Question

CH,CH:CH.CН,ОН + НBr CH&CH2CH2CH2Br + H2O H2SO4 1-Butanol Hydrobromic Acid 1-Bromobutane Вр 118° C Den 1.49 Вр 101.6° С Den 0.810 MW 80.9 Den 1275 (a)which is the limiting reagent, (b) Calculate the theoretical yield, and (e) percent yield of the above reaction if you were to react 48 mmol of 1-Butanol with 66 mmol of aqueous hydrobromic acid and you obtained 3.7g of 1-Bromobutane? SHOW ALL CALCULATIONS (a) Theoretical Yield: (b) Limiting Rcagent: (c) Percent Yield:

Answer 1

(a)

As per the reaction, the ratio of the number of moles of 1-butanol to HBr is 1:1.

Number of mmol of HBr (66) is more than the number of mmol of 1-butanol (48).

Hence, HBr is excess reagent and 1-butanol is limiting reagent.

(b)

As per the reaction, the ratio of the number of moles of 1-butanol to number of moles of 1-bromobutane is 1:1.

48 mmol 1-butanol will react to give $48 \ mmol \ 1-butanol \times \frac{1 \ mmol \ 1-bromobutane}{1 \ mmol \ 1-butanol}=48 \ mmol \ 1-bromobutane$

The molar mass of 1-bromobutane is 137 g/mol.

Theoretical yield $=48 \ mmol \times 137 \ g/mol \times \frac{1 \ mol}{1000 \ mmol}=6.58 \ g$

(c)

3.7 g of 1-bromobutane is obtained.

Percent yield $=\frac{3.7 \ g}{6.58 \ g} \times 100 =56.3 \ \%$