NEETChemistryHydrocarbons
Benzene is treated with propanoyl chloride in the presence of anhydrous AlCl ₃ to yield compound P . Compound P is then warmed with an alkaline solution of iodine. What is the expected outcome of the second step?
Options
- AA yellow precipitate of iodoform is formed along with sodium benzoate.
- BA yellow precipitate of iodoform is formed along with sodium phenylacetate.
- CElectrophilic substitution occurs on the aromatic ring to yield an iodinated product.
- DNo yellow precipitate is formed because P lacks a methyl group adjacent to the carbonyl carbon.
Correct answer
D. No yellow precipitate is formed because P lacks a methyl group adjacent to the carbonyl carbon.
Step-by-step solution
The first step is a Friedel-Crafts acylation. Benzene reacts with propanoyl chloride ( CH ₃ CH ₂ COCl ) in the presence of anhydrous AlCl ₃ to form propiophenone ( C ₆ H ₅ COCH ₂ CH ₃ ), which is compound P . The second step involves warming compound P with an alkaline solution of iodine ( I ₂/ NaOH ), which is the reagent for the haloform (iodoform) test. The iodoform test is positive only for compounds containing a methyl ketone group ( -COCH ₃ ) or a secondary alcohol group oxidizable to a methyl ketone ( -CH(OH