Q.Suggest a reagent for conversion of ethanol to ethanal.

The key idea is the oxidation of a primary alcohol to an aldehyde. Ethanol ($\text{CH}_3\text{CH}_2\text{OH}$) must lose two hydrogen atoms to become ethanal ($\text{CH}_3\text{CHO}$).

Reasoning:

1. A mild oxidizing agent is needed to stop the reaction at the aldehyde stage, preventing over-oxidation to ethanoic acid. …

Why This Approach Works

The conversion of ethanol to ethanal is a classic example of controlled oxidation of a primary alcohol. Ethanol has the structure $CH_3CH_2OH$, and ethanal is $CH_3CHO$ — the difference is the loss of two hydrogen atoms (one from the hydroxyl group and one from the adjacent carbon). This is an elimination of $H_2$ in the presence of an oxidising agent.

The challenge is that ethanal itself is easily oxidised further to ethanoic acid ($CH_3COOH$). So the reagent must be strong enough to oxidise the alcohol but mild enough to stop at the aldehyde stage. This is where the choice of reagent becomes critical.

Step-by-Step Reasoning

1. Identify the functional group change.

   Ethanol is a primary alcohol ($-CH_2OH$ group). Oxidation of a primary alcohol can give either an aldehyde or a carboxylic acid, depending on conditions. The target is ethanal, an aldehyde.

2. Recall the standard reagent for aldehyde formation.

   The most common laboratory reagent for this specific conversion is acidified potassium dichromate ($K_2Cr_2O_7$ in dilute $H_2SO_4$). The orange dichromate ion ($Cr_2O_7^{2-}$) is reduced to green $Cr^{3+}$ as it oxidises the alcohol.

   The reaction is:

Here $[O]$ represents the oxygen from the oxidising agent.

3. Control the reaction to stop at the aldehyde.

   To prevent further oxidation to ethanoic acid, the reaction must be distilled as soon as ethanal forms. Ethanal has a lower boiling point ($21^\circ C$) than ethanol ($78^\circ C$) and water, so it can be distilled off immediately. This is a key practical detail — the reagent alone isn't enough; the method matters.

4. Consider a milder, more selective reagent.

   In modern organic chemistry, pyridinium chlorochromate (PCC) in anhydrous dichloromethane ($CH_2Cl_2$) is preferred. PCC is a chromium(VI) reagent that oxidises primary alcohols to aldehydes without over-oxidation, because it works in the absence of water (which is needed for further oxidation). The reaction is: …

Method: Controlled Oxidation of Primary Alcohols

This is an oxidation reaction (not electrophilic addition), but it's the standard exam method for ethanol → ethanal.

Reagent

Acidified potassium dichromate ($K_2Cr_2O_7 / H_2SO_4$) under distillation conditions.

Why this reagent works

• Ethanol is a primary alcohol.
• Controlled oxidation stops at the aldehyde stage (ethanal) if the product is distilled off immediately — before it can be further oxidised to ethanoic acid.

Step-by-step method

1. Set up distillation apparatus — a round-bottom flask with a condenser and receiving flask.
2. Add ethanol to the flask.
3. Add acidified $K_2Cr_2O_7$ (orange colour) dropwise with gentle heating.
4. Distil the reaction mixture — ethanal (boiling point $\approx 21^\circ C$) vaporises and is collected in the receiver.
5. Observe colour change: orange $Cr_2O_7^{2-}$ → green $Cr^{3+}$.

Chemical equation …

Common Mistakes: Ethanol → Ethanal (Electrophilic Addition Context)

Students often confuse this conversion because it looks like an addition reaction but is actually an oxidation. Here are the most frequent errors and how to avoid them.

✗ Mistake 1: Suggesting an Electrophilic Addition Reagent

The error: Students see "ethanol" (an alkene? No — it's an alcohol) and think of reagents like $\text{Br}_2/\text{CCl}_4$ or $\text{H}_2\text{SO}_4$.

Why it's wrong:

Ethanol ($\text{CH}_3\text{CH}_2\text{OH}$) is a saturated alcohol, not an alkene. Electrophilic addition requires a $\pi$ bond (double/triple bond). Ethanol has none.

How to avoid:

• First identify functional group: Is it an alkene, alkyne, or alcohol?
• Ethanol → alcohol → oxidation, not addition.
• Memorise: Alcohols undergo oxidation; alkenes undergo addition.

✗ Mistake 2: Using $\text{KMnO}_4$ or $\text{K}_2\text{Cr}_2\text{O}_7$ in Acidic Medium

The error: Suggesting acidified $\text{K}_2\text{Cr}_2\text{O}_7$ or $\text{KMnO}_4$ — which is correct for oxidation — but forgetting that excess reagent or harsh conditions over-oxidise ethanol to ethanoic acid ($\text{CH}_3\text{COOH}$).

Why it's wrong:

• Acidified $\text{K}_2\text{Cr}_2\text{O}_7$ (hot, excess) gives ethanoic acid, not ethanal.
• Ethanal is an intermediate that gets further oxidised.

How to avoid:

• Use mild oxidising agents only.
• The correct reagent: Pyridinium chlorochromate (PCC) or Collins reagent ($\text{CrO}_3\cdot 2\text{pyridine}$).
• Alternatively: Catalytic dehydrogenation (Cu, 573 K) — but that's less common in exam short-answer.

✗ Mistake 3: Writing $\text{CH}_3\text{CHO}$ as Product but Wrong Reagent

The error: Writing "ethanol → ethanal" but giving $\text{NaBH}_4$ or $\text{LiAlH}_4$ (reducing agents).

Why it's wrong:

• $\text{NaBH}_4$ and $\text{LiAlH}_4$ reduce carbonyls to alcohols — they do the reverse reaction.
• Ethanol to ethanal is oxidation (loss of $\text{H}_2$).

How to avoid:

• Remember the oxidation ladder: $\text{Alcohol} \xrightarrow{[\text{O}]} \text{Aldehyde} \xrightarrow{[\text{O}]} \text{Carboxylic acid}$
• Reducing agents go down the ladder; oxidising agents go up.

✗ Mistake 4: Confusing Reagent Names (PCC vs PDC vs Jones)

The error: Writing "PCC" but meaning Jones reagent ($\text{CrO}_3/\text{H}_2\text{SO}_4$) — which over-oxidises.

Why it's wrong:

• PCC (pyridinium chlorochromate) in anhydrous $\text{CH}_2\text{Cl}_2$ stops at aldehyde.
• Jones reagent ($\text{CrO}_3$, $\text{H}_2\text{SO}_4$, acetone) goes to carboxylic acid.

How to avoid: …