Question

Identify product B in following reaction.
Propanone $\xrightarrow{\text{Ba(OH)}_2}A\xrightarrow[-\text{H}_2\text{O}]{\Delta}$ B

• A. 2-Methylpentan-3-one
• B. 4-Hydroxy-4-methylpentan-2-one
• C. 4-Methylpent-3-en-2-one
• D. 2-Methylpent-2-en-4-one

Hint:

To quickly predict aldol condensation products without drawing out the entire mechanism, write the two starting molecules side by side. Draw the carbonyl oxygen (=O) facing the two $\alpha$-hydrogens ($\text{-CH}_2\text{-}$) of the second molecule, remove $\text{H}_2\text{O}$, and join them directly with a double bond!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question presents a two-step organic reaction sequence starting with propanone (acetone). We need to determine the structural IUPAC name of the final product B.

Step 2: Key Formula or Approach:
Treating an aldehyde or ketone containing $\alpha$-hydrogens with a barium hydroxide base ($\text{Ba(OH)}_2$) initiates a classic Aldol Condensation reaction. The first step forms a $\beta$-hydroxy ketone intermediate (A), which undergoes subsequent thermal dehydration ($\Delta$) to produce an $\alpha,\beta$-unsaturated carbonyl compound (B).

Step 3: Detailed Explanation:
Let's trace out the chemical mechanism step-by-step:
1.

Formation of Intermediate A (Aldol Addition): One molecule of propanone ($\text{CH}_3\text{COCH}_3$) loses an $\alpha$-hydrogen to the base, forming a nucleophilic enolate ion. This ion attacks the electrophilic carbonyl carbon of a second propanone molecule:
$$(\text{CH}_3)_2\text{C=O} + \text{CH}_3\text{COCH}_3 \xrightarrow{\text{Ba(OH)}_2} (\text{CH}_3)_2\text{C(OH)-CH}_2\text{-CO-CH}_3$$ The IUPAC name for this intermediate molecule (A) is 4-hydroxy-4-methylpentan-2-one (diacetone alcohol).
2.

Formation of Product B (Dehydration): Heating ($\Delta$) this intermediate eliminates a molecule of water ($-\text{H}_2\text{O}$) from the neighboring adjacent positions (the $-\text{OH}$ group from C4 and an $\alpha$-hydrogen from C3). This creates a stable, conjugated carbon-carbon double bond:
$$(\text{CH}_3)_2\text{C(OH)-CH}_2\text{-CO-CH}_3 \xrightarrow[\Delta]{-\text{H}_2\text{O}} (\text{CH}_3)_2\text{C=CH-CO-CH}_3$$ Let's find the IUPAC name for this product framework:
Longest carbon chain containing the functional groups has 5 carbons (

pentane).
Numbering from right to left to prioritize the ketone carbonyl gives it the lowest locant index: C2 for the ketone group (-2-one).
This numbering places the double bond starting at position C3 (-3-en-) and a methyl substituent branch at C4 (4-methyl).
Putting it all together yields

4-Methylpent-3-en-2-one (commonly known as mesityl oxide). This perfectly aligns with option (C).

Step 4: Final Answer:
Product B is 4-Methylpent-3-en-2-one, which corresponds to option (C).