He well being of modern society is unimaginable without a myriad of products of industrial organic s — Organic Chemistry Chemistry Question
Towards Green Chemistry: The E-factor
He well being of modern society is unimaginable without a myriad of products of industrial organic synthesis, from pharmaceutical combating diseases or relieving pain, to synthetic dyestuffs for aesthetic appeal. The flip side of the coin is that many of these processes generate substantial amounts of waste. The solution is not less chemistry bur alternative, cleaner technologies that minimize the waste. In order to evaluate the environmental (un)friendliness of a process, the terms “atom utilization” and “the E-factor” were introduced. The atom utilization is obtained by dividing the molar mass of the desired product by the sum of the molar masses of all substances produced according to the reaction equations. The E-factor is the amount (in kg) of by-products per kg of the product. Methyl methacrylate is an important monomer for transparent materials (Plexiglas).
Figure 1: Methyl methacrylate synthesis.
5.1 Calculate the atom utilization and the E-factor for both processes. The classical and a process for methyl methacrylate manufacture are shown in Figure 1.
Model Answer
5.1 The relative molecular mass of methyl methacrylate = 100
The relative molecular mass of NH4HSO4 = 115
Classical route: Atom utilization = 100/(100 + 115) = 0.47 or 47 %
E-factor= 115/100 = 1.15
Classical route: Atom utilization = 115/115 = 1 or 100 %
E-factor= 0/100 = 0
Another example is the manufacture of ethane oxide (see Figure 2). The classical route produces calcium chloride. Moreover, 10 % of the ethane is converted into 1,2-ethanediol by hydrolysis. In the modern direct route a silver catalyst is applied. Here, 15 % of the ethene is oxidized to carbon dioxide and water.
Figure 2: Ethene oxide synthesis.
5.2 Calculate the atom utilization and E-factor for both processes.
Model Answer
5.2 Classical chlorohydrin route: Atom utilization = 44/173 = 0.25 or 25 %
Modern petrochemical route: Atom utilization = 44/44 = 1 or 100 %
Classical route: E-factor = 133.4/39.6 = 3.37
product: 44 – (10 % of 44) = 39.6
by-products: 111 + (10 % of 62) + (18 – 10 % of 18) =
= 111 + 6.2 + 16.2 = 133.4
Modern route: E-factor = 18.6 / 37.4 = 0.50
product: 44 – (15 % of 44) = 37.4
by-products: 2 CO2 + 2 H2O per mole of C2H4 (15 %)
2 × 15 % of 18 = 5.4