Question

8. If the reaction shown was carried out by allowing 7.50 mL of 2,2,3-trimethyl-3-pentano (MW=130.23 g/mol, d 0.857 g/mL) to react with catalytic acid to generate 3,4,4-trimethyl-2-pentene (MW=112.21 g/mol) and water, what is the theoretical yield (in grams) of the alkene product? (15 pts) OH H₂O H3PO4 + 4, 60 min 3,4,4-trimethyl-2-pentene 2,2,3-trimethyl-3-pentano 7. 50mL 15% 857 - 8.75 - 10 150.23 knool 112.215

Answer 1

We know  

Mass = density * volume

Number of moles = mass in gram / Molar mass

Mass of 2,2,3-trimethyl-3-pentanol = 7.50 ml * 0.857 g/ml = 6.4275 g

Number of moles of 2,2,3-trimethyl-3-pentanol = 6.4275 g / 130.23 g/mol = 0.04935 mol

From reaction, 1.0 mol of 2,2,3-trimethyl-3-pentanol produces 1.0 mol of 3,4,4-trimethyl-2-pentene then 0.04935 mol of 2,2,3-trimethyl-3-pentanol will produce 0.04935 mol of 3,4,4-trimethyl-2-pentene.

Theoretical yield of 3,4,4-trimethyl-2-pentene = 0.04935 mol * 112.21 g/mol = 5.54 g