Chemistry Alcohols & Phenols MDCAT 2011
PMDC Verified Question 94 of 95
An alcohol is converted into an aldehyde with same number of carbon atoms in the presence of \(\text{K}_2\text{CrO}_4\)/\(\text{H}_2\text{SO}_4\). the alcohol is
A
\(\text{CH}_3\text{CH}(\text{OH})\text{CH}_3\)
B
\(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}\)
C
\((\text{CH}_3)_3\text{COH}\)
D
\((\text{CH}_3)_2\text{CHOH}\)
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option B: \(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}\)
Concept:

Oxidation of primary alcohols yields aldehydes (with the same number of carbons) if the oxidizing agent is mild or distilled off quickly. Secondary alcohols yield ketones.

Formula:

$$ \text{R-CH}_2\text{OH} + [O] \longrightarrow \text{R-CHO} + \text{H}_2\text{O} $$

Solution:

  • We must identify the primary alcohol among the options.


  • \(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}\) (1-propanol) is a primary alcohol and oxidizes to propanal.


Why other options are incorrect:

Options A and D represent isopropyl alcohol, which is a secondary alcohol that oxidizes to acetone. Option C is a tertiary alcohol, which resists oxidation under these specific conditions.

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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