A volatile hydrocarbon Z has an open chain, and its vapour density is 34…

(a) Molar mass. $M = 2 \times 34 = 68\ \text{g mol}^{-1}$.

(b) Hydrogen uptake. $n(\text{Z}) = \dfrac{1.36\ \text{g}}{68\ \text{g mol}^{-1}} = 0.0200\ \text{mol}$ and $n(\text{H}_2) = \dfrac{896\ \text{mL}}{22400\ \text{mL mol}^{-1}} = 0.0400\ \text{mol}$. One mole of Z takes up $2$ moles of $\text{H}_2$, so it has $2$ $\pi$ bonds. An open chain with two $\pi$ bonds fits $\text{C}_n\text{H}_{2n-2}$, and $14n - 2 = 68$ gives $n = 5$: Z is $\text{C}_5\text{H}_8$.

Which compound? Hydrogenation keeps the carbon skeleton, so Z has the skeleton of 2-methylbutane. The student's diene has that skeleton, but it has no hydrogen on an $sp$ carbon, so it cannot react with sodium. The only alkyne with this skeleton and a terminal $\equiv\text{CH}$ is 3-methylbut-1-yne, $(\text{CH}_3)_2\text{CHC}\equiv\text{CH}$. The guess fails; Z is 3-methylbut-1-yne.

(c) Hydrogen with sodium. Each molecule gives up its one acidic hydrogen: $\text{RC}\equiv\text{CH} + \text{Na} \rightarrow \text{RC}\equiv\text{C}^-\text{Na}^+ + \tfrac{1}{2}\text{H}_2$. So $n(\text{H}_2) = \dfrac{0.0200\ \text{mol}}{2}$, and the volume is $0.0100\ \text{mol} \times 22400\ \text{mL mol}^{-1} = 224\ \text{mL}$.

(d) Hydration. Water adds by Markovnikov's rule: the OH goes to carbon 2. The enol changes to the ketone 3-methylbutan-2-one, $\text{CH}_3\text{COCH(CH}_3)_2$, with $M = 86\ \text{g mol}^{-1}$. Mass $= 0.0200\ \text{mol} \times 86\ \text{g mol}^{-1} = 1.72\ \text{g}$.

(e) Bromine. Two moles of $\text{Br}_2$ add per mole of Z: $2 \times 0.0200\ \text{mol} \times 160\ \text{g mol}^{-1} = 6.40\ \text{g}$.