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I would like to ask about the rate of hydrolysis for 4-hydroxyphenyl acetate (A), N-(4-hydroxyphenyl)acetamide (B), 4-hydroxyphenyl chloroacetate (C), and methyl 4-hydroxybenzoate (D). I know that hydrolysis depends on the positive charge of the carbonyl carbon and also the substituent around it (e.g. electron withdrawing).

Therefore, from the picture above I can see that compound C is the most reactive since it has a phenol and a chlorine as its substituents. Compound B is the least reactive due to nitrogen being the weakest electronegatively as compared to oxygen (I think?).

I am confused between compounds A and D because they are isomers of each other. The oxygen and phenol group are on the same substituent in compound A and are in two different substituent in compound D. Does this affect anything? or can I say that the strength of the electron withdrawing groups are additive, meaning compound A and D are similar in terms of reactivity?

  • $\begingroup$ A and D are not similar. When the methyl is attached to the carbonyl it shifts electron density towards the carbonyl carbon. What does the hydroxyl-substituted benzene do when attached to the carbonyl carbon? Draw some resonance structures for D where you use the OH oxygen lone pair. What effect does that have on the carbonyl carbon? $\endgroup$
    – ron
    Commented Mar 30, 2015 at 0:50
  • 1
    $\begingroup$ hi there, thanks for the reply. I have drawn the resonance structure of the hydroxyl-substituted benzene and realised that the negative charge is right beside the carbonyl carbon (stabilizing it? less reactive?). but still this doesnt help me understand whether compound A or D is more prone to hydrolysis. Both the methoxy group in compound D and the phenol + oxygen group in compound A are electron withdrawing. therefore both compound A and D have 1 group that stabilize the carbonyl carbon and 1 group that makes it more reactive? what does that mean in terms of reactivity? $\endgroup$
    – Joash
    Commented Mar 30, 2015 at 1:40
  • $\begingroup$ Yes, stabilizing it, less reactive. Methoxy and phenoxy are roughly the same; methyl is less electron donating to the carbonyl carbon than the hydroxyphenyl. $\endgroup$
    – ron
    Commented Mar 30, 2015 at 2:07
  • $\begingroup$ Thank you, that really helped me out and i understand it now. Therefore the answer to this question is C>A>D>B from the most reactive to the least. $\endgroup$
    – Joash
    Commented Mar 30, 2015 at 2:33
  • $\begingroup$ That is how I see it. $\endgroup$
    – ron
    Commented Mar 30, 2015 at 2:36

1 Answer 1



When a carbonyl compound reacts with a nucleophile we form a tetrahedral intermediate. Since the carbonyl carbon becomes $\ce{sp^3}$ hybridized in the tetrahedral intermediate there is no significant resonance stabilization - the various tetrahedral intermediates have (approximately) the same energy. However, strong resonance effects can occur in the starting carbonyl compound, potentially stabilizing the carbonyl compound.

In other words, while a substituent attached to the carbonyl carbon can stabilize the starting carbonyl compound, it should have little effect on the relative energetic position of the tetrahedral intermediate.

If we place the carbonyl compound and the tetrahedral intermediate on a reaction coordinate we can see that anything that stabilizes the carbonyl compound will increase the activation energy for reaction, slowing down the reaction rate.

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Below are the resonance structures we can draw for our carbonyl compound.

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If structures II and III contribute significantly to the overall description of the compound, then we would expect that carbonyl compound to be more stable, lower in energy, than a carbonyl compound where structures II and III contribute less.


Three of your compounds are esters, one is an amide. We know that resonance structure III is more important in an amide than in an ester (due to electronegativity differences, positive charge on nitrogen is better than positive charge on oxygen). Therefore the amide (B) will react the slowest.

Now to separate the three esters. You are right that ester C will be the most reactive. This is because the inductively electron withdrawing chlorine substituent destabilizes resonance structure II.

In compound D we can draw stabilizing resonance structures like the following; the para-hydroxy substituted aromatic ring is very effective at moving electron density towards the electron deficient carbonyl carbon thereby stabilizing it.

enter image description here

The methyl group in compound A also donates electron density to the carbonyl carbon, but not nearly as effectively as the aromatic ring.

Therefore our overall order of carbonyl reactivity is C > A > D > B


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