Question

Route A: 1-chloropropane +Mg -------Et2O----> propylmagnesium chloride --------i.Et2O ii.H2O----> 2-methylheptan-4-ol ----------NaOCl---------> 2-methylheptan-4-one

Route B: 1-chloro-2-methylpropane +Mg ---------Et2O-----> isobutylmagnesium chloride ---------i.Et2O ii.H2O----> 2-methylheptan-4-ol ----------NaOCl---------> 2-methylheptan-4-one

Assume both approach were similar yielding: what would make one approach superior over the other if this experiment was being done in a small lab on the ~5g scale? What about if you had to do this reaction on the 100kg scale in a pilot chemical plant?

Answer 1

At 5 g scale 1-chloropropane, will be better to use because it is more reactive than 1-chloro-2-methylpropane, because of the presence of 2-methyl group. All other conditions are similar for both the rotes.

But at 100 kg scale, 1-chloro-2-methylpropane, should be used because its boiling point is higher around 68-70^oC compare to that of 1-chloropropane with boiling point 47-48^oC. For large scale handling of reactants and reagent should be given importance because this is an exothermic reaction and working with low boiling material could be hazardous, considering the other steps are very similar.