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I tried to find Aldol products for two molecules of propanal.
While doing the mechanism, I observed the formation of a $2$° carbanion, so I did a $1,2$-hydride shift to convert that into a 1° carbanion (as $1$° carbanions are more stable than $2$° carbanions) but, while continuing with the $1$° carbanion, I get the wrong result for the product (hex-3-enal), and the $2$° carbanion forms the actual product (2-methylpent-2-enal). Why is this so?

Below attached is the image of my mechanism for reference showing both the 1° and 2° carbanion reactions:

enter image description here

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    $\begingroup$ Your equilibrium to a primary carbanion does not take place, only the secondary carbanion is formed. The carbanion formed is better drawn as the enolate anion. $\endgroup$
    – Waylander
    Aug 12, 2023 at 5:15
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    $\begingroup$ It is an alpha carbanion not a secondary. The left column is correct. Since the carbanion reacts with the alpha carbon the enol structure might not be correct either. The structure is a resonance hybrid with dispersed negative charge. Aldol reactions are reversible equilibria and can make a mess if not run carefully. $\endgroup$
    – jimchmst
    Aug 12, 2023 at 7:11

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You are comparing an isolated carbanion to a carbanion next to the carbonyl. The latter is much more stable and easily formed because of the partial +charge on the carbonyl carbon and resonance with the double bond. The alpha hydrogens are much more acidic. This is the property of carbonyl compounds such as aldehydes, ketones, esters, anhydrides, even nitriles, nitro compounds and the 'onium salts that enables aldol type reactions. These reactions take several chapters in most Organic Chem texts. [Get one, I recommend Morrison and Boyd]

Stability of carbanions: Your very first equations get you in trouble! There is no equilibrium between the primary and alpha carbanions; the energy difference is too large. The alpha carbanion has partial negative charge on the carbon and oxygen by resonance. [There is chemical equilibrium between the keto and enol forms of the molecule]. The primary carbanion is difficult to form but is stable and does not rearrange because the sp3 hybridization exposes the electron pair and it is strongly bonded to its counter ion. There must be no nearby Lewis acids for it to react with. If there were the carbanion would not have formed [your case above]. This lack of rearrangement extends to other electron rich transition states such as E2 and SN2. However, carbanions that are resonance stabilized lack the strong ionic or ion-paired bonding and can be multidentate.

Carbocations are the opposite. They are generally formed in a manner that solvates the anion leaving an intense +charge on the carbon. The nearest Lewis base is on an adjacent carbon atom. The electron pair will migrate in accordance with its basicity and the stability of the new carbocation with some geometric considerations.

Conclusion: Isolated Carbanions are difficult to form but their ionic bonds are stable if no Lewis acids are present; they do not rearrange. Conjugated carbanions as in carbonyl compounds are much more readily formed and can be polydentate. Carbocations are formed poorly solvated and subject to attack by adjacent electron pairs causing rearrangement to a more stable carbocation.

Finally these are my ideas on this. Study some texts for some specific examples and further discussion. look up the difference in acidity between alkanes and various carbonyl compounds.

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  • $\begingroup$ Thanks for the answer. This is the book you're referring to, right? I think I've seen it in my school's library, I'll check it out. $\endgroup$ Aug 12, 2023 at 11:39
  • $\begingroup$ The older editions might be just as good in describing mechanisms. $\endgroup$
    – jimchmst
    Aug 12, 2023 at 17:37
  • $\begingroup$ My library has the 7th edition, it's good. Thank you so much. $\endgroup$ Aug 14, 2023 at 13:34

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