The tests performed on compound X and their inferences are:
| Test | Inference |
| (a) 2,4-DNP test | Coloured precipitate |
| (b) Iodoform test | Yellow precipitate |
| (c) Azo-dye test | No dye formation |
Compound 'X' is:
–COCH3 is present it will show both 2, 4-DNP & iodoform test.
Due to steric inhibition of resonance. I.P of ‘N’ is not involved in delocalization so coupling reaction will not take place.
Let's analyze the given tests step by step to identify compound X:
The 2,4-Dinitrophenylhydrazine (2,4-DNP) test gives a colored precipitate (usually orange or yellow). This test is positive for carbonyl compounds (aldehydes and ketones). So, compound X contains a carbonyl group (C=O).
The iodoform test gives a yellow precipitate. This test is positive for compounds that have a methyl ketone group (CH3-C=O) or alcohols that can be oxidized to methyl ketones (like CH3CH(OH)-R). Since we already know it has a carbonyl, it is likely a methyl ketone (CH3-C=O).
The azo-dye test is used to detect aromatic primary amines. No dye formation means compound X does not contain an aromatic primary amine group (-NH2 attached to benzene ring).
From the tests:
Now, looking at the options (structures not fully visible, but based on common knowledge):
The compound must be a methyl ketone without an aromatic primary amine. Among typical options, it is likely acetophenone (C6H5COCH3), which is a methyl ketone and gives positive 2,4-DNP and iodoform tests, but has no amine group (so negative azo-dye test).
Confirming with the options provided (though images are not clear here), the correct structure should be the one with a benzene ring attached to a carbonyl and a methyl group, i.e., acetophenone.
2,4-DNP Test: Reacts with carbonyls to form hydrazones (colored precipitates).
Iodoform Test: For CH3CO- or CH3CH(OH)- groups; reaction involves halogenation and hydrolysis to form CHI3 (yellow precipitate).
Azo-Dye Test: For aromatic primary amines; involves diazotization and coupling to form a colored dye.