The stronger base (propylamine) holds a proton more tightly than the weaker base (benzylamine). Thus, the propylammonium ion is less acidic (larger pK_a) than the benzylammonium ion (smaller pK_a).

To calculate pK_b: K_a · K_b = 10 – 14, pK_a + pK_b = 14 and pK_b = 14 – pK_a.

At pH = 7.3, virtually 100% of the pyrimidine molecules are in the neutral form.

The compounds in parts (b) and (d) can’t be prepared by reduction of a nitrile.

Step 1: Addition of hydroxide.

Step 2: Ring opening.

Step 3: Proton transfer.

Step 4: Addition of hydroxide.

Step 5: Elimination of amine.

Step 6: Proton transfer.

Either method of nitro group reduction (SnCl₂, H₂) can be used in all parts of this problem; both methods are shown.

Coupling takes place between N,N-dimethylaniline and a benzenediazonium salt to yield the desired product.

Thiazole contains six π electrons. Each carbon contributes one electron, nitrogen contributes one electron, and sulfur contributes two electrons to the ring π system. Both sulfur and nitrogen have lone electron pairs in sp² orbitals that lie in the plane of the ring.

4.1% of histidine molecules have the imidazole nitrogen in the protonated form at the physiological pH

Reaction at C3 is favored over reaction at C2 or C4. The positive charge of the cationic intermediate of reaction at C3 is delocalized over three carbon atoms, rather than over two carbons and the electronegative pyridine nitrogen as occurs in reaction at C2 or C4.

The side chain nitrogen atom of N,N-dimethyltryptamine is more basic than the ring nitrogen atom because its lone electron pair is more available for donation to a Lewis acid. The aromatic nitrogen electron lone pair is part of the ring π electron system.

Positive charge can be stabilized by the nitrogen lone-pair electrons in reaction at both C2 and C3. In reaction at C2, however, stabilization by nitrogen destroys the aromaticity of the fused benzene ring. Reaction at C3 is therefore favored, even though the cationic intermediate has fewer resonance forms, because the aromaticity of the six-membered-ring is preserved.

The IR spectrum (pair of bands at around 3300 cm^–1) shows that B is a primary amine, and the ¹H NMR spectrum shows a 9-proton singlet, a one-proton quartet, and a 3-proton doublet. An absorption due to the amine protons is not visible.

Hofmann elimination is an E2 elimination, in which the two groups to be eliminated must be 180° apart. The product that results from this elimination geometry is the Z isomer.

The indicated nitrogen is most basic because it is more electron-rich. The electrons of the other nitrogens are part of the fused-ring π system and are less available for donation to a Lewis acid.

Steps 1, 6: Protonation.

Steps 2, 7: Nucleophilic addition.

Steps 3, 8: Proton transfer.

Steps 4, 9: Loss of water.

Steps 5, 10: Loss of proton.

The mechanism consists of the nucleophilic addition of ammonia, first to one of the ketones, and then to the other, with loss of two equivalents of water.

Steps 1, 6: Protonation.

Steps 2, 7: Nucleophilic addition of hydroxylamine.

Steps 3, 8: Proton transfer.

Steps 4, 9: Loss of water.

Steps 5, 10: Loss of proton.

This mechanism is virtually identical to the mechanism illustrated in the previous problem and involves two nucleophilic additions to carbonyl groups, with loss of water. These addition-eliminations are nucleophilic acyl substitution reactions.

Step 1: Nucleophilic addition.

Step 2: Cyclization.

Step 3: Elimination of lysine.

Step 4: Elimination of the other lysine.

Step 5: Tautomerization.

Steps 1 – 3: E1 elimination (protonation, loss of HOCH₃, deprotonation).

Steps 4 – 6: S_N2 substitution (protonation, substitution, deprotonation).

Step 7: E1 elimination of carbamate.

Steps 8 – 9: Conjugate addition of DNA (addition, deprotonation).

Notice that five of the nine steps are either protonations or deprotonations.

Step 1: Addition of NH3.

Step 2: Elimination of –OH.

Step 3: Addition of –CN.

Step 4: Hydrolysis of nitrile.

The mechanism of acid-catalyzed nitrile hydrolysis is shown in Problem 20.25.

Step 1: Conjugate addition of hydrazine.

Step 2: Proton transfer.

Step 3: Nucleophilic acyl substitution, forming the cyclic amide.

Step 4: Proton transfer.

Hofmann rearrangement (the mechanism is shown in Section 24.6) of an α-hydroxy amide produces a carbinolamine intermediate that expels ammonia to give an aldehyde.

Step 1: Conjugate addition of amine.

Step 2: Proton transfer.

Step 3: Nucleophilic addition of amine.

Step 4: Proton transfer.

Step 5: Elimination of methanol.

Step 1: S_N2 displacement of Br^– by amine.

Step 2: Deprotonation.

Step 3: Conjugate addition of alcohol.

Step 4: Proton transfer.

Step 1: Nucleophilic addition of the amine to the protonated aldehyde.

Step 2: Proton transfer.

Step 3: Loss of water.

Step 4: Electrophilic aromatic substitution.

Step 5: Loss of proton.

The “a” nitrogen is most basic because its electron pair is most available to Lewis acids.

The “c” nitrogen is the least basic because the lone-pair electrons of the pyrrole nitrogen are part of the ring π electron system.

In (e), the amine reacts with the Grignard reagent and inactivates it.

(a) NH₃, NaBH₄;
(b) excess CH₃I;
(c) Ag₂O, H₂O, heat;
(d) RCO₃H
(e) (CH₃)2NH.

Step (e) is an S_N2 ring opening of the epoxide by nucleophilic substitution of the amine at the primary carbon.

Oxazole is an aromatic 6 π electron heterocycle. Two oxygen electrons and one nitrogen electron are in p orbitals that are part of the π electron system of the ring, along with one electron from each carbon. An oxygen lone pair and a nitrogen lone pair are in sp² orbitals that lie in the plane of the ring. Since the nitrogen lone pair is available for donation to acids, oxazole is more basic than pyrrole.

Protonation occurs on oxygen because an O-protonated amide is stabilized by resonance.

(a) CH₂=CHCH₂Cl, AlCl₃;
(b) Hg(OAc)₂, H₂O; NaBH₄;
(c) CrO₃, H₃O⁺;
(d) CH₃NH₂, NaBH₄.

The last step of the synthesis is a reductive amination of a ketone that is formed by oxidation of the corresponding alcohol. The alcohol results from the Grignard reaction between cyclopentylmagnesium bromide and propylene oxide.

The synthesis in (a) is achieved by a reductive amination reaction. Reactions in (b) include formation of an acid chloride, esterification, and reduction of the nitro group.

We know the location of the –OH group of tropine because it is stated that tropine is an optically inactive alcohol. This hydroxyl group results from basic hydrolysis of the ester that is composed of tropine and tropic acid.

Tropilidene results from two cycles of Hofmann elimination on tropidene.

Step 1: Reduction of nitrile

Step 2: Nucleophilic addition.

Step 3: Dehydration.

Step 4: Reduction of double bond.

Step 1: Michael addition.

Step 2: Ester hydrolysis; decarboxylation.

Step 3: Reduction.

Step 4: Cyclization.

Step 5: Reduction.

When you see –CH₂NH₂, think of the reduction of a nitrile. The nitrile comes from substitution of a benzylic bromide by ^–CN.

The reactive intermediate is benzyne, which undergoes a Diels–Alder reaction with cyclopentadiene to yield the observed product.