Chemistry · Haloalkanes And Haloarenes · NEET
It is about two opposite needs. SN1 makes a carbocation first, so it needs a carbon that gives a STABLE cation. A tertiary carbon has 3 alkyl groups pushing electron density in, so its carbocation is very stable, so SN1 is easy. SN2 needs the nucleophile to hit the carbon from the back side. A tertiary carbon is crowded (3 bulky groups block the back), so SN2 is blocked. A primary carbon is open, so SN2 works well but its carbocation is unstable, so SN1 fails. Order for SN1: 3 > 2 > 1 > methyl. Order for SN2: methyl > 1 > 2 > 3 (exact reverse).
A polar protic solvent has O-H or N-H bonds (water, alcohols, ammonia). These form hydrogen bonds around the ions. In SN1, the slow step makes a carbocation and a leaving anion. Protic solvent surrounds and stabilises BOTH ions, which lowers the energy needed, so SN1 speeds up. A polar aprotic solvent (like acetone, DMSO, DMF) is polar but has no O-H to donate. It cannot cage the nucleophile, so the nucleophile stays 'naked' and reactive, which speeds up SN2. Simple rule: protic helps SN1, aprotic helps SN2.
No, and this is a common NEET trap. In SN1 the slow (rate-determining) step is only the leaving group leaving to form the carbocation. The nucleophile joins in the fast SECOND step, after the slow step is over. So SN1 rate = k[substrate] and does not include the nucleophile at all. In SN2, the nucleophile attacks IN the slow step, so a stronger nucleophile means a faster SN2 (rate = k[substrate][nucleophile]). Nucleophile strength = big deal for SN2, no effect on SN1 rate.
A good leaving group leaves easily and is a weak base (a stable anion). For the same carbon, the halide order is I⁻ > Br⁻ > Cl⁻ > F⁻. Iodide is largest, so the C-I bond is weakest and easiest to break, and iodide spreads its negative charge over a big size, so it is stable once free. Fluoride is small and holds charge tightly, so it leaves poorly. A better leaving group speeds up BOTH SN1 and SN2. This is exactly why an iodide reacts faster than a chloride (a real 2025 NEET question).
Steric hindrance means bulky groups getting in the way. In SN2 the nucleophile must reach the carbon from the back side, opposite the leaving group. Bulky alkyl groups (as in secondary and tertiary) block that path, so SN2 slows down or stops. In SN1 the carbon becomes a flat (planar) carbocation before the nucleophile arrives, so crowding is actually relieved and bulky groups even help by stabilising the cation. So: crowding hurts SN2, helps SN1.
Assertion (A): CH₃CH₂CH₂I (n-propyl iodide) undergoes SN2 reaction faster than CH₃CH₂CH₂Cl (n-propyl chloride). Reason (R): Iodine is a better leaving group because of its large size. Choose the correct answer.
The INCORRECT statement regarding chirality is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
The substrate (carbon type) is usually the biggest factor. Primary carbons go SN2, tertiary carbons go SN1, and secondary carbons can go either way depending on the nucleophile and solvent. Then solvent and nucleophile fine-tune the outcome.
Almost never on their own, because a primary carbocation is too unstable. Exceptions like allyl and benzyl halides can do SN1 because their carbocation is stabilised by resonance, even though the carbon is primary.
Water is a polar protic solvent, so it favours SN1. It hydrogen-bonds around and stabilises the carbocation and the leaving anion in the slow step.
Because the C-X bond breaks in the slow step of both mechanisms. In SN1 it breaks alone, in SN2 it breaks as the nucleophile attacks, but either way a weaker C-X bond and a more stable leaving anion lowers the energy barrier, so both speed up.
Some nucleophiles (like cyanide CN⁻ or nitrite NO₂⁻) can attach through two different atoms, giving different products. Which product forms depends on SN1 vs SN2 conditions, which is the next topic in this chapter.
