An aldol reaction is a reaction in organic chemistry where an enol or enolate ion reacts with a carbonyl compound (aldehyde or ketone) to form a β-hydroxy aldehyde or β-hydroxy ketone, followed by dehydration to give an enone. The reaction requires an α-hydrogen (hydrogen on the carbon adjacent to the carbonyl group) to form the enol/enolate.

Let's analyze the given transformation step by step:

Step 1: The reactant is CH₃COCH₂CH₂CH₂CHO. This molecule has two carbonyl groups: a ketone (COCH₃) and an aldehyde (CHO). Both have α-hydrogens:

• The ketone has α-hydrogens on the CH₂ group adjacent to it.
• The aldehyde has α-hydrogens on the CH₂ group adjacent to it.

So, it can undergo intramolecular aldol condensation. The enolate from one carbonyl can attack the other carbonyl. The most stable ring formed is 5- or 6-membered. Here, forming a 5-membered ring (by attack of enolate from aldehyde on ketone) or 6-membered ring (by attack of enolate from ketone on aldehyde) is possible. Typically, 5-membered ring formation is favored. So, one aldol reaction occurs intramolecularly to give a cyclic enone.

Step 2: After the first aldol, we get a cyclic enone (likely a 5-membered ring with a double bond). This product still has α-hydrogens? Actually, in the cyclic enone, the carbonyl might have α-hydrogens, but often such systems can undergo further aldol if conditions permit. However, in standard aldol conditions, the first condensation might be the major product, but sometimes multiple aldols can occur if there are multiple α-hydrogens.

But looking at the options and typical problems, the molecule CH₃COCH₂CH₂CH₂CHO can undergo two aldol reactions:

1. First, intramolecular aldol to form a 5-membered ring (cyclopentenone derivative).
2. This product has no α-hydrogen? Actually, the cyclopentenone might have α-hydrogens, but it is conjugated and usually stable. However, under strong base, it might undergo another aldol, but it's less common.

Actually, careful analysis: The given molecule is a 1,6-dicarbonyl compound. It undergoes intramolecular aldol to form a 5-membered ring (since 5 is favored over 6). The product is $C_6{H}_{8}O$ (after dehydration). This product has no α-hydrogen left? In cyclopentenone, the carbon adjacent to carbonyl is part of double bond, so no α-H. Therefore, only one aldol reaction occurs.

But wait: The molecule has two carbonyls, each with α-H. However, intramolecular reaction gives one cyclic product. So, only one aldol condensation happens.

However, some sources suggest that such molecules can give double aldol, but here it's not possible due to ring formation.

After checking standard knowledge: For HO−(CH₂)₅−CHO, it gives intramolecular aldol to form 5-membered ring. Similarly here, only one aldol occurs.

So, the number of aldol reactions is 1.

Final Answer: 1

Related Topics

• Aldol Condensation: A reaction between two carbonyl compounds with α-hydrogens, leading to β-hydroxy carbonyl or enone.
• Intramolecular Reactions: When a molecule reacts with itself to form cyclic structures.
• Carbonyl Chemistry: Reactions involving aldehydes and ketones, including nucleophilic addition.

Formulae

General aldol reaction: $R-\mathrm{CH}{-}_2-C=O+R\text{'}-\mathrm{CH}{-}_2-C=O\to R-\mathrm{CH}\left(\mathrm{OH}\right)-\mathrm{CH}{-}_2-C=O\to R-\mathrm{CH}=\mathrm{CH}-C=O$ (after dehydration)