[Winga]

To give acid bromides?

I think it can't, but I don't know the reason.
Is it due to the steric hinderance from alkyl group, -O-(R)?

[AWK]

Try to sketch mechanisms with PBr3 and PBr5. Treat PBr5 asi ionic PBr4(+)Br(-).

[Albert]

In my opinion, this can be an explaination.

[movies]

This is like the Hell-Vollhard-Zelinski reaction, isn't it?  In that case you need an acid though.  My guess is that an ester isn't nucleophilic enough.  Plus, when you have an acid then your byproducts is a phosphine oxide, which is more difficult to get to when your leaving group is an alkoxide.

Albert, the first step of your mechanism should have the lone pair on the carbonyl oxygen attacking the P, not the one that you have drawn.  I used to make that mistake all the time.  My organic prof. never let me live it down!

[Winga]

PBr3 can react with alcohols, acids and even phenols.
Alcohols, acids and phenols contain -OH group, that means PBr3 reacts with -OH group.

However, according to mechanism of the reaction, the first step is a lone pair of the single-bonded oxygen attacks to the electrophilic P of PBr3 and Br- leaves.

From the IR spectra of an acid and an ester, the C=O stretching of the ester is higher than that of the acid. This implies that the delocalization of the ester's -(O)-R lone pair is less available than the acid's -(O)-H lone pair. Therefore, the nucleophilicity of the ester's oxygen -(O)-R, should be higher than that of the acid's.

So, why esters can't react with PBr3? Or why PBr3 only reacts with -OH group?

[movies]

In esters, the oxygen of the carbonyl group is more nucleophilic.  Adding to the other oxygen gives you a cation that can't be stabilized by resonance.  It would be like reacting with an ether.  If the carbonyl oxygen attacks, then you can delocalize the positive charge onto the other oxygen by resonance.  In alcohols and phenols you don't have a choice, but in those cases you just need a proton to fall off, not an alkyl group.

[Winga]

In my opinion, this can be an explaination.

If I add excess Br-, I can still convert RCOOPBr2 into RCOBr.

By the way, are there 2 approaches describe the mechanism of alcohol + PBr3?

One is leaving the HOPBr2 (no deprotonated), another one is leaving -OPBr2 (deprotonated)?

[movies]

I think you probably want the protonated one to leave because then your leaving group is neutral instead of charged.