When o -phthalaldehyde is treated with base, o -(hydroxymethyl)benzoic acid is formed. Show the

When o -phthalaldehyde is treated with base, o...

Key Concepts

Cannizzaro Reaction

This reaction involves the base?induced disproportionation of non-enolizable aldehydes, where one molecule is reduced to an alcohol while another is oxidized to a carboxylate. It is a self-redox process that does not require an external oxidizing or reducing agent, and it is central to understanding how aldehydes without ?-hydrogens behave under strongly basic conditions.

Aldehyde Reactivity Under Basic Conditions

Aldehydes are electrophilic, and in the presence of a base, they undergo nucleophilic attack by hydroxide ions to form tetrahedral intermediates. This fundamental reactivity enables several base-catalyzed transformations, including formation of geminal diols and further rearrangements or redox processes, as seen in reactions like the Cannizzaro reaction.

Intramolecular Redox Processes

In intramolecular redox reactions, both oxidation and reduction occur within the same molecule. For example, in systems with two carbonyl groups positioned appropriately, the base-catalyzed mechanism can involve one carbonyl group being reduced to an alcohol while the other is oxidized to a carboxylate, highlighting the interplay of nearby reactive sites in organic molecules.

Hydride Transfer

A key step in many base-catalyzed reactions is the migration of a hydride ion from one carbon to another. This hydride transfer is essential in the Cannizzaro reaction mechanism, where the alkoxide intermediate formed from one aldehyde group donates a hydride to another carbonyl carbon, facilitating the simultaneous oxidation and reduction that leads to the final products.

Transcript

00:01 Okay, this problem is asking us to consider an aldol reaction catalyzed by acid.

00:04 So what we're normally accustomed to is using an aldol connoissation and reacting that with a base, but this is just demonstrating that we can actually do it in acidic conditions.

00:12 Okay, so what i'm going to start with is just a regular ketone.

00:16 I'm going to pronate that with hydronium.

00:19 Okay, so my oxygen is going to get protonated, and i'm going to move the electrons onto that oxygen to make water.

00:25 So i should end up with a protonated ketone like that, and then just water.

00:30 Oh, h2, or h2o.

00:32 Okay, so next up, i have my protonated ketone, and normally at this step, this is the step that we might consider if we're in basic conditions that we need to deprotonate the acidic hydrogen.

00:42 In this case, we're going to do the same because we have water.

00:45 Water is actually relatively basic.

00:47 It's going to deprotate my acidic hydrogen, which is attached to my alpha carbon.

00:51 Okay, so once i deprotonate the hydrogen, i'm going to have to move the electrons from that carbon hydrogen bond onto my single bond.

00:56 And when i accomplish that, this carbon has exceeded its octet, so i need to move.

01:00 The electrons up to that oxygen.

01:02 Okay, so after the result of this mechanism, i should get this in which i have my alcohol connected to an alken.

01:09 I'll redraw that.

01:10 So i have a vinyl alcohol, and then i have hydronium.

01:15 Okay, so i reformed my hydronium ion.

01:17 Okay, so this is the point where i want to react this nucleophile...

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