6D.36 - Nucleophilic Substitution Mechanisms

Why Primary Halogenoalkanes Are Attacked

A primary halogenoalkane has a polar carbon-halogen bond. The halogen atom is more electronegative than carbon, so the bonding pair is pulled towards the halogen:

$\ce{R-CH2}^{\delta+}\!-\ce{X}^{\delta-}$

The carbon bonded to the halogen is electron deficient enough to be attacked by a nucleophile. A nucleophile is an electron-pair donor. In this lesson the two nucleophiles are:

$\ce{K+}$ is a spectator ion in the hydroxide reaction. It helps identify the reagent, but it is not the species whose lone pair attacks the carbon.

Primary substrate focus

Here, the substrate is a primary halogenoalkane, such as $\ce{CH3CH2Br}$ or $\ce{CH3CH2CH2Cl}$. The carbon attached to the halogen is bonded to one other carbon atom and has relatively open access for the nucleophile. At this stage, use the one-step substitution pattern for primary halogenoalkanes: nucleophile attack and carbon-halogen bond breaking happen together.

For a singly negative nucleophile, the general substitution pattern is:

$\ce{Nu-}$ represents a singly negative nucleophile with a lone pair, such as hydroxide. Neutral ammonia instead gives a positively charged organic ion in its first step, as explained below.

Aqueous Hydroxide Forms An Alcohol

Aqueous potassium hydroxide converts a primary halogenoalkane into an alcohol. The ionic equation shows the species involved in the electron-pair movement:

For example:

If the equation is written using potassium hydroxide, the spectator ion appears in the salt:

Curly-arrow route

For the mechanism, the key electron movement is:

1. Put a dipole on the carbon-halogen bond: $\ce{C}^{\delta+}$ and $\ce{X}^{\delta-}$.
2. Draw a curly arrow from a lone pair on the oxygen of $\ce{OH-}$ to the electron-deficient carbon bonded to $\ce{X}$.
3. Draw a curly arrow from the $\ce{C-X}$ bond to the halogen atom.
4. Form the $\ce{C-O}$ bond as the $\ce{C-X}$ bond breaks.
5. Give the products as the alcohol and the halide ion.

The two curly arrows are simultaneous in this primary substitution mechanism. Do not draw a free carbocation intermediate for this primary halogenoalkane mechanism.

Worked example: bromoethane to ethanol

Question: Describe the mechanism for the reaction between bromoethane and aqueous potassium hydroxide.

Model answer route:

1. The $\ce{C-Br}$ bond in bromoethane is polar, with the carbon atom $\delta^{+}$ and bromine $\delta^{-}$.
2. A lone pair on the oxygen atom of $\ce{OH-}$ attacks the $\delta^{+}$ carbon.
3. The $\ce{C-Br}$ bond breaks and its electron pair moves to bromine.
4. Ethanol forms and bromide ion is released.

Chemical interpretation: hydroxide is acting as a nucleophile, not as a base, because its lone pair forms a new bond to carbon. The reaction is substitution because $\ce{OH}$ replaces $\ce{Br}$ on the carbon skeleton.

Applying the equation to an amount

The balanced substitution equation gives a 1:1 halogenoalkane:alcohol ratio. To find a maximum product mass, divide the starting mass by its relative molecular mass, transfer the amount using this ratio, then multiply by the product's relative molecular mass. For example, 2.50 g of bromoethane (relative molecular mass 108.9) with excess hydroxide gives at most $2.50/108.9=0.02296\,\mathrm{mol}$ of ethanol. Using ethanol's relative molecular mass 46.0, the maximum mass is $(2.50/108.9)\times46.0=1.06\,\mathrm{g}$ to three significant figures. This is a theoretical mass, assuming complete conversion and no product loss.

[DIAGRAM: asset_name: Hydroxide Substitution Mechanism; asset_slug: organic_sn2_hydroxide; recommended_method: deterministic_chart; description: Hydroxide substitution of bromoethane: oxygen lone pair attacks the carbon opposite bromine while the C–Br bond pair goes to bromine.]

Ammonia Forms A Primary Amine

Ammonia is a neutral nucleophile. Its nitrogen atom has a lone pair, so it can attack the electron-deficient carbon in a primary halogenoalkane. The first substitution step gives a positively charged alkylammonium ion.

The product of the lesson reaction is a primary amine, $\ce{R-CH2-NH2}$. To reach it, another ammonia molecule removes a proton from the alkylammonium ion:

Combining the two stages gives the common overall equation:

For bromoethane: