12 principles of green chemistry: Exercises — Chemistry deep dive

Before we start, one reminder of the two numbers we keep computing (both defined in the parent):

Atomic masses used throughout (g/mol): $H = 1, C = 12, N = 14, O = 16, N a = 23, C l = 35.5, B r = 80$ .

Level 1 — Recognition

Recall Solution L1.1

(a) Principle 6 — Energy Efficiency (ambient temperature & pressure). See Activation Energy and Reaction Rates. (b) Principle 5 — Safer Solvents/Auxiliaries. See Solvent Selection and Supercritical CO2. (c) Principle 7 — Renewable Feedstocks. See Renewable Feedstocks and Biomass. (d) Principle 10 — Design for Degradation.

Recall Solution L1.2

False. Yield only measures how much of the limiting reactant turned into product; atom economy asks how many of all the input atoms landed in the product. A reaction can convert 100 % of the limiting reactant yet still throw away half its atom-mass as byproduct salt. (This is the whole point of Atom Economy and Yield.)

Level 2 — Application

Recall Solution L2.1

What we do: find $M$ of the desired product and of every reactant.

$M (C_{2} H_{4}) = 2 (12) + 4 (1) = 28$
$M (C l_{2}) = 2 (35.5) = 71$
$M (product C_{2} H_{4} C l_{2}) = 28 + 71 = 99$

Why the sum of reactants is also 99: in an addition reaction all atoms enter the single product — nothing is expelled. $% AE = \frac{99}{28 + 71} \times 100 = \frac{99}{99} \times 100 = 100% .$ Addition reactions are inherently 100 % atom-economical.

Recall Solution L2.2

$M (C H_{3} C l) = 12 + 3 (1) + 35.5 = 50.5$
$M (N a O H) = 23 + 16 + 1 = 40$
$M (C H_{3} O H, desired) = 12 + 4 (1) + 16 = 32$

$% AE = \frac{32}{50.5 + 40} \times 100 = \frac{32}{90.5} \times 100 = 35.4% .$ The other $\approx 65%$ of the atom-mass becomes $N a C l$ waste.

Recall Solution L2.3

Convert to the same units: waste $= 1600 g$ , product $= 200 g$ . $E -factor = \frac{1600}{200} = 8 .$ An ideal green reaction has $E \to 0$ . A value of $8$ means 8 g of waste per gram of product — typical of fine-chemical/pharma processes, not green. See E-factor and Process Mass Intensity.

Level 3 — Analysis

Recall Solution L3.1

Why substitution loses: a substitution swaps a group — one incoming group takes a place, and the displaced group ( $C l^{-}$ , here as $N a C l$ ) leaves. That leaving group carries real mass out of the desired product, so it is guaranteed byproduct waste. Addition adds both reactants onto one molecule; nothing leaves, so no mass is lost. Prediction: $C_{2} H_{4} + H B r \to C_{2} H_{5} B r$ is an addition — both reactants fully incorporate. So AE should be high (100 %). Check: $M (C_{2} H_{4}) = 28$ , $M (H B r) = 81$ , product $M = 109 = 28 + 81$ → $% AE = \frac{109}{109} \times 100 = 100%$ . ✓

Recall Solution L3.2

Error: Pt is a catalyst — it is regenerated and does not appear in the balanced overall equation. It must be excluded from $\sum M_{reactants}$ . See Catalysis. Correct: $% AE = \frac{30}{28 + 2} \times 100 = \frac{30}{30} \times 100 = 100% .$ The student's wrong version would give $\frac{30}{225} \times 100 = 13.3%$ — a badly misleading number.

Recall Solution L3.3

Route A: $M (C_{2} H_{4}) = 28$ , $M (H_{2} O) = 18$ . Sum $= 46 = M (product)$ . $% AE_{A} = \frac{46}{46} \times 100 = 100% .$ Route B: $M (C_{2} H_{5} C l) = 64.5$ , $M (N a O H) = 40$ . Sum $= 104.5$ ; product $= 46$ . $% AE_{B} = \frac{46}{104.5} \times 100 = 44.0% .$ Route A wins. This is Principle 2 (Atom Economy) in action — and see the bar figure below.

Figure — 12 principles of green chemistry

Level 4 — Synthesis

Recall Solution L4.1

(a) Improvement factor $= \frac{77}{40} = 1.925 \approx 1.9 \times$ . Nearly double the input atoms become drug instead of waste. (b)

Principle 9 (Catalysis): stoichiometric reagents replaced by catalysts (e.g. HF/Raney-nickel/Pd steps that regenerate). See Catalysis.
Principle 8 (Reduce Derivatives): 6 steps → 3 steps means fewer temporary/protecting-group manipulations.
Principle 1 (Prevention): far less waste generated in the first place (and recovered acetic acid pushes AE toward 99 %).

Recall Solution L4.2

Proposal: use a direct catalytic esterification / addition-type coupling — e.g. acid + alcohol over a solid acid catalyst, or catalytic addition — instead of first converting the acid to the acyl chloride. Why greener:

The acyl-chloride route expels $H C l$ (a corrosive byproduct) — that is lost atom-mass and a hazard (violates Principle 3, Less Hazardous Synthesis).
A catalytic route (Principle 9) regenerates its catalyst and, being closer to an addition/condensation with water as the only byproduct, has higher atom economy (Principle 2).
Fewer aggressive reagents also serves Principle 12 (Inherently Safer).

Level 5 — Mastery

Worked example L5.1 — Full green audit, two routes

A pigment ( $M_{product} = 250 g/mol$ ) can be made two ways. Per mole of product:

	Route X	Route Y
$\sum M_{reactants}$ (balanced eqn)	$500$	$320$
Solvent + aqueous waste per mole product	$2000 g$	$150 g$
Temperature	$180^{\circ} C$	$25^{\circ} C$
Reagents	stoichiometric $C r O_{3}$ (toxic)	reusable enzyme catalyst

(a) Compute %AE for each route. (b) Compute the E-factor for each (treat all non-product mass — byproduct atoms + solvent/aqueous waste — as waste; assume 1 mol product = 250 g). (c) Give a verdict: which route is greener, citing at least four principles.

Recall Solution L5.1

(a) Atom economy (balanced-equation atoms only): $% AE_{X} = \frac{250}{500} \times 100 = 50%, % AE_{Y} = \frac{250}{320} \times 100 = 78.125% .$

(b) E-factor. Product mass $= 250 g$ . Route X waste = (byproduct atoms) + (solvent/aqueous). Byproduct atom-mass $= \sum M_{react} - M_{product} = 500 - 250 = 250 g$ . Add solvent $2000 g$ : total waste $= 2250 g$ . $E_{X} = \frac{2250}{250} = 9 .$ Route Y waste = $(320 - 250) + 150 = 70 + 150 = 220 g$ . $E_{Y} = \frac{220}{250} = 0.88 .$

(c) Verdict — Route Y is far greener:

Principle 2 (Atom Economy): 78 % vs 50 % → more atoms in product.
Principle 1 / E-factor: $E = 0.88$ vs $9$ → roughly 10× less waste. See E-factor and Process Mass Intensity.
Principle 6 (Energy): 25 °C vs 180 °C → much lower energy.
Principle 9 (Catalysis): reusable enzyme vs stoichiometric $C r O_{3}$ .
Principle 3 (Less Hazardous): avoids toxic chromium(VI).

See the comparison figure:

Recall Solution L5.2

What the naïve estimate captures: if the only waste were the byproduct atoms, then at 50 % AE waste-mass = product-mass, giving $E \approx 1$ . What it misses: AE counts only atoms in the balanced equation; it is blind to solvents, wash water, drying agents and workup losses. Route X dumps $2000 g$ of solvent/aqueous waste that AE never sees. That extra mass drives $E$ from $\approx 1$ up to $9$ . Lesson: AE and E-factor are complementary. AE is a design-of-the-molecule metric; E-factor is a reality-of-the-whole-process metric. A route can look decent on AE and still be filthy on E-factor. (Core message of Atom Economy and Yield + E-factor and Process Mass Intensity.)

Connections

12 principles of green chemistry (parent)
Atom Economy and Yield — the L2–L3 formula machinery
E-factor and Process Mass Intensity — the L5 waste audits
Catalysis — Principle 9, L3.2 & L4
Renewable Feedstocks and Biomass — Principle 7, L1.1(c)
Solvent Selection and Supercritical CO2 — Principle 5, L1.1(b)
Activation Energy and Reaction Rates — Principle 6, energy efficiency