Nucleophilic Aromatic Substitution (SNAr)

by James

Description: Aromatic rings containing electron withdrawing groups and a leaving group can undergo substitution reactions when treated with strong nucleophiles.

1-header

Notes: The reaction is facilitated by electron-withdrawing groups such as NO2. Common nucleophiles are alkoxides (RO-) hydroxide (HO-) amines, and sulphides (RS-). Common leaving groups are halides. Interestingly fluoride is an acceptable leaving group for this reaction (see mechanism section for an explanation).

Examples: 

2-examples

Notes: The first example shows an alkoxide nucleophile, with the alcohol as solvent. Heat increases the rate of reaction. The second example shows a primary amine as a nucleophile [extra detail – the amine is part of an amino acid here, and since amino acids are usually present in their “zwitterionic” form, a base (not shown) is generally added to deprotonate the nitrogen]. The third example has two possible products, but substitution on the ortho position will be favoured because this will lead to the more stable negative charge (see mechanism section below). The final example shows CN as the electron withdrawing group and a sulfide (NaSCH3) as nucleophile.

Mechanism: This reaction proceeds in two steps. In the first step, which is rate determining, because it leads to loss of aromaticity, the nucleophile attacks the aromatic ring, leading to formation of a negative charge (Step 1, arrows A and B). This is why this reaction is facilitated by electron withdrawing groups – because they help to stabilize this negative charge!  In the second step, the C–C π bond is restored, and the leaving group leaves (Step 2, arrows C and D).

3-mech-1

Why fluoride as a leaving group? I thought fluoride was a terrible leaving group? 

In SN1/SN2/E1/E2 reactions,  loss of a leaving group is part of the rate determining step, and fluoride is a poor leaving group because C-F bonds are strong. However, in the SNAr reaction, the rate limiting step is loss of aromaticity (Step 1). Step 2 (restoring aromaticity and loss of leaving group) is the fast step. Fluoride is a good choice here because it makes the carbon attached more electrophilic (electron-poor) due to its high electronegativity and this speeds up the rate of Step 1 (the rate-determining step).

Why does only the fluoride at the carbon “ortho” to the nitro group get displaced here, and not the “meta” fluoride? 

Good question! Let’s look at the resonance forms for ortho attack:
3-mech-2Notice how the second resonance form allows the negative charge to be placed on oxygen instead of carbon. Therefore this resonance form will have a large contribution to the hybrid and it will lower the activation energy towards attack, resulting in a faster reaction. Attack at the groups ortho and para to electron withdrawing groups will be favoured for this reason! 

What about attack at the carbon “meta” to the nitro group? Again, the resonance forms help to explain why it’s not as favoured:

3-mech-3

Notice how there are no resonance forms where the negative charge can be delocalized onto oxygen.

 

{ 2 comments… read them below or add one }

Trenten Witte

For the last diagram showing resonance you said “what about attack at the carbon ‘ortho’ to the nitro group” , but you meant to say meta.

Reply

James

Fixed! thank you

Reply

Leave a Comment