Question

In an SN2 reaction converting 1-butanol to 1-bromobutane what is the theoretical yield of 1-bromobutane

Starting material

25mL 8.7M sulfuric acid

7mL 1-butanol

10.4g Sodium Bromide

Answer 1

Answer – Given, [H₂SO₄] = 8.7 M , volume = 25 mL, volume of 1-butanol = 7 mL

Density of 1-butanol = 0.81 g/mL , mass of NaBr = 10.4 g

First we need to write the reaction

CH₃CH₂CH₂CH₂OH + NaBr + H₂SO₄ ------> CH₃CH₂CH₂CH₂Br + NaOH + H₂SO₄

We need to calculate the moles of each

Moles of H₂SO₄ = 8.7 M * 0.025 L = 0.218 moles

Moles of NaBr = 10.4 g / 102.894 g.mol^-1

                        = 0.101 moles

Mass of 1-butanol = 7 mL * 0.81 g/mL = 5.67 g

Moles of 1-butanol = 5.67 g / 74.12 g.mol^-1 = 0.0765 moles

We know all has mole ratio is 1:1 with each other

So lowest moles of 1-bromobutane formed from the 1-butanol, since 1-butanol is limiting reactant.

So theoretical yield of 1-bromobutane = 0.0765 moles * 137.018 g/mol

                                                              = 10.5 g