After breathing in a test tube that contains acidified K 2 Cr 2 O 7 , a man noticed the change in the colour…

Analysis

The question describes a classic chemical test used to identify specific functional groups in organic chemistry. We are looking for a substance that reacts with acidified potassium dichromate(VI) ($K_2Cr_2O_7$) to cause a color change from orange to green.

Concept Explanation

• The Reagent: Acidified potassium dichromate(VI) is a strong oxidizing agent.
• The Reaction: When it reacts with certain organic compounds, the dichromate ion ($Cr_2O_7^{2-}$, which is orange) is reduced to the chromium(III) ion ($Cr^{3+}$, which is green).
• The Target: This reaction occurs when the reagent oxidizes primary and secondary alkanols (alcohols). • Primary alkanols are oxidized to alkanals and then to alkanoic acids. • Secondary alkanols are oxidized to alkanones. • Tertiary alkanols do not react with acidified $K_2Cr_2O_7$ (no color change).
• The Context: The "man breathing into the test tube" refers to the breathalyzer test. Ethanol (an alkanol) present in the breath of someone who has consumed alcohol is oxidized by the acidified dichromate, causing the characteristic color change.

Step-by-Step Solution

1. Identify the oxidizing agent: Acidified $K_2Cr_2O_7$ is used to test for the presence of oxidizable alcohols.
2. Evaluate the options: • Alkanone: These are already oxidized products (ketones) and generally do not undergo further oxidation by dichromate. • Alkanal: While these can be oxidized to alkanoic acids, the breathalyzer test specifically targets the ethanol (alkanol) in the breath. • Alkanol: Primary and secondary alcohols are easily oxidized by acidified $K_2Cr_2O_7$, resulting in the reduction of chromium from the +6 oxidation state (orange) to the +3 oxidation state (green). • Alkanoic acid: These are the final products of the oxidation of primary alcohols and cannot be further oxidized by this reagent.
3. Conclusion: The color change from orange to green confirms the presence of an alkanol (specifically ethanol in the context of a breath test).

Correct Option

The correct option is C.

Exam Tip

Remember the mnemonic "Primary and Secondary Alcohols go Green". Tertiary alcohols are "stubborn"—they do not react with acidified $K_2Cr_2O_7$ or $KMnO_4$, so they remain orange. This is a common way to distinguish between different types of alcohols in the lab!