Different aldol products can be formed when butanone reacts with dilute sodium hydroxide (NaOH). The correct answer is '6' different aldol products. Let's break down the explanation into the following headings:

1. Introduction:
- Butanone, also known as methyl ethyl ketone (MEK), is a ketone with the molecular formula C4H8O.
- Dilute sodium hydroxide (NaOH) is a basic solution commonly used in organic chemistry reactions.

2. Aldol Reaction:
- The aldol reaction is a type of organic reaction that involves the condensation of an enol or enolate with a carbonyl compound.
- In this case, butanone (a carbonyl compound) will react with dilute NaOH to form aldol products.

3. Structure of Butanone (C4H8O):
- Butanone has a ketone functional group (C=O) attached to a four-carbon chain.
- The four-carbon chain provides different positions for the aldol reaction to occur.

4. Formation of Aldol Products:
- The aldol reaction involves the nucleophilic addition of the enolate ion (generated from NaOH) to the carbonyl carbon of butanone, followed by dehydration to form the final aldol product.
- In this case, the butanone molecule can react at two different positions (carbon 2 and carbon 3) to form four possible aldol products.

5. Isomerism:
- The aldol products can also exhibit isomerism due to the presence of a chiral center in butanone.
- The chiral center at carbon 2 can lead to the formation of two stereoisomers for each aldol product, resulting in a total of six different aldol products.

6. Conclusion:
- In summary, when butanone reacts with dilute NaOH, six different aldol products can be formed.
- These aldol products result from the reaction at two different positions of butanone and the possibility of stereoisomerism.
- The aldol reaction is an important organic transformation that allows the synthesis of complex molecules with multiple functional groups.