(a) Write the structure of the product(s) and the intermediate formed in the following reaction: Cl₂CH—COCl 1) Et₃N → ? 2)…

(a)

• Cl₂CH–C(=O)Cl has an acidic α-H because the conjugate base is stabilised by two Cl atoms and by the carbonyl group.
• Et₃N removes the α-proton: Cl₂CH–C(=O)Cl + Et₃N → [Cl₂C⁻–C(=O)Cl] + Et₃NH⁺.
• The enolate is represented by the resonance form: [Cl₂C⁻–C(=O)Cl] ↔ [Cl₂C=C(O⁻)–Cl].
• Collapse of the enolate reforms C=O and expels Cl⁻: [Cl₂C=C(O⁻)–Cl] → Cl₂C=C=O + Cl⁻.
• Overall step 1: Cl₂CH–COCl + Et₃N → Cl₂C=C=O + Et₃NH⁺Cl⁻. The intermediate that enters step 2 is dichloroketene, Cl₂C=C=O.
• In step 2, dichloroketene does not act as a Diels–Alder dienophile; it undergoes a thermal [2+2] cycloaddition ([π2s + π2a], allowed for ketenes) with one C=C bond of cyclopentadiene. The ketene C=C adds across that double bond to form a cyclobutanone; the C=O bond remains as a ketone carbonyl.
• The product is the cis-fused cyclobutanone in which the carbonyl carbon and the CCl₂ carbon form part of the four-membered ring. In IUPAC form: 7,7-dichlorobicyclo[3.2.0]hept-2-en-6-one. It has a bicyclo[3.2.0]hept-2-ene skeleton, a ketone at C-6, and two chlorine atoms on C-7.

(b)

The product is the allylic methyl ether (Z)-5-methoxy-2,2-dimethylhex-3-ene, (CH₃)₃C–CH=CH–CH(OMe)CH₃, and the four steps are read backwards as follows:

• Step 4 (NaOMe/MeOH) is an allylic substitution: methoxide displaces bromide from the allylic bromide by SN2, installing the OMe group without changing the carbon skeleton.
• Step 3 (NBS, ROOR, Δ) is a radical allylic bromination. The only allylic C–H available is on the ethyl side of the double bond (the other allylic carbon is the quaternary t-Bu carbon, which bears no H), so the product is (CH₃)₃C–CH=CH–CHBr–CH₃.
• Step 2 (Lindlar's catalyst, H₂) is a syn hydrogenation of the alkyne to the (Z)-alkene, (Z)-(CH₃)₃C–CH=CH–CH₂CH₃.
• Step 1 (NaNH₂, EtI) alkylates a terminal alkyne: deprotonation by NaNH₂ gives the acetylide, which undergoes SN2 with ethyl iodide. Therefore the alkyne was (CH₃)₃C–C≡C–CH₂CH₃, formed from the starting material A = 3,3-dimethyl-1-butyne (tert-butylacetylene), (CH₃)₃C–C≡CH.

Summary: A = (CH₃)₃C–C≡CH; intermediates: (i) (CH₃)₃C–C≡C–CH₂CH₃ (5,5-dimethylhex-2-yne); (ii) (Z)-(CH₃)₃C–CH=CH–CH₂CH₃ (5,5-dimethylhex-2-ene); (iii) (CH₃)₃C–CH=CH–CHBr–CH₃ (4-bromo-2,2-dimethylhex-3-ene); product (Z)-5-methoxy-2,2-dimethylhex-3-ene (the (Z)-geometry from the Lindlar step is retained through steps 3–4).

(c)

• Photon energy is E = hc/λ. Using h = 6.626×10⁻³⁴ J s and c = 2.998×10⁸ m s⁻¹: E(290 nm) = (6.626×10⁻³⁴ × 2.998×10⁸)/(290×10⁻⁹) = 6.85×10⁻¹⁹ J = 4.27 eV. E(450 nm) = (6.626×10⁻³⁴ × 2.998×10⁸)/(450×10⁻⁹) = 4.42×10⁻¹⁹ J = 2.76 eV.
• Jablonski diagram: put energy on the vertical axis. Place S₀ at the bottom. Place T₁ above S₀ at the energy corresponding to 450 nm. Place S₁ above T₁ at the energy corresponding to 290 nm. Draw a solid vertical arrow from S₀ to S₁ labelled 290 nm; this is the spin-allowed singlet–singlet absorption. Draw a dashed vertical arrow from S₀ to T₁ labelled 450 nm; this is the spin-forbidden singlet–triplet absorption. From S₁ draw short downward wavy arrows for vibrational relaxation, a straight downward arrow to S₀ for fluorescence, and a dashed arrow to T₁ for intersystem crossing. From T₁ draw an arrow to products for the photochemical reaction and a straight downward arrow to S₀ for phosphorescence.
• Answer: use 290 nm light. The 450 nm photon matches the S₀–T₁ gap, but direct S₀→T₁ absorption is spin-forbidden and therefore very weak. Efficient population of T₁ is achieved instead by the strong, spin-allowed S₀→S₁ excitation at 290 nm, followed by vibrational relaxation and fast intersystem crossing (ISC) to T₁, the state from which the reaction occurs. The excess energy (4.27 − 2.76 = 1.51 eV per photon) is dissipated as vibrational energy before reaction.

(d)(i)

• Pure rotational microwave spectroscopy requires a permanent electric dipole moment, μ ≠ 0, so that rotation changes the dipole and couples to the radiation field. The rotational selection rule is ΔJ = ±1.
• CH₄ is tetrahedral (T_d). The four C–H bond dipoles cancel exactly; μ = 0; inactive.
• H₂O is bent (C₂v). The O–H bond dipoles do not cancel; μ ≠ 0; active.
• NH₃ is pyramidal (C₃v). The N–H bond dipoles give a net dipole along the C₃ axis; μ ≠ 0; active.
• BCl₃ is trigonal planar (D₃h). The three B–Cl bond dipoles cancel in the plane; μ = 0; inactive.
• XeF₄ is square planar (D₄h). The four Xe–F bond dipoles cancel; μ = 0; inactive.
• Active molecules: H₂O and NH₃.

(d)(ii)

• For a rigid diatomic rotor, the rotational term in frequency units is F(J) = B J(J+1), with B = h/(8π²I). The transition J→J+1 has frequency ν = 2B(J+1), so the spacing between adjacent lines is 2B.
• The moment of inertia is I = μ r_e², where μ is the reduced mass. Isotopic substitution changes μ but, to an excellent approximation, leaves r_e unchanged.
• For ¹²C¹⁶O, using integer mass numbers: μ₁₂ = (12×16)/(12+16) = 192/28 = 48/7 u.
• For ¹³C¹⁶O: μ₁₃ = (13×16)/(13+16) = 208/29 u.
• Since B ∝ 1/I ∝ 1/μ, the line spacing is also proportional to 1/μ. Therefore: spacing(¹³CO)/spacing(¹²CO) = μ₁₂/μ₁₃ = (48/7)/(208/29) = 87/91 ≈ 0.956.
• Thus the spacing in ¹³CO is smaller by the factor 87/91, about 4.4%, because the heavier carbon increases the reduced mass and the moment of inertia, decreasing the rotational constant. The same ratio holds whether the spacing is reported in Hz or cm⁻¹.

(e)(i)

• Chemical shift in ppm is field independent: δ = (Δν/ν₀)×10⁶, where ν₀ is the spectrometer frequency and Δν is the offset from the reference in Hz.
• At 100 MHz, 1 ppm corresponds to 100 Hz. For δ = 3.0 ppm: Δν = 3.0×100 = 300 Hz.
• At 500 MHz, 1 ppm corresponds to 500 Hz. For the same δ = 3.0 ppm: Δν = 3.0×500 = 1500 Hz.
• The scalar coupling constant J is a property of the spin–spin interaction and, for first-order spectra, does not depend on the external field. Therefore J remains 4.5 Hz.
• Chemical shift in Hz: 1500 Hz; J: 4.5 Hz.

(e)(ii)

• n-Butyl phenyl ketone is C₆H₅COCH₂CH₂CH₂CH₃, formula C₁₁H₁₄O. Its nominal molecular mass is 11×12 + 14×1 + 16 = 162, so m/z 162 is the molecular ion [M]⁺•.
• m/z 120: 162 − 42 = 120. The loss of 42 u is C₃H₆. This is a McLafferty rearrangement: a γ-hydrogen from the butyl chain transfers to the carbonyl oxygen, and the Cα–Cβ bond cleaves, expelling propene. The charged fragment is [C₈H₈O]⁺•, represented as the enol radical cation PhC(OH)=CH₂⁺•, the acetophenone radical cation tautomer.
• m/z 105: 162 − 57 = 105. The loss of 57 u is an n-butyl radical, C₄H₉•. α-Cleavage of the carbonyl–alkyl bond gives the resonance-stabilised benzoyl cation [C₆H₅CO]⁺, C₇H₅O⁺. This ion may further lose CO to give m/z 77, but that fragment is not among the listed peaks.
• m/z 85: 162 − 77 = 85. The loss of 77 u is a phenyl radical, C₆H₅•. α-Cleavage of the carbonyl–phenyl bond gives the n-pentanoyl (valeryl) cation [CH₃CH₂CH₂CH₂CO]⁺, C₅H₉O⁺, an acylium ion stabilised as R–C≡O⁺ ↔ R–C=O⁺.
• Fragmentation: 162 = [M]⁺•; 120 = McLafferty loss of C₃H₆ to [C₈H₈O]⁺•; 105 = α-cleavage loss of C₄H₉• to [C₆H₅CO]⁺; 85 = α-cleavage loss of C₆H₅• to [CH₃CH₂CH₂CH₂CO]⁺.