The SN1 mechanism and stereochemistry

A different rate law

Shake 2-bromo-2-methylpropane with water and ethanol. It reacts and forms 2-methylpropan-2-ol. Now add more hydroxide. The rate does not change. The tertiary halide is not waiting for the nucleophile.

The rate depends only on the concentration of the haloalkane:

The reaction is first order. It is called $S_{\text{N}}1$, substitution, nucleophilic, unimolecular.

The $S_{\text{N}}1$ mechanism

It has two steps.

1. Ionisation (slow). The C-X bond breaks by itself, and the halide ion leaves. A carbocation forms. This step is the slowest, so it sets the rate. It is the rate-determining step.
2. Capture (fast). The nucleophile bonds to the positive carbon.

$(\text{CH}_{3})_{3}\text{C}\text{-}\text{Br}\ \rightarrow \ (\text{CH}_{3})_{3}\text{C}^{+}\ +\ \text{Br}^{-}$

$(\text{CH}_{3})_{3}\text{C}^{+}\ +\ \text{H}_{2}\text{O}\ \rightarrow \ (\text{CH}_{3})_{3}\text{C}\text{-}\text{OH}_{2}^{+}$, then the proton goes to the solvent to give the alcohol.

Look at the energy diagrams. $S_{\text{N}}2$ has a single hill. $S_{\text{N}}1$ has two hills and a valley between them, where the carbocation sits. The first hill is the higher one, so the first step is the slow one. When water is the nucleophile and the solvent, the reaction is called solvolysis, and a reaction with water is hydrolysis.

Why tertiary halides favour $S_{\text{N}}1$

The carbocation must be stable enough to form. A carbocation is stabilised by alkyl groups. They push electrons towards the positive carbon (the inductive effect), and the C-H bonds next to it overlap with the empty orbital (hyperconjugation). You met both in the lessons on organic basics. A tertiary cation has three alkyl groups, a secondary two and a primary one. So the order is tertiary, then secondary, then primary, then methyl.

Allylic and benzylic halides also react easily by $S_{\text{N}}1$. Their cations spread the charge over the double bond or the ring by resonance.

The solvent. A polar protic solvent, such as water, ethanol or aqueous acetone, stabilises the ions that form. It speeds up $S_{\text{N}}1$. The leaving group trend is the same as in $S_{\text{N}}2$: iodide is the best.

Rearrangement. A carbocation can change into a more stable one by moving a hydrogen atom, with its electron pair, from the neighbouring carbon. This is a 1,2-hydride shift. The secondary cation from 2-bromo-3-methylbutane has a tertiary C-H next door. The hydride moves, and a tertiary cation forms. The product from water is then mostly 2-methylbutan-2-ol, not 3-methylbutan-2-ol. Rearrangement is a sign of an $S_{\text{N}}1$ reaction. It never happens in $S_{\text{N}}2$.

Chirality and optical activity

To see what $S_{\text{N}}1$ does to a chiral halide, we need the language of mirror images.

A carbon with four different groups is a chiral centre. A molecule with one chiral centre and its mirror image cannot be placed on top of each other. They are enantiomers. They have the same melting point, boiling point and solubility in an ordinary solvent. They differ in one physical property: they turn the plane of polarised light by equal amounts in opposite directions.

• A compound that turns the light is optically active. The enantiomer that turns it clockwise is dextrorotatory, marked $(+)$. The other is laevorotatory, marked $(-)$.
• An equal mixture of the two is a racemic mixture. The turns cancel, so it is optically inactive.
• The specific rotation is $[\alpha] = \dfrac{\alpha}{l \times c}$, where $\alpha$ is the observed rotation in degrees, $l$ is the tube length in decimetres and $c$ is the concentration in g/mL. A pure enantiomer has a fixed value at a given temperature and wavelength.
• The signs $(+)$ and $(-)$ come from an experiment. The letters R and S come from the priority rules. There is no link between them.

The enantiomeric excess is the difference between the percentages of the two enantiomers. A racemic mixture has an excess of zero. A pure enantiomer has $100\ \%$.

Stereochemistry of substitution

$S_{\text{N}}2$: inversion. The nucleophile attacks from the back, so every molecule inverts. An optically pure halide gives an optically pure product of the opposite shape.

$S_{\text{N}}1$: racemisation. The carbocation is flat. The nucleophile can reach either face. If the two faces are equally open, the two enantiomers form in equal amounts, and the product is racemic. This is racemisation. In practice the leaving group has not quite left. The halide ion is still close to one face, so the other face is a little more open. Most $S_{\text{N}}1$ reactions give slightly more of the inverted product than of the retained one, which is partial racemisation. Retention of the shape can also happen, for example in some reactions with neighbouring groups, which is beyond this course.

$S_{\text{N}}2$. One step. Rate $k[\text{RX}][\text{Nu}]$. Substrates: methyl and primary. No intermediate. Inversion. Best in a polar aprotic solvent. No rearrangement.

$S_{\text{N}}1$. Two steps. Rate $k[\text{RX}]$. Substrates: tertiary, allylic and benzylic. A carbocation intermediate. Racemisation. Best in a polar protic solvent. Rearrangement is possible.

A secondary halide can use both routes. Set $k_2$ to zero and the line is flat, which is pure $S_{\text{N}}1$. Then raise $k_2$. At low nucleophile concentration the $S_{\text{N}}1$ part dominates, and at high concentration the $S_{\text{N}}2$ part takes over.

Practise with new numbers

Try it in Python

The program works out the enantiomeric excess for an $S_{\text{N}}2$ reaction, an ideal $S_{\text{N}}1$ reaction and an $S_{\text{N}}1$ reaction with partial racemisation. It also finds the rotation that a polarimeter would show.

Common mistakes

• Saying $S_{\text{N}}1$ gives exactly equal amounts every time. Often one enantiomer is slightly in excess.
• Mixing the sign of the rotation with R and S. They are not linked.
• Thinking that adding more nucleophile speeds up $S_{\text{N}}1$. The rate-determining step does not involve it.
• Forgetting rearrangement. A secondary cation next to a tertiary C-H can shift to the tertiary cation.

Remember these

• $S_{\text{N}}1$: two steps, a carbocation, a rate $= k[\text{RX}]$, racemisation, and favoured by tertiary, allylic and benzylic halides in polar protic solvents.
• Carbocation stability: tertiary, then secondary, then primary, then methyl.
• A chiral centre has four different groups. Enantiomers turn polarised light in opposite directions, and a racemic mixture does not turn it.
• $S_{\text{N}}2$ inverts the shape. $S_{\text{N}}1$ racemises it, with a small excess of inversion.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. (R)-2-chlorobutane reacts with a good nucleophile by a pure $S_{\text{N}}2$ pathway. Is the product a racemic mixture?

2. Why is the rate of hydrolysis of 2-bromo-2-methylpropane unchanged when hydroxide is added?

3. What does a polarimeter read for a racemic mixture?

4. A halide gives a product mixture with $70\ \%$ inversion and $30\ \%$ retention. What is the enantiomeric excess?

References

• NCERT, Chemistry Part II for Class XII, the chapter Haloalkanes and Haloarenes, on the $S_{\text{N}}1$ mechanism, optical activity and stereochemical aspects.
• OpenStax, Organic Chemistry, Chapter 11 (Reactions of Alkyl Halides: Nucleophilic Substitutions and Eliminations), the $S_{\text{N}}1$ section.
• Morrison and Boyd, Organic Chemistry, the chapters on nucleophilic aliphatic substitution and on stereochemistry.