2
$\begingroup$

enter image description here

from the reaction, the 3rd molecule (carboxylic acid) formed undergo neutralization with OH- (proton transfer to a base).

However, is it possible for the 3rd molecule to behave as the 1st molecule (ester) and undergo nucleophilic attack by the OH- as they both have an electrophilic centre (carbon)?

If yes, will this reaction form a cycle (never forms carboxylic salt). If no, may I ask why? Is it related to stability or other effects? Thank you very much

$\endgroup$
  • $\begingroup$ What would you expect the product of such a reaction to be? $\endgroup$ – LigninPauling May 25 at 13:03
  • $\begingroup$ @LigninPauling I think it may be a mixture of first 3 molecules in equilibrium as the OH in 2nd molecule seems to be a leaving group (regenrated) Sorry that I am a total beginner $\endgroup$ – 234ff May 25 at 13:08
  • $\begingroup$ If you chose the carboxylic acid as the starting compound, as opposed to the ester, would you not find that the nucleophile and the leaving group are the same? $\endgroup$ – LigninPauling May 25 at 13:24
  • $\begingroup$ Oh I see so can this be one of the reason why the first 2 steps are reversible (i.e. has chance to undergo the attack? Also, does this means the proton transfer step occurs by higher chance (lower energy level)? Thank you for answering $\endgroup$ – 234ff May 25 at 13:36
  • 1
    $\begingroup$ Yes indeed. For the first step, either the hydroxy group or the alkoxy group may leave, giving reactants or products, respectively. For the second step, it is reversible as the generated alkoxide ion may acts as a nucleophile to attack the carboxylic acid and give rise to the intermediate structure once more. $\endgroup$ – LigninPauling May 25 at 16:23
1
$\begingroup$

Because you are a beginner to organic chemistry, I'll tell you this: Both esterification (e.g., Fischer esterification) and de-esterification reactions are reversible. The question you had at the end is why the last step of base catalyzed de-esterification reaction is only forward reaction. As you have learned in general chemistry class, acid-base reaction is one of the fastest reactions in chemistry (not counting nuclear reactions). Thus, the last step is governed by preference for acid base reactions:

$$\ce{R-COOH + OH- -> R-CO2- + H2O}$$

The $\mathrm{p}K_\mathrm{a}$ of $\ce{R-COOH}$ is about $4.7$ ($\mathrm{p}K_\mathrm{a}$ of $\ce{CH3-COOH}$ is $4.75$) while $\mathrm{p}K_\mathrm{a}$ of $\ce{H2O}$ is $15.7$. Therefore, forward reaction is about $10^{10}$ times faster than backward reaction.

Also remember, this reaction is in aqueous medium, therefore there are much more $\ce{H2O}$ molecule available than $\ce{R-OH}$ molecules. Thus, going back to ester reaction is much slower than becoming acid (governed by Le Chatelier principle).

| improve this answer | |
$\endgroup$
  • $\begingroup$ This means in every step of reactions I should first consider position of equilibrium (pKa) of acid-base reactions other mechanisms (e.g. nucleophilic attack) since it occurs much easier? Thank you for answering $\endgroup$ – 234ff May 27 at 2:58
  • $\begingroup$ It is depends. Think, your reactant is a strong base as well as a good nucleophile. For example, consider Grignard reaction.Grignard reagent is a strong base, but under goes nucleophilic reactions with ketone, etc. But, if you hace trace of water or prtic solvent, it react with them acting as a base. But, even if $\alpha$-H in ketone is relatively acetic Grignard reaction prefers nucleophilic attack rether than abstracting $\alpha$-H. See, it's a little complicated. But, you'd be okay when you got enough experience. :-) $\endgroup$ – Mathew Mahindaratne May 27 at 4:00

Not the answer you're looking for? Browse other questions tagged or ask your own question.