Amines and Nitrogen Compounds Questions and Answers

Answer: B, C

The official key accepts B and C.

In the half-chair isochroman ring of X, a coupling of only 3 Hz between H_a and H_b means the two hydrogens are gauche (one axial, one equatorial). If the two phenyls were trans, both would sit equatorial and both hydrogens axial, giving a large anti coupling (about 8 to 10 Hz). So X is the cis-3,4-diphenylisochroman.

Methyl iodide converts the NMe₂ group into NMe₃⁺. On heating, the hydroxyl oxygen displaces it by an intramolecular S_N2 reaction at the benzylic CH₂, which closes the six-membered ring. That carbon is not a stereocentre, and neither of the two stereocentres is touched, so the relative configuration of the amino alcohol carries straight into X.

Working out the configuration at each carbon of the drawings, the amino alcohols in which the phenyl bond and the OH bond are the same kind (both hashed in B, both wedged in C) place the two phenyl groups on the same face of the new ring, which is the cis isomer. B and C are mirror images of each other, so both give X (as its two enantiomers). A and D have one hashed and one wedged bond, which gives the trans isomer, with the large coupling.

Answer: A, C

The official key accepts A and C.

The question states the rule: chymotrypsin cuts on the carboxyl side of Phe, Tyr and Trp. In Val-Phe-Leu-Met-Tyr-Pro-Gly-Trp-Cys that gives three cuts, after Phe, after Tyr and after Trp:

\[ \begin{aligned} & \text{Val-Phe} \mid \text{Leu-Met-Tyr} \\ & \mid \text{Pro-Gly-Trp} \mid \text{Cys} \end{aligned} \]

The tripeptides formed are Leu-Met-Tyr (A) and Pro-Gly-Trp (C). B, Phe-Leu-Met, is not a fragment: the chain is cut after Phe, not before it, so Phe stays with Val. D, Tyr-Pro-Gly, is not a fragment either, because the cuts fall after Tyr and after Trp, not before Tyr.

In real chymotrypsin chemistry the Tyr-Pro bond is cleaved poorly because the next residue is proline; the key follows the simple rule given in the stem.

Answer: A, C, D

The sets in A, C and D give aniline from benzoyl chloride; set B does not.

(A) benzoyl chloride and NH₃ give benzamide, and Br₂ with NaOH is the Hofmann degradation, which loses the carbonyl carbon and gives aniline. (D) NH₂OH gives the hydroxamic acid, TsCl makes its tosylate, and NaOH on heating is the Lossen rearrangement to phenyl isocyanate, which water converts to aniline. (C) NaN₃ gives benzoyl azide, which on warming in water undergoes the Curtius rearrangement through the isocyanate to aniline.

(B) diazomethane is the Arndt-Eistert homologation and leads to phenylacetic acid, which keeps the carbonyl and has one more carbon. The final key accepts either A, C, D or A, D alone, because the Curtius step needs heating that the set does not state.

Answer: B, D

Statements (B) and (D) are correct; (A) and (C) are incorrect.

First sequence. Acetic anhydride converts aniline to acetanilide, which protects the amino group. Nitration with conc. HNO₃ and conc. H₂SO₄ at 288 K gives mainly p-nitroacetanilide (P), a mono-nitro compound. Hydrolysis then gives p-nitroaniline, and NaNO₂/HCl at 273-278 K diazotises it to p-nitrobenzenediazonium chloride (Q).

Second sequence. The cumene process (\(\mathrm{O_2/H_3O^+}\)) gives phenol and acetone. Phenol with NaOH/CO₂ (Kolbe-Schmitt reaction) followed by acid gives salicylic acid (S). Coupling of the diazonium salt Q with S in aqueous NaOH gives an azo compound (T).

(A) A diazonium salt heated with ethanol is reduced (the diazo group is replaced by H), giving nitrobenzene and acetaldehyde. The aldehyde formed is acetaldehyde, which is aliphatic, so the statement is incorrect.

(B) S still has a free phenolic OH with its para position open, so it condenses with phthalic anhydride (conc. H₂SO₄) to give a phthalein dye: correct.

(C) P has only one nitro group (and the acetamido group): incorrect.

(D) T is an azo dye with an extended conjugated system, which is coloured: correct.

Answer: B, D

The official key accepts B and D.

B is the Cadogan-Sundberg indole synthesis: triethyl phosphite deoxygenates the nitro group of the o-nitrostyrene to a nitrene, which inserts into the alkene to close the five-membered ring; the propenyl group puts the methyl on C-2, giving 2-methylindole. D is the Gassman indole synthesis: N-chloroaniline reacts with the β-keto sulfide MeSCH₂COMe, a [2,3] shift and cyclisation give 2-methyl-3-(methylthio)indole, and Raney nickel removes the sulfur to give 2-methylindole.

A is wrong: o-nitrobenzaldehyde with acetone and base is the Baeyer-Drewsen reaction, which leads to indigo. C is wrong: o-nitrotoluene with diethyl oxalate and base is the Reissert synthesis, and reduction gives ethyl indole-2-carboxylate, which still carries the ester.