Q2(b). Cannizzaro in D₂O; E2 in rings; 1,2- vs 1,4-addition

Introduction. (i) is the Cannizzaro reaction: benzaldehyde has no α-H, so strong base makes it disproportionate. (ii) tests the anti-periplanar requirement of E2 in a cyclohexane. (iii) compares direct (1,2) and conjugate (1,4) addition to α,β-unsaturated ketones.

(i) Cannizzaro reaction in D₂O

1. OD⁻ adds to the C=O of one aldehyde molecule to give the tetrahedral anion PhCH(OD)O⁻ (in strong base partly the dianion).
2. This anion transfers its C–H as a hydride directly to the carbonyl carbon of a second aldehyde (RDS; the reaction is third or fourth order overall, second order in aldehyde).
3. The first molecule becomes benzoic acid, which is deprotonated; the second becomes the benzyl alkoxide, which takes D⁺ from the solvent.

2 PhCHO →PhCOO⁻ Na⁺ + PhCH₂OD Key observation: the benzyl alcohol carries no deuterium on carbon (PhCH₂OD, not PhCHDOD). The hydrogen that reaches carbon came from the other aldehyde molecule, not from the solvent, which proves a direct intermolecular hydride transfer. After work-up in H₂O the products are sodium benzoate (benzoic acid on acidification) and benzyl alcohol, PhCH₂OH.

(ii) Which can undergo E2: Y or Z?

E2 needs the H and the leaving Br anti-periplanar, which on a ring means trans-diaxial. So a β-H must be trans to Br. Z (all-cis): with both CH₃ equatorial, the Br on C-2 (cis to both neighbours) must be axial. The H atoms on C-1 and C-3 are trans to Br and axial, anti-periplanar to it. Z undergoes E2 readily.

Figure: Z in its stable chair: both CH₃ equatorial, Br axial, two axial H anti to Br.

Y: Br is trans to both methyl groups, so the H atoms on C-1 and C-3 are cis to Br. In the stable chair all three groups are equatorial; after a ring flip all three are axial, but the C-1 and C-3 hydrogens are then equatorial. In neither chair is any β-H anti to Br, so Y cannot undergo E2 (it reacts only very slowly, by other paths).

Scheme: Loss of the C-1 or C-3 hydrogen gives the two enantiomers of the same alkene.

(iii) Direct and conjugate addition

Scheme: M = 2-hydroxy-2-methylbut-3-enenitrile; N = 4-methyl-4-(phenylsulfanyl)pentan-2-one

M. At 5–10°C cyanide attacks the carbonyl carbon of but-3-en-2-one: the C=O carbon carries the larger positive charge, so 1,2-addition is fastest, and at low temperature the reversible cyanohydrin does not return to the starting materials. M is the kinetic product. (On warming to about 80°C the 1,4-product, 4-oxopentanenitrile, accumulates, because conjugate addition keeps the strong C=O bond.)

N. Thiophenol is a soft nucleophile and adds reversibly; it goes to the soft β-carbon of mesityl oxide (conjugate addition), even though that carbon is tertiary. The enol formed tautomerises to the ketone: N = 4-methyl-4-(phenylsulfanyl)pentan-2-one.

Conclusion. (i) Sodium benzoate and PhCH₂OD, with no C–D bond, which proves direct hydride transfer. (ii) Only Z undergoes E2: its axial Br has two anti-periplanar axial H atoms; in Y every β-H is cis to Br. (iii) M = the cyanohydrin (kinetic 1,2-addition); N = the thiol conjugate adduct (1,4-addition).

Application. The crossed Cannizzaro step is used industrially to make pentaerythritol (alkyd resins, explosives such as PETN); thia-Michael additions are the basis of thiol–ene polymer curing and of drug binding to cysteine residues.