SN1, SN2, E1, E2 Mechanisms — Mnemonic
Target: mnemonic for SN1 SN2 E1 E2 reaction mechanisms
Why is this hard to memorize?
Choosing between SN1, SN2, E1, and E2 is one of the biggest challenges in organic chemistry. The choice depends on four factors: substrate structure (1°/2°/3°), nucleophile strength, base strength, and solvent polarity. SN2 favours 1° substrates with strong nucleophiles in polar aprotic solvents. SN1 favours 3° substrates in polar protic solvents. E2 favours strong, bulky bases. E1 competes with SN1 at high temperatures. JEE Advanced tests this decision tree extensively.
Classic mnemonics you should know
"1° → SN2. 3° → SN1/E1. 2° → depends on nucleophile/base"
Primary substrates undergo SN2 (backside attack is easy, no steric hindrance). Tertiary substrates undergo SN1 or E1 (stable 3° carbocation forms). Secondary is the grey zone — depends on other factors.
"SN2: Strong Nu⁻, 1° substrate, Polar Aprotic solvent, Inversion of configuration"
SN2 needs: strong nucleophile (CN⁻, I⁻, RS⁻), unhindered substrate (methyl or 1°), polar aprotic solvent (DMSO, DMF, acetone). Result: 100% inversion (Walden inversion). Rate = k[substrate][Nu⁻].
"E2: Strong base, anti-periplanar, one step. E1: Weak base, carbocation, two steps."
E2: strong base rips off β-H while leaving group leaves — single concerted step, requires anti-periplanar geometry. E1: leaving group leaves first (carbocation), then any base removes H — two steps. E1 competes with SN1; E2 competes with SN2.
The complete list
- SN1: 3° substrate, weak Nu, polar protic, racemization
- SN2: 1° substrate, strong Nu, polar aprotic, inversion
- E1: 3° substrate, weak base, high T, Zaitsev product
- E2: strong bulky base, anti-periplanar, Zaitsev/Hofmann
- Rate: SN1/E1 = k[RX], SN2/E2 = k[RX][Nu/Base]
- 2° substrates: SN2 with strong Nu, SN1 with weak Nu
Frequently asked questions
How do I decide between SN1 and SN2?
Check substrate first: methyl/1° → SN2. 3° → SN1. For 2°: strong nucleophile → SN2, weak nucleophile → SN1. Then check solvent: polar aprotic favours SN2, polar protic favours SN1. Rate law confirms: SN2 is second-order, SN1 is first-order.
When does elimination (E) beat substitution (SN)?
Three factors push toward elimination: (1) Strong, bulky base (tBuO⁻) → E2. (2) High temperature → E1/E2 (entropy favours elimination). (3) 3° substrate with any base → E dominates (SN2 impossible due to steric hindrance, so if conditions don't favour SN1, elimination wins). In general: heat + bulky base = elimination.
What is the Zaitsev rule?
In E1 and E2 (with non-bulky base), the major product is the MORE substituted alkene — the one with more alkyl groups on the C=C. This is because more substituted alkenes are more stable (hyperconjugation). Exception: with bulky bases (tBuO⁻), Hofmann product (less substituted alkene) forms due to steric preference for the less hindered β-H.
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