Question

An organic compound [A] having molecular formula C₃H₈O gives turbidity with Lucas reagent within five minutes of mixing. On heating compound [A] with Cu at 573K, compound [B] is formed. Compound [B] does not reduce Fehling’s solution but when heated with iodine and sodium hydroxide, it gives a yellow precipitate of compound [C]. Identify the compounds [A], [B] and [C].

Write the chemical reaction for the conversion of compound [B] to compound [C].

Answer 1

Compound [A]: Propan-2-ol (isopropyl alcohol), CH₃-CH(OH)-CH₃. It is a secondary alcohol, which is why it reacts with Lucas reagent to produce turbidity within 5 minutes.

Compound [B]: Propanone (acetone), CH₃COCH₃. Formed by the catalytic dehydrogenation of propan-2-ol over heated copper at 573 K. As a ketone, it does not reduce Fehling’s solution.

\[\ce{CH3CH(OH)CH3->[Cu, 573 K][]CH3COCH3 + H2}\]

Compound [C]: Iodoform (CHI₃). This is the yellow precipitate formed when propanone (a methyl ketone) reacts with iodine and sodium hydroxide in the iodoform reaction.

Chemical reaction for the conversion of compound [B] to compound [C].

The reaction of propanone with iodine and sodium hydroxide is as follows:

\[\ce{CH3COCH3 + 3I2 + 4NaOH ->[Δ][]\underset{Yellow ppt}{CHI3} + CH3COONa + 3NaI + 3H2O}\]