3.3.3.1 - Nucleophilic Substitution

Why halogenoalkanes are attacked

Halogenoalkanes contain a polar carbon-halogen bond because halogens are more electronegative than carbon. The shared electron pair is pulled towards the halogen, so the carbon becomes slightly positive, $\delta^{+}$, and the halogen becomes slightly negative, $\delta^{-}$. That $\delta^{+}$ carbon is electron deficient, so it attracts species that can donate electrons.

The three nucleophiles in this topic all have a lone pair ready to donate:

• $\ce{OH-}$
• $\ce{CN-}$
• $\ce{NH3}$

$\ce{OH-}$ and $\ce{CN-}$ are negatively charged, so they are strongly attracted to the $\delta^{+}$ carbon. $\ce{NH3}$ is neutral, but the nitrogen atom still has a lone pair, so it can act as a nucleophile too.

When substitution happens, the halogen leaves with the bonding pair of electrons as a halide ion, $\ce{X-}$. This is why halide ions are described as leaving groups.

The diagram below shows the polar $\ce{C-Br}$ bond in bromoethane and the three lone-pair donors that can attack the electron-deficient carbon.

[DIAGRAM: asset_name: 3.3.1 - Nucleophilic Substitution - Diagram 1; asset_slug: 3.3.1 - Nucleophilic Substitution - Diagram 1; recommended_method: retained_png; description: Bromoethane shown as H3C-CH2-Br, with the CH2 carbon labelled δ+ and Br labelled δ-; place :OH-, :C≡N-, and :NH3 nearby as the three lone-pair donors used in nucleophilic substitution.]

The core mechanism pattern

Substitution replaces the halogen while retaining the rest of the carbon framework.

For the primary halogenoalkane examples here, substitution by $\ce{OH-}$ and $\ce{CN-}$ follows the same one-step pattern:

1. The nucleophile uses its lone pair to attack the $\delta^{+}$ carbon.
2. At the same time, the carbon-halogen bond breaks.
3. The halogen leaves as $\ce{X-}$.

In a curly-arrow mechanism, one arrow starts at the nucleophile's lone pair and points to the carbon atom. The second starts at the C-X bond and points to the halogen. Curly arrows always show the movement of an electron pair.

With hydroxide ions, the halogen is replaced by $\ce{OH}$, so an alcohol forms:

$\ce{CH3CH2Br}$ + $\ce{OH-}$ → $\ce{CH3CH2OH}$ + $\ce{Br-}$

Hydrolysis is carried out by heating with aqueous sodium or potassium hydroxide. Some ethanol can be added to help the poorly water-soluble halogenoalkane mix with the solution; hot predominantly ethanolic hydroxide instead favours elimination.

With cyanide ions, the halogen is replaced by $\ce{CN}$, so a nitrile forms:

$\ce{CH3CH2Br}$ + $\ce{CN-}$ → $\ce{CH3CH2CN}$ + $\ce{Br-}$

Use ethanolic potassium cyanide and heat under reflux for this substitution. The attacking lone pair is on carbon in $\ce{CN-}$, giving a C–C bond and a nitrile; drawing attack through nitrogen gives a different connectivity. Cyanide is acutely toxic, and contact with acid can release toxic hydrogen cyanide. This synthesis belongs in a controlled laboratory, with the specified risk controls.

This is a very useful synthetic step because the carbon in the -$\ce{CN}$ group becomes part of the product. Cyanide substitution therefore lengthens the carbon chain by one carbon atom.

Worked example: starting with 1-bromopropane, $\ce{CH3CH2CH2Br}$, substitution by $\ce{CN-}$ gives $\ce{CH3CH2CH2CN}$. The product has four carbons in total, so its name is butanenitrile.

The diagram below compares hydroxide and cyanide substitution side by side so the shared electron-flow pattern is easy to see.

[DIAGRAM: asset_name: 3.3.1 - Nucleophilic Substitution - Diagram 2; asset_slug: 3.3.1 - Nucleophilic Substitution - Diagram 2; recommended_method: retained_png; description: Side-by-side one-step nucleophilic substitution mechanisms for bromoethane. Left: :OH- attacking the δ+ carbon in H3C-CH2-Br, with one curly arrow from the oxygen lone pair to carbon and one from the C-Br bond to Br, giving H3C-CH2-OH + Br-. Right: the same electron-flow pattern for :C≡N- attacking through carbon to give H3C-CH2-C≡N + Br-.]

Once you understand that shared pattern, you do not need to memorise two unrelated mechanisms. The electron movement is the same; only the attacking nucleophile and the product change.

Why ammonia is slightly different

Ammonia starts in the same way: the lone pair on nitrogen attacks the $\delta^{+}$ carbon, and the C-X bond breaks. However, there is one extra step because ammonia is a neutral molecule rather than a negatively charged ion.

After the first step, the nitrogen has formed four bonds, so the immediate organic product is positively charged. For bromoethane, that intermediate is $\ce{CH3CH2NH3+}$.