The first transformation is two successive dehydrohalogenations; the second opens the lactone by two methyl additions and then closes a six-membered cyclic ether. Both schemes must be accounted for within this part.

For the dichloride, write $R=\mathrm{cyclohexyl}$. In the product-forming path, tert-butoxide removes a hydrogen from the methyl group of $\mathrm{R}-\mathrm{CC}{\mathrm{l}}_2-\mathrm{C}{\mathrm{H}}_3$. Draw arrows from the base lone pair to H, from C–H into the C–C bond, and from C–Cl to Cl. This E2 elimination gives $\mathrm{R}-\mathrm{C}(\mathrm{Cl})=\mathrm{C}{\mathrm{H}}_2$. A second base-induced elimination removes a vinyl hydrogen and the remaining chloride, forming $\mathrm{R}-\mathrm{C}\equiv \mathrm{CH}$. The second elimination is more demanding because a vinylic bond is involved, consistent with the strong-base, heated conditions. This route explains the displayed product without asserting that no other elimination regioisomer can ever form.

For the lactone, a methyl equivalent from methylmagnesium iodide attacks the ester carbonyl carbon, with its pi pair moving to oxygen. Collapse of the tetrahedral intermediate reforms C=O and cleaves the acyl–oxygen bond, opening the ring to the magnesium alkoxide corresponding to $\mathrm{HO}(\mathrm{C}{\mathrm{H}}_2{)}_4\mathrm{C}(=\mathrm{O})\mathrm{C}{\mathrm{H}}_3$. The newly formed ketone is more susceptible to a second methyl addition; its pi pair again moves to oxygen. After water workup, the product is the diol $\mathrm{HO}(\mathrm{C}{\mathrm{H}}_2{)}_4\mathrm{C}(\mathrm{OH})(\mathrm{C}{\mathrm{H}}_3{)}_2$.

Under acid and heat, protonate its tertiary OH, then move the C–O pair onto oxygen as water leaves. The resulting $\mathrm{HO}(\mathrm{C}{\mathrm{H}}_2{)}_4{\mathrm{C}}^{+}(\mathrm{C}{\mathrm{H}}_3{)}_2$ is trapped intramolecularly by the remaining primary-alcohol oxygen. Its lone pair forms the closing O–C bond, producing a six-membered cyclic oxonium ion. Deprotonation yields 2,2-dimethyloxane, also called 2,2-dimethyltetrahydropyran. Ring closure retains four chain methylenes; it does not form a five-membered ring. The two methyl groups both derive from the two equivalents of Grignard reagent.

The most acidic sites are: acetic-acid O–H, $\mathrm{p}{K}_a\approx 4.8$; phenol O–H, $\approx 10$; ethanal alpha-CH${}_3$, $\approx 17$–$20$; propyne terminal ≡C–H, $\approx 25$. These are ordinary approximate organic-chemistry estimates; exact acidities depend on solvent and reference convention.

Acetic acid gives a carboxylate in which negative charge is delocalized over two oxygens through equivalent resonance structures. Phenol gives a phenoxide with delocalization into the aromatic system, but its important contributors are less equivalent and do not stabilize the charge as effectively as a carboxylate.

For ethanal, removal of a methyl hydrogen gives the enolate ${}^{-}\mathrm{C}{\mathrm{H}}_2-\mathrm{CH}=\mathrm{O}\leftrightarrow \mathrm{C}{\mathrm{H}}_2=\mathrm{CH}-{\mathrm{O}}^{-}$. The aldehydic hydrogen is not the preferred acidic site: removing it does not give this alpha-carbon/oxygen delocalization. For propyne, loss of the terminal hydrogen gives an acetylide with a lone pair in an $sp$ orbital. Its greater s character stabilizes negative charge relative to an ordinary $s{p}^2$ or $s{p}^3$ carbanion; the methyl hydrogens are much less acidic.

The two carbon skeletons are allene, $\mathrm{C}{\mathrm{H}}_2=\mathrm{C}=\mathrm{C}{\mathrm{H}}_2$, and propyne, $\mathrm{C}{\mathrm{H}}_3-\mathrm{C}\equiv \mathrm{CH}$. “Linear” describes the three-carbon skeleton; it does not mean every hydrogen lies on the same line.

For allene, use the contributor ${}^{2-}\mathrm{C}=\mathrm{C}={\mathrm{C}}^{2-}$. A terminal lone pair attacks a water hydrogen while the O–H pair returns to oxygen, giving $\mathrm{H}{\mathrm{C}}^{-}=\mathrm{C}={\mathrm{C}}^{2-}$ and hydroxide. Repeat at the still-dianionic end, then protonate each remaining terminal carbanion. The four successive proton transfers givewith two hydrogens placed on each terminal carbon in this branch.

For propyne, the resonance contributor ${}^{3-}\mathrm{C}-\mathrm{C}\equiv {\mathrm{C}}^{-}$ makes the alternative electron flow clear. Three successive protonations of the singly bonded terminal carbon convert its three lone pairs into C–H bonds; protonation of the acetylide end gives $\mathrm{C}{\mathrm{H}}_3-\mathrm{C}\equiv \mathrm{CH}$. These are formal protonation pathways for the multiply charged anion, not a claim that the highly charged resonance contributors are separately isolable species. Each transfer produces $\mathrm{O}{\mathrm{H}}^{-}$; overallAllene has one proton resonance integrating to 4H, ideally a singlet because its four hydrogens are chemically equivalent. Proton-decoupled carbon NMR has two resonances, in ratio two terminal carbons to one central carbon: approximately $75$ and $210$ ppm, respectively. The unusually downfield central cumulene carbon is not a carbonyl.

Propyne has two proton resonances, integrating 3H near $1.8$–$2.0$ ppm and 1H near $2$–$3$ ppm. If the small four-bond coupling is resolved, they are a methyl doublet and an acetylenic quartet; if unresolved, each can look like a singlet. Proton-decoupled carbon NMR has three resonances: one methyl carbon near $5$ ppm and two distinct sp carbons around $65$–$85$ ppm. These are approximate predictions, not supplied measured shifts.

The intended ordinary-carbide comparison is energetic: the smaller, more strongly polarizing $\mathrm{M}{\mathrm{g}}^{2+}$ can stabilize a different balance of lattice energy and carbon-anion bonding from the larger $\mathrm{C}{\mathrm{a}}^{2+}$. Stabilization of the highly charged ${\mathrm{C}}_3^{4-}$ group competes with acetylide formation; electron counting alone does not require the same carbide stoichiometries for both metals. Ionic size and charge density provide a qualitative explanation, not a complete phase-stability calculation.

The literal claim that calcium has only one carbide is too broad. Primary calculations and high-pressure synthesis have established additional calcium carbides, including $\mathrm{C}{\mathrm{a}}_2{\mathrm{C}}_3$ and $\mathrm{C}{\mathrm{a}}_2\mathrm{C}$; known calcium-carbide chemistry also includes $\mathrm{Ca}{\mathrm{C}}_6$. Thus the conventional comparison must be limited to the familiar salt-like compounds and conditions intended here, rather than treated as a prohibition on other calcium–carbon compositions.^[ref]

The extra electron associated with trivalent lanthanum weakens the C–C bond by occupying states with C–C antibonding character. Calcium gives the ten-valence-electron acetylide description $\mathrm{C}{\mathrm{a}}^{2+}[{\mathrm{C}}_2{]}^{2-}$, with formal bond order three, consistent with the short $119.2$ pm distance.

A fully ionic count for $\mathrm{La}{\mathrm{C}}_2$ gives $\mathrm{L}{\mathrm{a}}^{3+}[{\mathrm{C}}_2{]}^{3-}$. The eleventh valence electron then enters a ${\pi }_g^{\ast }$ orbital, giving the simple estimate $3-1/2=2.5$. This explains the longer $130.3$ pm bond. In the actual solid that electron is not necessarily localized on an isolated ${\mathrm{C}}_2^{3-}$ radical: metal–carbon mixing and band delocalization make “triple bond plus partial antibonding occupation” more appropriate than an exact isolated-ion bond order. Bond length alone does not determine a unique numerical bond order.

The ions are acetylide ${\mathrm{C}}_2^{2-}$ and linear tricarbide ${\mathrm{C}}_3^{4-}$. Charge balance gives $\mathrm{M}{\mathrm{g}}^{2+}[{\mathrm{C}}_2{]}^{2-}$ and $(\mathrm{M}{\mathrm{g}}^{2+}{)}_2[{\mathrm{C}}_3{]}^{4-}$.

The acetylide has ten valence electrons, two more than neutral ${\mathrm{C}}_2$. They fill the p-derived sigma bonding orbital, giving the formal configuration ${\pi }_u^4{\sigma }_g(2p{)}^2$ above the cancelling $2s$ pair. Its bond order is three, with the Lewis structure ${}^{-}\mathrm{C}\equiv {\mathrm{C}}^{-}$ and a lone pair at each end; each carbon is approximately $sp$ hybridized.

The tricarbide has $3(4)+4=16$ valence electrons and is isoelectronic with carbon dioxide. The cumulene contributor ${}^{2-}\mathrm{C}=\mathrm{C}={\mathrm{C}}^{2-}$ has a central $sp$ carbon, two sigma bonds and two perpendicular pi bonds. The end carbons each have two lone pairs and can be described with an $s{p}^2$ localized basis. Its two C–C bonds have formal order two in this model; the central C–C–C angle is $180^{\circ }$. Other resonance contributors help describe its protonation chemistry.

The elementary valence-MO model gives $\boxed{\text{bond order}({\mathrm{C}}_2)=2}$. Each carbon supplies four $n=2$ valence electrons, so fill eight electrons in the ordering appropriate to the lighter second-row diatomics:The two ${\pi }_u$ orbitals are degenerate, as are the two ${\pi }_g^{\ast }$ orbitals. The $2s$ bonding and antibonding occupations cancel; four net p-bonding electrons remain, giving $(6-2)/2=2$. Mixing between sigma orbitals of compatible symmetry places the occupied ${\pi }_u$ pair below the empty p-derived ${\sigma }_g$ orbital in this diagram. All occupied orbitals are paired.

The omitted $1s$ orbitals form filled ${\sigma }_{g}1s$ and ${\sigma }_u^{\ast }1s$ levels. Their contributions to the formal bond order cancel, and their low energies and small spatial extent make them largely core-like. Omitting them simplifies this valence diagram without claiming they do not exist or have exactly no effect on molecular energies.

The product is carbonic acid, $\mathrm{HO}-\mathrm{C}(=\mathrm{O})-\mathrm{OH}$. The oxygen lone pair Homo of water donates into the carbon-centred component of the carbon dioxide ${\pi }^{\ast }$ LUMO. Draw an electron-pair arrow from the water oxygen to carbon and a second from one C=O pi bond onto its oxygen:The first step bends the original linear electrophile and gives carbon three oxygen neighbours; carbon is approximately trigonal planar in the resulting carbonate framework. Proton transfer from the oxonium oxygen to the negatively charged oxygen produces the neutral acid. In real water, solvent molecules relay that proton rather than requiring a strained isolated intramolecular transfer. The forward mechanism does not assert that most dissolved carbon dioxide is present as carbonic acid at equilibrium.

The central carbon in carbon dioxide has two bonding directions and no lone pair, whereas the central oxygen in ozone has two bonding directions and one lone pair. In the VSEPR model, carbon dioxide therefore has a linear arrangement; ozone has an approximately trigonal-planar electron-domain arrangement but a bent atomic skeleton. Multiple bonding occupies one direction for this geometrical count. The extra central-oxygen lone pair, rather than a different number of atomic nuclei, creates the difference.

The carbon spectra are A → 5, B → 8, C → 6, and D → 7. Number each acyclic acid or ester chain from its carbonyl carbon. The assignments below account for every carbon-13 NMR environment.

For A, both ketone carbonyl carbons have $\delta =208.3$ ppm and all four methylene carbons have $\delta =36.5$ ppm. Molecular symmetry and rapid conformational averaging explain why six carbons give only two resonances.

For B, written $\mathrm{HOOC}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}-\mathrm{C}{\mathrm{H}}_3$, assign $\delta ({\mathrm{C}}_1)=172.8$, $\delta ({\mathrm{C}}_2)=118.1$, $\delta ({\mathrm{C}}_3)=147.4$, $\delta ({\mathrm{C}}_4)=129.7$, $\delta ({\mathrm{C}}_5)=140.8$, and $\delta ({\mathrm{C}}_6)=18.1$ ppm. The alternation among vinylic chemical shifts reflects polarization of the conjugated chain, especially the deshielded carbon beta to the carboxyl group.

For C, written $\mathrm{HOOC}-\mathrm{C}\equiv \mathrm{C}-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}{\mathrm{H}}_3$, assign the carbonyl at $158.5$, the adjacent sp carbon at $72.9$, the other sp carbon at $92.6$, and the terminal methyl at $13.4$ ppm. The two methylene resonances are $20.7$ and $21.0$ ppm. A plausible assignment is ${\mathrm{C}}_4=20.7$ and ${\mathrm{C}}_5=21.0$ ppm, but their ordering is tentative from the supplied shift list alone; a correlation experiment would resolve it.

For D, written $\mathrm{C}{\mathrm{H}}_3\mathrm{CH}=\mathrm{CHC}(=\mathrm{O})\mathrm{OCH}=\mathrm{C}{\mathrm{H}}_2$, the carbonyl is at $163.2$ ppm. The crotonyl beta-CH, alpha-CH and methyl carbons are at $148.0$, $121.7$ and $18.2$ ppm, respectively. The O-bound vinyl CH is at $141.3$ ppm and its terminal $\mathrm{C}{\mathrm{H}}_2$ is at $97.4$ ppm. The two sides of the ester oxygen therefore produce quite different vinylic environments.