbon oxygen bond breaking bimolecular mechanism.

In the last one it is unimolecular, and acid catalysed and acyl carbon oxygen bond breaking mechanism.

In esters, Hydrolysis can occur via breaking of two possible bonds. If the ester is $\ce{R'-CO-O-R}$, then the $\ce{CO-O}$ bond is the acyl carbon oxygen bond, whereas the $\ce{O-R}$ bond is alkyl carbon oxygen bond. Either of these two breaks in the process of hydrolysis.

If you take usual hydrolysis of esters, they can occur in both acid and base catalysed reaction medium condition, and in that case the $\ce{R'-CO}$ part of the ester goes to the acid, and the $\ce{-O-R}$ goes to alcohol. So, this is the acyl carbon oxygen bond breaking and bimolecular as the $\ce{OH-}$(in base catalysed) and $\ce{H2O}$ (in acid actalysed) attacks at the rate determining step. So, $A_{AC^2}$ and $B_{AC^2}$ is easily seen.

In an ester, where the attack of $\ce{OH-}$ or $\ce{H2O}$ at the carbonyl carbon is highly sterically hindered (e.g. in mesitoic ester) the reaction passes through alkyl carbon oxygen bond breaking, and thus the $\ce{R'-CO-O}$ part of ester goes to acid, whereas only the R part goes to alcohol, and they are also bimolecular generally, as in the slow step, both the nucleophile attacking at the alkyl carbon remains present along with the molecule. That's why $A_{AL^2}$ and $B_{AL^2}$ are also seen but it is slightly rare.

But for showing $A_{AC^1}$, the molecule doesn't have any other chance rather than forming an acyl carbocation(as, the carbocation intermidiate follows unimolecular reactions). But if it forms acyl acrbocation in acid medium, the lone pair on the adjacent oxygen is already donated to $\ce{H+}$, so, there is no extra stabilising factor rather than any highly electron donating group, like tert butyl or phenyl attached to carbonyl carbon. But if you have those, the molecule generally prefers the alkyl bond breaking to avoid steric hindrance.So, $A_{AC^1}$ is very very rare to happen.

But $A_{AL^1}$ can be seen where R is tert butyl group. because, it can get stabilisation by forming a carbocation of its own, and the remaining portion of molecule can get stabilisation through resonance.