Q.An aromatic organic compound 'A' with molecular formula $C_8H_8O$ gives positive DNP and iodoform tests. It neither reduces Tollens' reagent nor does it decolourise bromine water. Write the structure of 'A'.

Concept and Intuition

The iodoform reaction (I$_2$ + NaOH) converts a methyl ketone (or acetaldehyde) or a secondary

alcohol of the type $\mathrm{CH_3{-}CH(OH){-}R}$ into the corresponding carboxylate plus

iodoform, $\mathrm{CHI_3}$, seen as a characteristic pale yellow precipitate. The key structural

requirement is a methyl group directly on the carbonyl carbon, or on the carbinol carbon of a

secondary alcohol (which is first oxidised in situ by the hypoiodite formed from I$_2$/NaOH).

Step-by-Step Solution

1. All four given structures are isomers of $\mathrm{C_4H_8O}$, so molecular formula alone cannot distinguish them — the test is structural.
2. (A) $\mathrm{CH_3CH_2CH_2CHO}$ (butanal): the carbonyl carbon is attached to a propyl group, not a methyl group — no iodoform reaction.
3. (B) $\mathrm{CH_2{=}CHCH(OH)CH_3}$: the carbinol carbon carries a methyl group directly ($\mathrm{CH_3{-}CH(OH){-}}$) — this is oxidised in situ to a methyl ketone and gives a positive iodoform test.
4. (C) and (D) are enol ethers (no free carbonyl or $\mathrm{CH_3CH(OH)}$ group present, and …

Concept and Intuition

The haloform reaction is diagnostic for methyl ketones ($\mathrm{CH_3CO-}$ groups, or alcohols that can be oxidised to them): triiodomethylation of the methyl group followed by cleavage under basic conditions gives iodoform ($\mathrm{CHI_3}$, a characteristic yellow solid) and the carboxylate salt of the rest of the molecule. Because the reaction is entirely localised on the terminal methyl carbon of the ketone, any other functional groups elsewhere in the molecule — here, the ring's C=C double bond — are left completely unaffected.

Step-by-Step Solution

1. Starting material: cyclohexene ring bearing $-\mathrm{C(=O)CH_3}$ on the alkene (sp²) carbon — i.e. 1-acetylcyclohexene.
2. $\mathrm{NaOI}$ (from $\mathrm{I_2 + NaOH}$) progressively iodinates the acidic $\alpha$-hydrogens of the methyl group, ultimately forming $R\mathrm{-CO-CI_3}$.
3. Hydroxide attacks the carbonyl carbon, and $\mathrm{CI_3^-}$ (which becomes $\mathrm{CHI_3}$ on protonation, i.e. iodoform) leaves as the carbanion, cleaving the C–C bond between the carbonyl carbon and the trihalomethyl carbon. …

Concept and Intuition

The industrial/lab route to chloroform ($CHCl_3$) from ethanol via bleaching powder proceeds through a sequence: generate chlorine in situ, use it to oxidize the alcohol to an aldehyde, then further chlorinate that aldehyde, and finally hydrolyze the chlorinated aldehyde with base to release the chloroform. Getting the identity of each intermediate right (acetaldehyde, not acetic acid) and knowing that the final hydrolysis step genuinely happens (not a step that fails) is the crux of this question.

Step-by-Step Solution

1. Statement (i): Bleaching powder, $CaOCl_2$, reacts with water: $CaOCl_2 + H_2O \rightarrow Ca(OH)_2 + Cl_2$. This is textbook-standard and TRUE.
2. Statement (ii): The $Cl_2$ generated oxidizes ethanol: $CH_3CH_2OH + Cl_2 \rightarrow CH_3CHO + 2HCl$ — the product is acetaldehyde, not acetic acid. So statement (ii) is FALSE.
3. Statement (iii): Excess chlorine then chlorinates the acetaldehyde at the methyl group: $CH_3CHO + 3Cl_2 \rightarrow CCl_3CHO$ (chloral) $+ 3HCl$. This is TRUE. …

The iodoform test detects a methyl ketone (CH₃CO-) directly, or a secondary alcohol of the form CH₃CH(OH)-, which iodine/base first oxidises to a methyl ketone before cleaving it to give yellow CHI₃. The key requirement: the methyl group must sit directly on the carbon bearing the OH (or the carbonyl) — not on a neighbouring carbon.

1. (i) 1-(3,4-dimethoxyphenyl)ethan-1-ol, Ar-CH(OH)-CH₃. The carbinol carbon carries a methyl group directly → CH₃CH(OH)- pattern. Ring substituents (methoxy groups) don't interfere with the test. Positive.

2. (ii) Ph-CH=C(CH₃)-CH(OH)-CH₃. The terminal carbon is -CH(OH)-CH₃ — a methyl directly on the carbinol carbon, regardless of what's further down the chain. Positive.

3. (iii) Ph-CH(OH)-C(CH₃)=CH-CH�3. The carbinol carbon (attached to Ph) is bonded to Ph, OH, H, and the alkene carbon — no methyl group directly on it (the methyl sits on the next carbon, part of the double bond). Negative.

4. (iv) 2-methylcyclopentan-1-one. The carbonyl carbon (C1) is flanked by two ring carbons, not a methyl group; the methyl substituent is on the adjacent C2 (the α-carbon), not on the carbonyl carbon itself — so there is no CH₃CO- unit. Negative. …