Three tests were performed on an unknown organic compound.
| Test reagent | Test result |
|---|---|
| 2,4-DNPH reagent | orange ppt |
| Tollens’ reagent | no change |
| alkaline \(\mathrm{I_2(aq)}\) | yellow ppt |
What is the organic compound tested?
Correct Answer: \( \boxed{\mathrm{B}} \)
The orange precipitate with 2,4-DNPH confirms the presence of a carbonyl group (aldehyde or ketone).
No reaction with Tollens’ reagent shows that the compound is not an aldehyde, so it must be a ketone.
The yellow precipitate with alkaline \(\mathrm{I_2}\) is a positive iodoform test, indicating the presence of a \(\mathrm{CH_3CO-}\) group (a methyl ketone).
Of the four structures, only Structure B contains a methyl ketone group.
Therefore, the correct answer is (B).
\(\mathrm{HOCH_2CHO}\) is heated under reflux with an excess of acidified \(\mathrm{K_2Cr_2O_7}\) until there is no further reaction.
What is the final product of this reaction?
Correct Answer: \( \boxed{\mathrm{C}} \)
Acidified potassium dichromate oxidises both functional groups present.
The final product is ethanedioic acid:
\(\mathrm{HOCH_2CHO \xrightarrow[\text{reflux}]{K_2Cr_2O_7/H^+} HOOCCOOH}\)
Therefore, the correct answer is (C).
An organometallic lithium compound, \(\mathrm{RLi}\), contains the nucleophile \(\mathrm{R^-}\).
\(\mathrm{CH_3CH_2Li}\) reacts with pentan-2-one. The mechanism is nucleophilic addition. The first step produces an anion which is then protonated to form the final product.
Which organic product is formed?
Correct Answer: \( \boxed{\mathrm{B}} \)
The ethyl group, \(\mathrm{CH_3CH_2^-}\), attacks the carbonyl carbon of pentan-2-one.
After protonation, the carbonyl oxygen becomes a hydroxyl group, producing a tertiary alcohol.
The carbon bearing the \(\mathrm{OH}\) group is attached to:
This corresponds to the structure shown in B.
Therefore, the correct answer is (B).
