Question

The compound of molecular formula C\textsubscript{5H\textsubscript{10}O(A) reacts with Tollen's reagent to give silver mirror but does not undergo aldol condensation. The compound A is}

• A. 3-pentanone
• B. 2,2-dimethylpropanal
• C. 3-hydroxy-2-pentene
• D. 3-methylbutanal
• E. 3-methyl-2-butanone

Hint:

Aldol condensation requires $\alpha$-H to form the enolate ion. If an aldehyde has no $\alpha$-H (like benzaldehyde or 2,2-dimethylpropanal), it's a prime candidate for Cannizzaro's reaction!

Answer 1

The Correct Option is B

Concept: This problem involves identifying a compound based on two classic chemical tests: the Tollen's test and the Aldol condensation requirement.
• Tollen's Reagent Test: A positive result (silver mirror) indicates the presence of an aldehyde group ($-CHO$). Ketones do not react.
• Aldol Condensation: For a compound to undergo aldol condensation, it must possess at least one alpha-hydrogen ($\alpha$-H). These are hydrogen atoms attached to the carbon immediately adjacent to the carbonyl group.

Step 1: Filter by the Tollen's test. Compounds (A) and (E) are ketones (pentanone and butanone), so they are eliminated. Compound (C) is an enol/alcohol. This leaves (B) 2,2-dimethylpropanal and (D) 3-methylbutanal as the possible aldehydes.

Step 2: Check for alpha-hydrogens. - (D) 3-methylbutanal: The $\alpha$-carbon is a $-CH_2-$ group. It has two $\alpha$-hydrogens and will undergo aldol condensation. - (B) 2,2-dimethylpropanal: The $\alpha$-carbon is a quaternary carbon (attached to three methyl groups). It has zero alpha-hydrogens. Structure: $(CH_3)_3C-CHO$. Because it is an aldehyde with no $\alpha$-H, it reacts with Tollen's but fails the Aldol condensation (it would undergo Cannizzaro reaction instead).