(a) The following molecule shows the rigid or fluxional behaviour at higher temperature or in the presence of a base. Justify the…

(a)

The molecule printed with the question is chlorido(η³-allyl)(triphenylphosphane)palladium(II), [Pd(η³-C₃H₅)(Cl)(PPh₃)] — a π-allyl palladium(II) complex. Palladium is square-planar; the η³-allyl group occupies two coordination sites, while the remaining two sites carry Cl⁻ and PPh₃. The allyl fragment has two terminal CH₂ groups (C1, C3 — bearing H_A/H_A′ and H_B/H_B′) and a central CH (C2, H_x). Terminal protons pointing towards the central H are the syn protons (H_A, H_A′); those pointing away are the anti protons (H_B, H_B′). A labelled structure is shown below.

¹H NMR evidence. At low temperature the complex is rigid on the NMR timescale: syn and anti protons are inequivalent, so distinct signals appear — H_x (central, 1H) at ≈5.5–6.0 δ, the syn protons H_A/H_A′ (2H) downfield at ≈3.5–4.5 δ, and the anti protons H_B/H_B′ (2H) upfield at ≈2.0–3.0 δ, with characteristic couplings to H_x (J(anti–H_x) ≈ 12–14 Hz, larger than J(syn–H_x) ≈ 6–7 Hz). Because one allyl terminus is trans to PPh₃ and the other trans to Cl, the two ends are themselves slightly inequivalent, and the terminal protons may also show ³¹P coupling. At higher temperature or in the presence of a base, the complex becomes fluxional: the η³-allyl undergoes rapid η³ ⇌ η¹ interconversion (slippage to an η¹ intermediate, rotation about the remaining C–C bond, and re-coordination), which exchanges the syn and anti environments (apparent allyl rotation). When this syn–anti exchange becomes fast on the NMR timescale, the separate syn and anti resonances broaden, coalesce and average (signal averaging), so all four terminal allyl protons ultimately appear equivalent. An added base or donor ligand (e.g., an amine, pyridine or excess PPh₃) attacks palladium to give a five-coordinate species that slips to the σ-bonded η¹-allyl, so it accelerates the exchange even at lower temperature. The temperature- and base-dependent ¹H NMR pattern (splitting when rigid; averaging when fluxional) is therefore direct evidence of the rigid ⇌ fluxional behaviour of the π-allyl complex.

(b)

The photochemical reaction is a radical chain process. Initiation is Br₂ + hν → 2Br•; if I_a is the rate of photon absorption, the primary production rate of Br atoms is 2I_a. Propagation consists of Br• + H₂ → HBr + H• (k₂) and H• + Br₂ → HBr + Br• (k₃). The product HBr can also react with Br•, H• + HBr → H₂ + Br• (k₋₁), which inhibits the chain. Termination is 2Br• + M → Br₂ + M (k₄); because [M] is the constant concentration of the third body, the effective termination constant is k₄[M].

Steady-state for H• gives d[H•]/dt = k₂[Br•][H₂] − k₃[H•][Br₂] − k₋₁[H•][HBr] = 0, hence [H•] = k₂[Br•][H₂]/(k₃[Br₂] + k₋₁[HBr]). The net HBr formation rate is d[HBr]/dt = k₂[Br•][H₂] + k₃[H•][Br₂] − k₋₁[H•][HBr] = 2k₂k₃[Br•][H₂][Br₂]/(k₃[Br₂] + k₋₁[HBr]). For Br•, d[Br•]/dt = 2I_a − k₂[Br•][H₂] + k₃[H•][Br₂] + k₋₁[H•][HBr] − 2k₄[M][Br•]² = 0. Adding the H• steady-state equation gives 2I_a = 2k₄[M][Br•]², so [Br•] = (I_a/k₄[M])^1/2 exactly; when HBr is small the rate simplifies. Therefore d[HBr]/dt ≈ 2k₂[H₂](I_a/k₄[M])^1/2, showing the required square-root dependence on absorbed intensity. The square-root law is therefore a direct consequence of second-order radical termination competing with first-order photon production. With appreciable HBr, the rate is reduced by the factor 1/(1 + k₋₁[HBr]/(k₃[Br₂])).

Each absorbed photon dissociates one Br₂ molecule into two Br atoms, but the observed quantum yield of the H₂–Br₂ reaction at ordinary temperature is very low, only about 0.01, and it decreases further as HBr accumulates. Thus most Br atoms recombine without completing a chain cycle. It is low compared with very long chain reactions because many Br atoms recombine in the encounter pair or gas cage, terminate by 2Br• + M, while H atoms are consumed by the back reaction with HBr; above all, Br• + H₂ → HBr + H• is strongly endothermic (about +70 kJ mol⁻¹) and very slow at ordinary temperature, so the chain hardly propagates.

(c)

The Langmuir isotherm rests on five assumptions: adsorption is confined to a monolayer; the surface contains a fixed number of equivalent sites; adsorbed molecules do not interact laterally; adsorption and desorption are elementary steps that reach dynamic equilibrium; and the enthalpy of adsorption is independent of coverage. Let θ be the fractional coverage and P the gas pressure. The rate of adsorption is proportional to the pressure and the number of vacant sites, k_a P(1−θ); the rate of desorption is proportional to occupied sites, k_d θ. At equilibrium, k_a P(1−θ) = k_d θ. Rearranging gives θ = KP/(1+KP), where K = k_a/k_d. Equivalently, P/θ = 1/K + P, which is linear in P. The linear form is useful experimentally because a plot of P/θ against P gives a straight line whose intercept gives 1/K.

At low pressure, KP is much less than 1, so θ ≈ KP and the amount adsorbed is directly proportional to pressure; this is the Henry-law region where the Langmuir model is most reliable. As pressure rises, KP becomes large, θ approaches unity, and the surface approaches monolayer saturation. At high pressure the simple Langmuir form often fails because real adsorption is not limited to one ideal layer: surfaces are heterogeneous, adsorbate-adsorbate interactions become important, and multilayer adsorption can occur. In such cases the amount adsorbed may continue to increase beyond monolayer saturation, or the apparent K may vary with coverage. BET, Freundlich, or Temkin isotherms then provide better descriptions. Thus Langmuir adsorption correctly explains the initial linear uptake and monolayer saturation, but it cannot account for the non-ideal, multilayer behaviour that dominates at high pressure.