Q.Suggest a reagent for the following conversion. The starting material is a secondary allylic alcohol, pent-3-en-2-ol (CH3-CH(OH)-CH=CH-CH3), and the product is the corresponding alpha,beta-unsaturated ketone, pent-3-en-2-one (CH3-CO-CH=CH-CH3); the carbon-carbon double bond is retained unchanged and only the -CH(OH)- group is oxidised to a >C=O group.

Alcohol Oxidation: Why the Reactions Work the Way They Do

Alcohol oxidation is a fundamental reaction in organic chemistry, and understanding why it proceeds as it does is crucial for Indian board exams (Class 12, JEE, NEET). Let's break it down step-by-step.

1. The Core Idea: Loss of Hydrogen

Oxidation in organic chemistry means loss of hydrogen (or gain of oxygen). For alcohols, this happens at the carbon bearing the –OH group.

• Primary alcohol ($R-CH_2OH$): Has two hydrogens on the carbon attached to –OH.
• Secondary alcohol ($R_2CHOH$): Has one hydrogen on that carbon.
• Tertiary alcohol ($R_3COH$): Has zero hydrogens on that carbon.

Key insight: The number of hydrogens on the carbon with –OH determines if and how far oxidation can go.

2. Why Primary Alcohols Give Aldehydes (Then Carboxylic Acids)

Step 1: Aldehyde formation

When a primary alcohol ($R-CH_2OH$) is oxidized, the first product is an aldehyde ($R-CHO$).

Why? The oxidizing agent (like $K_2Cr_2O_7$ / $H_2SO_4$ or $PCC$) removes two hydrogens:

• One from the –OH group
• One from the carbon atom

The carbon–oxygen bond becomes a double bond ($C=O$), forming the aldehyde.

But why stop here? The aldehyde still has one hydrogen on the carbonyl carbon. If a strong oxidant is present, it can remove that hydrogen too.

Step 2: Carboxylic acid formation

With excess strong oxidant (e.g., $K_2Cr_2O_7$ / $H_2SO_4$, heat), the aldehyde is further oxidized to a carboxylic acid ($R-COOH$).

Why does this happen? The aldehyde's carbonyl carbon is electrophilic (partially positive). Water (from the reaction medium) adds to it, forming a gem-diol intermediate. The oxidant then removes two more hydrogens, giving the acid.

Exam tip: To stop at the aldehyde, use a mild oxidant like PCC (pyridinium chlorochromate) in anhydrous conditions — no water means no gem-diol formation.

3. Why Secondary Alcohols Give Ketones (and Stop)

A secondary alcohol ($R_2CHOH$) has only one hydrogen on the carbon with –OH. Oxidation removes:

• One hydrogen from –OH
• One hydrogen from the carbon

This forms a ketone ($R_2C=O$).

Why does it stop here? The ketone has no hydrogen on the carbonyl carbon. Without that hydrogen, further oxidation (under normal conditions) is impossible — you'd need to break a $C-C$ bond, which requires much harsher conditions.

Key result: Secondary alcohols cannot be oxidized further than ketones under standard conditions.

4. Why Tertiary Alcohols Do NOT Oxidize

A tertiary alcohol ($R_3COH$) has zero hydrogens on the carbon bearing –OH.

What happens if you try? The oxidant cannot remove any hydrogen from that carbon. The only possible reaction would be breaking a $C-C$ bond, which doesn't happen under normal oxidation conditions.

Result: Tertiary alcohols are resistant to oxidation under mild to moderate conditions. They require strong heating with powerful oxidants (like $K_2Cr_2O_7$ / $H_2SO_4$, heat) to break carbon–carbon bonds — this is destructive oxidation, not useful for synthesis.

5. The "Why" in One Table