Q.What products will be formed on reaction of propanal with 2-methylpropanal in the presence of NaOH? Write the name of the reaction also.

The key idea is that this is a crossed (mixed) aldol condensation, not a Cannizzaro reaction — Cannizzaro requires an aldehyde with NO alpha-hydrogens, but both propanal ($\text{CH}_3\text{CH}_2\text{CHO}$) and 2-methylpropanal ($(\text{CH}_3)_2\text{CHCHO}$) have alpha-hydrogens, so both are enolisable.

In the presence of NaOH, either aldehyde can form an enolate that attacks the carbonyl carbon of either aldehyde (itself or the other), so a mixture of four beta-hydroxy aldehydes (aldols) forms — two from a single aldehyde reacting with itself, and two crossed products. …

Reaction

In dilute NaOH each aldehyde forms an enolate that can add to the carbonyl carbon of either aldehyde. Propanal $\mathrm{CH_3CH_2CHO}$ has an $\alpha\text{-}CH_2$; 2-methylpropanal $\mathrm{(CH_3)_2CHCHO}$ has a single $\alpha\text{-}CH$. Four aldol (β-hydroxy aldehyde) adducts form.

1. Propanal + propanal (self)

3-hydroxy-2-methylpentanal, which dehydrates to $\mathrm{CH_3CH_2CH{=}C(CH_3)CHO}$ — 2-methylpent-2-enal.

2. Enolate of propanal + 2-methylpropanal (crossed)

3-hydroxy-2,4-dimethylpentanal, which dehydrates to $\mathrm{(CH_3)_2CHCH{=}C(CH_3)CHO}$ — 2,4-dimethylpent-2-enal.

3. Enolate of 2-methylpropanal + propanal (crossed)

3-hydroxy-2,2-dimethylpentanal.

4. 2-Methylpropanal + 2-methylpropanal (self)

3-hydroxy-2,2,4-trimethylpentanal. …

Method: Crossed Cannizzaro Reaction

This is a Cannizzaro reaction — a disproportionation of aldehydes without α-hydrogen atoms, in the presence of concentrated alkali.

Why this applies here

• Propanal ($\mathrm{CH_3CH_2CHO}$) has α-hydrogens → normally undergoes aldol, not Cannizzaro.
• 2-Methylpropanal ($\mathrm{(CH_3)_2CHCHO}$) also has α-hydrogens → same issue.

Correction: Neither aldehyde lacks α-hydrogens. Therefore, a Cannizzaro reaction does not occur under these conditions. Instead, a mixed aldol condensation takes place.

Correct Method: Mixed Aldol Condensation (Crossed Aldol)

Steps

1. Identify the reactive sites

   Both aldehydes have α-hydrogens. In the presence of $\mathrm{NaOH}$, the α-carbon of each aldehyde forms an enolate.

2. Formation of enolates

   • Propanal enolate: $\mathrm{CH_3CH_2CHO \xrightarrow{OH^-} CH_3CH=CH(O^-)}$
   • 2-Methylpropanal enolate: $\mathrm{(CH_3)_2CHCHO \xrightarrow{OH^-} (CH_3)_2C=CH(O^-)}$

3. Nucleophilic attack

   Each enolate can attack the carbonyl carbon of the other aldehyde, giving four possible products (self + crossed).

4. Major product

   The crossed product is favoured when one aldehyde is more electrophilic (less hindered).

   • 2-Methylpropanal is more sterically hindered → propanal enolate attacks 2-methylpropanal preferentially.

   Reaction:

   $\mathrm{CH_3CH_2CH=CH(O^-) + (CH_3)_2CHCHO \rightarrow (CH_3)_2C(OH)CH(CH_3)CHO}$

   (after protonation) …

Here are the common mistakes students make on this question, along with how to avoid each.

1. Mistake: Forgetting the Reaction Name

The Mistake: Students often write the mechanism correctly but fail to name the reaction, or they incorrectly label it as "Aldol Condensation" (which requires an $\alpha$-hydrogen on both aldehydes).

How to Avoid:

• Memorize the trigger: When you see two different aldehydes (without $\alpha$-hydrogens) reacting with concentrated NaOH, the reaction is Crossed Cannizzaro Reaction.
• Key difference: Aldol condensation needs $\alpha$-H. Cannizzaro needs no $\alpha$-H on the aldehyde that gets reduced.

2. Mistake: Assuming Both Aldehydes React Equally

The Mistake: Students think both propanal and 2-methylpropanal undergo Cannizzaro simultaneously, producing a messy mixture of four products.

Why this is wrong: Propanal has $\alpha$-hydrogens (it can enolize), so it cannot undergo Cannizzaro. Only 2-methylpropanal (pivalaldehyde) has no $\alpha$-H and can undergo the reaction.

How to Avoid:

• Check for $\alpha$-hydrogens first. Draw the structure:
  • Propanal: $\text{CH}_3\text{CH}_2\text{CHO}$ → has $\alpha$-H (on the carbon next to CHO).
  • 2-methylpropanal: $(\text{CH}_3)_3\text{CCHO}$ → no $\alpha$-H (the alpha carbon is quaternary).
• Rule: In a crossed Cannizzaro, the aldehyde without $\alpha$-H gets reduced to alcohol; the one with $\alpha$-H (if present) gets oxidized to acid.

3. Mistake: Writing Cannizzaro Products (an Alcohol + an Acid) …