Q7(c). SeO₂/H₂O₂, Pb(OAc)₄, OsO₄, periodate
Write the products (X) and (Y) of the following reactions:
(i) [structure] → (H₂O₂, SeO₂) (X)(ii) [structure] → (Pb(OAc)₄) (X)
(iii) [structure] → (OsO₄, Aq. Na₂SO₃) (X)(iv) [structure] → ((1) OsO₄, (2) NaSO₃H) (X) → (HIO₃) (Y)
Comment on the stereochemistry of product (X) in each case.
Answer in brief
(i) (±)-Tartaric acid (anti addition, racemic); (ii) cis,cis-mucononitrile (both C=C Z); (iii) cis-cyclopentane-1,2-diol (syn, meso); (iv) X = cis-3-methyl-3-phenylcyclopropane-1,2-diol (syn, from the face anti to Ph; achiral), Y = 2-methyl-2-phenylpropanedial.
Key points
- OsO₄ adds syn to the strained, reactive cyclopropene C=C from the face anti to the larger phenyl group (the methyl face is less hindered). Bisulfite cleaves the osmate. X = cis-3-methyl-3-phenylcyclopropane-1,2-diol, both OH…
- periodic acid forms a cyclic periodate ester with the cis-diol and cleaves the C1–C2 bond; both carbinol carbons become CHO. Y = 2-methyl-2-phenylpropanedial, CH₃C(Ph)(CHO)₂. The stereocentres of X disappear.
The full model answer
Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.
Introduction. (i) and (iii) are two ways of adding two OH groups to a C=C, anti (via an epoxide) and syn (via an osmate ester); (ii) is an oxidative ring cleavage that keeps the cis geometry of the ring bonds; (iv) is a syn dihydroxylation followed by glycol cleavage.
Reading of the question: “NaSO₃H” is sodium hydrogen sulfite (NaHSO₃); “HIO₃” is read as periodic acid, HIO₄, the reagent that cleaves 1,2-diols.
(i) Maleic acid with H₂O₂/SeO₂
Scheme: Syn epoxidation and anti ring opening: net anti dihydroxylation.
SeO₂ and H₂O₂ form peroxyselenous acid, an electrophilic O-transfer agent like a peracid. It epoxidises the C=C (syn); in the aqueous acidic medium water opens the epoxide from the back side (anti). Net anti addition of two OH to a cis alkene gives the racemic (±)-tartaric acid, (2R,3R) + (2S,3S). (Syn reagents, OsO₄ or cold KMnO₄, give meso-tartaric acid from maleic acid.)
(ii) o-Phenylenediamine with Pb(OAc)₄
Scheme: Oxidative cleavage of the ring between the two N-bearing carbons.
Lead tetraacetate oxidises both amino groups (through the quinone di-imine and a dinitrene-like stage), and the C1–C2 bond between them breaks: each former C–N becomes a C#N. The product is cis,cis-mucononitrile. The two C=C bonds were ring bonds, which are necessarily cis; the cleavage does not touch them, so both double bonds are Z. (On heating or with iodine it isomerises to the cis,trans isomer.)
(iii) Cyclopentene with OsO₄, then aqueous Na₂SO₃
Scheme: Both C–O bonds form on the same face in one concerted step.
OsO₄ adds to the C=C in a concerted [3+2] cycloaddition, giving a cyclic osmate(VI) ester; sulfite reduces osmium and cleaves the Os–O bonds without touching the C–O bonds. Both OH are on the same face: cis-cyclopentane-1,2-diol, an achiral meso compound (1R,2S).
(iv) 3-Methyl-3-phenylcyclopropene: OsO₄, then periodate
Scheme: The diol carbons become two CHO groups on the quaternary carbon.
- X: OsO₄ adds syn to the strained, reactive cyclopropene C=C from the face anti to the larger phenyl group (the methyl face is less hindered). Bisulfite cleaves the osmate. X = cis-3-methyl-3-phenylcyclopropane-1,2-diol, both OH cis to CH₃ and trans to Ph. A mirror plane passes through C-3, Me and Ph, so X is achiral (meso); C-3 is pseudo-asymmetric. The minor diastereomer has both OH cis to Ph.
- Y: periodic acid forms a cyclic periodate ester with the cis-diol and cleaves the C1–C2 bond; both carbinol carbons become CHO. Y = 2-methyl-2-phenylpropanedial, CH₃C(Ph)(CHO)₂. The stereocentres of X disappear.
Conclusion. (i) (±)-Tartaric acid (anti addition, racemic); (ii) cis,cis-mucononitrile (both C=C Z); (iii) cis-cyclopentane-1,2-diol (syn, meso); (iv) X = cis-3-methyl-3-phenylcyclopropane-1,2-diol (syn, from the face anti to Ph; achiral), Y = 2-methyl-2-phenylpropanedial.
Application. Syn (OsO₄) versus anti (peracid, then hydrolysis) dihydroxylation is the standard way to choose the relative configuration of 1,2-diols, as in the synthesis of meso- and (±)-tartaric acids and of sugar-like polyols.
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All parts of Question 7
- 7(a) Structure of C₈H₄O₃ from IR, NMR and mass spectrum: phthalic anhydride15 marks
- 7(b) Borane: acid reduced, ester kept; δ-lactone15 marks
- 7(c) SeO₂/H₂O₂, Pb(OAc)₄, OsO₄, periodate20 marks