Stereochemistry of Nucleophilic Substitution Reactions

So far, we have looked at nucleophilic substitution in terms of which bonds break and form, and how fast those steps happen. But there is another question we have not yet answered: what happens to the three-dimensional arrangement of groups around the carbon under attack? Does the product molecule look exactly like the reactant, or has something flipped? To explore this, we first need a few tools from stereochemistry.

What is Optical Activity?

When ordinary light passes through a Nicol prism, it becomes plane-polarised light (light that vibrates in only one plane). The prism is named after William Nicol (1768-1851), the Scottish physicist who developed the first device capable of producing plane-polarised light. If you then pass this polarised beam through a solution of certain compounds, something interesting happens: the plane of vibration rotates. The compounds that cause this rotation are called optically active compounds, and the angle of rotation is measured using an instrument called a polarimeter.

The direction of rotation tells us what type we are dealing with:

• Dextrorotatory (from the Greek for “right-rotating”): rotates the plane clockwise. Labelled $d$- or with a ($+$) sign.
• Laevorotatory (“left-rotating”): rotates the plane anticlockwise. Labelled $l$- or with a ($-$) sign.

A pair of compounds that rotate light by the same angle but in opposite directions are called optical isomers, and the phenomenon itself is optical isomerism.

Chirality: Why Some Molecules Rotate Light

The mirror-image test

Think about your left and right hands. They look alike, but if you place one on top of the other (keeping both palms down), they do not match up. Your thumbs point in opposite directions. Objects like this, whose mirror images cannot be perfectly overlapped, are said to be chiral (from the Greek word for “hand”), and the property is called chirality.

Molecules follow the same rule. A molecule whose mirror image cannot be superimposed on the original is chiral and will be optically active. A molecule whose mirror image can be superimposed on the original is achiral and will not rotate polarised light.

The asymmetric carbon: a quick test for chirality

In 1874, the Dutch scientist J. Van’t Hoff and the French scientist C. Le Bel independently proposed that the four bonds around a carbon atom point towards the corners of a tetrahedron. They realised that if all four substituents on such a carbon are different, the molecule and its mirror image will never be superimposable. A carbon that carries four distinct groups is called an asymmetric carbon or stereocentre. Van’t Hoff (1852-1911) went on to receive the very first Nobel Prize in Chemistry in 1901, awarded for his groundbreaking work on chemical dynamics and osmotic pressure in solutions.

This idea built on Louis Pasteur’s observation in 1848 that certain crystal forms exist as mirror-image pairs, and that their solutions rotate polarised light by equal amounts in opposite directions. Pasteur suspected the difference lay in the three-dimensional arrangement of atoms within the molecule, what we now call configuration.

Propan-2-ol vs butan-2-ol: seeing chirality in action

Consider propan-2-ol ($CH_3CH(OH)CH_3$). The central carbon carries $CH_3$, $CH_3$, $H$, and $OH$. Two of these groups are identical (both are methyl). If you build a model and then build its mirror image, rotating the mirror image by 180° lets it overlap perfectly with the original. The molecule is achiral.

Now look at butan-2-ol ($CH_3CH(OH)CH_2CH_3$). Carbon-2 carries $CH_3$, $C_2H_5$, $H$, and $OH$: all four groups are different. When you try to overlap the mirror image onto the original, they refuse to match. Butan-2-ol is chiral.

Other common chiral molecules include 2-chlorobutane, 2,3-dihydroxypropanal ($OHC{-}CHOH{-}CH_2OH$), bromochloroiodomethane ($BrClCHI$), and 2-bromopropanoic acid ($CH_3CHBr{-}COOH$). In each case, one carbon has four different groups attached.

Enantiomers: Mirror-Image Partners

Stereoisomers that are related as non-superimposable mirror images are called enantiomers. For instance, molecules A and B in the propan-2-ol diagram (Fig. 6.5) are mirror images, but since propan-2-ol is achiral they are actually the same molecule. Molecules D and E in the butan-2-ol diagram (Fig. 6.6), however, are true enantiomers because they cannot be superimposed.

Enantiomers share all ordinary physical properties: same melting point, same boiling point, same density, same refractive index, same solubility. The only measurable difference between them is the direction in which they rotate plane-polarised light. If one enantiomer is dextrorotatory ($+$), the other is laevorotatory ($-$), and both rotate light by exactly the same angle.

One important caution: the sign of optical rotation ($+$ or $-$) does not tell you the actual arrangement of atoms in space (the absolute configuration). Two different compounds with the same configuration at their stereocentre can have opposite signs of rotation.

Racemic Mixtures

When you mix equal quantities of both enantiomers, the clockwise rotation caused by one is exactly cancelled by the anticlockwise rotation of the other. The result is zero net rotation. Such a 50:50 mixture is called a racemic mixture (or racemic modification) and is written with the prefix $dl$- or ($\pm$). For example, ($\pm$)-butan-2-ol is the racemic form of butan-2-ol.

The process of converting a pure enantiomer into a racemic mixture is called racemisation.

Worked Example: Spotting Chiral and Achiral Molecules

Example 6.8: Identify which molecules in each pair are chiral and which are achiral.

How to solve this: For each molecule, find the tetrahedral carbons and check whether any of them carries four different groups.

• Pair (i): The first molecule has a carbon bonded to $CH_3$, $H$, $Br$, and $OH$: four different groups, so it is chiral. The second has $H$, $Br$, $Br$, and $CH_3$: two groups are the same ($Br$ appears twice), so it is achiral.
• Pair (ii): Check the candidate carbons carefully. If any carbon carries two identical groups (for instance, two $CH_3$ groups), the molecule is achiral.
• Pair (iii): $CH_3CHBrCH_2CH_3$ (2-bromobutane) has carbon-2 with $CH_3$, $H$, $Br$, and $C_2H_5$: chiral. $CH_3CH_2CH_2CH_2Br$ (1-bromobutane) has no carbon with four different groups: achiral.

Retention of Configuration

When a reaction takes place at a site away from the stereocentre, leaving all four bonds of the asymmetric carbon intact, the spatial arrangement of groups around that carbon stays exactly the same. This is called retention of configuration.

Consider what happens when ($-$)-2-methylbutan-1-ol is heated with concentrated $HCl$:

The $-OH$ on carbon-1 is replaced by $-Cl$, but the asymmetric carbon is carbon-2, and none of its bonds break. The product therefore has the same configuration as the reactant.

Notice, though, that the sign of optical rotation has changed from ($-$) to ($+$). This is not a contradiction: the sign of rotation depends on the whole molecule, not just on the stereocentre’s arrangement. Two different compounds can share the same configuration yet show opposite signs of rotation.

Three Possible Outcomes at a Stereocentre

When a reaction does break a bond directly at the asymmetric carbon (for example, replacing a leaving group $X$ with a nucleophile $Y$), there are three possible stereochemical outcomes:

1. Retention: If only product A forms, with $Y$ taking up the same spatial position that $X$ occupied, the configuration is preserved.
2. Inversion: If only product B forms, with $Y$ on the opposite side from where $X$ was, the configuration has been flipped (like an umbrella turning inside out in the wind).
3. Racemisation: If a 50:50 mixture of A and B forms, the product is optically inactive because the rotations cancel out.

Which of these outcomes actually happens depends entirely on the mechanism of the reaction.

$S_N2$ Reactions Give Inversion

In the $S_N2$ mechanism, the nucleophile approaches from the back side of the carbon, directly opposite the leaving group. As the new bond forms and the old bond breaks simultaneously, the three remaining groups swing to the other side, like an umbrella flipping. The result is a product with inverted configuration.

A concrete example: when ($-$)-2-bromooctane reacts with sodium hydroxide via $S_N2$, the product is ($+$)-octan-2-ol. The $-OH$ group sits in the position opposite to where the $-Br$ was originally.

Key takeaway: $S_N2$ reactions on optically active substrates always produce inversion of configuration.

$S_N1$ Reactions Give Racemisation

In the $S_N1$ mechanism, the leaving group departs first (slow step) to form a carbocation. This carbocation is $sp^2$-hybridised and therefore flat (planar), with an empty $p$-orbital sticking out above and below the plane. The original chirality is lost the moment the carbocation forms, because a flat species has no “handedness.”

When the nucleophile comes in during the fast second step, it can attack from either face of this flat carbocation with equal probability. Attack from one face produces the retained product; attack from the other produces the inverted product. Since both faces are equally accessible, a roughly 50:50 mixture forms: a racemic mixture.

For example, hydrolysis of optically active 2-bromobutane through $S_N1$ gives ($\pm$)-butan-2-ol, a racemic product.

Key takeaway: $S_N1$ reactions on optically active substrates always produce racemisation.

Quick Note on Naming Carbon Positions

In discussions of haloalkane reactions, two Greek-letter labels come up often:

• $\alpha$-carbon: the carbon atom directly bonded to the halogen (or other functional group).
• $\beta$-carbon: the carbon atom next to the $\alpha$-carbon.

These labels are useful shorthand when describing which bonds break in substitution and elimination reactions.