Consider the above sequence of reactions. 151 g of 2-bromopentane is made to react. Yield of major product P is 80 % whereas Q is 100 %.
Mass of product Q obtained is________ g.
(Given molar mass in g mol⁻¹ H: 1, C: 12, O: 16, Br: 80 )

Solution

Chemistry ma’am · 4 steps

Exactly 1 mol of bromide goes in. Carry the 80% through the alkene, then weigh the dibromide.

1. Moles of starting bromide. 2-Bromopentane is C₅H₁₁Br, with molar mass 60 + 11 + 80 = 151 g mol⁻¹. So 151 g is exactly 1 mol.
2. Find P. Alcoholic KOH removes HBr. By Saytzeff's rule the more substituted alkene, pent-2-ene (C₅H₁₀), is the major product P. At 80% yield you get 0.8 mol of P.
3. Find Q. Br₂ adds across the double bond to give 2,3-dibromopentane, C₅H₁₀Br₂, with molar mass 60 + 10 + 160 = 230 g mol⁻¹. The yield is 100%, so 0.8 mol of P gives 0.8 mol of Q.

   m_Q = 0.8 mol × 230 g mol⁻¹ = 184 g

4. So. The mass of Q obtained is 184 g.

The common slip: 230. Ignoring the 80% yield of the first step gives 1 × 230 = 230 g, not 184.

The idea to keep. In a multistep sequence, multiply the moles by each step's fractional yield in turn. Convert to grams only at the end, using the molar mass of the final product.

Answer from the NTA final answer key. Solution by PrepWiser.