---
title: "Total Synthesis: Strategy and Classic Routes"
book: "University Chemistry — Year 3"
subject: chemistry
language: en
chapter: 30
exercises: 12
source: https://one-course.com/books/chemistry/4/en/chapter/30-total-synthesis-strategy-and-classic-routes
license: CC-BY-NC-SA-4.0
credit: "One Chemistry Book, One Course (one-course.com)"
---

# Chapter 30 — Total Synthesis: Strategy and Classic Routes

The anticancer drug paclitaxel was first isolated from the bark of the Pacific yew, a slow-growing tree; stripping the bark killed the tree, and the supply was far short of what patients needed. Chemists answered in two ways: total syntheses, which showed that the molecule could be built from simple starting materials but were far too long for production, and a [semisynthesis](#def-b3-total-synthesis-total) from a related compound extracted from the needles of a common yew, which is renewable. This chapter is about how such routes are judged and designed: how yields multiply, why convergence pays, what protecting groups cost, which bonds to make first, and three landmark syntheses read with those tools.

**You already know.**

The Year 2 volume treated retrosynthetic analysis (target molecule, disconnection, synthon, synthetic equivalent, functional-group interconversion), orthogonal protecting groups, the aldol reaction, the Robinson annulation, the Wittig reaction and iminium chemistry; the Year 1 volume protecting groups and yields. [Chapter 28](https://one-course.com/books/chemistry/4/en/chapter/28-asymmetric-synthesis#ch-b3-asymmetric-synthesis), [Chapter 26](https://one-course.com/books/chemistry/4/en/chapter/26-pericyclic-reactions#ch-b3-pericyclic) and [Chapter 21](https://one-course.com/books/chemistry/4/en/chapter/21-homogeneous-catalysis-in-industry#ch-b3-homogeneous-catalysis) supplied the modern key steps.

![Bark and needles of a Pacific yew, the first source of paclitaxel. Photograph: JOE BLOWE, CC BY-SA 2.0.](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-total-synthesis/img-0aef26a104df.jpg)

*Bark and needles of a Pacific yew, the first source of paclitaxel. Photograph: JOE BLOWE, CC BY-SA 2.0.*

## 30.1 Measuring a synthesis

**Definition 30.1 (Total synthesis).**

A *total synthesis* makes a complex molecule, often a natural product, from simple commercial compounds. A *formal synthesis* makes an intermediate of an earlier total synthesis, completing the route on paper. A *semisynthesis* starts from a natural compound already close to the target.

**Definition 30.2 (Route architecture).**

In a *linear synthesis* each step acts on the product of the one before; in a *convergent synthesis* separate fragments are made in parallel and joined late. The *longest linear sequence* (LLS) is the largest number of steps from any starting material to the target. The *overall yield* is the amount of target obtained divided by the amount of the limiting starting material.

**Proposition 30.3 (Overall yield).**

For a sequence of steps of yields $y_1, \dots, y_n$, the [overall yield](#def-b3-total-synthesis-architecture) is $Y = \prod_i y_i$.

**Proof.** Each step converts the amount it receives into $y_i$ times as much product; the amount reaching the target is the starting amount multiplied by every factor in turn. ∎

**Proposition 30.4 (Convergence).**

A route of $n = 2m + 1$ steps of equal yield $y$, built as two branches of $m$ steps joined by one coupling step, has the [overall yield](#def-b3-total-synthesis-architecture) $y^{m+1}$ along its [longest linear sequence](#def-b3-total-synthesis-architecture), against $y^{2m+1}$ for the linear route of the same steps.

**Proof.** Each branch is made separately, on as much starting material as needed: the material of each branch passes through $m$ steps and then the coupling, $m + 1$ factors $y$. In the linear route the material passes through all $2m + 1$. For $m = 10$ and $y = 0.90$: 31 % against 11 %. ∎

![Overall yield against the number of steps for three yields per step (solid: linear routes), and for convergent routes at 90 % per step (two equal branches joined in a last step; dashed). Twenty steps at 90 % leave 12 %; the same steps arranged convergently leave about three times more.](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-total-synthesis/fig-332d6695b17b.svg)

*[Overall yield](#def-b3-total-synthesis-architecture) against the number of steps for three yields per step (solid: linear routes), and for convergent routes at 90 % per step (two equal branches joined in a last step; dashed). Twenty steps at 90 % leave 12 %; the same steps arranged convergently leave about three times more.*

![Route trees: each circle is a compound, each arrow a step. A linear route passes all its material through every step; a convergent one makes two fragments side by side and couples them, so that its longest linear sequence is shorter.](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-total-synthesis/fig-98b89bb18d9b.svg)

*Route trees: each circle is a compound, each arrow a step. A linear route passes all its material through every step; a convergent one makes two fragments side by side and couples them, so that its [longest linear sequence](#def-b3-total-synthesis-architecture) is shorter.*

**Definition 30.5 (Economies and ideality).**

*Step economy* is the aim of reaching a target in as few steps as possible; *protecting-group economy*, that of using as few protection and deprotection steps as possible. The *ideality* of a route is the fraction of its steps that build the skeleton or make a necessary change of oxidation level: $(\text{construction steps} + \text{strategic redox steps})/(\text{total steps})$.

**Proposition 30.6 (Ideality of a route).**

A one-pot route that only forms skeletal bonds has [ideality](#def-b3-total-synthesis-economy) 100 %; each protection, deprotection, functional-group interconversion or non-strategic redox step lowers it.

**Proof.** Directly from the definition: such steps add to the denominator without adding to the numerator. ∎

## 30.2 Protecting groups and their economy

Each protecting group costs at least two steps, the reagents to put it on and take it off, and the material lost in both. A route of eight steps at 90 % keeps 43 % of its material; adding two protections and two deprotections at 95 % each brings it to 35 %. Protecting groups are chosen orthogonal, so that each can be removed without touching the others; better still, the order of steps is arranged so that the reactive group is not present, or not yet revealed, when it would interfere: protecting-group-free syntheses are a goal, not a curiosity.

**Method 30.7 (Reading a total synthesis).**

1. Identify the target’s rings, stereocentres and sensitive groups.
2. Find the key disconnections and the reactions that make them in the forward direction ( [strategic bonds](#def-b3-total-synthesis-strategic-bond) ).
3. Count the steps and find the [longest linear sequence](#def-b3-total-synthesis-architecture) ; note the convergence.
4. Note how each stereocentre is set (substrate control, auxiliary, catalyst, [chiral pool](https://one-course.com/books/chemistry/4/en/chapter/28-asymmetric-synthesis#def-b3-asymmetric-synthesis-chiral-pool) ).
5. List the protecting groups and other non-ideal steps; compute the [overall yield](#def-b3-total-synthesis-architecture) and the [ideality](#def-b3-total-synthesis-economy) .

**Method 30.8 (Counting steps).**

1. A step is one operation ending in an isolated (or at least worked-up) product; a one-pot sequence without isolation counts as one step.
2. Count all steps for the total; for the LLS, follow the longest path from a commercial compound to the target.
3. Multiply the yields along the LLS for the [overall yield](#def-b3-total-synthesis-architecture) .

## 30.3 Strategic bonds and key ring-forming steps

**Definition 30.9 (Strategic bond).**

A *strategic bond* of a target is a bond whose disconnection simplifies the target most, usually by splitting a ring system or removing several stereocentres at once, and which a reliable reaction can form.

Ring-forming reactions that make two bonds or several stereocentres at once are the workhorses: the Diels–Alder reaction (two C–C bonds, up to four stereocentres, stereospecifically, [Chapter 26](https://one-course.com/books/chemistry/4/en/chapter/26-pericyclic-reactions#ch-b3-pericyclic)), the Robinson annulation (a six-membered ring onto a ketone), [ring-closing metathesis](https://one-course.com/books/chemistry/4/en/chapter/21-homogeneous-catalysis-in-industry#def-b3-homogeneous-catalysis-metathesis) ([Chapter 21](https://one-course.com/books/chemistry/4/en/chapter/21-homogeneous-catalysis-in-industry#ch-b3-homogeneous-catalysis)) for medium and large rings, and cascades.

**Definition 30.10 (Cascade reaction).**

A *cascade reaction* is a sequence of transformations in which each step creates the functionality for the next, without isolating intermediates or adding reagents.

**Definition 30.11 (Biomimetic synthesis).**

A *biomimetic synthesis* imitates the route by which a natural product is thought to be made in the organism.

The cyclisation of squalene oxide to lanosterol in cells, four rings and seven stereocentres in one enzymatic cascade, inspired the polyene cyclisations of William Johnson, in which a cation started at one end of a carefully designed polyene closes ring after ring through chair-like transition states, setting the *trans* ring fusions of the steroids.

## 30.4 Landmark routes

**Definition 30.12 (Mannich reaction).**

The *Mannich reaction* is the addition of an enol or enolate to an iminium ion, made in situ from an aldehyde and a primary or secondary amine, giving a $\beta$-amino carbonyl compound.

In 1917 Robert Robinson made tropinone, the bicyclic core of atropine and cocaine, in one pot from three simple components in water, imagining that the plant might build it the same way: $\ce{C4H6O2 + CH3NH2 + C5H6O5 -> C8H13NO + 2CO2 + 2H2O}$ (succinaldehyde, methylamine and acetonedicarboxylic acid give tropinone).

![Robinson’s tropinone synthesis. Two Mannich reactions, the first intermolecular and the second closing the ring, join the three components; the two carboxylic acid groups that made the central CH2 groups enolisable then leave as carbon dioxide. In the bicyclic product the N-methyl bridge spans the seven-membered ring.](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-total-synthesis/fig-83a328e32496.svg)

*Robinson’s tropinone synthesis. Two [Mannich reactions](#def-b3-total-synthesis-mannich), the first intermolecular and the second closing the ring, join the three components; the two carboxylic acid groups that made the central $\ce{CH2}$ groups enolisable then leave as carbon dioxide. In the bicyclic product the N-methyl bridge spans the seven-membered ring.*

**Proposition 30.13 (Tropinone in one pot).**

The one-pot tropinone synthesis is two [Mannich reactions](#def-b3-total-synthesis-mannich) followed by two decarboxylations.

**Argued.** Methylamine and one aldehyde group of succinaldehyde give an iminium ion; an enol of acetonedicarboxylic acid adds to it (first Mannich). The secondary amine formed condenses with the second aldehyde group into a cyclic iminium ion, which the enol on the other side of the ketone attacks (second Mannich), closing the bicycle. The two $\beta$-keto acid groups then lose $\ce{CO2}$, as $\beta$-keto acids do on mild warming. ∎

Elias Corey’s synthesis of prostaglandin $\mathrm F_{2\alpha}$ (1969) made retrosynthetic analysis explicit. A bicyclo[2.2.1]heptenone built by a Diels–Alder reaction carries the relative configuration of the cyclopentane’s substituents; a [Baeyer–Villiger oxidation](https://one-course.com/books/chemistry/4/en/chapter/27-radicals-carbenes-and-rearrangements#def-b3-radicals-carbenes-baeyer) ([Chapter 27](https://one-course.com/books/chemistry/4/en/chapter/27-radicals-carbenes-and-rearrangements#ch-b3-radicals-carbenes)) and an iodolactonisation turn it into the Corey lactone, from which the two side chains are attached by olefinations. The lactone has served for many prostaglandin drugs since.

![Corey’s retrosynthesis of prostaglandin F_2, in outline (double arrows: “is made from”). The two side chains are disconnected first; the cyclopentane with its four stereocentres comes from a lactone, the lactone from a bicyclic ketone, and the ketone from a Diels–Alder reaction that sets their relative configuration.](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-total-synthesis/fig-269724cbb81b.svg)

*Corey’s retrosynthesis of prostaglandin $\mathrm F_{2\alpha}$, in outline (double arrows: “is made from”). The two side chains are disconnected first; the cyclopentane with its four stereocentres comes from a lactone, the lactone from a bicyclic ketone, and the ketone from a Diels–Alder reaction that sets their relative configuration.*

Oseltamivir, the influenza drug, is made industrially from shikimic acid, a chiral-pool compound extracted from star anise and later produced by fermentation: its six-membered ring and stereocentres are already in place. The route converts the ring diol into an epoxide, opens it with azide, closes an aziridine and opens that with azide again, setting the two nitrogen atoms *trans*; acetylation and reduction finish the job. Because sodium azide and organic azides are toxic and potentially explosive on a large scale, azide-free routes were later developed.

![Oseltamivir from shikimic acid, key stages only: the natural ring and its stereocentres are kept, and two nitrogen atoms are introduced trans to each other by successive openings of an epoxide and an aziridine with azide.](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-total-synthesis/fig-baea324844c3.svg)

*Oseltamivir from shikimic acid, key stages only: the natural ring and its stereocentres are kept, and two nitrogen atoms are introduced *trans* to each other by successive openings of an epoxide and an aziridine with azide.*

## 30.5 From discovery to process

A route that delivers milligrams for biological testing is rarely the one used for tonnes. Process chemists scout routes for cost, safety and robustness: they remove chromatography in favour of crystallisation, replace hazardous reagents and cryogenic temperatures, shorten the sequence, check the thermal safety of every step by calorimetry, and set specifications for impurities. These choices meet the measures of [Chapter 31](https://one-course.com/books/chemistry/4/en/chapter/31-green-chemistry-and-industrial-processes#ch-b3-green-industrial).

**In the lab — From one gram to one kilogram.**

A step run on one gram in a round-bottomed flask is repeated at ten grams in a jacketed reactor with an overhead stirrer and a temperature probe in the liquid, adding the reactive reagent at a controlled rate while watching the internal temperature: the heat released grows with the volume, but the area that removes it grows only as the square of the size. Only when the heat flow and the work-up (phase separations, filtrations, drying) behave is the step run at a kilogram.

**Safety.**

![](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-total-synthesis/fig-8f09793f6441.svg)

![](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-total-synthesis/fig-ce8aa75a0f1b.svg)

![](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-total-synthesis/fig-584c0be6faf1.svg)

Sodium azide: fatal if swallowed, very toxic to aquatic life; with acids it releases hydrazoic acid, a toxic and explosive gas, and with heavy metals (copper and lead in drains) it forms explosive azides. Waste is destroyed chemically, never poured away.

**History — Robinson, Woodward, Corey.**

Robert Robinson received the 1947 Nobel Prize in Chemistry for his work on alkaloids and other plant products; his tropinone synthesis of 1917 is still quoted for its economy. Robert Woodward’s syntheses of quinine, cholesterol, chlorophyll and, with Albert Eschenmoser, vitamin $\mathrm B_{12}$ made [total synthesis](#def-b3-total-synthesis-total) an art (1965 prize). Elias Corey formalised retrosynthetic analysis and received the 1990 prize for the theory and methodology of organic synthesis.

## 30.6 Exercises

**Exercise 30.1 ★.**

Compute the [overall yield](#def-b3-total-synthesis-architecture) of 10- and 20-step linear routes at 90 % and at 80 % per step.

**Solution of Exercise 30.1.**

At 90 %: $0.9^{10} = 35$ % and $0.9^{20} = 12$ %. At 80 %: $0.8^{10} = 11$ % and $0.8^{20} = 1.2$ %.

**Exercise 30.2 ★.**

A route makes fragment A in five steps and fragment B in three, couples them, and needs four more steps to the target. Give the total number of steps and the [longest linear sequence](#def-b3-total-synthesis-architecture).

**Solution of Exercise 30.2.**

$5 + 3 + 1 + 4 = 13$ steps in all; LLS $5 + 1 + 4 = 10$.

**Exercise 30.3 ★.**

Give the product of the [Mannich reaction](#def-b3-total-synthesis-mannich) of acetone, formaldehyde and dimethylamine.

**Solution of Exercise 30.3.**

4-(Dimethylamino)butan-2-one, $\ce{CH3COCH2CH2N(CH3)2}$: the enol of acetone adds to the iminium ion $\ce{CH2=N+(CH3)2}$.

**Exercise 30.4 ★.**

Give a Diels–Alder disconnection of 4-acetylcyclohexene, and the forward reaction.

**Solution of Exercise 30.4.**

Disconnect the two ring bonds allylic to the C=C: buta-1,3-diene and but-3-en-2-one (methyl vinyl ketone), which react on heating.

**Exercise 30.5 ★★.**

A 21-step linear route runs at 90 % per step. Compute its [overall yield](#def-b3-total-synthesis-architecture) and that of a convergent version with two 10-step branches and one coupling.

**Solution of Exercise 30.5.**

Linear: $0.9^{21} = 11$ %. Convergent: $0.9^{11} = 31$ %, almost three times as much.

**Exercise 30.6 ★★.**

Route A has 12 steps, of which 7 make skeletal bonds and 1 is a strategic reduction; route B has 9 steps, 7 of them skeletal. Compute their idealities.

**Solution of Exercise 30.6.**

A: $(7 + 1)/12 = 67$ %. B: $7/9 = 78$ %.

**Exercise 30.7 ★★.**

An eight-step route at 90 % per step needs two protecting groups, each put on and taken off at 95 %. Compute the [overall yield](#def-b3-total-synthesis-architecture) with and without them.

**Solution of Exercise 30.7.**

Without: $0.9^8 = 43$ %. With four extra steps at 95 %: $43\,\% \times 0.95^4 = 35$ %.

**Exercise 30.8 ★★.**

A triol has a primary, a secondary and a tertiary hydroxy group. Propose protecting groups that allow each to be revealed in turn, and their removal conditions.

**Solution of Exercise 30.8.**

Primary: a bulky silyl ether (*tert*-butyldiphenylsilyl), removed with fluoride. Secondary: a 4-methoxybenzyl ether, removed by oxidation with DDQ. The tertiary alcohol, hindered, can often be left free, or protected as a small silyl ether cleaved by mild acid. Each condition leaves the other groups untouched.

**Exercise 30.9 ★★.**

Why was paclitaxel produced by [semisynthesis](#def-b3-total-synthesis-total) rather than by any of its total syntheses?

**Solution of Exercise 30.9.**

The total syntheses needed tens of steps with [overall yields](#def-b3-total-synthesis-architecture) far below 1 %; a compound with the full ring system and most stereocentres, extracted from the needles of the European yew, a renewable source, is converted in a few steps.

**Exercise 30.10 ★★★.**

2-Methylcyclohexane-1,3-dione reacts with methyl vinyl ketone in a Robinson annulation. Identify the Michael and aldol steps and give the bicyclic enone formed (the Wieland–Miescher ketone).

**Solution of Exercise 30.10.**

Michael addition: the enolate of the dione (C2, between the carbonyls) adds to the $\beta$-carbon of methyl vinyl ketone, giving a triketone. Aldol: the methyl ketone’s enolate attacks a ring carbonyl intramolecularly, and dehydration gives the cyclohexenone: the Wieland–Miescher ketone, a bicyclic enedione with an angular methyl group.

**Exercise 30.11 ★★★.**

Show that the convergent route of [Proposition 30.4](#prop-b3-total-synthesis-convergent) keeps $y^{-m}$ times more material than the linear one, and evaluate this factor for $m = 5$ and $m = 15$ at 90 %.

**Solution of Exercise 30.11.**

$y^{m+1}/y^{2m+1} = y^{-m}$: at 90 %, 1.7 for $m = 5$ and 4.9 for $m = 15$. The longer the route, the more convergence pays.

**Exercise 30.12 ★★★.**

In a polyene cyclisation, why do chair-like transition states give *trans* ring fusions, and why does the geometry (*E* or *Z*) of each double bond matter?

**Solution of Exercise 30.12.**

Each ring closes with the cation and the attacking double bond in a chair-like arrangement; addition across each double bond is anti, and the substituents take pseudo-equatorial positions, which makes every ring fusion *trans*. An *E* double bond places its two chain continuations so that the *trans* fusion results; a *Z* one would give the *cis* fusion: the geometry of the polyene encodes the stereochemistry of the product.

## 30.7 Problem: Tropinone in One Pot

**Problem 30.1.**

Weekend problem — tropinone in one pot: the three components and the balanced equation, the two Mannich reactions, a comparison with a long linear route, and the stereochemistry of tropinone and its reduction products

Data of the problem: the one-pot synthesis, run in buffered water near pH 5, gives tropinone in 42 % yield; an older linear route needs 14 steps averaging 75 %.

**Part I — Components and equation.**

1. Give the formulas of succinaldehyde, methylamine, acetonedicarboxylic acid and tropinone.
2. Write the balanced equation.
3. Compute the molar masses of the reactants and of tropinone.
4. Compute the atom economy.
5. Why is acetonedicarboxylic acid used instead of acetone?
6. Why does buffering near neutral pH help?

**Part II — Mechanism.**

7. Write the formation of the first iminium ion.
8. Show the first [Mannich reaction](#def-b3-total-synthesis-mannich) .
9. How does the second iminium ion form?
10. Show the second, ring-closing [Mannich reaction](#def-b3-total-synthesis-mannich) .
11. Why do the two carboxylic acid groups leave as $\ce{CO2}$ ?
12. Count the C–C and C–N bonds formed.

**Part III — One pot against fourteen steps.**

13. Compute the [overall yield](#def-b3-total-synthesis-architecture) of the linear route.
14. Compute the ratio of the one-pot yield to it.
15. Compute the [ideality](#def-b3-total-synthesis-economy) of the one-pot route.
16. What else besides yield favours the one-pot route?
17. Why is the one-pot synthesis called biomimetic?
18. Name another reaction class of this chapter that makes several bonds at once.

**Part IV — Stereochemistry.**

19. Is tropinone chiral? Find its [symmetry element](https://one-course.com/books/chemistry/4/en/chapter/4-symmetry-and-point-groups#def-b3-point-groups-operation) .
20. Why are the two bridgehead carbons stereocentres nonetheless?
21. Reduction of the ketone gives two alcohols, tropine and pseudotropine. What is their relationship?
22. Are tropine and pseudotropine chiral?
23. Why does a hydride approach the carbonyl group preferentially from one face?
24. State the result: the ratio of the one-pot yield to the linear route’s [overall yield](#def-b3-total-synthesis-architecture) .

**Solution of Problem 30.1.**

**1.** $\ce{C4H6O2}$, $\ce{CH5N}$, $\ce{C5H6O5}$, $\ce{C8H13NO}$.

**2.** $\ce{C4H6O2 + CH5N + C5H6O5 -> C8H13NO + 2CO2 + 2H2O}$.

**3.** 86.0, 31.0 and $146.0\,\mathrm{g}/\mathrm{mol}$; tropinone $139.0\,\mathrm{g}/\mathrm{mol}$.

**4.** $139.0/(86.0 + 31.0 + 146.0) = 0.53$: 53 %.

**5.** Its central $\ce{CH2}$ groups, each between two carbonyl groups, enolise readily in water; acetone itself is too weakly nucleophilic, and the activating carboxyl groups leave afterwards.

**6.** The iminium ions need free amine (too much acid protonates it), and the enol and the iminium formation need some acid: a pH near neutrality balances them.

**7.** $\ce{CH3NH2}$ adds to one aldehyde group; loss of water gives the iminium ion $\ce{R-CH=N+H-CH3}$.

**8.** An enol carbon of acetonedicarboxylic acid attacks the iminium carbon, making a C–C bond and a secondary amine.

**9.** The secondary amine condenses with the second aldehyde group of the same chain, giving a cyclic (five-membered) iminium ion.

**10.** The enol on the other side of the ketone attacks it, closing the six-membered ring and the bicyclic skeleton.

**11.** Each is a $\beta$-keto acid, which loses $\ce{CO2}$ through a cyclic six-membered transition state, giving the enol of the ketone.

**12.** Two C–C bonds and two C–N bonds.

**13.** $0.75^{14} = 0.018$: 1.8 %.

**14.** $42/1.78 = 24$.

**15.** 100 %: one step, all bond construction.

**16.** One operation, water as solvent, cheap components, no protecting groups and almost no waste besides carbon dioxide and water.

**17.** The plant builds the tropane skeleton from an amino acid-derived cyclic iminium ion and an acetate-derived unit, through the same kind of Mannich chemistry.

**18.** Cascades (and ring-forming [cycloadditions](https://one-course.com/books/chemistry/4/en/chapter/26-pericyclic-reactions#def-b3-pericyclic-families) such as the Diels–Alder reaction).

**19.** No: a [mirror plane](https://one-course.com/books/chemistry/4/en/chapter/4-symmetry-and-point-groups#def-b3-point-groups-mirror) passes through N, its methyl group, C3 and the carbonyl oxygen.

**20.** Each has four different groups, but they are mirror images of each other in the molecule’s own plane of symmetry: a meso-type compound.

**21.** Diastereomers, differing in the orientation of the OH on C3 relative to the nitrogen bridge.

**22.** No: the [mirror plane](https://one-course.com/books/chemistry/4/en/chapter/4-symmetry-and-point-groups#def-b3-point-groups-mirror) is kept in both.

**23.** The two faces of the carbonyl group are not equivalent: one faces the nitrogen bridge, the other the two-carbon bridge, which hinder differently.

**24.** The one-pot route gives about 24 times the [overall yield](#def-b3-total-synthesis-architecture) of the 14-step linear route.