Question

03 O-Al H-Al-H H-Al-H H-Al-H OH

In #5 (carboxylic acid), why doesn't the AlH4(-) attack the carbonyl carbon and proceed to the tetrahedral intermediate...why does it instead attack the hydroxy group to form O-AlH3 and then does an H(-) attack on the carbonyl carbon?

On the other hand, in #1 (ester), the ALH4(-) attacks the carbonyl carbon...tetrahedral intermediate, OCH3 leaving group, etc. (this is the mechanism I expected #5 to be)

Answer 1

So, you are in confusion with two different types of mechanism instead of having the same type of starting materials.

Well as you know, the carboxylic acid proton is much more labile and susceptible towards bases. LiAlH₄ is nothing but hydride donor (an H^- , a base itself). So it will preferentially take up the proton instead of acting as nucleophile in first step.

The 2nd reason is that OH^- is bad leaving group unlike OMe^- . The weaker the Y^- is as a base (or the stronger H-Y is as an acid) the better a leaving group will be. Methanol is slightly more acidic than water. Their pKa values, in water, are 15.5 and 15.7, respectively. So it is rather better to render the OH^- group into a good leaving group by OAlH₃^- .

Hope the reason is clear to you.