The product of the reaction between ethanol and excess acidified K$_2$Cr$_2$O$_7$ is

Analyze the question

Ethanol is a primary alcohol. Acidified potassium dichromate(VI), K₂Cr₂O₇/H⁺, is a strong oxidising agent. The key phrase is "excess", which tells you how far the oxidation goes.

Concept: oxidation of primary alcohols

A primary alcohol is oxidised in two stages:

1. Primary alcohol → aldehyde (partial oxidation)
2. Aldehyde → carboxylic acid (complete oxidation)

• If the aldehyde is distilled off as it forms, you stop at the aldehyde.
• With excess oxidising agent (usually under reflux), the oxidation continues to the carboxylic acid.

Step-by-step solution

Step 1: Ethanol is oxidised to ethanal.

CH₃CH₂OH → [O] → CH₃CHO + H₂O

Step 2: With excess K₂Cr₂O₇/H⁺, ethanal is oxidised further to ethanoic acid.

CH₃CHO + [O] → CH₃COOH

Overall:

CH₃CH₂OH + 2[O] → CH₃COOH + H₂O

Colour change: the orange Cr₂O₇²⁻ is reduced to green Cr³⁺.

Checking the other options

• A. CH₂=CH₂ (ethene) is formed by dehydration of ethanol with conc. H₂SO₄ at 170 °C, not by oxidation.
• B. CH₃OCH₃ (methoxymethane) is an isomer of ethanol. It is not formed by oxidation.
• D. CH₃CH₃ (ethane) would need reduction, not oxidation.

Correct answer: C. CH₃COOH (ethanoic acid)

Exam tip: Remember "excess oxidant = carboxylic acid". For primary alcohols:

• Mild or limited oxidant with distillation gives the aldehyde.
• Excess oxidant with reflux gives the carboxylic acid.

Also, secondary alcohols give ketones, and tertiary alcohols resist oxidation.