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:
  $\ce{(CH3)3CBr} &gt; \ce{CH3Br} &gt; \ce{CH3CH2Br} &gt; \ce{(CH3)2CHBr}$.
  
  Provide an explaination for this order of reactivity.


But my doubt is how is this order $\ce{(CH3)3CBr} &gt; \ce{CH3Br} &gt; \ce{CH3CH2Br} &gt; \ce{(CH3)2CHBr}$ 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 $(1)&gt;(4)&gt;(3)&gt;(2)$.
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)