I am currently studying aldehydes and ketones and their reactions. I've noticed that acid catalysed reactions are represented with reversible arrows, while base catalysed reactions are shown with regular arrows. Why is this so? Is this because the addition of the nucleophilic base stabilises the electrophilic carbon? Or is there some other reason?

Update: Sorry, "all" would not be appropriate here. In my text-book, it is mentioned only once. Nucleophilic substitution by weak nucleophiles is carried out in a weakly acidic medium, while a basic medium is used for strong nucleophiles. The corresponding mechanism use the convention I've spoken about. I can't see the use of reversible arrows elsewhere. I wanted to know why is the nucleophilic addition reversible (as per my book) in a weakly acidic medium.

My text book (not a standard one): Nootan ISC Chemistry XII

The pictures of the reactions (I'm not able to rotate them for some reason) Acid catalysed reaction Base catalysed reaction

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    $\begingroup$ Can you give us examples of the sort of reactions you have been studying? $\endgroup$
    – bon
    Commented Jul 22, 2016 at 20:21
  • $\begingroup$ @bon Sorry for the late reply. I wrongly used the word "all". I've seen it only once. $\endgroup$ Commented Jul 22, 2016 at 20:46
  • $\begingroup$ @ambigram_maker- any chance of you taking a picture/drawing out the actual reactions in your book... it might be easier for you to get a useful answer if the question is specific. (By weak and strong nucleophiles I'm imagining you're talking about something like hydroxide vs something like a Grignard? ) $\endgroup$
    – NotEvans.
    Commented Jul 22, 2016 at 20:59
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    $\begingroup$ Given the mechanisms in your post, I don't think it's appropriate to use the reversible arrows, and I don't believe the mechanism is quite accurate. In the acid-catalyzed case, the first step should really be protonation of $\ce{O}$ on the carbonyl group; in addition, negatively charged intermediates should not be present in acidic media. $\endgroup$ Commented Jul 23, 2016 at 2:00

1 Answer 1


enter image description here

This is a picture from wiki (Simple aldol reaction). Both Acid and base catalyzed reactions are reversible up to aldol intermediate. Dehydration is the only irreversible step. So this statement " I've noticed that all acid catalysed reactions are represented with reversible arrows, while base catalysed reactions are shown with regular arrows. " is incorrect.

Single arrow means that the step is irreversible. Reversible arrow means that reaction can go both forward and backward. Usually reversible steps are happening before the irreversible step and a happening faster than irreversible step.

This is easy to demonstrate with a pool table that nave a small obstacle near the hole. Reversible reactions are balls that move over the the surface of the table. They can get from any point to any other point of the table. Sometimes they get on that small obstacle near the hole and roll back to the table (they were not able to overcome activation barrier). Irreversible reactions are represented by balls that were able to overcome the barrier and now ended up in the hole. Now they have no chance to go back to the table on their own.

What makes reaction irreversible vs reversible. From thermodynamics perspective irreversible reactions are reactions with a large change in $\Delta$G. From organic chemists perspective reversible reactions are typically those that include protonation/deprotonation steps. Irreversible reactions are those where a strong bond like C-C or O-P forms.

Reversibility is a subject to conditions. 2H -> H2 (recombination of radicals) is effectively irreversible at room temperature, but is reversible at 3000K.

Why do we need to account for reversible reactions? (1) To understand the mechanism. (2) To know how to increase concentration of the reagents right before the first irreversible step to accelerate the overall reaction.

tl;dr: in reactions you are asking about formation of C-C bond is effectively irreversible step.

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    $\begingroup$ Im not sure this really answers why acid catalysed mechanisms have reversible steps whilst base catalysed ones have unidirectional steps in some cases... $\endgroup$
    – NotEvans.
    Commented Jul 22, 2016 at 20:13
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    $\begingroup$ $\ce{C-C}$ bonds can form in both acid and base catalysed mechanisms (aldol reaction for example). This doesn't actually answer the question. $\endgroup$
    – bon
    Commented Jul 22, 2016 at 20:20
  • $\begingroup$ @NotNicolaou because the assumption is incorrect, that's why. $\endgroup$
    – sixtytrees
    Commented Jul 22, 2016 at 20:33

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