Animated Solution for Chemistry - Organic Chemistry: Comprehension Passage

This problem is a beautiful symphony of classic organic reactions, testing your ability to track moles, stoichiometry, and percentage yields across multiple sequential steps. Let's break down this chemical journey into two distinct pathways.

From Carbide to Aromatic Ring

We begin with the hydrolysis of magnesium carbide, ${\text{Mg}}_2{\text{C}}_3$. Unlike calcium carbide which yields ethyne, magnesium carbide contains the allylide ion (${\text{C}}_3^{4-}$), which upon hydrolysis gives propyne ($\text{P}$).

Given $4.0\text{ g}$ of propyne (molar mass $40\text{ g/mol}$), we start with exactly $0.1\text{ mol}$ of $\text{P}$.

Next, propyne is treated with sodamide (${\text{NaNH}}_2$) followed by methyl iodide ($\text{MeI}$). The strong base deprotonates the terminal alkyne, forming a nucleophilic acetylide ion that undergoes an $S_{N}2$ reaction with methyl iodide to form But-2-yne ($\text{Q}$).

Applying the $75\%$ yield, the moles of $\text{Q}$ become $0.1\times 0.75=0.075\text{ mol}$.

Now comes a critical stoichiometry trap. But-2-yne is passed through a red hot iron tube at $873\text{ K}$, undergoing cyclic trimerization to form hexamethylbenzene ($\text{R}$).

Because three moles of reactant form one mole of product, we must divide the moles of $\text{Q}$ by $3$ before applying the $40\%$ yield.

Moles of $\text{R}=\left(\frac{0.075}{3}\right)\times 0.40=0.01\text{ mol}$.

The molar mass of hexamethylbenzene is $162\text{ g/mol}$. Therefore, the mass $x=0.01\times 162=1.62\text{ g}$.

Hydration and Condensation

Returning to our $0.1\text{ mol}$ of propyne ($\text{P}$), the bottom pathway begins with the Kucherov reaction using ${\text{Hg}}^{2+}/{\text{H}}^{+}$. This adds water across the triple bond following Markovnikov's rule, yielding an enol that rapidly tautomerizes to acetone ($\text{S}$).

With a $100\%$ yield, we have $0.1\text{ mol}$ of acetone.

Acetone is then heated with barium hydroxide, triggering an Aldol condensation. Two molecules of acetone condense to form mesityl oxide ($\text{T}$).

Again, stoichiometry dictates we divide the moles by $2$. Applying the $80\%$ yield, the moles of $\text{T}=\left(\frac{0.1}{2}\right)\times 0.80=0.04\text{ mol}$.

Haloform Reaction

Finally, mesityl oxide is treated with sodium hypochlorite ($\text{NaOCl}$), initiating the haloform reaction. This cleaves the methyl ketone group, yielding a carboxylate salt and chloroform.

The problem states that product $\text{U}$ decolourises Baeyer's reagent, confirming it is the fragment containing the carbon-carbon double bond. Assuming an implicit acidic workup (standard in such problems), $\text{U}$ is 3-methylbut-2-enoic acid (${\text{C}}_5{\text{H}}_8{\text{O}}_2$), with a molar mass of $100\text{ g/mol}$.

Applying the $80\%$ yield, the moles of $\text{U}=0.04\times 0.80=0.032\text{ mol}$.

The mass $y=0.032\times 100=3.20\text{ g}$.