Aldol Condensation - Overview, Reaction, Types, Conclusion, FAQs

Aldol Condensation Reaction

The Aldol condensation reaction is an important organic reaction in which aldehydes or ketones containing at least one $\boldsymbol{\alpha}$-hydrogen react in the presence of a base (or acid) to form $\boldsymbol{\beta}$-hydroxy aldehydes or $\boldsymbol{\beta}$-hydroxy ketones (aldols), which may further lose water to give $\boldsymbol{\alpha}, \boldsymbol{\beta}$-unsaturated carbonyl compounds.
Step 1: Aldol addition

$2 \mathrm{CH}_3 \mathrm{CHO} \xrightarrow{\text { dil. } \mathrm{NaOH}} \mathrm{CH}_3 \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_2 \mathrm{CHO}$

Step 2: Condensation (dehydration)

$\mathrm{CH}_3 \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_2 \mathrm{CHO} \xrightarrow{\Delta} \mathrm{CH}_3 \mathrm{CH}=\mathrm{CHCHO}+\mathrm{H}_2 \mathrm{O}$

Here:

• Acetaldehyde undergoes self-condensation.
• The product formed is crotonaldehyde.

Aldol Condensation Mechanism

Taking acetaldehyde as an example:

Step 1: Formation of Enolate Ion
The base removes an $\alpha$-hydrogen from acetaldehyde to form an enolate ion.

$\mathrm{CH}_3 \mathrm{CHO}+\mathrm{OH}^{-} \rightarrow \mathrm{CH}_2 \mathrm{CHO}^{-}+\mathrm{H}_2 \mathrm{O}$
The enolate ion is resonance stabilized.

Step 2: Nucleophilic Attack
The enolate ion attacks the carbonyl carbon of another acetaldehyde molecule.

$\mathrm{CH}_2 \mathrm{CHO}^{-}+\mathrm{CH}_3 \mathrm{CHO} \rightarrow \mathrm{CH}_3 \mathrm{CH}\left(\mathrm{O}^{-}\right) \mathrm{CH}_2 \mathrm{CHO}$
This step forms an alkoxide ion.

Step 3: Protonation
The alkoxide ion gains a proton from water to form aldol ( $\beta$-hydroxy aldehyde).

$\mathrm{CH}_3 \mathrm{CH}\left(\mathrm{O}^{-}\right) \mathrm{CH}_2 \mathrm{CHO}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{CH}_3 \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_2 \mathrm{CHO}+\mathrm{OH}^{-}$
The product formed is 3-hydroxybutanal.

Step 4: Dehydration (Condensation)
On heating, the aldol loses a molecule of water to form an $\alpha, \beta$-unsaturated aldehyde.

$\mathrm{CH}_3 \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_2 \mathrm{CHO} \xrightarrow{\Delta} \mathrm{CH}_3 \mathrm{CH}=\mathrm{CHCHO}+\mathrm{H}_2 \mathrm{O}$
The final product formed is crotonaldehyde.

Crossed Aldol Condensation

The condensation reaction which occurs between two aldehyde molecules or ketone in protic solvents like water or alcohol triggers a cross-linked aldol reaction. When the condensate is between two different carbonyl compounds, it is called aldol condensation crossing. When both aldehydes contain alpha hydrogens, both can form carbanions and can act as carbanion receptors. A combination of four products was therefore made with a small amount of processing.

If one of the aldehydes does not contain alpha hydrogen then it can only act as a carbanion receptor. In that case, only two products were made. A typical substrate of an aldol reaction crosses a fragrant aldehyde, which has no alpha position. In addition, the dehydration of the first condensation product accelerates leading to the formation of α, β - an unrefined ketone and prevents the formation of retro-aldol from occurring.

$\mathrm{RCH}_2 \mathrm{CHO}+\mathrm{R}^{\prime} \mathrm{CHO} \xrightarrow{\text { dil. } \mathrm{NaOH}} \mathrm{RCH}(\mathrm{OH}) \mathrm{CH}\left(\mathrm{R}^{\prime}\right) \mathrm{CHO}$

On heating:

$\mathrm{RCH}(\mathrm{OH}) \mathrm{CH}\left(\mathrm{R}^{\prime}\right) \mathrm{CHO} \xrightarrow{\Delta} \mathrm{RCH}=\mathrm{CH}\left(\mathrm{R}^{\prime}\right) \mathrm{CHO}+\mathrm{H}_2 \mathrm{O}$

Example of Crossed Aldol Condensation
Reaction between acetaldehyde and benzaldehyde:
Step 1: Aldol Formation

$\mathrm{CH}_3 \mathrm{CHO}+\mathrm{C}_6 \mathrm{H}_5 \mathrm{CHO} \xrightarrow{\text { dil. } \mathrm{NaOH}} \mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_2 \mathrm{CHO}$

Step 2: Dehydration

$\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_2 \mathrm{CHO} \xrightarrow{\Delta} \mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}=\mathrm{CHCHO}+\mathrm{H}_2 \mathrm{O}$

The final product formed is cinnamaldehyde.

Aldol Condensation of Cyclohexanone

The aldol condensation of cyclohexanone occurs because cyclohexanone contains α-hydrogen atoms. In the presence of a dilute base such as NaOH , two molecules of cyclohexanone undergo condensation to form a $\beta$-hydroxy ketone, which on dehydration gives an $\alpha, \beta$-unsaturated ketone.

Reaction
Step 1: Aldol Addition

$2 \mathrm{C}_6 \mathrm{H}_{10} \mathrm{O} \xrightarrow{\text { dil. } \mathrm{NaOH}} \mathrm{C}_{12} \mathrm{H}_{20} \mathrm{O}_2$

This forms 2-(1-hydroxycyclohexyl)cyclohexanone.

Step 2: Dehydration
On heating, the $\beta$-hydroxy ketone loses water to form an unsaturated ketone.

$\mathrm{C}_{12} \mathrm{H}_{20} \mathrm{O}_2 \xrightarrow{\Delta} \mathrm{C}_{12} \mathrm{H}_{18} \mathrm{O}+\mathrm{H}_2 \mathrm{O}$

The final product formed is 2-cyclohexylidenecyclohexanone

Mechanism

The aldol condensation reaction of cyclohexanone is a smart process of action.

Step 1: (Carbanion formation)

In the first step of the cyclohexanone aldol reaction, carbanion is formed as normal as a decrease in normal aldol. Base (Hydroxide ion) releases Alpha-Hydrogen for cyclohexanone. This molecule contains two Alpha-Carbons (Carbon presenting close to the active carbon-binding group) so here any Alpha-Hydrogen can be released. It leads to the formation of carbanions. It also deals with the enrichment of Enolate-carbanion

Step 2: (Electrophilic Center Attack)

In this step, the carbanion formed in the previous step invades the Electrophilic Center of another cyclohexanone molecule. That is why it is called Intermolecular aldol condensation

Step 3: (Acid workup)

A small amount of acid is added here to convert the oxygen produced into its hydroxy form and the product produced is called Aldol's cyclohexanone product.

Step 4: (Dehydration)

The aldol product made of cyclohexanone enters the body at high temperatures and forms a condensation product (Water molecule is removed)

Acyloin Condensation

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Some Solved Examples

Question 1: Which of the following compounds do not undergo aldol condensation?

Solution:

Aldol condensation takes place only if $\alpha$ hydrogen is present. In the reaction, aldehydes and ketones with 1 or more than 1 alpha hydrogen react in presence of dilute alkali to form aldol and ketol respectively.

Hence, the correct answer is option (ii) and (iv)

Question 2:

Which of the following aldehydes will not undergo aldol condensation in the presence of NaOH?

A. Formaldehyde
B. Acetaldehyde
C. Benzaldehyde
D. Butanal

Solution:

Aldol condensation requires $\alpha-\mathrm{H}$.
Benzaldehyde has no $\alpha-\mathrm{H} \rightarrow$ cannot form enolate $\rightarrow$ does not undergo aldol condensation.

Hence, the correct answer is option (c)

Question 3: Predict the major product of the reaction:

$\mathrm{CH}_3 \mathrm{CHO}+\mathrm{C}_6 \mathrm{H}_5 \mathrm{CHO} \xrightarrow{\mathrm{NaOH}} ?$

A. $\mathrm{CH}_3 \mathrm{CH}(\mathrm{OH}) \mathrm{C}_6 \mathrm{H}_5$
B. $\mathrm{CH}_3 \mathrm{CH}=\mathrm{C}\left(\mathrm{C}_6 \mathrm{H}_5\right) \mathrm{OH}$
C. $\mathrm{CH}_3 \mathrm{CH}=\mathrm{CHC}_6 \mathrm{H}_5$
D. $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}=\mathrm{CHCHO}$

Solution:

Acetaldehyde has $\alpha-\mathrm{H} \rightarrow$ forms enolate.
Benzaldehyde has no $\alpha-\mathrm{H} \rightarrow$ acts as electrophile.
Enolate attacks benzaldehyde $\rightarrow \beta$-hydroxy aldehyde $\rightarrow$ dehydrates:

$\mathrm{CH}_3 \mathrm{CH}=\mathrm{CHC}_6 \mathrm{H}_5$

Hence, the correct answer is option (c)

Question 4: Which of the following compounds is formed when ethyl benzoate undergoes acyloin condensation in the presence of sodium metal and dry ether followed by hydrolysis?

(A) Benzoin
(B) Benzil
(C) 2-Hydroxy-1,2-diphenylethanone
(D) 1,2-Diphenylethane-1,2-diol

Solution:

In acyloin condensation, two ester molecules combine in the presence of sodium metal to form an αhydroxy ketone (acyloin).

$2 \mathrm{C}_6 \mathrm{H}_5 \mathrm{COOC}_2 \mathrm{H}_5 \xrightarrow{\mathrm{Na} / \text { dry ether }} \mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}(\mathrm{OH}) \mathrm{COC}_6 \mathrm{H}_5$

The product formed is $\mathbf{2}$-hydroxy-1,2-diphenylethanone, which is an $\boldsymbol{\alpha}$-hydroxy ketone.

Hence, the correct answer is option (C)