Chemistry of the Carbonyl Group II & III Notes

Important notes regarding imine formation:
- Imine formation usually requires acid catalysis.
- The initial step (nucleophilic addition) is rate determining (slow) below pH4 (amine is protonated so not a good nucleophile at pH levels below this).
- Acid catalysis is needed for elimination of water (acid protonates OH to turn it into a good leaving group) - this step is slow above pH6 (limited acid (H+) to protonate)

The optimal reaction rate for imine formation occurs at pH 5-6, but either side of this pH the reaction proceeds slowly. This is a sign of change in rate determining step.
This type of reaction (and its mechanism) is general for a range of different primary amines:

The imines that are formed can be easily hydrolysed in aqueous acid (or neutral conditions $[H_2O]$ )
Draw the mechanism for the hydrolysis of imines:
Imines in which the nitrogen atom carries a group bearing an electronegative group are relatively stable and require forcing acid or base catalysis for their hydrolysis - why?
- Delocalisation of imine C=N decreases the electrophilicity of the carbon of the imine making it less susceptible to nucleophilic attack

Compare acid catalysed imine formation and acetal formation up to iminium / oxonium formation
We require acid catalysis for the addition of the alcohol (a relatively poor nucleophile) but not the amine (a good nucleophile). Apart from that the mechanisms are similar up to iminium / oxonium ion formation.
Why does an oxonium react with another equivalent of an alcohol, but the iminium generates the imine?

The iminium ion carries a proton but the oxonium ion does not.
This means that iminium ions act as acids, readily losing a proton to become an imine:
Oxonium ions cannot lose $H^+$ so act as electrophiles, adding another alcohol to become an acetal.

Secondary amines react with aldehydes and ketones to generate enamines
For example, pyrrolidine reacts with isobutyraldehyde under acid catalysis to make an enamine
Mechanism? Similar to imine formation, apart from the iminium ion has no N-H proton to lose in this instance and so loses one of the C-H protons next to the C=N to give an enamine

Carboxylic acid derivatives react in a different fashion to aldehydes and ketones - they typically give the product of substitution
For example; if we take a carboxylic acid in the presence of methanol and an acid we substitute the OH of the acid for an OMe group and make an ester:
This type of reaction is common for the following carboxylic acid derivatives

For example: acid chlorides and anhydrides react with alcohols to make esters
How does this type of transformation proceed - consider a mechanism
The first step is nucleophilic addition - to form a tetrahedral intermediate
This tetrahedral intermediate contains a good leaving group ( $Cl^-$ ) which is eliminated to generate the product ester
The mechanism for the formation of an ester from an anhydride is similar
The first step is nucleophilic addition - to form a tetrahedral species.
In the presence of pyridine (a base) deprotonation occurs
The resultant tetrahedral intermediate is unstable as contains a good leaving group ( $MeCO_2^-$ ) which is eliminated to generate the product ester
Key to mechanism: nucleophilic addition generates a tetrahedral intermediate; a leaving group $[(X^-)$ -usually something whose conjugate base has a pKa of less ~20] is eliminated from this intermediate to form the product
Why are the tetrahedral intermediates unstable in these cases?

Remember that we have previously seen that the tetrahedral alkoxide formed from addition of a Grignard to an aldehyde or ketone is stable:
But the tetrahedral intermediate formed from addition to an acid chloride is unstable? Why?

Once the nucleophile has added to the carbonyl, the stability of the tetrahedral intermediate depends upon how good the groups attached to the $sp^3$ carbon atom are at leaving with the negative charge - this is called leaving group ability
In the case below we have three choices as leaving groups; $Cl^-$ , $Me^-$ , or $OEt^-$ .
Only $Cl^-$ leaves - why? - This is the best leaving group.
- We can use $pKaH$ as a useful guide to leaving group ability
The best leaving group is the conjugate base of the strongest acid; the leaving group with the lowest $pKaH$ leaves
$Cl^-$ leaves from the tetrahedral intermediate in the reaction of an acid chloride with an alcohol to generate the ester product.
In the reaction with an anhydride the carboxylate leaves
By the same mechanism, acid chlorides react with carboxylates to form anhydrides:
- And amines react with acid chlorides to form amides: