1. Identifying the Compounds and Reactions

The reaction scheme starts with an alkyne P, which undergoes two different reduction reactions.

• Pathway 1 (P → Q → R):

  • P → Q: The reagent is $H_2, Pd/BaSO_4, S$ (Lindlar's catalyst), which reduces an alkyne to a cis-alkene. The yield is 80%.
  • Q → R: The reagent is $Br_2/CCl_4$, which is an electrophilic addition of bromine to an alkene. The product R is given as 2,3-dibromobutane (from its eventual formation), which means Q must be but-2-ene. Since Q is formed via Lindlar's catalyst, it is cis-but-2-ene. Therefore, P must be but-2-yne.
    • P: But-2-yne ($CH_3-C \equiv C-CH_3$)
    • Q: cis-But-2-ene ($cis-CH_3-CH=CH-CH_3$)
    • R: racemic-2,3-Dibromobutane ($CH_3-CH(Br)-CH(Br)-CH_3$). The addition of $Br_2$ to a cis-alkene is an anti-addition, resulting in a racemic mixture.

• Pathway 2 (P → S → T → U):

  • P → S: The reagent is $Na/liq. NH_3$ (Birch reduction for alkynes), which reduces an alkyne to a trans-alkene. The yield is 100%.
    • S: trans-But-2-ene ($trans-CH_3-CH=CH-CH_3$)
  • S → T: The reagent is $Br_2/CCl_4$. The anti-addition of $Br_2$ to a trans-alkene gives a meso compound.
    • T: meso-2,3-Dibromobutane ($CH_3-CH(Br)-CH(Br)-CH_3$)
  • T → U: The reagent is Alc. KOH, $\Delta$. This causes double dehydrohalogenation (E2 elimination) to form an alkyne.
    • U: But-2-yne ($CH_3-C \equiv C-CH_3$), which is the same as the starting material P.

2. Molar Mass Calculations

Using the given atomic masses (C=12, H=1, O=16) and assuming Br=80 g/mol (a common approximation in JEE problems, which is consistent with integer calculations based on the given mass of R).

• Molar mass of P and U ($C_4H_6$): $4 \times 12 + 6 \times 1 = 54$ g/mol.
• Molar mass of R and T ($C_4H_8Br_2$): $4 \times 12 + 8 \times 1 + 2 \times 80 = 48 + 8 + 160 = 216$ g/mol.

3. Stoichiometric Calculation for Pathway 1 (P → Q → R)

We are given that the mass of R produced is x = 108 g. We work backwards to find the moles of P required.

1. Moles of R: $\text{Moles of R} = \frac{\text{Mass of R}}{\text{Molar Mass of R}} = \frac{108 \text{ g}}{216 \text{ g/mol}} = 0.5 \text{ mol}$

2. Moles of Q required: The reaction $Q \rightarrow R$ has a 50% yield. The stoichiometry is 1:1. $\text{Moles of Q required} = \frac{\text{Moles of R produced}}{\text{Yield}} = \frac{0.5 \text{ mol}}{0.50} = 1.0 \text{ mol}$

3. Moles of P required ($n_P$): The reaction $P \rightarrow Q$ has an 80% yield. The stoichiometry is 1:1. $n_P = \text{Moles of P required} = \frac{\text{Moles of Q produced}}{\text{Yield}} = \frac{1.0 \text{ mol}}{0.80} = 1.25 \text{ mol}$

So, 1.25 moles of P are needed to produce 108 g of R.

4. Stoichiometric Calculation for Pathway 2 (P → S → T → U)

The problem statement does not specify the amount of P used for the second pathway. A standard and logical assumption in such problems is that the two pathways are to be compared starting with the same initial amount of reactant.

Assumption: The moles of P used for the second pathway ($n'_P$) are the same as for the first pathway. So, $n'_P = n_P = 1.25$ mol.

1. Moles of S produced: The reaction $P \rightarrow S$ has a 100% yield. $\text{Moles of S} = n'_P \times 1.00 = 1.25 \text{ mol} \times 1.0 = 1.25 \text{ mol}$

2. Moles of T produced: The reaction $S \rightarrow T$ has a 50% yield. $\text{Moles of T} = \text{Moles of S} \times 0.50 = 1.25 \text{ mol} \times 0.5 = 0.625 \text{ mol}$

3. Moles of U produced: The reaction $T \rightarrow U$ has an 80% yield. $\text{Moles of U} = \text{Moles of T} \times 0.80 = 0.625 \text{ mol} \times 0.8 = 0.5 \text{ mol}$

4. Mass of U produced (y): $y = \text{Mass of U} = \text{Moles of U} \times \text{Molar Mass of U}$ $y = 0.5 \text{ mol} \times 54 \text{ g/mol} = 27 \text{ g}$

The calculated value for y is 27.00.

5. Conclusion and Answer Verification

The calculated value for y is 27. The stored correct answer is 3.20. There is a large discrepancy.

The calculation leading to y=27 is based on the most direct and logical interpretation of the problem statement, namely that the same initial amount of starting material (P) is used for both reaction pathways. Without this assumption, the problem is not solvable as there is no other information provided to link the two pathways.

The stored answer of 3.20 cannot be obtained through any standard chemical or stoichiometric principle from the data given. It is highly likely that the question as stated is flawed, either containing a typo in the given values (e.g., the mass of R or the yields) or is missing a crucial piece of information connecting the two reaction sequences. For instance, if the mass of R produced (x) were 12.8 g instead of 108 g, the calculated value of y would be 3.2 g. Given the information provided, the derived answer is 27.