1.(a)(i) pKₐ value of cyclopentadiene is almost similar to water. Explain. 5 marks 1.(a)(ii) Rate of hydrogen exchange reaction…

(a)(i) pKₐ of cyclopentadiene close to that of water

Cyclopentadiene (pKₐ ≈ 16) is about as acidic as water (15.7) although it is only a hydrocarbon. Loss of H⁺ from its sp³ CH₂ gives the cyclopentadienyl anion, which is planar, cyclic and fully conjugated with 6π electrons (4n+2, n = 1), so it is aromatic. This aromatic stabilisation of the conjugate base makes ionisation easy. Neutral cyclopentadiene itself is not aromatic, because the sp³ carbon breaks the ring conjugation.

(a)(ii) H–D exchange about 6000 times faster in A than in B

Exchange proceeds through the carbanion (enolate) formed at the ring carbon α to the benzoyl group. A is cyclopropyl phenyl ketone and B is cyclopropenyl phenyl ketone, as read from the drawing.

In A the carbanion lies on a cyclopropane carbon and is stabilised by the C=O group. In B the same carbanion lies in the planar cyclopropene ring and forms a cyclopropenyl anion with 4π electrons. This is antiaromatic (Breslow), so it is strongly destabilised and forms far more slowly. A therefore exchanges about 6000 times faster. By contrast the cyclopropenyl cation (2π) is aromatic, so the electron count, not the three-membered ring, decides the stability.

(b)(i) IUPAC name with stereodescriptors

The drawn compound has a five-carbon chain: CHO (C1), C2 bearing OH, C3 bearing Cl, and a terminal vinyl group (C4=C5). Its name is (2R,3R)-3-chloro-2-hydroxypent-4-enal.

CIP priorities: at C2, OH > C3 (Cl-bearing) > CHO > H; at C3, Cl > C2 (O-bearing) > C4 (vinyl) > H. In the drawing the H on each centre points toward the viewer, so the sense seen on paper is reversed. Both centres then come out R.

(b)(ii) Order of dimerisation tendency of the radicals

A is Ph₃C•, B is tris(4-fluorophenyl)methyl and C is Ph₂C•–tBu. The order of increasing tendency to dimerise is B < A < C.

Dimerisation of trityl radicals occurs at the para position of one ring (Gomberg-type quinoid dimer), not at the crowded central carbon. In B the para positions carry fluorine, which blocks this path, so B is the most persistent. A has three aryl rings and open para positions, so it dimerises reversibly. C has only two aryl rings for delocalisation and a small, reactive centre, so it dimerises most readily. The relative size of the steric and delocalisation effects is an interpretation, not a measured fact.

(c) Solvent effect on MeI + NaN₃

The reaction is an SN2 substitution, CH₃I + N₃⁻ → CH₃N₃ + I⁻. In DMF, a polar aprotic solvent, the Na⁺ ion is strongly solvated through the oxygen of DMF, but the azide anion is not, leaving a "naked", highly nucleophilic N₃⁻. In methanol, a protic solvent, N₃⁻ is hydrogen-bonded and caged by solvent molecules, so its energy is lowered and its nucleophilicity falls. More energy is needed to strip this shell in the transition state, so the activation energy is higher and the reaction is slower. DMF therefore gives the much faster reaction.

(d) Photochemical cyclisation of the cyclodeca-1,3-dienes

A and B are ten-membered ring conjugated dienes; A has one trans double bond (E,Z) and B has both cis (Z,Z), as read from the drawing. Under light, a 4π system (butadiene unit) undergoes a disrotatory electrocyclic ring closure, because the photochemical 4n case is disrotatory (thermal is conrotatory).

Disrotatory closure of the Z,Z diene B gives a cis-fused bicyclo[6.2.0]dec-9-ene (cyclobutene fused to the ring). The E,Z diene A, closing disrotatorily, gives the trans-fused bicyclo[6.2.0]dec-9-ene. The ring-fusion stereochemistry in each product follows directly from the geometry of the starting diene.

(e)(i) 2-Chloro-6-methylcyclohexanone with MeO⁻

This is a Favorskii rearrangement. Methoxide removes the α′-hydrogen at C6 (the side without chlorine, bearing the methyl). The enolate displaces Cl⁻ from C2 to form a bicyclic cyclopropanone. MeO⁻ attacks the C=O and the three-membered ring opens at the bond that leaves the more stable carbanion, which is the secondary carbon (former C2), not the tertiary one (former C6). The ring has contracted, and the ester group stays on the methyl-bearing carbon. The product is methyl 1-methylcyclopentanecarboxylate (A).

(e)(ii) Reaction that evolves N₂

The answer is (B) von Richter reaction. A nitroarene heated with cyanide (KCN, aqueous ethanol) gives the carboxylic acid at the position meta to the original nitro group (cine substitution), and the nitro nitrogen is lost as N₂. The Fischer indole synthesis loses NH₃, and the Stobbe and Bischler–Napieralski reactions do not release N₂.