SN1 versus SN2: The Complete Comparison
Four factors decide which pathway operates. Check them in a fixed order and the prediction is reliable every time.
BSc & MSc · Organic Chemistry · Concept
The two mechanisms side by side
| Feature | SN2 | SN1 |
|---|---|---|
| Steps | One, concerted | Two, via carbocation |
| Kinetics | Second order, rate = k[RX][Nu] | First order, rate = k[RX] |
| Stereochemistry | Inversion at the carbon | Racemisation, often with slight inversion excess |
| Substrate preference | Methyl > primary > secondary | Tertiary > secondary > primary |
| Nucleophile | Strong nucleophile required | Strength irrelevant to rate |
| Solvent | Polar aprotic | Polar protic |
| Rearrangement | Never | Possible, via the cation |
Why the substrate dominates
The two mechanisms respond to substitution in opposite directions, which is why substrate structure decides most cases on its own.
SN2 requires the nucleophile to reach the carbon from the side opposite the leaving group. Every additional alkyl group blocks that approach, so rate falls steeply from methyl to tertiary. This is pure steric hindrance.
SN1 requires a carbocation to form, and carbocation stability rises with substitution through hyperconjugation and induction. So the same change that kills SN2 helps SN1.
The other three factors
Nucleophile
A strong nucleophile pushes toward SN2 because it participates in the rate-determining step. In SN1 the nucleophile attacks after the slow step, so its strength does not appear in the rate law at all — a point worth stating explicitly when asked why concentration has no effect.
Leaving group
Both mechanisms need a good leaving group, so this rarely discriminates. The general rule is that a weak base is a good leaving group, because it is stable carrying the negative charge. Iodide leaves better than bromide, which beats chloride; hydroxide is a poor leaving group unless protonated first.
Solvent
Polar protic solvents hydrogen-bond to the nucleophile, reducing its reactivity, while stabilising the developing carbocation. Both effects favour SN1. Polar aprotic solvents dissolve the salt but leave the anion relatively bare and highly reactive, favouring SN2. This is one of the cleanest cause-and-effect chains in the topic.
Allylic and benzylic substrates
These are the standard exception. A primary allylic or benzylic halide can undergo SN1 readily, because the resulting cation is stabilised by resonance with the adjacent π system. A question giving a primary substrate that nonetheless reacts by SN1 is almost always testing this.
A working decision order
- Classify the substrate. Methyl or primary points to SN2; tertiary points to SN1.
- Check for allylic or benzylic stabilisation, which overrides the substrate class.
- For secondary, look at the nucleophile: strong pushes SN2, weak pushes SN1.
- Confirm with the solvent: aprotic supports SN2, protic supports SN1.
- State the stereochemical consequence, because it usually carries its own mark.
Frequently asked questions
Why does SN2 give inversion?
The nucleophile must attack opposite the leaving group, so the three remaining bonds flip through the plane like an umbrella in wind. Inversion is a geometric necessity of the mechanism, not a separate rule.
Why is SN1 racemisation rarely complete?
The leaving group does not depart instantly. While it lingers near the cation it partially blocks that face, so attack from the opposite side is slightly favoured, giving a small excess of the inverted product.
Can a tertiary substrate ever undergo SN2?
Essentially never. The steric barrier to backside attack is prohibitive, and a strong base with a tertiary substrate gives elimination instead.
How do I tell substitution from elimination?
Basicity and temperature. A strong base that is also bulky favours elimination, and higher temperature always favours elimination because it is entropically favoured.
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