[schafer]

Hey,
I am currently studying organic reactions, and was hoping someone could help me understand why carbonyl and carboxyl react so differently to alcohol.
Carbonyl + Alcohol (ROH):
1. The O in alcohol is attracted to the carbon of the carbonyl (Nucleophilic attack)
2. Alcohol binds to C, causing the double bond of carbonyl to break

Carboxyl+Alcohol:
1. The O in alcohol is attracted to the carbon of the carboxyl (Nucleophilic attack)
2. as a result, the bond between the carbon and the hydroxyl group is broken (-OH is released).

Both reactions started the same with a nucleophilic attack, but ended differently.
How come in the first one the strong double bond is broken, and not C-H?
How come in the second reaction it wasn't the double bond to break, but the C-OH?

Thanks a lot!

[Hunter2]

First reaction gives an Acetal or Ketal. This goes easily back to Alcohol and Aldehyd or Keton.

In second reaction you get an Ester by releasing of water what is thermodynamic more stable, although this reaction goes in an equilibrium.

Its the rule of Erlenmeyer that two single O-Bonding to one carbon is unstable and it try to get back the carbonyl.

[OrgXemProf]

Here's an alternative way to view the situation:

Both aldehyde/ketone C=O groups and also C=O groups in acyl derivatives (like CO₂H) are electrophilic. Both undergo nucleophilic attack by, e. g, Nu:^- (like an alkoxide ion), upon the electrophilic carbonyl carbon atom. As you pointed out, the C=O group opens to Nu-C-O^-.

The difference lies in the fate of the Nu-C-O^- anion.

With aldehydes and ketones, the group attached to the Nu-C-O^- carbon atom is either hydrogen (aldehyde) or an alkyl/aryl group (ketone), neither of which are suitable leaving groups.

However, with acyl derivatives (like CO₂H), the group attached to the Nu-C-O^- can function as a leaving group (OH^- in CO₂H, or possibly neutral OH₂ if acid is present).

Thus, aldehyde and ketone C=O groups undergo addition of the elements of Nu-H across the C=O double bond, whereas acyl C=O group undergo a two-step addition-elimination sequence.

So, with an aldehyde or ketone as substrate and alkoxide as nucleophile, the reaction results in initial formation of a hemiacetal via addition of the elements of RO-H across the C=O double bond.

With RCO₂H as substrate, addition of Nu:^- (like alkoxide ion) occurs in the first step, which is followed by elimination of OH in the second step thereby resulting in net substitution of OH by Nu to afford RC(=O)Nu. In the case of alkoxide ion as nucleophile, the reaction results in ester formation.

Hope the foregoing is clear. It's much easier to see when written out in typical organic reaction mechanistic fashion, curly arrows and all.

[schafer]

Thank you very much!