Nucleophilic substitution and the SN2 mechanism

A halogen is replaced

Warm bromoethane with aqueous potassium hydroxide. After a time, you find ethanol and potassium bromide.

$\text{CH}_{3}\text{CH}_{2}\text{Br}\ +\ \text{OH}^{-}\ \rightarrow \ \text{CH}_{3}\text{CH}_{2}\text{OH}\ +\ \text{Br}^{-}$

The hydroxide ion took the place of the bromide ion. This is a substitution reaction. The incoming species had a pair of electrons to give. It is a nucleophile, which means "nucleus loving". It attacks the carbon with the small positive charge, the one that holds the halogen. The halogen leaves with the electron pair of the old bond. It is the leaving group.

We write the general reaction as follows. Nu stands for any nucleophile.

$\text{Nu}^{-}\ +\ \text{R}\text{-}\text{X}\ \rightarrow \ \text{R}\text{-}\text{Nu}\ +\ \text{X}^{-}$

The carbon is the electrophilic centre, which means it is short of electrons. The bond to the halogen breaks in the process.

What nucleophiles do

The same reaction with different nucleophiles gives a whole family of compounds. This is why haloalkanes are so useful.

• Aqueous $\text{KOH}$ (or moist $\text{Ag}_{2}\text{O}$) gives an alcohol, $\text{R}\text{-}\text{OH}$.
• A sodium alkoxide, $\text{R}'\text{ONa}$, gives an ether, $\text{R}\text{-}\text{O}\text{-}\text{R}'$ (Williamson synthesis).
• Alcoholic $\text{KCN}$ gives a nitrile (alkyl cyanide), $\text{R}\text{-}\text{CN}$.
• $\text{AgCN}$ gives an isocyanide, $\text{R}\text{-}\text{NC}$.
• $\text{KNO}_{2}$ gives an alkyl nitrite, $\text{R}\text{-}\text{O}\text{-}\text{N}=\text{O}$.
• $\text{AgNO}_{2}$ gives a nitroalkane, $\text{R}\text{-}\text{NO}_{2}$.
• Excess $\text{NH}_{3}$ in ethanol, in a sealed tube, gives an amine, $\text{R}\text{-}\text{NH}_{2}$.
• A silver carboxylate, $\text{R}'\text{COOAg}$, gives an ester, $\text{R}'\text{COOR}$.
• $\text{NaI}$ in dry acetone gives an iodoalkane, $\text{R}\text{-}\text{I}$ (Finkelstein).

Ambident nucleophiles. The cyanide ion and the nitrite ion have two atoms that can give a pair of electrons. They are ambident nucleophiles. Ionic $\text{KCN}$ attacks through carbon and gives the nitrile. Silver cyanide has covalent character. The silver ion holds the carbon end, so nitrogen attacks and the isocyanide forms. In the same way, ionic $\text{KNO}_{2}$ gives the nitrite ester, by attack through oxygen, and $\text{AgNO}_{2}$ gives the nitro compound, by attack through nitrogen.

With ammonia the reaction does not stop. The amine formed is a nucleophile too, and it goes on to give secondary and tertiary amines and a quaternary salt. A large excess of ammonia keeps the first product the main one.

The rate of the halide: a test you can see

Shake a few drops of three haloalkanes with a warm solution of silver nitrate in ethanol. A precipitate of silver halide forms as the halide ion leaves. The chloride gives a white precipitate slowly, the bromide gives a cream one faster, and the iodide gives a yellow one almost at once. The weaker the C-X bond, the faster it breaks. The order of reactivity of the halogens as leaving groups is I, then Br, then Cl, then F. This test is used in A-Level and IB courses.

The $S_{\text{N}}2$ mechanism

$S_{\text{N}}2$ stands for substitution, nucleophilic, bimolecular. The reaction has one step. The nucleophile comes in from the side of the carbon opposite to the leaving group. The new bond forms as the old one breaks. In the middle, the transition state has the nucleophile and the leaving group partly bonded to the carbon, and the other three groups lie flat.

After the reaction the three groups have turned over, like an umbrella blown inside out. This is the inversion of configuration, also called the Walden inversion. If the carbon was chiral, the product has the opposite spatial arrangement. For (R)-2-bromobutane and hydroxide, the product is (S)-butan-2-ol. Here the label changes from R to S because OH in the product has the top priority that Br had in the reactant. In other reactions the priorities can change, and the label may stay the same, so always work out the label.

The rate depends on both reactants:

The reaction is second order overall. Doubling either concentration doubles the rate.

Choose class 1 and double the nucleophile. The rate doubles. Now change the class. The methyl halide reacts about thirty times as fast as the primary one, and the tertiary halide hardly reacts at all. The next section says why.

What makes $S_{\text{N}}2$ fast or slow

The substrate. The nucleophile must reach the back of the carbon. Methyl and primary carbons have little in the way. A secondary carbon has two alkyl groups close to the path, and a tertiary carbon has three. This steric hindrance cuts the rate. The order is methyl, then primary, then secondary, then tertiary. The numbers in the simulation are typical textbook values for illustration. The tertiary halides do not react by $S_{\text{N}}2$ in practice.

The leaving group. A good leaving group leaves as a stable, weakly basic ion. The order is iodide, then bromide, then chloride, then fluoride. Fluoride is a poor leaving group because the C-F bond is strong.

The nucleophile. A stronger nucleophile gives a faster reaction. A negative ion is usually a better nucleophile than its neutral partner: $\text{OH}^-$ is better than $\text{H}_2\text{O}$.

The solvent. Polar aprotic solvents, such as acetone, dimethyl sulphoxide (DMSO) and dimethylformamide (DMF), favour $S_{\text{N}}2$. They dissolve the salt and leave the anion free. A polar protic solvent, such as water or ethanol, wraps the anion in hydrogen bonds and slows it down.

Practise with new numbers

Try it in Python

The program defines the $S_{\text{N}}2$ and $S_{\text{N}}1$ rate laws and compares what happens when concentrations change. It also shows how long a reaction takes for different classes of halide.

Common mistakes

• Saying $S_{\text{N}}2$ happens in two steps. It is one step, with no intermediate.
• Forgetting that the rate depends on the nucleophile. In $S_{\text{N}}2$ it does. In $S_{\text{N}}1$ it does not.
• Thinking inversion always changes the label R to S. The shape inverts. The label changes only if the priorities allow it.
• Putting a tertiary halide first in the $S_{\text{N}}2$ order. It is last, because of steric hindrance.

Remember these

• A nucleophile has a pair of electrons to give. The leaving group takes the old bond pair with it.
• $S_{\text{N}}2$ has one step, a rate $= k[\text{RX}][\text{Nu}]$, and inversion of configuration.
• $S_{\text{N}}2$ speed: methyl, then primary, then secondary, then tertiary. Leaving group: I, then Br, then Cl, then F.
• Polar aprotic solvents favour $S_{\text{N}}2$. Ambident ions: $\text{KCN}$ gives the nitrile, $\text{AgCN}$ gives the isocyanide, $\text{KNO}_{2}$ gives the nitrite, $\text{AgNO}_{2}$ gives the nitro compound.

Check yourself

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

1. In an $S_{\text{N}}2$ reaction the concentration of the nucleophile is tripled and that of the haloalkane is unchanged. What happens to the rate?

2. Which reacts faster by $S_{\text{N}}2$, 1-bromobutane or 2-bromo-2-methylpropane? Why?

3. What product forms when bromoethane is heated with silver cyanide?

4. Why does 1-iodobutane give a precipitate with silver nitrate faster than 1-chlorobutane?

References

• NCERT, Chemistry Part II for Class XII, the chapter Haloalkanes and Haloarenes, on nucleophilic substitution reactions.
• OpenStax, Organic Chemistry, Chapter 11 (Reactions of Alkyl Halides: Nucleophilic Substitutions and Eliminations), the $S_{\text{N}}2$ section.
• Clayden, Greeves and Warren, Organic Chemistry, the chapter on nucleophilic substitution at saturated carbon.