12 principles of green chemistry: Worked examples — Chemistry deep dive

Before anything: the one formula this whole page turns, stated in plain words.

See Atom Economy and Yield for the yield-vs-AE distinction and E-factor and Process Mass Intensity for the waste-side twin measure.

The scenario matrix

Every AE problem you will ever meet falls into one of these cells. Each row is a case class; the last column names the worked example that covers it.

#	Case class	What makes it special	Covered by
1	Addition (combine, nothing ejected)	AE hits the maximum, 100%	Example 1
2	Substitution (an atom is kicked out)	AE drops — byproduct carries mass away	Example 2
3	Elimination / decomposition (ONE reactant → several products)	Degenerate: $\sum M_{reactants}$ is a single molecule	Example 3
4	Catalytic reaction	The catalyst must be excluded from the sum — the trap	Example 4
5	Stoichiometric coefficients $> 1$	Must multiply molar masses by coefficients	Example 5
6	Two competing routes to the SAME product	Compare AE to choose the greener route	Example 6
7	Real-world word problem (mass given, not moles)	Translate grams → the same formula still works	Example 7
8	Limiting / extreme values (AE → 0 and AE → 100)	Sanity-check the boundaries of the formula	Example 8

Two edge behaviours worth stating up front, because they bound every answer:

Example 1 — Cell 1: Addition (the 100% case)

Worked example Bromine adds across an alkene

$C_{2} H_{4} + B r_{2} \to C_{2} H_{4} B r_{2}$ Find the atom economy.

Forecast: In an addition, nothing leaves the flask — every atom of both reactants ends up in the single product. Guess the AE before reading on. (Guess: 100%?)

List molar masses. $M (C_{2} H_{4}) = 2 (12) + 4 (1) = 28$ ; $M (B r_{2}) = 2 (80) = 160$ . Why this step? AE needs masses, and mass is (atoms $\times$ their atomic weight).
Sum the reactants. $\sum M_{reactants} = 28 + 160 = 188$ . Why this step? The denominator is all incoming atom-mass.
Identify the desired product's mass. $M (C_{2} H_{4} B r_{2}) = 2 (12) + 4 (1) + 2 (80) = 188$ . Why this step? The numerator is only the product we want.
Divide. $% AE = \frac{188}{188} \times 100 = 100%$ .

Verify: Product mass (188) equals reactant sum (188) → conservation of mass holds AND there is no byproduct, so AE must be exactly 100%. Consistent. ✓

Example 2 — Cell 2: Substitution (an atom is ejected)

Worked example Hydrolysis of bromomethane

$C H_{3} B r + N a O H \to C H_{3} O H + N a B r$ We want the methanol, $C H_{3} O H$ . Find the AE.

Forecast: Something leaves the product this time — the $N a$ and $B r$ pair off into salt. So AE must be below 100%. Will it be closer to 50% or closer to 20%? Guess first.

Molar masses. $M (C H_{3} B r) = 12 + 3 + 80 = 95$ ; $M (N a O H) = 23 + 16 + 1 = 40$ . Why? Same reason — denominator needs every reactant's mass.
Sum reactants. $\sum M = 95 + 40 = 135$ .
Desired product mass. $M (C H_{3} O H) = 12 + 4 + 16 = 32$ . Why only this? $N a B r$ ( $M = 103$ ) is the unwanted byproduct — it does not count in the numerator.
Divide. $% AE = \frac{32}{135} \times 100 \approx 23.7%$ .

Verify: Byproduct fraction $= \frac{103}{135} \times 100 \approx 76.3%$ . The two must sum to 100%: $23.7 + 76.3 = 100$ . ✓ The salt really is carrying away most of the atom-mass — that is why substitutions are usually poor on atom economy.

Example 3 — Cell 3: Decomposition (degenerate, ONE reactant)

Example 4 — Cell 4: The catalyst trap

Figure — 12 principles of green chemistry

Worked example Catalytic hydrogenation of ethene

$C_{2} H_{4} + H_{2} Pt C_{2} H_{6}$ The platinum sits above the arrow. Find the AE — and decide whether Pt goes in the denominator.

Forecast: Addition-type (H adds on), so intuition says 100%. But there's a Pt. Does the catalyst's mass count? Look at the figure: Pt goes in and comes out unchanged — the red loop.

Decide what counts as a reactant. Pt is regenerated — it appears on both sides and cancels out of the net balanced equation. So it is excluded. Why this step? AE counts only species in the stoichiometric balanced equation; a catalyst that returns unchanged is not consumed, so it adds no atoms to the sum.
Molar masses of the real reactants. $M (C_{2} H_{4}) = 28$ ; $M (H_{2}) = 2$ .
Sum. $\sum M = 28 + 2 = 30$ .
Product mass. $M (C_{2} H_{6}) = 2 (12) + 6 = 30$ .
Divide. $% AE = \frac{30}{30} \times 100 = 100%$ .

Verify: No byproduct, reactant sum = product mass = 30 → AE must be 100%. ✓ The trap sprung: if you had wrongly added, say, $M (P t) = 195$ to the denominator, you'd get $30/225 \approx 13.3%$ — absurdly low and wrong, because the Pt you added on the left comes back on the right. See Catalysis.

Example 5 — Cell 5: Coefficients bigger than 1

Worked example Combustion-style balance — ammonia oxidation (Ostwald step)

$4 N H_{3} + 5 O_{2} \to 4 N O + 6 H_{2} O$ We want nitric oxide, $N O$ . Find the AE.

Forecast: Now every molecule has a coefficient. You must weigh all four $N H_{3}$ and all five $O_{2}$ . Miss a coefficient and your denominator is too small → AE too high. Guess: above or below 60%?

Molar masses of one molecule each. $M (N H_{3}) = 14 + 3 = 17$ ; $M (O_{2}) = 32$ ; $M (N O) = 14 + 16 = 30$ .
Multiply by coefficients for the reactant sum. $\sum M_{reactants} = 4 (17) + 5 (32) = 68 + 160 = 228.$ Why this step? The coefficient tells you how many moles react; four moles of $N H_{3}$ weigh $4 \times 17$ . Forgetting it is the #1 error here.
Desired product mass — also coefficient-weighted. We make $4$ mol $N O$ : $4 (30) = 120$ . Why weight the numerator too? Because we want all the useful product mass produced by that same balanced equation.
Divide. $% AE = \frac{120}{228} \times 100 \approx 52.6%$ .

Verify: Waste is the water, $6 (18) = 108$ . Check conservation: $120 + 108 = 228 =$ reactant sum. ✓ AE $= 52.6%$ ; water fraction $= 108/228 \times 100 \approx 47.4%$ ; they sum to 100%. ✓

Example 6 — Cell 6: Two routes, pick the greener one

Worked example Making methyl acetate (

$C H_{3} C O O C H_{3}$ , $M = 74$ ) two ways Route A (substitution): $C H_{3} C O C l + C H_{3} O H \to C H_{3} C O O C H_{3} + H C l$ Route B (addition, catalytic): $C H_{3} C O O H + C H_{3} O H H^{+} C H_{3} C O O C H_{3} + H_{2} O$ Compute AE for each and say which is greener. (In Route B the $H^{+}$ is a catalyst — exclude it.)

Forecast: Both make the SAME product ( $M = 74$ ). The one whose byproduct is lighter wins. $H C l$ ( $M = 36.5$ ) vs $H_{2} O$ ( $M = 18$ ) — guess which route has the higher AE.

Route A reactant sum. $M (C H_{3} C O C l) = 59 + ...$ compute: $12 + 3 + 12 + 16 + 35.5 = 78.5$ ; $M (C H_{3} O H) = 32$ . Sum $= 78.5 + 32 = 110.5$ .
Route A AE. $\frac{74}{110.5} \times 100 \approx 67.0%$ .
Route B reactant sum. $M (C H_{3} C O O H) = 60$ ; $M (C H_{3} O H) = 32$ ; $H^{+}$ excluded (catalyst). Sum $= 92$ .
Route B AE. $\frac{74}{92} \times 100 \approx 80.4%$ . Why compare? Same numerator (74), smaller waste in B (water, 18) → smaller denominator waste → higher AE. B is the greener route.

Verify: Route A byproduct $H C l$ : $110.5 - 74 = 36.5 = M (H C l)$ . ✓ Route B byproduct $H_{2} O$ : $92 - 74 = 18 = M (H_{2} O)$ . ✓ B wins as forecast.

Example 7 — Cell 7: Real-world word problem (grams given)

Worked example A plant runs the Example 2 reaction on scale

A factory feeds 190 g of $C H_{3} B r$ and the stoichiometric matching $N a O H$ into a hydrolysis reactor to make methanol. Ignoring yield losses, what mass of methanol is the maximum atom-economical output, and what is the AE?

Forecast: 190 g of $C H_{3} B r$ is exactly 2 moles ( $M = 95$ ). Since AE is a mass fraction, the AE percentage is identical to Example 2 — a scale-up doesn't change AE. Guess the methanol mass.

Moles of $C H_{3} B r$ . $\frac{190}{95} = 2$ mol. Why? Word problems give grams; convert to moles so stoichiometry (1:1 here) applies.
Moles of methanol (1:1). $2$ mol $C H_{3} O H$ .
Mass of methanol. $2 \times 32 = 64$ g.
AE (unchanged by scale). Still $\frac{32}{135} \times 100 \approx 23.7%$ . Why unchanged? AE is a ratio of molar masses, independent of how many moles you run.

Verify: Total reactant mass $= 2 (95) + 2 (40) = 270$ g. Useful product $= 64$ g. $64/270 \times 100 \approx 23.7%$ — matches the per-mole AE exactly. ✓

Example 8 — Cell 8: The limiting extremes

Worked example Bounding cases — how low and how high can AE go?

(i) A reaction where a huge protecting group ( $M = 300$ ) is attached then discarded to reveal a tiny target ( $M = 15$ , e.g. a $C H_{3}$ unit), from a reactant sum of $315$ . Find AE. (ii) A pure isomerisation $A \to B$ where $A$ rearranges into $B$ with the same formula ( $M = 100$ ). Find AE.

Forecast: (i) tiny product, giant waste → AE near the 0% wall. (ii) nothing leaves at all → AE at the 100% wall. These are the boundaries every other example lives between.

Case (i). $% AE = \frac{15}{315} \times 100 \approx 4.76%$ . Why so low? The protecting group carried away almost all the mass — this is exactly why Principle 8 (reduce derivatives) exists.
Case (ii). Isomerisation: reactant and product are the same atoms rearranged, so $M (B) = M (A) = 100$ . $% AE = \frac{100}{100} \times 100 = 100%$ . Why exactly 100%? No byproduct is possible when the formula is unchanged — every atom stays.

Verify: (i) $4.76% < 100%$ and $> 0$ — inside the walls. ✓ (ii) equals the upper wall exactly. ✓ No computed AE ever crossed 0% or 100%, confirming the bounds stated in the matrix.

Connections

Green Chemistry & Sustainability (parent)
Atom Economy and Yield — the yield-vs-AE contrast underpinning every cell
E-factor and Process Mass Intensity — the waste-side twin of these numbers
Catalysis — Example 4's exclusion rule
Renewable Feedstocks and Biomass — Principle 7 context
Solvent Selection and Supercritical CO2 — Principle 5 context
Activation Energy and Reaction Rates — why catalytic routes (Ex 4, 6) run milder