---
title: "Stereochemistry in Depth"
book: "University Chemistry — Year 1"
subject: chemistry
language: en
chapter: 16
exercises: 12
source: https://one-course.com/books/chemistry/2/en/chapter/16-stereochemistry-in-depth
license: CC-BY-NC-SA-4.0
credit: "One Chemistry Book, One Course (one-course.com)"
---

# Chapter 16 — Stereochemistry in Depth

Crush a few caraway seeds, then a spearmint leaf: two quite different smells. The main odorant of each is carvone, $\ce{C10H14O}$, the same atoms joined in the same order — but the two carvones are mirror images of each other, as a left hand is of a right hand, and the receptors of the nose, themselves built of mirror-asymmetric molecules, tell them apart. Medicines, flavours and the molecules of life are all built in three dimensions, and a structural formula that ignores the third dimension misses half the story. This chapter gives the tools to draw molecules in space, to name each arrangement without ambiguity, to follow the rotations that molecules perform constantly, and to measure the one physical property that tells mirror images apart.

**You already know.**

The school volume (grade 12) introduced [chiral](#def-b1-stereochemistry-in-depth-stereocentre) molecules, the asymmetric carbon, [enantiomers](#def-b1-stereochemistry-in-depth-enantiomers), $Z$/$E$ isomers and the [conformations](#def-b1-stereochemistry-in-depth-isomers) of ethane. The tetrahedral arrangement of four bonds around carbon and the $\sigma$/$\pi$ description of single and double bonds come from [Chapter 3](https://one-course.com/books/chemistry/2/en/chapter/3-lewis-structures-resonance-and-vsepr#ch-b1-lewis-resonance-vsepr).

![Caraway seeds and spearmint leaves. Caraway owes its smell mainly to (S)-carvone, spearmint to its mirror image, (R)-carvone.](https://one-course.com/images/onecourse/chapters/chemistry-2/b1-stereochemistry-in-depth/img-a25c80581a43.jpg)

*Caraway seeds and spearmint leaves. Caraway owes its smell mainly to (*S*)-carvone, spearmint to its mirror image, (*R*)-carvone.*

## 16.1 Isomers and representations

**Definition 16.1 (Constitution, configuration, conformation).**

Two compounds of the same formula whose atoms are joined in a different order are *constitutional isomers*. Compounds with the same constitution whose atoms differ only in their arrangement in space are *stereoisomers*. The arrangement that can be changed only by breaking bonds is the *configuration* of a stereoisomer; an arrangement obtained by rotation about single bonds, without breaking any, is a *conformation*.

At room temperature, rotations about single bonds take place billions of times a second: [conformations](#def-b1-stereochemistry-in-depth-isomers) interconvert and cannot be isolated, while [configurations](#def-b1-stereochemistry-in-depth-isomers) are stable and give distinct compounds.

**Definition 16.2 (Representations).**

In the *Cram representation* two bonds are drawn in the plane of the page, a solid wedge points towards the reader and a hashed wedge away. A *Newman projection* views a molecule along one C–C bond: the front carbon is a point where its three other bonds meet, the rear carbon a circle from whose rim its bonds emerge. In a *Fischer projection* the carbon chain is vertical, horizontal bonds point towards the reader and vertical bonds away; each crossing is a stereogenic carbon.

![Left: Newman projections of ethane along its C–C bond, staggered (rear bonds between the front ones) and eclipsed (rear bonds hidden behind the front ones, drawn slightly offset). Right: the same molecule, (R)-lactic acid, as a Fischer projection and in Cram representation (the vertical bonds of the Fischer cross point away from the reader, the horizontal ones towards the reader).](https://one-course.com/images/onecourse/chapters/chemistry-2/b1-stereochemistry-in-depth/fig-3d6f136cc9dd.svg)

*Left: [Newman projections](#def-b1-stereochemistry-in-depth-representations) of ethane along its C–C bond, staggered (rear bonds between the front ones) and eclipsed (rear bonds hidden behind the front ones, drawn slightly offset). Right: the same molecule, (*R*)-lactic acid, as a [Fischer projection](#def-b1-stereochemistry-in-depth-representations) and in [Cram representation](#def-b1-stereochemistry-in-depth-representations) (the vertical bonds of the Fischer cross point away from the reader, the horizontal ones towards the reader).*

**Method 16.3 (From Cram to Newman).**

1. Choose the bond to look along and the end nearer to the eye.
2. Draw the three bonds of the front carbon from the centre, keeping their order (clockwise or anticlockwise) as seen from the eye.
3. Draw the rear carbon’s bonds from the circle, with their order as seen from the same side.
4. Check: rotating the rear carbon by $60^\circ$ passes from staggered to eclipsed; by $120^\circ$ , from one [staggered conformation](#def-b1-stereochemistry-in-depth-conformations) to the next.

## 16.2 Configuration: the CIP rules

**Definition 16.4 (CIP rules, descriptors).**

The *CIP rules* (Cahn–Ingold–Prelog) rank the four substituents of a [stereogenic centre](#def-b1-stereochemistry-in-depth-stereocentre): higher atomic number first; on a tie, compare the sets of atoms attached next, and so on outwards; a double bond counts as two single bonds to duplicated atoms. With the lowest-ranked substituent pointing away from the eye, the sequence $1 \to 2 \to 3$ turning clockwise defines the descriptor (*R*), and anticlockwise (*S*). For a double bond with two substituents on each carbon, (*Z*) means the two higher-ranked ones are on the same side and (*E*) on opposite sides.

**Method 16.5 (Assigning (R) or (S)).**

1. Rank the four substituents by the [CIP rules](#def-b1-stereochemistry-in-depth-cip) .
2. If the lowest-ranked one points away from you (hashed wedge, or a vertical bond of a [Fischer projection](#def-b1-stereochemistry-in-depth-representations) ), read the turn of $1 \to 2 \to 3$ directly.
3. If it points towards you (solid wedge, horizontal Fischer bond), read the turn and reverse the descriptor.
4. Exchanging any two substituents reverses the descriptor: a quick check.

In lactic acid, $\ce{OH} > \ce{COOH} > \ce{CH3} > \ce{H}$ (O; then C(O,O,O) beats C(H,H,H)). In the [Fischer projection](#def-b1-stereochemistry-in-depth-representations) above, H is horizontal (towards us) and $\ce{OH} \to \ce{COOH} \to \ce{CH3}$ turns anticlockwise: reversed, the centre is (*R*).

## 16.3 Chirality and its consequences

**Definition 16.6 (Stereogenic centre, chirality).**

An object is *chiral* if it cannot be superposed on its mirror image, *achiral* otherwise; the property is *chirality*. A *stereogenic centre* is an atom at which exchanging two substituents gives a stereoisomer; a tetrahedral carbon with four different substituents is one.

A molecule with a plane of symmetry or a centre of symmetry is [achiral](#def-b1-stereochemistry-in-depth-stereocentre); a molecule with neither is [chiral](#def-b1-stereochemistry-in-depth-stereocentre) in practice. The exact criterion, in terms of symmetry operations, is given in the Year 3 volume.

**Definition 16.7 (Enantiomers, diastereomers, meso compound, racemic mixture).**

Two [stereoisomers](#def-b1-stereochemistry-in-depth-isomers) that are mirror images of each other are *enantiomers*; [stereoisomers](#def-b1-stereochemistry-in-depth-isomers) that are not are *diastereomers*. A *meso compound* has [stereogenic centres](#def-b1-stereochemistry-in-depth-stereocentre) but is [achiral](#def-b1-stereochemistry-in-depth-stereocentre). An equimolar mixture of two enantiomers is a *racemic mixture*.

**Proposition 16.8 (Number of stereoisomers).**

A molecule with $n$ [stereogenic centres](#def-b1-stereochemistry-in-depth-stereocentre) (and no other stereogenic unit) has at most $2^n$ [stereoisomers](#def-b1-stereochemistry-in-depth-isomers), in at most $2^{n-1}$ pairs of [enantiomers](#def-b1-stereochemistry-in-depth-enantiomers); meso forms make the number smaller.

**Proof.** Each centre is (*R*) or (*S*) independently: $2^n$ combinations. The mirror image changes every descriptor, so the combinations pair up into [enantiomers](#def-b1-stereochemistry-in-depth-enantiomers), $2^{n-1}$ pairs. If a combination is its own mirror image after a rotation of the molecule (a meso form), its pair collapses into one compound. ∎

Tartaric acid, $\ce{HOOC-CH(OH)-CH(OH)-COOH}$, has two equivalent centres: (*R,R*) and (*S,S*) are [enantiomers](#def-b1-stereochemistry-in-depth-enantiomers), while (*R,S*) has a mirror plane between the two carbons and is meso: three [stereoisomers](#def-b1-stereochemistry-in-depth-isomers), not four.

**Proposition 16.9 (Properties of enantiomers).**

Two [enantiomers](#def-b1-stereochemistry-in-depth-enantiomers) have the same melting and boiling points, solubilities in [achiral](#def-b1-stereochemistry-in-depth-stereocentre) solvents, spectra and reactivity towards [achiral](#def-b1-stereochemistry-in-depth-stereocentre) reagents; they differ in their action on polarised light and in their interactions with other [chiral](#def-b1-stereochemistry-in-depth-stereocentre) molecules. [Diastereomers](#def-b1-stereochemistry-in-depth-enantiomers) differ in all their properties.

**Proof.** Every property that depends only on distances and angles inside the molecule, or between it and [achiral](#def-b1-stereochemistry-in-depth-stereocentre) partners, is the same for an object and its mirror image, since a reflection preserves distances and angles. A [chiral](#def-b1-stereochemistry-in-depth-stereocentre) partner (a receptor, a [chiral](#def-b1-stereochemistry-in-depth-stereocentre) reagent, the helical path of polarised light described below) makes pairs that are diastereomeric, no longer mirror images, which therefore differ. [Diastereomers](#def-b1-stereochemistry-in-depth-enantiomers) are not related by any reflection: their internal distances differ. ∎

![The two carvones, drawn on the same skeleton: the isopropenyl group on C5 points towards the reader (solid wedge) in one and away from the reader (hashed wedge) in the other, the hydrogen of C5 (not drawn) on the opposite side. They are mirror images of each other: enantiomers ().](https://one-course.com/images/onecourse/chapters/chemistry-2/b1-stereochemistry-in-depth/fig-fc58dac59927.svg)

*The two carvones, drawn on the same skeleton: the isopropenyl group on C5 points towards the reader (solid wedge) in one and away from the reader (hashed wedge) in the other, the hydrogen of C5 (not drawn) on the opposite side. They are mirror images of each other: [enantiomers](#def-b1-stereochemistry-in-depth-enantiomers) ([Exercise 16.4](#exo-b1-stereochemistry-in-depth-4)).*

**History — A tetrahedral carbon.**

![](https://one-course.com/images/onecourse/chapters/chemistry-2/b1-stereochemistry-in-depth/img-e19167e924bf.jpg)

Jacobus Henricus van ’t Hoff proposed, as a young chemist, that the four bonds of a carbon atom point to the corners of a tetrahedron. A carbon with four different substituents then exists as two mirror-image arrangements, which explained why some compounds come in two forms that differ only in the way they rotate polarised light. The idea is the starting point of every drawing in this chapter. (Photograph published in 1911, public domain, Wikimedia Commons.)

## 16.4 Conformations

**Definition 16.10 (Torsion angle and conformations).**

For four atoms A–B–C–D, the *torsion angle* $\theta$ about B–C is the angle between the planes ABC and BCD, read on a [Newman projection](#def-b1-stereochemistry-in-depth-representations) along B–C. A [conformation](#def-b1-stereochemistry-in-depth-isomers) is an *eclipsed conformation* when the bonds of the front and rear carbons are aligned in projection, a *staggered conformation* when they alternate. In butane, viewed along C2–C3, the staggered [conformation](#def-b1-stereochemistry-in-depth-isomers) with the two methyl groups at $180^\circ$ is the *anti conformation*, those at $\pm 60^\circ$ are *gauche conformations*. In cyclohexane, the strain-free puckered ring is the *chair*; each carbon carries one *axial* bond, parallel to the ring’s axis, and one *equatorial* bond, roughly in the ring’s mean plane. The *ring flip* converts one chair into the other, every axial bond becoming equatorial and the reverse.

![Torsional energy of butane about its central bond (solid; = 0 when the methyl groups eclipse each other) and of ethane (dashed), computed from experimental torsional potentials. Ethane: barrier 12.2\, kJ/ mol. Butane: anti most stable; gauche 2.8\, kJ/ mol higher, at = 62; barriers 15.2\, kJ/ mol (methyl eclipsing hydrogen) and 16.5\, kJ/ mol (methyl eclipsing methyl).](https://one-course.com/images/onecourse/chapters/chemistry-2/b1-stereochemistry-in-depth/fig-119126b559e9.svg)

*Torsional energy of butane about its central bond (solid; $\theta
= 0$ when the methyl groups eclipse each other) and of ethane (dashed), computed from experimental torsional potentials. Ethane: barrier $12.2\,\mathrm{kJ}/\mathrm{mol}$. Butane: anti most stable; gauche $2.8\,\mathrm{kJ}/\mathrm{mol}$ higher, at $\theta = 62^\circ$; barriers $15.2\,\mathrm{kJ}/\mathrm{mol}$ (methyl eclipsing hydrogen) and $16.5\,\mathrm{kJ}/\mathrm{mol}$ (methyl eclipsing methyl).*

The barriers are small compared with the thermal energy available in collisions ($RT = 2.5\,\mathrm{kJ}/\mathrm{mol}$ at room temperature, and much more in the tail of the distribution, [Chapter 9](https://one-course.com/books/chemistry/2/en/chapter/9-reaction-mechanisms-elementary-steps-and-approximations#ch-b1-elementary-steps)): rotation is fast, and butane is a mixture of anti and gauche molecules that cannot be separated.

**Method 16.11 (Drawing a chair).**

1. Draw two parallel lines slanting slightly downwards to the right, offset; join their ends with a V at the left and an inverted V at the right to close the six-membered ring.
2. [Axial](#def-b1-stereochemistry-in-depth-conformations) bonds are vertical: up on the carbons that point up, down on those that point down, alternating round the ring.
3. [Equatorial](#def-b1-stereochemistry-in-depth-conformations) bonds are parallel to the ring bonds once removed, and point outwards.
4. To flip the ring, draw the other [chair](#def-b1-stereochemistry-in-depth-conformations) ; a substituent [axial](#def-b1-stereochemistry-in-depth-conformations) on one is [equatorial](#def-b1-stereochemistry-in-depth-conformations) on the other.

![Left: the chair of cyclohexane with its six axial bonds (orange, alternately up and down) and six equatorial bonds (blue). Right: the two chairs of methylcyclohexane, interconverted by the ring flip; the methyl group is equatorial in one, axial in the other.](https://one-course.com/images/onecourse/chapters/chemistry-2/b1-stereochemistry-in-depth/fig-63a0ca9ff47f.svg)

*Left: the [chair](#def-b1-stereochemistry-in-depth-conformations) of cyclohexane with its six [axial](#def-b1-stereochemistry-in-depth-conformations) bonds (orange, alternately up and down) and six [equatorial](#def-b1-stereochemistry-in-depth-conformations) bonds (blue). Right: the two [chairs](#def-b1-stereochemistry-in-depth-conformations) of methylcyclohexane, interconverted by the [ring flip](#def-b1-stereochemistry-in-depth-conformations); the methyl group is [equatorial](#def-b1-stereochemistry-in-depth-conformations) in one, [axial](#def-b1-stereochemistry-in-depth-conformations) in the other.*

**Proposition 16.12 (Equatorial substituents are preferred).**

In a monosubstituted cyclohexane the [chair](#def-b1-stereochemistry-in-depth-conformations) with the substituent [equatorial](#def-b1-stereochemistry-in-depth-conformations) is the more stable. For a methyl group, the axial position adds two gauche interactions with ring carbons, each worth that of butane ($2.8\text{ to }2.9\,\mathrm{kJ}/\mathrm{mol}$), about $5.7\,\mathrm{kJ}/\mathrm{mol}$ in all: an estimate of the energy difference.

**Proof.** Viewed along the C1–C2 bond, an [axial](#def-b1-stereochemistry-in-depth-conformations) methyl on C1 is gauche to C3 of the ring; viewed along C1–C6, gauche to C5: two gauche butane-like interactions, which are also seen as the crowding of the methyl with the [axial](#def-b1-stereochemistry-in-depth-conformations) hydrogens on C3 and C5 (the 1,3-diaxial interactions). An [equatorial](#def-b1-stereochemistry-in-depth-conformations) methyl is anti to C3 and C5. Each gauche interaction costs what it costs in butane, $2.8\,\mathrm{kJ}/\mathrm{mol}$ in the torsional potential of the figure. ∎

With $\Delta E = 5.7\,\mathrm{kJ}/\mathrm{mol}$, the ratio of [equatorial](#def-b1-stereochemistry-in-depth-conformations) to [axial](#def-b1-stereochemistry-in-depth-conformations) [chairs](#def-b1-stereochemistry-in-depth-conformations) at $25\,{}^{\circ}\mathrm{C}$ is $\exp(\Delta E/RT) = 10$: about $91\,\%$ [equatorial](#def-b1-stereochemistry-in-depth-conformations) ([Exercise 16.9](#exo-b1-stereochemistry-in-depth-9)). Large groups, such as tert-butyl, lock the ring in the [chair](#def-b1-stereochemistry-in-depth-conformations) where they are [equatorial](#def-b1-stereochemistry-in-depth-conformations).

## 16.5 Optical activity

**Definition 16.13 (Optical activity, specific rotation).**

A substance shows *optical activity* if it rotates the plane of polarisation of linearly polarised light passing through it. Seen by an observer facing the light, a *dextrorotatory* substance, written $(+)$, rotates it clockwise, a *laevorotatory* one, $(-)$, anticlockwise. The *specific rotation* is $[\alpha] = \alpha/(\ell c)$, where $\alpha$ is the measured angle in degrees, $\ell$ the path length in decimetres and $c$ the mass concentration in $\mathrm{g}/\mathrm{mL}$; it depends on the wavelength (usually the yellow sodium line), the temperature and the solvent.

![A polarimeter. The polariser lets through light vibrating in one plane; the chiral sample rotates that plane by ; the analyser is turned until the light is extinguished again, which measures .](https://one-course.com/images/onecourse/chapters/chemistry-2/b1-stereochemistry-in-depth/fig-2f99af1f3fb3.svg)

*A polarimeter. The polariser lets through light vibrating in one plane; the [chiral](#def-b1-stereochemistry-in-depth-stereocentre) sample rotates that plane by $\alpha$; the analyser is turned until the light is extinguished again, which measures $\alpha$.*

**Proposition 16.14 (Biot’s law).**

For a dilute solution of several optically active solutes, the angle of rotation is additive, $\alpha = \ell\sum_i [\alpha]_i c_i$ (*Biot’s law*). Two [enantiomers](#def-b1-stereochemistry-in-depth-enantiomers) have opposite [specific rotations](#def-b1-stereochemistry-in-depth-optical-activity); a [racemic mixture](#def-b1-stereochemistry-in-depth-enantiomers) does not rotate light.

**Proof.** Each solute contributes independently in dilute solution, in proportion to the number of molecules crossed, that is to $\ell c_i$; this is the experimental content of the law. A reflection changes clockwise into anticlockwise: the mirror-image molecule rotates the plane by the opposite angle, and in a [racemic mixture](#def-b1-stereochemistry-in-depth-enantiomers) the two contributions cancel. ∎

For a mixture of two [enantiomers](#def-b1-stereochemistry-in-depth-enantiomers) at total concentration $c$, the fraction $x$ of the $(-)$ form gives $\alpha = \ell c[\alpha]_{(-)}(x - (1 - x)) =
\ell c[\alpha]_{(-)}(2x - 1)$: measuring $\alpha$ gives the composition. Optical rotation is not linked in any simple way to the descriptor: an (*R*) compound may be $(+)$ or $(-)$; spearmint’s (*R*)-carvone is [laevorotatory](#def-b1-stereochemistry-in-depth-optical-activity).

## 16.6 Exercises

**Exercise 16.1 ★.**

Classify each pair as [constitutional isomers](#def-b1-stereochemistry-in-depth-isomers), [enantiomers](#def-b1-stereochemistry-in-depth-enantiomers), [diastereomers](#def-b1-stereochemistry-in-depth-enantiomers) or two [conformations](#def-b1-stereochemistry-in-depth-isomers) of one compound: butan-1-ol and butan-2-ol; (*R*)- and (*S*)-butan-2-ol; (*Z*)- and (*E*)-but-2-ene; the anti and [gauche conformations](#def-b1-stereochemistry-in-depth-conformations) of butane; (*R,R*)- and (*R,S*)-tartaric acid.

**Solution of Exercise 16.1.**

[Constitutional isomers](#def-b1-stereochemistry-in-depth-isomers); [enantiomers](#def-b1-stereochemistry-in-depth-enantiomers); [diastereomers](#def-b1-stereochemistry-in-depth-enantiomers) ([stereoisomers](#def-b1-stereochemistry-in-depth-isomers) that are not mirror images); two [conformations](#def-b1-stereochemistry-in-depth-isomers) of one compound; [diastereomers](#def-b1-stereochemistry-in-depth-enantiomers) ((*R,S*) is the meso form).

**Exercise 16.2 ★.**

Rank by the [CIP rules](#def-b1-stereochemistry-in-depth-cip): $\ce{-OH}$, $\ce{-NH2}$, $\ce{-CH3}$, $\ce{-H}$; then $\ce{-CH2OH}$, $\ce{-CHO}$, $\ce{-COOH}$, $\ce{-CH(CH3)2}$; then $\ce{-Cl}$, $\ce{-CH2Cl}$, $\ce{-CCl3}$, $\ce{-Br}$.

**Solution of Exercise 16.2.**

$\ce{-OH}$ $>$ $\ce{-NH2}$ $>$ $\ce{-CH3}$ $>$ $\ce{-H}$ (O, N, C, H). $\ce{-COOH}$ (O,O,O) $>$ $\ce{-CHO}$ (O,O,H) $>$ $\ce{-CH2OH}$ (O,H,H) $>$ $\ce{-CH(CH3)2}$ (C,C,H). $\ce{-Br}$ $>$ $\ce{-Cl}$ $>$ $\ce{-CCl3}$ (Cl,Cl,Cl) $>$ $\ce{-CH2Cl}$ (Cl,H,H): the first atom decides before its neighbours.

**Exercise 16.3 ★.**

How many [stereoisomers](#def-b1-stereochemistry-in-depth-isomers) have 2,3-dichlorobutane, 2-bromo-3-chlorobutane and pentane-2,4-diol? Identify the meso forms.

**Solution of Exercise 16.3.**

2,3-Dichlorobutane: two equivalent centres, (*R,R*), (*S,S*) and the meso (*R,S*): three. 2-Bromo-3-chlorobutane: two different centres, four [stereoisomers](#def-b1-stereochemistry-in-depth-isomers), no meso form. Pentane-2,4-diol: three, (*2R,4S*) being meso.

**Exercise 16.4 ★.**

Rank the four substituents of C5 of carvone by the [CIP rules](#def-b1-stereochemistry-in-depth-cip) and check the descriptors given in the figure of the two carvones (solid wedge: the isopropenyl group towards the reader, H away).

**Solution of Exercise 16.4.**

On C5: isopropenyl (C,C,C, the double bond counting twice) $>$ C6 $>$ C4 $>$ H. C6 and C4 both carry (C,H,H); one step further C6 leads to the carbonyl carbon (O,O,C) and C4 to C3 (C,C,H): C6 wins. With the isopropenyl group in front and H behind, the order isopropenyl $\to$ C6 $\to$ C4 turns clockwise as drawn: (*R*), spearmint carvone; the hashed one is (*S*).

**Exercise 16.5 ★★.**

Draw the [Newman projections](#def-b1-stereochemistry-in-depth-representations) of butane along C2–C3 at $\theta = 0$, 60, 120 and $180^\circ$ and read their energies on the curve. What fraction of the molecules would be gauche if the gauche and anti energies were equal?

**Solution of Exercise 16.5.**

$\theta = 0$: methyls eclipsed, $16.5\,\mathrm{kJ}/\mathrm{mol}$; $60^\circ$: gauche, about $2.9\,\mathrm{kJ}/\mathrm{mol}$ (minimum 2.8 at $62^\circ$); $120^\circ$: methyl eclipsing H, $15.2\,\mathrm{kJ}/\mathrm{mol}$; $180^\circ$: anti, 0. With equal energies, two [gauche conformations](#def-b1-stereochemistry-in-depth-conformations) for one anti: two thirds gauche.

**Exercise 16.6 ★★.**

Draw the [Fischer projection](#def-b1-stereochemistry-in-depth-representations) of (*S*)-alanine, $\ce{H2N-CH(CH3)-COOH}$, with the acid group at the top. On which side is the amino group?

**Solution of Exercise 16.6.**

Chain vertical with $\ce{COOH}$ on top and $\ce{CH3}$ at the bottom; ranks $\ce{NH2} > \ce{COOH} > \ce{CH3} > \ce{H}$. Putting $\ce{NH2}$ on the left and H on the right, $\ce{NH2} \to \ce{COOH} \to \ce{CH3}$ turns clockwise as seen; H is horizontal (towards the reader), so the descriptor is reversed: (*S*). The amino group is on the left.

**Exercise 16.7 ★★.**

Show that *cis*-1,2-dimethylcyclohexane has one methyl [axial](#def-b1-stereochemistry-in-depth-conformations) and one [equatorial](#def-b1-stereochemistry-in-depth-conformations) in both [chairs](#def-b1-stereochemistry-in-depth-conformations), whereas the *trans* isomer has a [chair](#def-b1-stereochemistry-in-depth-conformations) with both [equatorial](#def-b1-stereochemistry-in-depth-conformations). Which isomer is more stable? Use the estimate of the chapter.

**Solution of Exercise 16.7.**

*cis*: on adjacent carbons one bond up and one down among the [axial](#def-b1-stereochemistry-in-depth-conformations) and equatorial positions gives axial–equatorial in both [chairs](#def-b1-stereochemistry-in-depth-conformations). *trans*: diequatorial or diaxial. In the diequatorial [chair](#def-b1-stereochemistry-in-depth-conformations) the two methyls are gauche to each other (one interaction); the *cis* isomer has that interaction plus an [axial](#def-b1-stereochemistry-in-depth-conformations) methyl, about $5.7\,\mathrm{kJ}/\mathrm{mol}$ more. The *trans* isomer is the more stable.

**Exercise 16.8 ★★.**

A solution of a [chiral](#def-b1-stereochemistry-in-depth-stereocentre) compound at $0.100\,\mathrm{g}/\mathrm{mL}$ in a $2.00\,\mathrm{dm}$ tube rotates light by $+13.2^\circ$. Compute $[\alpha]$. What would a $1.00\,\mathrm{dm}$ tube give? And the enantiomer at the same concentration?

**Solution of Exercise 16.8.**

$[\alpha] = 13.2/(2.00 \times 0.100) = +66.0^\circ$. A $1.00\,\mathrm{dm}$ tube: $+6.60^\circ$. The enantiomer: $-13.2^\circ$ in the $2.00\,\mathrm{dm}$ tube.

**Exercise 16.9 ★★.**

Compute, with $\Delta E = 5.7\,\mathrm{kJ}/\mathrm{mol}$ and the Boltzmann ratio $\exp(\Delta E/RT)$, the fraction of methylcyclohexane [chairs](#def-b1-stereochemistry-in-depth-conformations) with an [equatorial](#def-b1-stereochemistry-in-depth-conformations) methyl at $25\,{}^{\circ}\mathrm{C}$ and at $-50\,{}^{\circ}\mathrm{C}$.

**Solution of Exercise 16.9.**

$25\,{}^{\circ}\mathrm{C}$: $\exp(5700/(8.314 \times 298.15)) = 10.0$, fraction $10.0/11.0 = 91\,\%$. $-50\,{}^{\circ}\mathrm{C}$: $\exp(5700/(8.314
\times 223.15)) = 21.6$, $96\,\%$.

**Exercise 16.10 ★★★.**

Explain why ethane has a single kind of [staggered conformation](#def-b1-stereochemistry-in-depth-conformations) and butane two (anti and gauche), and why gauche is less stable. Using the energies of the figure, estimate the anti : gauche population ratio of butane at $25\,{}^{\circ}\mathrm{C}$ (two [gauche conformations](#def-b1-stereochemistry-in-depth-conformations), one anti).

**Solution of Exercise 16.10.**

In ethane all six hydrogens are alike: every [staggered conformation](#def-b1-stereochemistry-in-depth-conformations) is the same. In butane the rear methyl can stand opposite the front one (anti) or beside it ($\pm 60^\circ$, gauche), where the two methyls crowd each other. Populations: anti 1, gauche $2\exp(-2.8/2.48) = 0.65$: about $61\,\%$ anti and $39\,\%$ gauche.

**Exercise 16.11 ★★★.**

Consider 2,3,4-trihydroxypentanedioic acid, $\ce{HOOC-CH(OH)-CH(OH)-CH(OH)-COOH}$. Count its [stereogenic centres](#def-b1-stereochemistry-in-depth-stereocentre), show that the central carbon is stereogenic only when the two outer ones have opposite descriptors (pseudo-asymmetric), and count the [stereoisomers](#def-b1-stereochemistry-in-depth-isomers) (meso included).

**Solution of Exercise 16.11.**

C2 and C4 are stereogenic. C3 carries H, OH and two branches that are identical in constitution; they differ only when C2 and C4 have opposite descriptors, and only then is C3 stereogenic (pseudo-asymmetric). Isomers: (*2R,4R*) and (*2S,4S*), a pair of [enantiomers](#def-b1-stereochemistry-in-depth-enantiomers) (C3 not stereogenic); (*2R,4S*) with the two arrangements at C3, two meso forms. Four in all.

**Exercise 16.12 ★★★.**

A sample of one enantiomer of a compound ($[\alpha] = +66.0^\circ$ for the pure compound in the conditions used) has partly racemised: at $0.050\,\mathrm{g}/\mathrm{mL}$ in a $1.00\,\mathrm{dm}$ tube it gives $+2.31^\circ$. Compute the fractions of the two [enantiomers](#def-b1-stereochemistry-in-depth-enantiomers).

**Solution of Exercise 16.12.**

$[\alpha]_{\text{sample}} = 2.31/(1.00 \times 0.050) = +46.2^\circ$, and $2x - 1 = 46.2/66.0 = 0.700$: $x = 0.850$, $85\,\%$ of the $(+)$ enantiomer and $15\,\%$ of the $(-)$.

## 16.7 Problem: Menthol, the Cool Molecule

**Problem 16.1.**

Weekend problem — the three stereocentres of menthol, its eight stereoisomers, the all-equatorial chair, Biot’s law, and the percentage of (−)-menthol in a partly racemised sample

Menthol, 2-isopropyl-5-methylcyclohexan-1-ol, gives mint its cooling feel. The natural compound is $(-)$-menthol, of [configuration](#def-b1-stereochemistry-in-depth-isomers) (*1R,2S,5R*). A laboratory measures with a polarimeter (sodium line, $25\,{}^{\circ}\mathrm{C}$, ethanol, $\ell = 1.00\,\mathrm{dm}$): its reference sample of pure $(-)$-menthol at $c = 0.0500\,\mathrm{g}/\mathrm{mL}$ gives $\alpha =
-2.46^\circ$; a sample to be tested, at the same concentration, gives $-1.97^\circ$.

**Part I — Stereocentres.**

1. Draw the constitution of menthol and mark the stereogenic carbons.
2. How many [stereoisomers](#def-b1-stereochemistry-in-depth-isomers) can there be? How many pairs of [enantiomers](#def-b1-stereochemistry-in-depth-enantiomers) ?
3. Why is there no meso form?
4. Rank the substituents of C1 by the [CIP rules](#def-b1-stereochemistry-in-depth-cip) .
5. Rank the substituents of C2.
6. Rank the substituents of C5.
7. Write the [configuration](#def-b1-stereochemistry-in-depth-isomers) of the enantiomer of $(-)$ -menthol.

**Part II — The [chair](#def-b1-stereochemistry-in-depth-conformations).**

8. Draw a [chair](#def-b1-stereochemistry-in-depth-conformations) of cyclohexane and place the three substituents of $(-)$ -menthol all [equatorial](#def-b1-stereochemistry-in-depth-conformations) . Check that this corresponds to the relative arrangement (1,2- *trans* and 1,5- *cis* ) of menthol.
9. What happens to the three substituents in the flipped [chair](#def-b1-stereochemistry-in-depth-conformations) ?
10. Using the estimate of the chapter for one [axial](#def-b1-stereochemistry-in-depth-conformations) methyl, explain why menthol is almost entirely in one [chair](#def-b1-stereochemistry-in-depth-conformations) .
11. Neomenthol differs from menthol only at C1. Is it an enantiomer or a diastereomer of menthol?
12. In the best [chair](#def-b1-stereochemistry-in-depth-conformations) of neomenthol, which group is [axial](#def-b1-stereochemistry-in-depth-conformations) ?
13. Why do menthol and neomenthol have different boiling points while the two [enantiomers](#def-b1-stereochemistry-in-depth-enantiomers) of menthol do not?

**Part III — Optical rotation.**

14. Compute $[\alpha]$ of the reference sample. Compare with the range $-45^\circ$ to $-51^\circ$ given by a handbook for $(-)$ -menthol.
15. What would be $[\alpha]$ of $(+)$ -menthol? Of racemic menthol?
16. Compute $[\alpha]$ of the tested sample.
17. Write [Biot’s law](#prop-b1-stereochemistry-in-depth-biot) for a mixture of the two [enantiomers](#def-b1-stereochemistry-in-depth-enantiomers) , with $x$ the fraction of $(-)$ -menthol.
18. Express $x$ as a function of the measured and reference rotations.
19. Could a [chiral](#def-b1-stereochemistry-in-depth-stereocentre) impurity of another compound distort the measurement? Give a check.

**Part IV — The composition.**

20. Compute the fraction of $(-)$ -menthol in the tested sample.
21. A second supplier’s sample gives $-2.40^\circ$ in the same conditions. Compute its composition.
22. Which sample is closer to natural menthol?
23. With a $2.00\,\mathrm{dm}$ tube, what angle would the tested sample give, and why is a longer tube better?
24. State the percentage of $(-)$ -menthol in the tested sample, to three significant figures.

**Solution of Problem 16.1.**

**1.** Cyclohexane ring: C1 carries OH, C2 the isopropyl group, C5 the methyl; C1, C2 and C5 are stereogenic. **2.** $2^3 = 8$, four pairs of [enantiomers](#def-b1-stereochemistry-in-depth-enantiomers). **3.** No arrangement has a mirror plane: the three substituents are all different. **4.** $\ce{OH}$ $>$ C2 (C,C,H) $>$ C6 (C,H,H) $>$ H. **5.** C1 (O,C,H) $>$ isopropyl (C,C,H) $>$ C3 (C,H,H) $>$ H. **6.** C6 and C4 both (C,H,H); next, C6 leads to C1 (O,C,H), C4 to C3 (C,H,H): C6 $>$ C4 $>$ $\ce{CH3}$ $>$ H. **7.** (*1S,2R,5S*). **8.** In a [chair](#def-b1-stereochemistry-in-depth-conformations), two [equatorial](#def-b1-stereochemistry-in-depth-conformations) bonds on adjacent carbons (C1, C2) point one up and one down: *trans*; on carbons 1 and 3 of the ring (C1 and C5, through C6) both [equatorial](#def-b1-stereochemistry-in-depth-conformations) bonds point the same way: *cis*. That is the arrangement of menthol. **9.** All three become [axial](#def-b1-stereochemistry-in-depth-conformations). **10.** In the flipped [chair](#def-b1-stereochemistry-in-depth-conformations) the methyl alone would cost about $5.7\,\mathrm{kJ}/\mathrm{mol}$ and the larger isopropyl group more; the OH adds a smaller cost of its own, and the [axial](#def-b1-stereochemistry-in-depth-conformations) OH and methyl, on carbons 1 and 3 and on the same face, would also crowd each other. That is well over $12\,\mathrm{kJ}/\mathrm{mol}$, a ratio of more than $\exp(12000/2479) \approx 130$ to one: menthol is almost entirely in the [all-equatorial](#def-b1-stereochemistry-in-depth-conformations) [chair](#def-b1-stereochemistry-in-depth-conformations). **11.** A diastereomer: only one of three centres differs. **12.** With isopropyl and methyl [equatorial](#def-b1-stereochemistry-in-depth-conformations), the OH is [axial](#def-b1-stereochemistry-in-depth-conformations). **13.** [Diastereomers](#def-b1-stereochemistry-in-depth-enantiomers) differ in all their properties (here the hydrogen bonding of an [axial](#def-b1-stereochemistry-in-depth-conformations) or [equatorial](#def-b1-stereochemistry-in-depth-conformations) OH); [enantiomers](#def-b1-stereochemistry-in-depth-enantiomers) differ only towards [chiral](#def-b1-stereochemistry-in-depth-stereocentre) partners. **14.** $[\alpha] = -2.46/(1.00 \times 0.0500) = -49.2^\circ$, inside the handbook range.

**15.** $+49.2^\circ$; $0$. **16.** $-1.97/0.0500 = -39.4^\circ$. **17.** $\alpha = \ell c[\alpha]_{(-)}\bigl(x - (1 - x)\bigr) =
\ell c[\alpha]_{(-)}(2x - 1)$. **18.** $x = \frac12(1 + \alpha/\alpha_{\text{ref}})$ at equal $\ell$ and $c$. **19.** Yes: any other optically active compound adds its own term ([Biot’s law](#prop-b1-stereochemistry-in-depth-biot)). A check of chemical purity by another method (chromatography) is needed before reading the rotation as a composition. **20.** $x = \frac12(1 + 1.97/2.46) = 0.900$. **21.** $x = \frac12(1 + 2.40/2.46) = 0.988$. **22.** The second ($98.8\,\%$). **23.** $-3.94^\circ$; the same reading error is then a smaller fraction of the angle. **24.** **$90.0\,\%$ of $(-)$-menthol** (and $10.0\,\%$ of its enantiomer).