See image — Alcohols Phenols and Ethers Chemistry Question

💡 Solution & Explanation

Concept: CrO3 in pyridine (also known as Collins reagent or the Sarett-Collins conditions) is a mild oxidizing agent selective for primary alcohols, oxidizing them to aldehydes without over-oxidizing to carboxylic acids. Secondary alcohols are oxidized to ketones. However, the key feature of CrO3/pyridine under cold conditions is that it is a mild oxidant that preferentially oxidizes the more reactive primary alcohol to an aldehyde and stops there. Step 1: Identify the substrate. The starting material is CH3-CH(OH)-CH2-CH2-OH, which is 1,3-butanediol (4-hydroxy-2-butanol / 1-hydroxy-3-butanol). It contains two hydroxyl groups: a secondary OH on C2 (CH3-CH(OH)-) and a primary OH on C4 (-CH2-OH). Step 2: Determine selectivity of CrO3/pyridine (cold). CrO3 in pyridine is a mild oxidant. Under cold conditions, primary alcohols are oxidized to aldehydes (not further to carboxylic acids). Secondary alcohols are oxidized to ketones. Both OH groups can be oxidized. Step 3: Apply oxidation to both groups. The primary alcohol (-CH2-OH) is oxidized to an aldehyde (-CHO). The secondary alcohol (CH3-CH(OH)-) is oxidized to a ketone (CH3-C(=O)-). This gives the product: CH3-C(=O)-CH2-C(=O)-H, which is 4-oxopentanal or more precisely 3-oxobutanal (methylglyoxal type) — specifically the product is CH3-C(=O)-CH2-CHO. Step 4: Match to options. Option (b) is CH3-C(=O)-CH2-C(=O)-H, which corresponds to oxidation of both the secondary OH to a ketone and the primary OH to an aldehyde. This matches the expected product under CrO3/pyridine conditions where both alcohols are oxidized. Why other options fail: - Option (a): Only the primary alcohol is oxidized to aldehyde; secondary OH remains unchanged. CrO3/pyridine oxidizes both alcohols. - Option (c): The primary alcohol is over-oxidized to a carboxylic acid. CrO3/pyridine under cold conditions does not over-oxidize aldehydes to carboxylic acids. - Option (d): Only the primary alcohol is oxidized to a carboxylic acid (over-oxidation), and secondary OH is unchanged. This does not match CrO3/pyridine selectivity. Therefore, the correct answer is B.