Question

When tert-butanol is placed in the presence of excess equimolar amounts of chloride and bromide ions in aqueous acid experimental data indicates 45% tert-butyl chloride and 55% tert-butyl bromide is obtained. On the other hand, when n-butanol is placed in the presence of excess equimolar amounts of chloride and bromide ions in aqueous acid experimental data indicates 16% of n-butyl chloride and 84% n-butyl bromide is obtained. Now when n-butanol is placed in the presence of excess equimolar amounts of chloride ions and bromide ions in acid dimethyl sulfoxide the experimental data provides an 84% n-butyl chloride and 16% n-butyl bromide as the outcome.

1.) write balanced chemical equations for all three reactions (hint: account for mole ratios!).

2.) draw complete mechanisms for all three reactions.

3.) calculate the theoretical yield for the chlorides and bromides if each reaction was started with 5.55ml of the corresponding alcohol, using an aqueous acid media.

Sample: tert
Theoretical g of chloride
Theoretical g of bromide

Sample: n-
Theoretical g of chloride
Theoretical g of bromide

Writing and balancing the equations is what is really throwing me off

Answer 1

The points to be remember

a) the nucleophilicity of bromide ion in aqueous solution (polar protic solvent) is more than the nucleophilicity of chloride ion as the chloride ion gets more solvated

b) the nucleophilicity of bromide ion in DMSO (polar aprotic solvent) is less than the nucleophilicity of chloride ion

so if we are perfoming any reaction in aqeous medium we will get more of bromo product while in DMSO (or DMF) we will get more of chloro product.

However the difference in percentage in first reaction (t-butyl alcohol) is not much as it undergoes substitution by SN1 mechanism in which there is formation of Carbocation (the rate determining step) and the effect of strenght of nucleophile is not much.

So we obtaine 45:55 ratio of the two products

while in case of n-butanol which undergoes substitution by SN2 the effect of nucleophile is highly remarkable as the attack of nucleophile on substrate is the rate determining step

1) the balanced chemical reaction will be

t-BuOH + 0.45Cl- + 0.55Br- --> 0.45 t-BuCl + 0.55 t-BuBr

n-BuOH + 0.16Cl- + 0.84Br- ---> 0.16n-BuCl + 0.84 nBuBr

nBuOH + 0.84Cl- + 0.16Br- --> 0.84nBuCl + 0.16nBuBr

2)

3) the theorectical yield will be calculated as

The volume of t-butanol = 5.55ml

The density of t-butanol = 0.775 g /mL

so mass of t-butanol = Density X volume = 0.775 x 5.55 grams = 4.30 grams

Moles of t-butanol = mass / Molecular weight = 4.30 / 74.12 = 0.058 moles

so moles of t-butyl chloride formed = 0.45 x 0.058 = 0.0261

Mass of t-butyl chloride formed = 0.0261 X molecular weight = 0.0261 x 92.57 = 2.42 grams (theoretical yield)

Moles of t-butyl bromide formed = 0.55 x 0.058 = 0.0319

Mass of t-butyl bromide formed = 0.0319 x mol wt = 137.02 X 0.0319 = 4.37 grams (theoretical yield)

Second reaction

The density of n-butanol = 0.81 g /mL

so mass of t-butanol = Density X volume = 0.81 x 5.55 grams = 4.49 grams

Moles of n-butanol = mass / Molecular weight = 4.49 / 74.12 = 0.0606 moles

so moles of butyl chloride formed = 0.16 x 0.0606 = 0.0097

Mass of butyl chloride formed = 0.0097 X molecular weight = 0.0097 x 92.57 = 0.898 grams (theoretical yield)

Moles of butyl bromide formed = 0.84 x 0.0606 = 0.0509

Mass of butyl bromide formed = 0.0509 x mol wt = 137.02 X 0.0509= 6.97 grams (theoretical yield)

Third reaction

The density of n-butanol = 0.81 g /mL

so mass of t-butanol = Density X volume = 0.81 x 5.55 grams = 4.49 grams

Moles of n-butanol = mass / Molecular weight = 4.49 / 74.12 = 0.0606 moles

so moles of butyl chloride formed = 0.84 x 0.0606 = 0.0509

Mass of butyl chloride formed = 0.0509 X molecular weight = 0.0509 x 92.57 = 4.71 grams (theoretical yield)

Moles of butyl bromide formed = 0.16 x 0.0606 = 0.00969

Mass of butyl bromide formed = 0.00969 x mol wt = 137.02 X 0.00969= 1.32 grams (theoretical yield)