Question

[1].

CH3 (CH2 3OH HBr H2O C4H9Br

(a) Calculate the theoretical yield of 1-bromobutane in volume and mass when 72.0 mmol of 1-butanol react.

$CH_3CH_2(OH)CH_3+HBr\rightarrow \space H_2O + CH_3CH_2CHBrCH_3$

(b) Calculate the theoretical yield of 2-bromobutane in volume and mass when 72.0 mmol of 2-butanol react.

Answer 1

Reaction:

1)CH₃(CH₂)₃-OH + HBr ---------->   CH₃(CH₂)₃-Br + H₂O

G.Mol.Wt. 74                                              G. Mol. Wt. 137              : 1 mole = 1000 mmoles

74     G.Mol.Wt. of n-Butanol = 137       G. Mol. Wt.   of n-Butyl bromide

0.072 G.Mol.Wt. of n-Butanol    = 0.072x 137   = 0.133 mols of n-Butylbromidex x1000   ( Theoretical yield )      

                                                      74              = 133   mmole              

           Density of 1-Bromobutane =1.275

   Density = Mass / Volume                                                            

              Volume = 133/ 1.275      = 104.3 mmol (Theoretical yield)                                     

2)CH₃(CH₂)₂CHOH-CH₃ + HBr ---------->  )CH₃(CH₂)₂CHBr-CH₃ + H₂O

Theoretical yield of the 2-bromobuane will be same as that of 1-Bromobutane as shown above.133 mmol

Density = Mass / Volume ,   as Density of 2-bromobutane is 1.255

    Volume = Mass / Density

               = 133 /1.255 = 105.9 mmol (Theoretical yield)