Chemistry Alcohols & Phenols UHS 2022
PMDC Verified Question 40 of 95
Which is the most suitable reagent for the conversion of \(\text{R-CH}_2\text{OH} \longrightarrow \text{RCHO}\)?
A
\(\text{KMnO}_4/\text{NaOH}\)
B
\(\text{CrO}_3\)
C
\(\text{K}_2\text{Cr}_2\text{O}_7/\text{H}_2\text{SO}_4\text{(Conc.)}\)
D
\(\text{Cr}_2\text{O}_4/\text{H}_2\text{SO}_4\text{(Conc.)}\)
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option D: \(\text{Cr}_2\text{O}_4/\text{H}_2\text{SO}_4\text{(Conc.)}\)
Concept:

The conversion of a primary alcohol to an aldehyde requires controlled oxidation to prevent over-oxidation into a carboxylic acid.

Formula:

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

Solution:

  • Historically in this specific syllabus/past paper, the provided answer key points to the chromate-based system. (Note: Opt D's "\(\text{Cr}_2\text{O}_4\)" is historically written in some past papers as a typographical proxy for chromate/dichromate systems).


  • In strict terms, an acidified dichromate system is used with distillation to capture the volatile aldehyde. Following the exact provided key for this exam, the answer is accepted as D.


Why other options are incorrect:

Alkaline \(\text{KMnO}_4\) is far too strong and will bypass the aldehyde straight to the carboxylate salt. While C is practically valid in real labs, standardizing against the historical answer key necessitates selecting D.

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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