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

5. (a) Bases of DNA and RNA; stability

Purines (a fused pyrimidine + imidazole ring; they bond to the sugar through N9):

• adenine (A, 6-aminopurine), in DNA and RNA;
• guanine (G, 2-amino-6-oxopurine), in DNA and RNA.

Pyrimidines (a single six-membered ring; they bond to the sugar through N1):

• cytosine (C, 4-amino-2-oxopyrimidine), in both;
• thymine (T, 5-methyluracil), in DNA only;
• uracil (U, 2,4-dioxopyrimidine), in RNA only.
• Base pairing: A=T (or A=U) is held by two hydrogen bonds; G≡C by three.

Why DNA is more stable than RNA:

• 2′-OH: ribose carries a 2′-OH. In base, or with metal ions, the 2′-O⁻ attacks the neighbouring 3′-phosphorus intramolecularly, giving a 2′,3′-cyclic phosphate and cutting the chain. 2′-Deoxyribose has no such group, so the DNA backbone is roughly 10⁵ times more resistant to hydrolysis.
• Double helix: DNA is double-stranded, with complementary H-bonding and extensive π-stacking inside the helix that shields the bases. RNA is mostly single-stranded and flexible.
• Thymine instead of uracil: cytosine can deaminate to uracil. Using T in DNA lets repair enzymes recognise any U as damage. The 5-CH₃ group also adds hydrophobic stacking stability.
• Consequence: DNA is suited to long-term storage of genetic information, whereas RNA is a short-lived messenger or catalyst.

(b) P, Q and R

P = (E)-2-methylhepta-2,5-diene, (CH₃)₂C=CH–CH₂–CH=CH–CH₃.

• Na/NH₃(l) reduces the internal alkyne by the dissolving-metal route: electron transfer, protonation, a second electron transfer, then protonation of the trans vinyl anion. This gives the trans alkene.
• The isolated trisubstituted C=C is not reduced.

Q = 2-methylheptane-2,3,5,6-tetrol, (CH₃)₂C(OH)–CH(OH)–CH₂–CH(OH)–CH(OH)–CH₃.

• OsO₄ adds syn to each C=C through a cyclic osmate ester, which NaHSO₃/H₂O cleaves.
• Each diol pair is therefore cis-added, and Q is a mixture of diastereomers.

R = acetone + malondialdehyde (OHC–CH₂–CHO) + acetaldehyde.

• HIO₄ cleaves each vicinal diol through a cyclic periodate ester.
• Cleaving C2–C3 gives acetone plus an aldehyde; cleaving C5–C6 gives an aldehyde plus acetaldehyde. The central C3–C5 fragment becomes propanedial.
• If only 1 equivalent of OsO₄ is used: the more electron-rich trisubstituted alkene is dihydroxylated first, and HIO₄ then gives acetone + (E)-pent-3-enal.

(c) Photoreduction of α-keto esters by alcohols

• Primary process: n→π* excitation, then intersystem crossing, gives the triplet keto ester. Its half-filled n orbital on O abstracts the H on the carbinol carbon of the alcohol, producing a ketyl radical and an α-hydroxyalkyl radical.

(i) C₆H₅COCO₂Et + CH₃CH(OH)C₂H₅: the radicals are PhC•(OH)CO₂Et and •C(OH)(CH₃)C₂H₅.

• Cross-coupling gives ethyl 2,3-dihydroxy-3-methyl-2-phenylpentanoate, PhC(OH)(CO₂Et)–C(OH)(CH₃)C₂H₅, as a diastereomer mixture.
• Because the alcohol radical is a hindered tertiary radical, ketyl–ketyl coupling to diethyl 2,3-diphenyltartrate (meso + dl) competes. The alcohol radical then transfers an H and becomes butan-2-one.

(ii) CH₃COCO₂Et + CH₃OH: the radicals are CH₃C•(OH)CO₂Et and •CH₂OH.

• The small •CH₂OH radical couples efficiently, giving ethyl 2,3-dihydroxy-2-methylpropanoate, CH₃C(OH)(CO₂Et)CH₂OH, as the main photoaddition product.
• Minor products are diethyl 2,3-dimethyltartrate and HCHO.
• These are intermolecular H-abstraction (photoreduction) reactions. They are not Norrish type I (α-cleavage) or type II fragmentations.

(d) Which has the higher λmax?

(i) A (4-methylacetophenone), about 252 nm.

• The para-CH₃ extends the aryl-ketone chromophore.
• In B the ortho-CH₃ forces the C=O out of the ring plane (steric inhibition of resonance), so B absorbs at a lower wavelength (about 242 nm) and less intensely.

(ii) B (1,2-dicyclohexylideneethane).

• Both are s-trans dienes with four ring residues.
• A: 214 + 4×5 = 234 nm. B: 214 + 4×5 + 2×5 (two exocyclic C=C) = 244 nm.

(iii) B, the trans (E)-penta-1,3-diene, but only marginally (about 223.5 nm against 223 nm, with higher ε).

• Both calculate as about 222 nm.
• In the cis isomer the CH₃ crowds the diene, slightly reducing planarity and intensity.

(iv) A (4-nitroaniline), about 380 nm with ε ≈ 13 500.

• Para donor (NH₂) and acceptor (NO₂) give a strong intramolecular charge-transfer band.
• In the meta isomer the groups cannot conjugate directly, so its band is weaker and lies at a shorter wavelength.

(v) B.

• A (enone C=C in the ring): 215 + 2 β-substituents (2×12) = 239 nm.
• B (ring-fusion enone): 215 + α-CH₃ (10) + 2 β ring residues (24) + 1 exocyclic C=C (5) = 254 nm.
• The isolated isopropenyl C=C is not conjugated and is ignored.

(e) ¹H NMR signals (a < b < c = increasing δ)

I, anthracene (D₂h): 3 signals.

• Ha: H-2,3,6,7 (4H), about 7.45.
• Hb: H-1,4,5,8 (4H), about 8.0.
• Hc: H-9,10 (2H), about 8.4, the most deshielded because they sit in the ring-current field of both outer rings.

II, cyclohexa-1,3-diene (C₂ᵥ): 3 signals.

• Ha: CH₂ at C-5,6 (4H), about 2.1 (allylic).
• Hb: H-1,4 (2H), about 5.8.
• Hc: H-2,3 (2H), about 5.9. Hb and Hc overlap closely as a multiplet.

III, norbornadiene (C₂ᵥ): 3 signals.

• Ha: C-7 bridge CH₂ (2H), about 2.0. The two H-7 are equivalent by the molecule's symmetry.
• Hb: bridgehead H-1,4 (2H), about 3.6 (doubly allylic).
• Hc: vinylic H-2,3,5,6 (4H), about 6.75.
• Order of increasing δ: I Ha < Hb < Hc; II Ha ≪ Hb < Hc; III Ha < Hb < Hc.