= Solution
\b[The two carbon skeletons are <allene>, $\mathrm{CH_2=C=CH_2}$, and <propyne>, $\mathrm{CH_3-C\equiv CH}$.] “Linear” describes the three-carbon skeleton; it does not mean every hydrogen lies on the same line.
For allene, use the contributor $\mathrm{^{2-}C=C=C^{2-}}$. A terminal lone pair attacks a water hydrogen while the O–H pair returns to oxygen, giving $\mathrm{HC^-=C=C^{2-}}$ and hydroxide. Repeat at the still-dianionic end, then protonate each remaining terminal carbanion. The four successive proton transfers give
$$
\mathrm{C_3^{4-}\to HC_3^{3-}\to H_2C_3^{2-}\to H_3C_3^-\to H_4C_3},
$$
with two hydrogens placed on each terminal carbon in this branch.
For propyne, the resonance contributor $\mathrm{^{3-}C-C\equiv C^-}$ 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 $\mathrm{CH_3-C\equiv CH}$. 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 $\mathrm{OH^-}$; overall
$$
\mathrm{Mg_2C_3+4H_2O\longrightarrow 2Mg(OH)_2+C_3H_4}.
$$
Allene has \b[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 \b[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.
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