Aldehydes Ketones And Carboxylic Acids

Step 1: Requirement
The iodoform test is given by compounds containing either a (methyl ketone) group or a (methyl carbinol) group.
Step 2: Analysis
- Acetophenone ( ), Acetaldehyde ( ), and Ethylmethyl ketone ( ) all contain the group. - n-butyl alcohol ( ) lacks both required groups.
Step 3: Conclusion
n-butyl alcohol does not answer the iodoform test.
Final Answer: (A)

Concept: Kolbe–Schmitt reaction.

Step 1: Phenol reacts with base to form phenoxide ion.

Step 2: Phenoxide reacts with CO (under pressure).

Step 3: Forms sodium salicylate.

Step 4: Acidification gives salicylic acid.
Conclusion:
Best method = Kolbe reaction

Concept: Different reducing agents give different products depending on their strength.

Step 1: Option (a): Primary alcohol aldehyde → valid reaction.

Step 2: Option (b): Ester gives primary alcohol, not aldehyde.

Step 3: Option (c): Ester (controlled) gives aldehyde → valid.

Step 4: Option (d): Acid chloride gives aldehyde → valid.
Conclusion:
Only option (b) does not give propionaldehyde.

Concept: Hydrazine reacts with aldehydes to form azines when aldehyde is in excess.

Step 1: Benzaldehyde reacts with NH NH (hydrazine).

Step 2: Two molecules of aldehyde condense with one hydrazine molecule.

Step 3: Product formed: azine (C H CH=N–N=CHC H ).
Conclusion:
Required reagent = Hydrazine

Concept: Cannizzaro reaction occurs only in aldehydes without -hydrogen.

Step 1: Acetaldehyde (CH CHO) has -hydrogen.

Step 2: Hence it undergoes aldol condensation and haloform reaction.

Step 3: Cannizzaro reaction does NOT occur.
Conclusion:
Incorrect statement = (B)

Step 1: Understanding the Concept:
Tollens' test is positive for aldehydes and -hydroxy ketones (which can tautomerize to aldehydes).

Step 2: Detailed Explanation:
Glucose (aldehyde) gives positive Tollens'. Fructose ( -hydroxy ketone) tautomerizes to glucose and gives positive Tollens'. Sucrose (no free aldehyde or -hydroxy ketone) gives negative. Acetophenone (aromatic ketone) gives negative. Hexanal (aldehyde) gives positive. So pair (B) both give positive.

Step 3: Final Answer:
Glucose, fructose

Step 1: Understanding the Concept:
Tollens' test is for aldehydes. Oxime formation indicates a carbonyl compound (aldehyde or ketone).

Step 2: Detailed Explanation:
No silver mirror → not an aldehyde. Oxime formation → carbonyl group present. So compound must be a ketone. C H O ketone is propanone (acetone, CH COCH ).

Step 3: Final Answer: