(a) An organic compound A (molecular formula $C_3H_6O$) does not react with Tollen's reagent but on reduction gives compound B ($C_3H_8O$). Compound B, on treatment with HBr, gives Bromide C, which on treatment with alcoholic KOH gives Alkene D ($C_3H_6$). Identify compounds A, B

Question

(a) An organic compound A (molecular formula $C_3H_6O$) does not react with Tollen's reagent but on reduction gives compound B ($C_3H_8O$). Compound B, on treatment with HBr, gives Bromide C, which on treatment with alcoholic KOH gives Alkene D ($C_3H_6$). Identify compounds A, B, C and D with essential chemical reactions. [3] (b) Write down the Gattermann-Koch reaction with chemical equation. [1] OR In following chemical equations, write down the products obtained with their names and chemical formulae in the blank spaces - (a) $CH_3COCl \xrightarrow{H_2,\ Pd/BaSO_4,\ S,\ boiling\ xylene}$ ____ (b) Phthalimide $\xrightarrow{KOH,\ CH_3I}$ then $\xrightarrow{H_3O^+}$ benzene-1,2-dicarboxylic acid (with -COOH, -COOH groups) + ____ (c) $HCHO \xrightarrow{50\%\ NaOH,\ \Delta} HCOONa +$ ____ (d) $(CH_3COO)_2Ca \xrightarrow{\Delta} CaCO_3 +$ ____ [1x4=4]

Accepted Answer

>

> [!TLDR]
> A (C₃H₆O, no Tollens test) is the ketone acetone; successive reduction, HBr substitution and dehydrohalogenation give B, C and D.
>
> **(a) Identification of A, B, C, D:**
>
> Since A ($C_3H_6O$) does NOT give a positive Tollens' test, A is not an aldehyde but a ketone: **A = Acetone, $CH_3COCH_3$**.
>
> On reduction (e.g. with $NaBH_4$ or $H_2/Ni$), the ketone is reduced to a secondary alcohol, $B$ ($C_3H_8O$): **B = Isopropyl alcohol (propan-2-ol), $CH_3CH(OH)CH_3$**.
> $CH_3COCH_3 \xrightarrow{[H]} CH_3CH(OH)CH_3$
>
> B, on treatment with HBr, undergoes substitution of -OH by -Br to give bromide C: **C = 2-Bromopropane, $CH_3CHBrCH_3$**.
> $CH_3CH(OH)CH_3 + HBr ightarrow CH_3CHBrCH_3 + H_2O$
>
> C, on treatment with alcoholic KOH, undergoes dehydrohalogenation (elimination) to give alkene D ($C_3H_6$): **D = Propene, $CH_3CH=CH_2$**.
> $CH_3CHBrCH_3 \xrightarrow[\Delta]{alc.\ KOH} CH_3CH=CH_2 + KBr + H_2O$
>
> **(b) Gattermann-Koch reaction:** Benzene is treated with carbon monoxide and hydrogen chloride gas in the presence of anhydrous $AlCl_3$ (with $CuCl$) under pressure, introducing a formyl (-CHO) group directly onto the ring to give benzaldehyde (a special Friedel-Crafts formylation):
> $C_6H_6+CO+HCl \xrightarrow[	ext{pressure}]{anhyd.\ AlCl_3/CuCl} C_6H_5CHO+HCl$
>
> **OR (fill in the blanks with products, names and formulae):**
>
> **(a)** $CH_3COCl \xrightarrow{H_2,\ Pd/BaSO_4\ (poisoned),\ S,\ boiling\ xylene}$ — this is **Rosenmund reduction**, which reduces an acid chloride to an aldehyde:
> $CH_3COCl ightarrow CH_3CHO$ (**Acetaldehyde**) $+ HCl$
> …

Concept understanding — Tollens Test

The Tollens Test: The Silver Mirror That Spots an Aldehyde

The Intuition: Why Aldehydes Are Special

The Precise Statement

How the Reagent Is Made

The Procedure in the Lab

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