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A mechanism leaves its signature on a stereocentre: SN2 back-side attack inverts it, SN1 through a flat carbocation racemizes it, and a reaction that breaks no bond to it retains it; read backwards, the outcome is evidence for the mechanism.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
You will predict inversion, racemization or retention from a mechanism, follow an enantiomeric excess through SN2, match observations to mechanisms, pick out reactions that keep a configuration, and explain why SN1 racemizes.
You can assign R and S, and you know the SN1 and SN2 rate laws. The stereochemistry of a product is the other half of the evidence for a mechanism.
Inversion turns a stereocentre into its mirror arrangement. Retention keeps it. Racemization turns a single enantiomer into a 50:50 mixture. Back-side attack is a nucleophile approaching from the side opposite the leaving group.
SN2 — inversion. The nucleophile attacks from the back, opposite the leaving group, in one step. As the new bond forms and the old one breaks, the other three groups flip through like an umbrella turned inside out in the wind. Every molecule is inverted.
SN1 — racemization. The leaving group departs first, leaving a carbocation whose carbon is sp2 and flat. The nucleophile can arrive from either face equally, so the two configurations form in equal amounts.
No bond to the centre broken — retention. A reaction at another carbon leaves the stereocentre exactly as it was.
A caution: inversion is about geometry. The R/S letter usually flips too, but not always, because the new group's priority may differ from the old one's.
Another way: table
The signatures.
| Mechanism | At the stereocentre |
|---|---|
| SN2 | inversion |
| SN1 | racemic mixture |
| reaction elsewhere | retention |
Chemists read these signatures backwards. A single enantiomer that gives a single inverted product points to SN2; one that gives a racemic product points to a flat intermediate, SN1. Combined with the rate law, the stereochemical outcome is some of the strongest evidence available for a mechanism. It also carries numbers: SN2 on a sample with an enantiomeric excess of 80% gives a product of 80% excess, in the opposite enantiomer; complete SN1 gives zero.
A reaction never changes stereochemical information. Breaking a bond to the centre can invert or scramble it.
SN1 gives inversion. Its flat intermediate gives a racemic mixture.
A racemic product has no stereocentre. It has one, in both configurations.
Inversion always flips the R/S letter. Priorities can change.
Hydroxide attacks from the back as bromide leaves.
One-step back-side attack.
The centre inverts; OH takes Br's place as priority 1.
Umbrella flip.
So the product is (R)-butan-2-ol, a single enantiomer.
Inversion, letter flipped.
What intermediate forms?
A flat carbocation.
From which faces can water attack?
Both equally, so the alcohol is a racemic mixture.
A single enantiomer undergoes a reaction at a different carbon that never breaks a bond to the stereocentre. What happens to the configuration of that stereocentre?
A sample of (R)-2-bromobutane with an enantiomeric excess of $60$% reacts completely with hydroxide by SN2. What is the enantiomeric excess of the butan-2-ol formed, in per cent?
Answer:
A chemist starts each reaction from a single enantiomer. Match each observation to the mechanism it points to.
| SN2: one-step back-side attack | SN1: through a flat carbocation | no bond to the stereocentre broke | |
|---|---|---|---|
| the product is entirely inverted | |||
| the product is racemic | |||
| the stereocentre is unchanged; the reaction happened two carbons away |
(S)-3-chlorobutan-1-ol, ClCH(CH3)CH2CH2OH, has its stereocentre at the carbon carrying Cl. Mark every reaction that is sure to keep that centre's configuration.
This task has no paper form; do it on a device.
(R)-3-bromo-3-methylhexane is dissolved in water and the product, 3-methylhexan-3-ol, shows almost no optical rotation. What explains that?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Each reaction starts from a single enantiomer. Give the stereochemical outcome at the stereocentre.
| outcome | |
|---|---|
| (R)-2-iodobutane with cyanide, SN2 | |
| (S)-3-chloro-3-methylheptane in ethanol, SN1 | |
| (R)-2-methylbutan-1-ol made into its ethanoate ester |
You can predict what a reaction does to a stereocentre. Tell someone why an SN1 product shows no optical rotation. Next unit: substitution and elimination in full.
8. Your turn: (R)-2-chloro-2-phenylbutane in water, by SN1, step 3