Chemistry Aldehydes & Ketones MDCAT 2013
PMDC Verified Question 87 of 94
Which group gives a yellow precipitate of triiodomethane when warmed with alkaline aqueous iodine?
A
An amide group, \( CH_3-CO-NH_2 \)
B
Ethyl ketone group, \( C_2H_5-CO-NH_2 \)
C
A primary alcohol group as in propanol, \( CH_3CH_2CH_2OH \)
D
Methyl ketone group, \( CH_3-CO- \)
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option D: Methyl ketone group, \( CH_3-CO- \)
Concept:

The formation of a yellow precipitate of triiodomethane (\( CHI_3 \)), also known as iodoform, is the definitive positive result of the Iodoform Test.

Solution:

  • The reagents for this test are Iodine (\( I_2 \)) in an alkaline solution (\( NaOH \)).


  • This specific reagent mixture cleaves and halogenates only those molecules possessing a specific structural motif: the methyl ketone group (\( CH_3-CO- \)).


  • The methyl group is fully iodinated to \( CI_3-CO-R \), which is then cleaved by the base to yield \( CHI_3 \) (iodoform).


Why other options are incorrect:

Amides do not undergo the haloform reaction. Propanol is a primary alcohol lacking the \( CH_3-CH(OH)- \) structure, so it cannot be oxidized to a methyl ketone.

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.

Previous Question All Aldehydes & Ketones MCQs Next Question

Want to solve full-length papers under timed exam conditions?

Practice with zero-scroll lockdown sprints, dynamic latency zone timers, live peer selection telemetry, and the automated Amber mistake recovery loop.

Enter Arena Sprint iOS App Android App