What exactly is hydrolysis reaction in organic chemistry?What mechanism does it follow?

According to my study material:



> 
> Hydrolysis is a special type of nucleophilic substitution ($\mathrm{S\_N1}$) where water acts as both nucleophile and a solvent molecule.
> 
> 
> 


Is this definition correct? If so how to solve the following problem (picked up from my textbook exercise which has no solution)?



> 
> When alkyl bromides (listed here) were subjected to hydrolysis in a
>  mixture of ethanol and water ($80\ \%\ \ce{C2H5OH} / 20~\%\ \ce{H2O}$)
>  at $55~\mathrm{^\circ C}$ the rates of reaction showed the following order:
>  $\ce{(CH3)3CBr} > \ce{CH3Br} > \ce{CH3CH2Br} > \ce{(CH3)2CHBr}$.
> 
> 
> Provide an explaination for this order of reactivity.
> 
> 
> 


But my doubt is how is this order $\ce{(CH3)3CBr} > \ce{CH3Br} > \ce{CH3CH2Br} > \ce{(CH3)2CHBr}$ possible?


I mean I know the first compound will form a tertiary carbocation. That is okay for $\mathrm{S\_N1}$ but how will the second compound undergo $\mathrm{S\_N1}$ (because methyl carbocation is very unstable)? Moreover if really $\mathrm{S\_N1}$ is taking place the fourth compound should form a more stable carbocation compared to the third compound. These things are confusing me. Is hydrolysis really $\mathrm{S\_N1}$ or something else?


According to me the order should have been $(1)>(4)>(3)>(2)$.




Your textbook is almost correct. Here is the corrected version of its statement:



> 
> Hydrolysis is a special type of nucleophilic substitution where water acts as both nucleophile and a solvent molecule.
> 
> 
> 


Note me omitting the $\mathrm{S\_N1}$ part. This is because a hydrolysis mechanism can be *any* nucleophilic substitution, whether $\mathrm{S\_N1, S\_N2, S\_N}$ for heavy atoms or $\mathrm{S\_NAr}$.


That also explains the given order of reactivity. *tert*-Butyl bromide is very fast at $\mathrm{S\_N1}$ reactions. Bromomethane very rapidly undergoes $\mathrm{S\_N2}$ reactions. The other two are in-between with bromoethane being just a tad better at $\mathrm{S\_N2}$ reactions than 2-bromopropane is at $\mathrm{S\_N1}$ reactions.


($\mathrm{S\_N1}$ is preferred in 2-bromopropane as the solvent is polar protic and it can stabilize the intermediate carbocation well)


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