Q.(a) An organic compound (A) of molecular formula $C_6H_6O$ gives violet colour with neutral ferric chloride. Compound (A) when heated with zinc dust gives a hydrocarbon (B). Also compound (A) reacts with phthalic anhydride and conc. $H_2SO_4$ to give compound (C) of molecular formula $C_{20}H_{14}O_4$. Identify (A), (B) and (C). Explain the reactions.

(a) Phenol → benzene (Zn-dust distillation) and → phenolphthalein (with phthalic anhydride/conc. $H_2SO_4$). (b) Copper (period 4, group 11) gives CuO and $Cu_2O$ at different temperatures, and $Cu(NO_3)_2$ with conc. $HNO_3$. (c) Benzaldehyde gives hydrobenzamide with $NH_3$ and benzoin with alcoholic KCN. (d) A 1:1 weak-acid/conjugate-base buffer has pH = pKa ≈ 4.87.

(a) Identification of A, B, C

A — Phenol, $C_6H_5OH$: The molecular formula $C_6H_6O$ and the characteristic violet colouration with neutral ferric chloride solution are the standard identifying tests for a phenolic $-OH$ group; phenol fits both exactly.

B — Benzene, $C_6H_6$: Distilling phenol with zinc dust is the classical "zinc-dust distillation" deoxygenation reaction, which removes the oxygen-containing functional group and reduces the compound to the parent hydrocarbon skeleton:

$C_6H_5OH + Zn \xrightarrow{\Delta} C_6H_6 + ZnO$

C — Phenolphthalein, $C_{20}H_{14}O_4$: Heating phenol with phthalic anhydride in the presence of concentrated sulfuric acid (Baeyer's classic synthesis) causes two molecules of phenol to condense with one molecule of phthalic anhydride, giving phenolphthalein and water:

$2C_6H_5OH + C_8H_4O_3 \xrightarrow{\text{conc. } H_2SO_4} C_{20}H_{14}O_4 + H_2O$

(Atom check: LHS $= 2(C_6H_6O) + C_8H_4O_3 = C_{20}H_{16}O_5$; RHS $= C_{20}H_{14}O_4 + H_2O = C_{20}H_{16}O_5$ — balances exactly, confirming the given formula for phenolphthalein, $C_{20}H_{14}O_4$.) Phenolphthalein is the familiar acid–base indicator, colourless in acid and pink/magenta in alkaline solution.

(b) Identification of A, B, C, D

A — Copper, Cu: The element is in Period 4 and Group 11 of the periodic table (the coinage-metal group, Cu/Ag/Au); the Group-11, Period-4 member is copper (atomic number 29). Copper's principal ore is copper pyrites (chalcopyrite), $CuFeS_2$, from which it is extracted — matching "extracted from its pyrite ore".

B and C — the two oxides of copper: Copper reacts with atmospheric oxygen differently depending on temperature:

• At a comparatively lower temperature, copper is oxidised fully to black cupric oxide: $2Cu + O_2 \xrightarrow{\Delta,\ \text{lower }T} 2CuO$ (B)
• At a higher temperature, the product is instead red cuprous oxide: $4Cu + O_2 \xrightarrow{\Delta,\ \text{higher }T} 2Cu_2O$ (C)

D — Copper(II) nitrate, $Cu(NO_3)_2$: Copper reacts with concentrated nitric acid with vigorous evolution of brown $NO_2$ gas (concentrated $HNO_3$ is reduced to $NO_2$, not $NO$):

$Cu + 4HNO_3(\text{conc.}) \rightarrow Cu(NO_3)_2 + 2NO_2\uparrow + 2H_2O$

(c) OR alternative — identification of A, B, C

A — Benzaldehyde, $C_6H_5CHO$ ($C_7H_6O$): the characteristic smell of bitter almonds is the classic identifying property of benzaldehyde.

B — Hydrobenzamide, $C_{21}H_{18}N_2$: benzaldehyde reacts with ammonia (not by simple addition, since the initial imine is unstable) in a 3:2 ratio to give hydrobenzamide, $(C_6H_5CH=N)_2CHC_6H_5$, with elimination of water:

$3C_6H_5CHO + 2NH_3 \rightarrow (C_6H_5CH=N)_2CHC_6H_5 + 3H_2O$

(Atom check for hydrobenzamide: three $C_6H_5$ groups contribute $C_{18}H_{15}$; the three connecting CH/CH= carbons contribute $C_3H_3$; plus 2 N atoms — total $C_{21}H_{18}N_2$, matching the given formula.)

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