Calculate Mass of Product U in Organic Reaction Pathway

To find the mass of product $\mathbf{U}$ ($y\text{ g}$), we analyze the step-by-step chemical transformations starting from $\mathbf{P}$:

1. Identification of Compound $\mathbf{P}$ and its Amount:

   • $\text{Mg}_2\text{C}_3$ undergoes hydrolysis to yield propyne ($\mathbf{P}$): $\text{Mg}_2\text{C}_3 + 4\text{H}_2\text{O} \rightarrow 2\text{Mg(OH)}_2 + \text{CH}_3-\text{C}\equiv\text{CH}$
   • Molar mass of propyne ($\text{C}_3\text{H}_4$) = $3 \times 12 + 4 \times 1 = 40\text{ g mol}^{-1}$.
   • Given mass of $\mathbf{P} = 4.0\text{ g}$.
   • Moles of $\mathbf{P} = \frac{4.0\text{ g}}{40\text{ g mol}^{-1}} = 0.1\text{ mol}$.

2. Conversion of $\mathbf{P}$ to $\mathbf{S}$ (Hydration):

   • Propyne ($\mathbf{P}$) reacts with $\text{Hg}^{2+}/\text{H}^+$ at $333\text{ K}$ via Markovnikov addition of water followed by tautomerization to yield acetone ($\mathbf{S}$): $\text{CH}_3-\text{C}\equiv\text{CH} + \text{H}_2\text{O} \xrightarrow[\text{333 K}]{\text{Hg}^{2+}/\text{H}^+} \text{CH}_3-\text{CO}-\text{CH}_3$
   • Percentage yield of this step = $100\%$.
   • Moles of $\mathbf{S}$ formed = $0.1\text{ mol} \times 1.00 = 0.1\text{ mol}$.

3. Conversion of $\mathbf{S}$ to $\mathbf{T}$ (Aldol Condensation):

   • Acetone ($\mathbf{S}$) undergoes self-aldol condensation in the presence of $\text{Ba(OH)}_2$ followed by dehydration on heating to form mesityl oxide ($\mathbf{T}$): $2\text{CH}_3-\text{CO}-\text{CH}_3 \xrightarrow[\text{heat}]{\text{Ba(OH)}_2} (\text{CH}_3)_2\text{C}=\text{CH}-\text{CO}-\text{CH}_3 + \text{H}_2\text{O}$
   • Stoichiometric ratio: $2\text{ moles of }\mathbf{S} \rightarrow 1\text{ mole of }\mathbf{T}$.
   • Theoretical yield of $\mathbf{T} = \frac{0.1\text{ mol}}{2} = 0.05\text{ mol}$.
   • Given percentage yield = $80\%$.
   • Actual moles of $\mathbf{T}$ obtained = $0.05\text{ mol} \times 0.80 = 0.04\text{ mol}$.

4. Conversion of $\mathbf{T}$ to $\mathbf{U}$ (Haloform Reaction):

   • Mesityl oxide ($\mathbf{T}$) contains a methyl ketone group ($-\text{CO}-\text{CH}_3$) and reacts with $\text{NaOCl}$ via the haloform reaction to form 3-methylbut-2-enoic acid ($\mathbf{U}$) which decolourises Baeyer's reagent due to the carbon-carbon double bond: $(\text{CH}_3)_2\text{C}=\text{CH}-\text{CO}-\text{CH}_3 \xrightarrow{\text{NaOCl}} (\text{CH}_3)_2\text{C}=\text{CH}-\text{COOH} + \text{CHCl}_3$
   • Stoichiometric ratio: $1\text{ mole of }\mathbf{T} \rightarrow 1\text{ mole of }\mathbf{U}$.
   • Theoretical yield of $\mathbf{U} = 0.04\text{ mol}$.
   • Given percentage yield = $80\%$.
   • Actual moles of $\mathbf{U}$ obtained = $0.04\text{ mol} \times 0.80 = 0.032\text{ mol}$.

5. Calculation of the Mass of Product $\mathbf{U}$ ($y$):

   • Molecular formula of 3-methylbut-2-enoic acid ($\mathbf{U}$) is $\text{C}_5\text{H}_8\text{O}_2$.
   • Molar mass of $\mathbf{U} = 5(12) + 8(1) + 2(16) = 60 + 8 + 32 = 100\text{ g mol}^{-1}$.
   • Mass of $\mathbf{U}$ ($y$) = $\text{moles of }\mathbf{U} \times \text{molar mass of }\mathbf{U}$: $y = 0.032\text{ mol} \times 100\text{ g mol}^{-1} = 3.20\text{ g}$

Thus, the value of $\mathbf{y}$ is $3.20$ (or $3.2$).