Chemistry · General Principles Of Organic Chemistry · NEET
In SN1 the slow (rate-determining) step is the leaving group breaking off to form the carbocation. The nucleophile only joins in the fast second step, after the slow step is over. Because rate is fixed by the slowest step, only the substrate concentration appears in the rate law: rate = k[substrate]. This is why we call it first order (SN1).
SN2 always gives inversion of configuration (called Walden inversion). The nucleophile attacks from the side opposite the leaving group, so the other three bonds flip like an umbrella turning inside out in the wind. If the starting carbon was a chiral centre, the product is the mirror-image configuration. SN1, by contrast, forms a flat carbocation that the nucleophile can attack from either face, giving a roughly 50:50 mix (racemisation).
SN1 needs a stable carbocation. A tertiary carbon is surrounded by three alkyl groups that push electrons in (+I effect and hyperconjugation), so its cation is stable and forms easily. A primary carbocation is very unstable, so tertiary avoids SN2 anyway because the three bulky groups block the back-side attack. A primary carbon is open and forms a poor cation, so it reacts by SN2. Secondary carbons can do either, depending on conditions.
Look at the carbon bearing the leaving group. Tertiary, allylic or benzylic (cation is resonance-stabilised) points to SN1. Methyl or primary points to SN2. Also check clues: a weak nucleophile in a polar protic solvent (like water or alcohol) favours SN1; a strong nucleophile in a polar aprotic solvent favours SN2. For NEET this fast rule solves most one-mark questions.
A polar protic solvent (has O-H or N-H, like water or ethanol) can surround and stabilise both the carbocation and the leaving anion through solvation. This lowers the energy needed to make the ions, so ionisation (the SN1 slow step) becomes easier. Polar aprotic solvents cannot hydrogen-bond to the nucleophile, so they leave it 'naked' and reactive, which instead speeds up SN2.
For the following reactions: (a) CH3CH2CH2Br + KOH (alcoholic) gives CH3CH=CH2 + KBr + H2O; (b) a bromo-dimethylcyclohexane + KOH gives the corresponding hydroxy compound (Br replaced by OH); (c) cyclohexene + Br2 gives 1,2-dibromocyclohexane. Which statement is correct?
Among the following, the reaction that proceeds through an electrophilic substitution is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Neither is always faster; it depends on the substrate. Tertiary substrates react much faster by SN1, while methyl and primary substrates react faster by SN2. The names refer to reaction order, not to speed.
SN1 causes racemisation because the flat carbocation intermediate can be attacked from both faces, giving a near 50:50 mixture of two mirror-image products. SN2 instead gives clean inversion of configuration.
SN2 needs a strong, concentrated nucleophile because the nucleophile takes part in the rate-determining step. SN1 works even with a weak nucleophile since the nucleophile joins only after the slow step.
Their carbon-halogen bond is partly double-bond in character and the carbon is sp2, so the C-X bond is strong and short. They cannot form a carbocation (no SN1) and the back side is blocked by the ring or double bond (no SN2).
Both SN1 and E1 share the same slow first step: forming the carbocation. After that, a nucleophile attacking the cation gives the SN1 product, while a base removing a beta-hydrogen gives the E1 (elimination) product. This is why the two often compete.
