(a) Write the structures of the bases present in DNA and RNA. Compare the stability of DNA and RNA. (10 marks) (b) Predict the…

(a) Bases and stability. DNA contains the four bases adenine, guanine, cytosine and thymine; RNA contains adenine, guanine, cytosine and uracil. Adenine is 6-aminopurine and guanine is 2-amino-6-oxopurine; cytosine is 4-amino-2-oxopyrimidine, thymine is 5-methyluracil and uracil is 2,4-dioxopyrimidine. The amino and carbonyl groups are arranged so that complementary base pairs form: in DNA, A–T has two hydrogen bonds and G–C has three; in RNA, A–U has two and G–C has three. DNA is more stable than RNA mainly because deoxyribose lacks the 2′-OH group. The 2′-OH of ribose makes RNA susceptible to base-catalysed hydrolysis: the 2′-oxygen can attack the adjacent phosphorus to give a 2′,3′-cyclic phosphate and strand scission. DNA also usually exists as a double helix with extensive base stacking and two-strand association, which adds thermodynamic stability; RNA is often single-stranded and more chemically reactive.

(b) P, Q and R. The starting enyne is (CH3)2C=CH–CH2–C≡C–CH3. Sodium in liquid ammonia reduces the alkyne by the dissolving-metal pathway to the trans alkene, while the pre-existing isolated alkene is not reduced. Therefore P is (CH3)2C=CH–CH2–CH=CH–CH3, with the newly formed double bond having E/trans geometry. OsO4 followed by NaHSO3/H2O gives syn dihydroxylation of the C=C bonds. With both alkenes available, Q is the tetraol CH3–C(OH)(CH3)–CH(OH)–CH2–CH(OH)–CH(OH)–CH3; the two hydroxyl groups added to each former double bond are syn, and Q is formed as a mixture of stereoisomers. Periodic acid cleaves vicinal diols through cyclic periodate esters. Cleavage of the C2–C3 diol gives acetone and an aldehyde at C3; cleavage of the C5–C6 diol gives acetaldehyde and an aldehyde at C5. The central CH2 group therefore remains as a dialdehyde. Hence R is a mixture of acetone, malonaldehyde, OHC–CH2–CHO, and acetaldehyde.

(c) Photochemical products. In both reactions the excited α-keto ester abstracts a hydrogen atom from the carbon of the alcohol. This gives a ketyl radical of the ester and an α-hydroxyalkyl radical; radical–radical coupling gives a cross-coupled α,β-dihydroxy ester. In (i), ethyl phenylglyoxylate, C6H5COCO2Et, reacts with butan-2-ol, CH3CH(OH)C2H5, to give C6H5C(OH)(CO2Et)C(OH)(CH3)C2H5, named ethyl 2,3-dihydroxy-3-methyl-2-phenylpentanoate. The product contains two new stereocentres and is obtained as a mixture of diastereomers. In (ii), ethyl pyruvate, CH3COCO2Et, reacts with methanol to give CH3C(OH)(CO2Et)CH2OH, ethyl 2,3-dihydroxy-2-methylpropanoate, as a racemate. Competing Norrish-type fragmentation or ester exchange may occur, but the principal photochemical addition product is the cross-coupled dihydroxy ester.

(d) UV λmax. The individual drawings for pairs (i)–(iv) are not available in the supplied text; assigning A or B for those four pairs would be conjecture. The Woodward–Fieser principle to apply is that the compound with the longer, more planar conjugated system, greater alkyl substitution at α/β positions, stronger auxochrome donation, or an exocyclic double bond has the higher λmax, whereas steric twisting of the enone lowers the wavelength. For the supplied decalin pair (v), A is expected to have the higher λmax. Both structures are cyclic enones, but in A the enone double bond carries the methyl substituent, giving an additional alkyl/hyperconjugative increment. In B the methyl is moved to a saturated position, so the conjugated enone is less substituted. Steric inhibition of resonance would reduce the shift if severe, but the dominant Woodward–Fieser effect is the methyl-substituted enone in A, so A absorbs at longer wavelength.

(e) 1H NMR signal count and order. Anthracene (I) has three signals by its high symmetry: H2,3,6,7, the outer β protons, about 7.4 ppm; H1,4,5,8, the outer α protons, about 7.6 ppm; and H9,10, the central protons, about 8.4 ppm. The order of increasing chemical shift is H2,3,6,7 < H1,4,5,8 < H9,10. 1,3-Cyclohexadiene (II) has four signals: the two terminal vinylic protons H1/H4, about 5.6 ppm; the two internal vinylic protons H2/H3, about 5.8 ppm; and two sets of diastereotopic allylic methylene protons, H5b/H6b about 2.55 ppm and H5a/H6a about 2.7 ppm. The increasing order is H5b/H6b < H5a/H6a < H1/H4 < H2/H3. 1,4-Methano-1,3-cyclohexadiene (III, norbornadiene) has three signals: the bridge methylene protons, about 1.8 ppm; the two equivalent bridgehead protons, about 3.0 ppm; and the four equivalent vinylic protons, about 5.8 ppm. The increasing order is bridge CH2 < bridgehead CH < vinylic CH. Thus the predicted numbers of signals are 3 for I, 4 for II and 3 for III, with the chemical-shift orders given above.