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Past exam of the chemistry course of the year 4 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 8 / a / iii by

Codex  0 2026-10-20

Past exam of the chemistry course of the year 4 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 8 / a / ii by

Codex  0 2026-10-20

Past exam of the chemistry course of the year 4 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 8 / a / i by

Codex  0 2026-10-20

Past exam of the chemistry course of the year 4 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 7 / f / Solution by

Codex  0 2026-10-20

With independent samples and finite variance, Monte Carlo integration has a root-mean-square stochastic error proportional to $N^{- 1/2}$ . The estimated standard error measures that random sampling uncertainty; it is not a deterministic bound.

For a sufficiently smooth one-dimensional integrand, the composite trapezoidal rule has error

I - T_{n} = - \frac{( b - a ) ^{3}}{12 n ^{2}} f^{''} (ξ)

for some

ξ \in (a, b)

, hence $O (n^{- 2})$ . It is normally much more efficient in one dimension. The stated continuity assumption alone guarantees convergence, but does not guarantee this second-order rate; bounded second derivatives are an additional assumption.

The dimensional advantage is different: a tensor grid with

N

total points in

d

dimensions has only about

N^{1/ d}

points per direction and typically error

O (N^{- 2/ d})

. The Monte Carlo exponent remains

- 1/2

, although its variance prefactor can still deteriorate with dimension. This explains why the stochastic method becomes attractive for high-dimensional integrals.

Past exam of the chemistry course of the year 4 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 7 / e / Solution by

Codex  0 2026-10-20

Most high-dimensional phase space has negligible equilibrium probability. Uniform sampling spends nearly all its work in these irrelevant regions and seldom visits the small subset that dominates the partition function or the numerator of an equilibrium average. The resulting estimator can have enormous variance.

Importance sampling places samples where the product of statistical weight and observable matters. Sampling the Boltzmann distribution, for example, converts a weighted phase-space integral into an equilibrium expectation. If a Markov chain Monte Carlo algorithm is used, adequate exploration of separate free-energy basins and allowance for autocorrelation remain necessary.

Past exam of the chemistry course of the year 4 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 7 / d / Solution by

Codex  0 2026-10-20

For a fixed-

N, V, T

canonical ensemble, sample the normalized Boltzmann distribution in phase space:

π (r, p) = Z^{- 1} e^{- β H (r, p)}, β = (k_{B} T)^{- 1} .

After integrating out the momenta for an ordinary separable Hamiltonian mechanics, configurational sampling uses

π (r) \propto e^{- β U (r)}

. The correct equilibrium distribution is ensemble-dependent; the Boltzmann form is the canonical answer, rather than a universal prescription for every possible ensemble.

Past exam of the chemistry course of the year 4 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 7 / c / Solution by

Codex  0 2026-10-20

The constant density is reasonable when the integrand has comparable magnitude throughout the interval. It is inefficient for a narrow peak because most samples contribute little while rare samples carry large weights.

For independent importance sampling,

Var (I_{N}) = \frac{1}{N} [\int_{a}^{b} \frac{f ( x ) ^{2}}{q ( x )} d x - I^{2}] .

A good density has support wherever

f \neq = 0

, is easy to sample and evaluate, and resembles

∣ f ∣

. Subject to normalization, $q \propto ∣ f ∣$ minimizes this variance. For nonnegative

f

, the ideal normalized choice makes every sample contribution equal, although its normalization is generally the unknown integral itself. Uniform weighting is therefore acceptable for a flat integrand, not universally suitable.

Past exam of the chemistry course of the year 4 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 7 / b / Solution by

Codex  0 2026-10-20

A weighting function must first be normalized to a probability density. The constant choice on

[a, b]

means

q (x) = \frac{1}{b - a}, I = E_{q} [\frac{f ( X )}{q ( X )}] = (b - a) E_{q} [f (X)] .

Equivalently, transform

x = a + (b - a) z

and use the already normalized

g (z) = 1

[0, 1]

. Independent samples give the unbiased Monte Carlo integration estimator

I_{N} = \frac{b - a}{N} j = 1 \sum N f (a + (b - a) U_{j}), U_{j} \sim Uniform (0, 1) .

(2)

function uniform_monte_carlo(f, a, b, N):
    require N >= 2
    total = 0
    total_squared = 0
    for j in 1,...,N:
        u = independent_uniform_0_1()
        value = f(a+(b-a)*u)
        total += value
        total_squared += value*value
    average = total/N
    sample_variance = (total_squared-N*average*average)/(N-1)
    integral = (b-a)*average
    standard_error = abs(b-a)*sqrt(sample_variance/N)
    return integral, standard_error

For a general positive normalized importance density

q

, replace each sample contribution by

f (X_{j}) / q (X_{j})

with

X_{j} \sim q

. Here the uniform choice is the special case of importance sampling that assigns equal weight to every region of equal length.

Past exam of the chemistry course of the year 4 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 7 / a / Solution by

Codex  0 2026-10-20

Both rules use a uniform partition with

Δ x = (b - a) / n

. An original implementation in pseudocode is:

function trapezoid(f, a, b, n):
    dx = (b-a)/n
    total = (f(a)+f(b))/2
    for j in 1,...,n-1:
        total += f(a+j*dx)
    return dx*total

function midpoint(f, a, b, n):
    dx = (b-a)/n
    total = 0
    for j in 0,...,n-1:
        total += f(a+(j+1/2)*dx)
    return dx*total

The trapezoidal rule gives each endpoint half weight; the composite midpoint rule uses one interior point per subinterval. Both converge for a continuous integrand on the closed interval.

The second equality in the printed trapezoidal hint is missing a factor $1/2$ . The average of the left and right rectangle sums is required. For

f = 1

, the uncorrected equality would return

2 (b - a)

instead of

b - a

Past exam of the chemistry course of the year 4 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 6 / c / Solution by

Codex  0 2026-10-20

The key sequence is deprotonation, anionic eight-pi electrocyclization, then elimination. Label the conjugated route from the lower cyclopentene to the right cyclopentene as

C_{1} = C_{2} - C_{3} - C_{4} = C_{5} - C_{6} = C_{7},

where

C_{3}

is the methylene joining the lower ring to the central alkene, and

C_{4}

carries the

C H_{2} OC (O) NM e_{2}

substituent.

potassium hexamethyldisilazide removes a proton at

C_{3}

. The resulting anion is delocalized over the seven-carbon array: its six pi-bond electrons plus the lone pair make an eight-electron system. Thermal electrocyclic reaction is disrotatory for this

4 n

electron count. Joining

C_{1}

and

C_{7}

forms the seven-membered central ring and puts the remaining anionic character at

C_{4}

, with double bonds at

C_{2} = C_{3}

and

C_{5} = C_{6}

That

C_{4}

anion then forms

C_{4} = C H_{2}

, expelling the O-dimethylcarbamate group from the attached methylene. This E1cb elimination supplies the exocyclic methylene and explains why no oxygen remains in the depicted product. The disrotatory closure, with the cyclopentene ends folded to avoid unfavorable overlap, gives the trans relationship between the two lower fused-ring bonds shown by the wedge and dash. A simple [3,3] rearrangement would not create the required bond between the two remote cyclopentene termini.

Past exam of the chemistry course of the year 4 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 6 / b / Solution by

Codex  0 2026-10-20

A is the kinetic, less-hindered-side silyl enol ether. Its double bond lies away from the methyl substituent. Forming B requires removal of a proton from the more hindered

α

position adjacent to the methyl-bearing carbon.

The base strengths alone do not determine the ratios. trimethylsilyl triflate is strongly electrophilic: oxygen silylation activates the ketone towards deprotonation, and irreversible trapping of the resulting silyl enol ether can make the product ratio reflect competing deprotonation rates. A weak amine therefore need not first produce a substantial equilibrium concentration of the free ketone enolate.

DBU is the strongest and comparatively accessible base, and distinguishes the two sites least strongly, giving the smallest A:B ratio. Triethylamine gives greater preference for the less-hindered site. The more sterically hindered diisopropylethylamine discriminates most strongly and gives the largest proportion of A. Its slightly higher conjugate-acid

p K_{a}

than that of triethylamine does not reverse this trend, because steric accessibility and the coupled silylation/deprotonation mechanism matter alongside basicity. The data do not establish a simple thermodynamic enolate distribution.

Past exam of the chemistry course of the year 4 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 6 / a / Solution by

Codex  0 2026-10-20

This is an Ireland–Claisen rearrangement. Number the ester carbonyl carbon

C_{0}

, its enolizable carbon

C_{α}

, and the three allylic ring carbons

C_{1} - C_{2} = C_{3}

, starting at the carbon bearing the ester oxygen.

lithium hexamethyldisilazide removes an

α

proton and trimethylsilyl chloride traps the oxygen of the resulting ester enolate, giving a silyl ketene acetal. Heating permits a six-electron, [3,3]-sigmatropic rearrangement through the array

C_{α} = C_{0} - O - C_{1} - C_{2} = C_{3} .

The electron-pair movements are

π (C_{α} = C_{0}) \to C_{0} = O

σ (O - C_{1}) \to π (C_{1} = C_{2})

, and

π (C_{2} = C_{3}) \to σ (C_{3} - C_{α})

. Acidic work-up cleaves the silyl ester to the carboxylic acid. The benzyl ether remains a protecting group; it is not the migrating bond.

The suprafacial reaction transfers the starting allylic stereochemistry to

C_{3}

, where the new side chain is delivered from the original oxygen face. In the product's printed orientation, the new C $_{3}$ –C $_{α}$ bond is behind the ring and the C $_{3}$ methyl group is toward the viewer. At that ring star, the Cahn-Ingold-Prelog priority rules orders the groups as side chain, alkenyl ring direction, saturated ring direction, methyl, giving $R$ .

The second star is

C_{α}

. Its priorities are carboxyl, substituted ring carbon,

C H_{2} C H_{2} OBn

, H. A chair-like transition state normally places the bulky side chain pseudoequatorial. However, its absolute configuration also depends on the E/Z geometry of the trapped silyl ketene acetal: opposite geometries give opposite configurations at this star under the same suprafacial ring transfer. The printed conditions do not state solvent or the ketene-acetal geometry. Thus the secure assignment is the ring-star

R

configuration and conditional

α

-star stereochemistry, rather than an unqualified absolute assignment to both stars. Drawing the assumed ketene-acetal geometry and its chair is necessary before selecting one

α

epimer.

Past exam of the chemistry course of the year 4 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 5 / f / ii by

Codex  0 2026-10-20

Past exam of the chemistry course of the year 4 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 5 / f / i by

Codex  0 2026-10-20

Past exam of the chemistry course of the year 4 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 5 / e / ii by

Codex  0 2026-10-20

Past exam of the chemistry course of the year 4 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 5 / e / i by

Codex  0 2026-10-20

Past exam of the chemistry course of the year 4 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 5 / d / ii by

Codex  0 2026-10-20

Past exam of the chemistry course of the year 4 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 5 / d / i by

Codex  0 2026-10-20