Chemistry Carboxylic Acids SZABMU 2024
PMDC Verified Question 5 of 94
Oxidation of Propanal in the presence of \( \text{K}_2\text{Cr}_2\text{O}_7 / \text{H}_2\text{SO}_4 \) produces
A
Propanol
B
Propanoic acid
C
Propanone
D
Propene
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option B: Propanoic acid
Concept:

Aldehydes are effortlessly oxidized to carboxylic acids under acidic dichromate conditions, preserving the carbon chain length.

Formula:

$$ \text{CH}_3\text{CH}_2\text{CHO} + [\text{O}] \longrightarrow \text{CH}_3\text{CH}_2\text{COOH} $$

Solution:

  • Propanal is a 3-carbon aldehyde.


  • The strong oxidizing agent (\( \text{K}_2\text{Cr}_2\text{O}_7 \)) converts the terminal \( -\text{CHO} \) group directly into a \( -\text{COOH} \) group.


  • The resultant 3-carbon acid is propanoic acid.


Why other options are incorrect:

Propanol requires reduction. Propanone is a ketone (requiring a secondary alcohol backbone). Propene requires dehydration/elimination.

Quality & Fidelity Assurance: Every question on BeambePrep is rigorously curated against the official PMDC syllabus with zero filler, zero out-of-syllabus content, and zero typos. When an authentic past paper originally contained a historical mistake or ambiguity from the examining board (such as UHS or NUMS), BeambePrep faithfully reflects the original paper while detailing the nuance and scientific consensus in the autopsy above.

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