Q.What is lanthanide contraction? Explain the cause of lanthanide contraction. Draw the structures of chloroxylenol and adenine. How are ethylamine and ethyl methyl amine distinguished by using nitrous acid? OR What is the action of the following reagents on ethanoic acid?

A combined question on lanthanide contraction, two named-compound structures, and a nitrous-acid test distinguishing 1°/2° amines (or, in the OR branch, ethanoic acid's reactions with three reagents, an AgCN/Wurtz-type identification, and Hoffmann bromamide degradation).

Lanthanide contraction: across the lanthanide series (La, Z=57, to Lu, Z=71), there is a slow but steady and cumulative decrease in atomic and ionic (usually $M^{3+}$) radii with increasing atomic number, even though electrons and protons are both being added — this is the lanthanide contraction.

Cause: as atomic number increases across the series, the added electrons enter the inner $4f$ subshell (not the outermost shell). The $4f$ orbitals are diffuse and have a complex, poorly-directed shape, so they shield (screen) the outer $5s$, $5p$, $6s$ electrons from the increasing positive nuclear charge very inefficiently (poor shielding). As a result, the effective nuclear charge experienced by the outer electrons increases steadily across the series, pulling them progressively closer to the nucleus — causing the observed steady contraction in size.

Structures: Chloroxylenol (4-chloro-3,5-dimethylphenol) — a benzene ring bearing $-OH$ at position 1, a $-Cl$ at position 4, and $-CH_3$ groups at positions 3 and 5 (the active antiseptic ingredient of Dettol). Adenine — a purine base: a fused bicyclic ring system (a six-membered pyrimidine ring fused to a five-membered imidazole ring), bearing an $-NH_2$ (amino) substituent on the six-membered ring, and forms one of the four nitrogenous bases of DNA/RNA.

Distinguishing ethylamine from ethyl methyl amine using nitrous acid: ethylamine (a primary amine, $C_2H_5NH_2$) reacts with nitrous acid ($NaNO_2$/HCl) via an unstable diazonium intermediate that decomposes immediately, evolving nitrogen gas with brisk effervescence and forming ethanol: $C_2H_5NH_2 + HNO_2 \rightarrow C_2H_5OH + N_2\uparrow + H_2O$. Ethyl methyl amine ($C_2H_5-NH-CH_3$, a secondary amine) instead forms a yellow, oily N-nitroso compound with NO gas evolved: $C_2H_5N(CH_3)H + HNO_2 \rightarrow C_2H_5-N(CH_3)-NO + H_2O$. The presence/absence of gas evolution and the appearance of a yellow oil thus distinguishes the two.

OR — Action of reagents on ethanoic acid:

1. $LiAlH_4$/$H_3O^+$: a strong reducing agent that reduces the carboxylic acid all the way to a primary alcohol: $CH_3COOH \xrightarrow{LiAlH_4} \xrightarrow{H_3O^+} CH_3CH_2OH$.
2. $PCl_3$, heat: converts the acid to the acid chloride: $3CH_3COOH + PCl_3 \xrightarrow{\Delta} 3CH_3COCl\ (\text{acetyl chloride}) + H_3PO_3$.
3. $P_2O_5$, heat: a strong dehydrating agent, converts two molecules of the acid into the acid anhydride: $2CH_3COOH \xrightarrow[\Delta]{P_2O_5} (CH_3CO)_2O\ (\text{acetic anhydride}) + H_2O$. Identify A and B: ethyl bromide reacts with silver cyanide (a covalent, N-attacking reagent, unlike ionic KCN which C-attacks) to give the ISOcyanide: …