[confusedstud]

When we add an unsymmetrical ketone to Br2 and HOAc, we get a brominated product with the more substituted alpha carbon getting the proton like this https://ibb.co/Zzj30Vr . Would this be because the rate determining step of the reaction be the bromination step, such that in the first step, the thermodynamic tautomer form (which takes longer to produce) has enough time to be generated and react with the Br2 in the subsequent step (Rate determining bromination step) ? Thereby making the major project the thermodynamic product.

On the other hand, when using a strong base on the unsymmetrical ketone, the product would be brominated in the less substituted alpha carbon like this https://ibb.co/gTmvj3P . Would this be because the rate determining step would be the formation of the enolate ion, and as such, the faster forming kinetic product which is formed from the less substituted would immediately react with the Br2 and not have enough time to convert to the thermodynamic product. And thereby making the major product the kinetic product?

[pgk]

Yes and no!
1). Bromination of ketones under acidic conditions:
Indeed, bromination occurs via the enol form. So, it is similar to the bromination of a double bond that follows the mechanism of the Markownikoff's rule and thus, the bromine anion ought to attack on the carbonyl carbon. But this does not happen because bromine anion prefers to attack on the hydroxyl proton with simultaneous carbonyl formation, due to both thermodynamic and kinetic reasons (simultaneous formation of hydrogen bromide).
2). Bromination of ketones under acidic basic conditions:
Indeed, bromination occurs via a carbanion formation and mainly via the formation of a primary carbanion, which is more stable and thus, it is thermodynamically favored than the secondary one. However, a more stable carbanion means a longer half-life and thus, bromination of a primary carbanion is kinetically favored, too.


*mod edit for clarity

[confusedstud]

Regarding the bromination under acidic conditions:
https://imgur.com/a/0T1iMjF

initially wouldn't most of the ketone exist in the keto form, then when acid is first added wouldn't the kinetic enol be formed faster, and, if it continues to react with the bromine in the kinetic enol form, wouldn't the less substituted brominated product be formed like this as the major product?

Which was why I had the initial impression that the subsequent bromination step was the slow step such that there is sufficient time for the kinetic enol to convert into the thermodynamic enol which then reacts irreversibly with the bromine to form the more substituted brominated product.

[pgk]

Acids catalyze the enol formation and remove keto-enol equilibrium to the right. But regardless if being formed faster, enol behaves as a double bond towards bromination and its π-electrons attack on the bromine molecule in a way that the now formed bromine anion can attack by its turn, to the most electropositive carbon, which is the carbonyl carbon. However and as explained above, the latter is interrupted by the labile hydrogen of the enol hydroxyl that is involved in enol-ketone tautomerism and finally leads to the bromination of the more substituted carbon with simultaneous formation of HBr.
This happens because the less substituted enol-double bond tends to isomerize to the more substituted one, under acidic conditions.

[confusedstud]

I see this makes a lot more sense now, for the base-catalyzed case like this: https://imgur.com/Wf23STG we would say that attacking the primary alpha carbon's hydrogen (HA) is more favored since its both kinetically more favorable as it has a more stable transition state as shown in the first image.

But would it also be more thermodynamically favorable since, https://imgur.com/6UKz0Y9, for A in the red alkene form it is more substituted and so more stable, but in the blue carbanion resonance form it is less stable since there are two alkyl donating group on the carbanion? So I don't really know how to compare which is more thermodynamically stable. But I presume as you've mentioned in the first reply that the primary carbanion is most stable, the blue resonance structure is the main contributor to B being more stable?

But in the final step, https://imgur.com/vTujkZH the product of A becomes the most stable as we take away the destabilization effects of the carbanion, then the more substituted alkene is more stable. So if the reaction is done at cold temperature, we get the kinetic enol and at higher temperatures, we convert them into the thermodynamic enol form?

 

[pgk]

Attention because the mechanism of ketone bromination under acidic conditions is different than the one, under basic conditions.

1). Under neutral and acidic conditions, keto-enol equilibrium is a tautomerism, meaning that the keto and enol forms are separate, distinguishable by chemical and spectral methods and sometimes, isolable.
In your example, you have three tautomers: a ketone, a less substituted enol and a more substituted enol that is more stable (due to a lower dipole moment).
As mentioned above, bromination occurs via the double bond and consequently, the more substituted enol that is more stable enol is brominated, in preference.
Once the enol is brominated and due to keto-enol equilibrium, Le Chatelier's principle, etc. the reaction goes on.

2). Under basic conditions, keto-enol anion is a conjugation product, meaning that there is a unique compound that derives from hybridization of orbitals, where the conjugation forms do not equally contribute therein. Consequently, bromination occurs with the most electronegative participant, which is the less substituted (carbanion forming) carbon.

PS 1: Conjugation under basic conditions, explains why enols have one pka only and not different pka values for the hydoxyl anion and the carbanion, respectively.

PS 2: The above is for your education. But if this is a question in exams, the answer is more simple.
1). Under acidic conditions:
Enol is a hard base that cannot react with bromine that is a soft Lewis acid, in contrast to the double bond that is asoft base and can.
Due to substitution by a hydroxyl group, the double bond is highly polar and consequently, bromination of the double bond mainly occurs to the most substituted double bond.
2). Under basic conditions:
Enol anion is a hard base that cannot react with bromine that is a soft Lewis acid, in contrast to the carbanion that is a soft base and can.
Bromination mainly occurs in the less substituted carbanion, which is the more stable.