3
$\begingroup$

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 $(1)>(4)>(3)>(2)$.

Improve this question
$\endgroup$
3
  • $\begingroup$ Do you call it a hydrolysis reaction? ... Because I just call it a substitution reaction... $\endgroup$
    – DHMO
    Commented Sep 27, 2016 at 13:31
  • $\begingroup$ @DHMO I stated what my textbook and study material wrote.I did'nt make up the definitions myself.See en.wikipedia.org/wiki/Solvolysis $\endgroup$
    – user14857
    Commented Sep 27, 2016 at 13:34
  • $\begingroup$ Hydrolysis is not necessarily SN1, it can also refer to a SN2 process (especially if there is OH-). Or it can be any process really. There are many different ways of hydrolysing esters, for example. However, in this particular case, it is weird. I would expect only SN1 reactivity under the stated conditions, so I don't know why MeBr is so high up. $\endgroup$
    – orthocresol
    Commented Sep 27, 2016 at 13:40

1 Answer 1

Reset to default
4
$\begingroup$

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)

Improve this answer
$\endgroup$
7
  • $\begingroup$ By the extension of your definition: In the solvolysis definition in Wikipedia too should the $\mathrm{S_N1}$ be removed? See en.wikipedia.org/wiki/Solvolysis $\endgroup$
    – user14857
    Commented Sep 27, 2016 at 13:43
  • $\begingroup$ @ZOZ Yes.$%MathJax comments hurray!$ $\endgroup$
    – Jan
    Commented Sep 27, 2016 at 13:45
  • $\begingroup$ Morever,I cannot understand how can water in this case act as nucleophile in 20 percent ethanol.Water is considered to be a weak nuceophile.Can it really initiate a $\mathrm{S_N2}$ ? $\endgroup$
    – user14857
    Commented Sep 27, 2016 at 13:45
  • $\begingroup$ @ZOZ Water isn’t that weak a nucleophile, tbh. $\endgroup$
    – Jan
    Commented Sep 27, 2016 at 13:46
  • $\begingroup$ Well by that logic if it were 80 % ethanol and 20 % water would ethanol have acted as the nucleophile? I mean the ethyl group exerts a $+I$ effect on the O atom.So would ethanol be nucleophile in that case? @Jan $\endgroup$
    – user14857
    Commented Sep 27, 2016 at 13:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.