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Volume of 1 M H2SO4 needed for the ammonia from a Kjeldahl estimation

Build the compound before you analyse it. The bench walks the printed reaction sequence in three stages - J, then K weighed against the stem's own 350 g/mol checksum, then L - and then stops and makes you commit: L carries two nitrogen atoms, so how many moles of ammonia come off per mole of L? The clock waits for your answer. Only then does it digest, distil and titrate, and the titre it reports is the one your commitment earns.

Sub-topics covered

• Friedel-Crafts acylation and where two methyl groups send the acyl
• The Finkelstein exchange, and why the iodide is made first
• Williamson ether synthesis from a phenoxide
• Selective reduction: NaBH4 takes the ketone and leaves the nitro group
• Alcohol to bromide with PBr3
• Nucleophilic substitution by excess ammonia
• Molar mass assembled atom by atom from given atomic masses
• Using a stated molar mass as a checksum on a proposed structure
• The Kjeldahl method and the nitrogens it cannot estimate
• Acid-base stoichiometry with a dibasic acid
• Molarity to volume, and answering in the unit asked for

What it teaches

• The stem prints one molar mass, 350 for K, and that number is a checksum rather than decoration: whatever structure is proposed for K has to weigh exactly 350 on the given atomic masses, which confirms the whole sequence before any arithmetic begins
• m-Xylene's two methyl groups are both o,p-directing, so the acyl group enters the one ring position activated by both of them
• The chloride is exchanged for iodide before the phenoxide step because iodide is the better leaving group - the Finkelstein exchange is there to serve the substitution that follows
• Sodium borohydride is chosen for its selectivity: it reduces the ketone to a secondary alcohol and leaves the aromatic nitro group completely untouched, which is exactly why the nitro group survives all the way to L
• L contains TWO nitrogen atoms - one amine and one nitro - and the entire question turns on that fact being noticed
• Kjeldahl's digestion converts nitrogen to ammonium only from amine-like nitrogen; nitro, azo, diazo and ring nitrogen are not reduced and are never estimated, so only one of L's two nitrogens is seen
• The method therefore under-reports a nitro compound: it finds about 4.9 per cent nitrogen in L where 9.8 per cent is actually present, and the gap is the nitro nitrogen
• Sulphuric acid is dibasic: 2 NH3 + H2SO4 gives (NH4)2SO4, so the moles of acid are half the moles of ammonia - a factor that is trivial to state and easy to drop

Simulation ID

• ADV-2026-P2-CHE-Q17 · revision 2

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