(a) An organic compound [A] with molecular formula $C_9H_{10}O$ forms an orange precipitate [B] with 2,4-DNP reagent. Compound [A] gives yellow precipitate [C] on heating with iodine in presence of sodium hydroxide along with a colourless compound [D]. The compound [A] does not r

Question

(a) An organic compound [A] with molecular formula $C_9H_{10}O$ forms an orange precipitate [B] with 2,4-DNP reagent. Compound [A] gives yellow precipitate [C] on heating with iodine in presence of sodium hydroxide along with a colourless compound [D]. The compound [A] does not reduce Tollen's reagent or Fehling's reagent nor it decolorise bromine water. On drastic oxidation with chromic acid, compound [A] gives a carboxylic acid [D] having molecular formula $C_7H_6O_3$. Deduce the structures of the compounds [A] to [D].

**OR**

(b) Predict the product when carboxylic acid reacts with the following reagents: (i) Alcohols in presence of mineral acids (ii) Ammonia at high temperatures (iii) $P_2O_5$ at high temperatures (iv) Na metal (v) Thionyl chloride

Accepted Answer

>

> [!TLDR]
> The 2,4-DNP test confirms a carbonyl, the positive iodoform test (with no Tollens'/Fehling's reduction) pins A down as a methyl ketone, and oxidative cleavage of that methyl ketone gives a benzoic-acid-type product D.
>
> Working through each clue for compound A ($C_9H_{10}O$):
>
> 1. **Forms an orange precipitate with 2,4-DNP (B):** confirms A is an **aldehyde or ketone** (has a $C=O$ group).
> 2. **Does NOT reduce Tollens'/Fehling's reagent, does not decolourise bromine water:** rules out an aldehyde (no $-CHO$) and rules out a $C=C$ double bond — A is a simple **saturated ketone**.
> 3. **Gives a yellow precipitate (iodoform, C) + a colourless compound on heating with $I_2$/NaOH:** the **iodoform test is positive**, which specifically requires a **methyl ketone** group, $CH_3-CO-$ (an ethyl or higher ketone, like propiophenone, would NOT give this test).
> 4. **Drastic oxidation with chromic acid gives a carboxylic acid D:** oxidative cleavage of an aryl methyl ketone ($Ar-CO-CH_3$) removes the terminal methyl carbon (as $CO_2$/via the same type of C–C cleavage as the haloform step), converting the ring-attached carbonyl carbon directly into $-COOH$ — i.e. $Ar-CO-CH_3 \xrightarrow{[O]} Ar-COOH$.
>
> Putting this together, A must be an **aryl methyl ketone** with formula $C_9H_{10}O$ — since $C_6H_5-CO-CH_3$ (acetophenone) is only $C_8H_8O$, A needs one more $CH_2$ of complexity, most simply as an extra ring methyl substituent: **4-methylacetophenone** (or the isomeric 2- or 3-methylacetophenone), $CH_3-C_6H_4-CO-CH_3$.
>
> **Structures:** A = $CH_3C_6H_4COCH_3$ (a methylacetophenone); B = its 2,4-DNP hydrazone (orange solid); C = $CHI_3$ (iodoform, yellow precipitate); D = the corresponding **methylbenzoic acid** ($CH_3C_6H_4COOH$, $C_8H_8O_2$), formed by oxidative loss of the acetyl group's terminal carbon.
>
> **⚠️ Honest discrepancy note:** the paper as transcribed states D's molecular formula as $C_7H_6O_3$ — but a formula with 3 oxygens and o

Concept understanding — Haloform (Iodoform) Reaction

Haloform (Iodoform) Reaction

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