Q.$CH_3CH_2OH$ can be converted into $CH_3CHO$ by ____________.

The key idea is Alcohol Oxidation — specifically, the controlled oxidation of a primary alcohol to an aldehyde without over-oxidising to the carboxylic acid.

1. $CH_3CH_2OH$ (ethanol) is a primary alcohol. To get $CH_3CHO$ (acetaldehyde), we need an oxidising agent that stops at the aldehyde stage.
2. Catalytic hydrogenation and $LiAlH_4$ are reducing agents — they would convert ethanol to ethane or keep it as alcohol, not oxidise it.
3. $KMnO_4$ is a strong oxidant — it would directly oxidise ethanol all the way to acetic acid ($CH_3COOH$), not stop at the aldehyde. …

The question asks for a method to convert ethanol into acetaldehyde. This is a classic example of partial oxidation of a primary alcohol. The key challenge is to stop the oxidation at the aldehyde stage, not push it all the way to the carboxylic acid (acetic acid, $CH_3COOH$).

Let's think about what each option does.

1. Catalytic hydrogenation (Option (i)) adds hydrogen across double or triple bonds. Ethanol has no such bonds — it's already saturated. Hydrogenation would not change it. This is a reduction process, not an oxidation. So this is wrong.

2. Treatment with $LiAlH_4$ (Option (ii)) is a powerful reducing agent. It reduces carbonyl compounds (aldehydes, ketones, acids) back to alcohols. Using it on ethanol would do nothing — ethanol is already at the alcohol oxidation level. This is the opposite of what we need. So this is wrong.

3. Treatment with pyridinium chlorochromate (PCC) (Option (iii)) is a mild oxidizing agent specifically designed for this job. PCC in anhydrous conditions (usually in dichloromethane, $CH_2Cl_2$) oxidizes primary alcohols to aldehydes and stops there. It does not have water present to hydrate the aldehyde, so no further oxidation to the carboxylic acid occurs. This is the correct choice. …

Method: Controlled Oxidation of Primary Alcohols

The concept here is that a primary alcohol ($\text{CH}_3\text{CH}_2\text{OH}$, ethanol) can be oxidised to an aldehyde ($\text{CH}_3\text{CHO}$, ethanal) only if the oxidising agent is mild and the reaction is stopped before further oxidation to a carboxylic acid occurs.

Steps to identify the correct reagent:

1. Recognise the functional group change

   • Starting material: Primary alcohol ($-\text{OH}$ on a terminal carbon).
   • Product: Aldehyde ($-\text{CHO}$).
   • This is a two-electron oxidation (loss of two hydrogen atoms).

2. Eliminate reducing agents

   • Catalytic hydrogenation (A) and $\text{LiAlH}_4$ (B) are reducing agents — they would convert an aldehyde back to an alcohol, not oxidise it.
   • ✗ (A) and (B) are incorrect.

3. Distinguish between strong and mild oxidants

   • $\text{KMnO}_4$ (D) is a strong oxidant. It will over-oxidise a primary alcohol all the way to a carboxylic acid ($\text{CH}_3\text{COOH}$), not stop at the aldehyde.
   • ✗ (D) is incorrect.

4. Identify the mild, selective oxidant …

Here are the common mistakes students make on this question, along with how to avoid each.

Mistake 1: Confusing Oxidation with Reduction

• The Error: Choosing (A) catalytic hydrogenation or (B) treatment with $LiAlH_4$.
• Why it happens: Students see a conversion and forget to check whether the carbon is gaining or losing oxygen/hydrogen. They might recall that $LiAlH_4$ is a strong "reagent" without remembering its function.
• How to avoid: Always check the oxidation state of the carbon attached to the -OH group.
  • In $CH_3CH_2OH$ (ethanol), the carbon has one bond to oxygen.
  • In $CH_3CHO$ (acetaldehyde), the carbon has a double bond to oxygen.
  • Moving from a single bond to a double bond with oxygen is oxidation (loss of hydrogen).
  • Catalytic hydrogenation ($H_2$/catalyst) and $LiAlH_4$ are reducing agents. They add hydrogen, which would turn an aldehyde back into an alcohol. They cannot perform the oxidation needed here.

Mistake 2: Overlooking the "Mild vs. Strong" Oxidizer

• The Error: Choosing (D) treatment with $KMnO_4$.
• Why it happens: Students know $KMnO_4$ is an oxidising agent, so they assume it will work. They forget that strong oxidisers do not stop at the aldehyde stage.
• How to avoid: Memorise the selectivity of common oxidising agents.
  • Mild oxidisers (like PCC – pyridinium chlorochromate) stop at the aldehyde.
  • Strong oxidisers (like acidic $KMnO_4$ or $K_2Cr_2O_7$) will over-oxidise a primary alcohol all the way to a carboxylic acid ($CH_3COOH$).
  • Therefore, $KMnO_4$ would give $CH_3COOH$, not $CH_3CHO$.

Mistake 3: Not Knowing the Specific Reagent (PCC)

• The Error: Not recognising option (C) as the correct answer. …