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Students learn that tertiary halides react by SN1 and primary halides by SN2. Then a question shows benzyl chloride, a primary halide, reacting quickly in water by SN1, and the rule seems to break. It does not: the rule was always about how stable the carbocation is, and a benzylic or allylic cation is unusually stable. This post explains why, shows how to rank these halides, works through the rearranged products that allylic cations give, and lists the comparisons examiners like.
Where is this in NCERT and the syllabus?
The current NCERT Class 12 Chemistry Part II chapter Haloalkanes and Haloarenes (2026-27 reprint) defines allylic and benzylic halides in its classification section, where the halogen is on an sp³ carbon next to a C=C or a benzene ring. In the SN1 section it says the rate depends on the slow C–X cleavage, that greater carbocation stability means a faster reaction, and that "allylic and benzylic halides show high reactivity towards the SN1 reaction" because the carbocation is stabilised through resonance. Its Example 6.7 ranks four benzylic bromides. The JEE (Main) 2026 syllabus lists "nature of C-X bond, mechanisms of substitution reactions" and "stability of carbocations", and the NEET (UG) 2026 syllabus has the same lines. Check your exam year's syllabus.
For the full SN1 and SN2 comparison, including stereochemistry, read our guide to SN1 and SN2 in haloalkanes. Here we stay with one question: why are these two families so fast by SN1?
Why does resonance make the cation stable?
In SN1, the halide leaves first and the carbocation forms in the slow step. Anything that lowers the energy of that cation lowers the energy needed to reach it, and the reaction speeds up.
Allyl cation. Remove Cl⁻ from CH₂=CH–CH₂Cl and you get CH₂=CH–CH₂⁺. The empty p orbital on the end carbon lines up with the π bond next to it, so you can draw a second structure, ⁺CH₂–CH=CH₂. The real ion is a blend of the two: the positive charge is shared by carbons 1 and 3.
Benzyl cation. Remove Cl⁻ from C₆H₅CH₂Cl and the empty orbital lines up with the ring's π system. You can draw the charge on the CH₂ carbon and on the two ortho carbons and the para carbon of the ring. Spread over four carbons, the charge is much easier to carry.
A charge shared over several atoms is a lower-energy charge. That is the whole reason.
Which halides are not allylic or benzylic?
Students often confuse four families. The test is where the halogen-bearing carbon sits.
Allylic: X on the carbon next to C=C, as in CH₂=CH–CH₂Cl. Fast SN1.
Vinylic: X on the C=C carbon itself, as in CH₂=CHCl. No SN1; the cation would be very unstable and the C–X bond has partial double-bond character.
Benzylic: X on a carbon attached to the ring, as in C₆H₅CH₂Cl. Fast SN1.
Aryl: X on the ring itself, as in C₆H₅Cl. NCERT notes that the phenyl cation is not stabilised by resonance, so SN1 is ruled out.
Worked example 1: ranking SN1 reactivity
Rank for SN1: (a) CH₃CH₂CH₂Cl, (b) CH₂=CHCH₂Cl, (c) C₆H₅CH₂Cl, (d) (C₆H₅)₂CHCl.
Draw each cation after Cl⁻ leaves.
(a) gives a primary alkyl cation with no resonance: least stable.
(b) shares its charge over two carbons.
(c) shares it over the CH₂ carbon and three ring positions.
(d) has two rings to share it, and is also secondary.
Order: (d) > (c) > (b) > (a). This matches the reasoning in NCERT Example 6.7, where the bromide with two phenyl groups on the reacting carbon is more reactive by SN1 than the one with a phenyl and a methyl.
Worked example 2: how do ring substituents change the rate?
Compare para-methoxybenzyl chloride, benzyl chloride and para-nitrobenzyl chloride.
The para carbon carries part of the positive charge in the benzyl cation. An OCH₃ group there can donate a lone pair by resonance and stabilise it further. An NO₂ group pulls electron density away and makes the cation less stable.
SN1 order: p-methoxy > unsubstituted > p-nitro.
Worked example 3: why do two products form?
Hydrolyse 3-chlorobut-1-ene, CH₂=CH–CHCl–CH₃, in water. Cl⁻ leaves to give a cation whose charge is shared between C3 and C1:
CH₂=CH–CH⁺–CH₃ and ⁺CH₂–CH=CH–CH₃
Water can attack either positive carbon. Attack at C3 gives but-3-en-2-ol; attack at C1 gives but-2-en-1-ol. This allylic rearrangement is a common trap: questions ask for "the products", and one answer is not enough.
Which mistakes cost marks?
Classifying by primary, secondary or tertiary alone. Check for an adjacent C=C or ring first.
Treating vinyl and aryl halides like allyl and benzyl. The halogen's position decides it.
Giving only one product from an allylic cation. Draw both resonance structures and attack both ends.
Forgetting the conditions. A fast SN1 needs a polar protic solvent such as water or ethanol and a weak nucleophile. With a strong nucleophile in a polar aprotic solvent, a primary benzylic halide can react by SN2 instead.
If you are unsure about the resonance ideas behind this, the Fermi Concept Library covers every concept in physics, chemistry, biology, mathematics and computing, so you can revise resonance and carbocations before ranking reactivities.
A practice routine
Draw all resonance structures of the benzyl cation and mark each carbon that carries charge.
Rank five halides for SN1 each day for a week, mixing allylic, vinylic, benzylic, aryl and plain alkyl examples.
Predict both products of hydrolysing 1-chlorobut-2-ene and check that they match those in example 3.
Explain in two lines why para-nitrobenzyl chloride is slower than benzyl chloride.
This week, redo example 1 from memory and say aloud why each halide sits where it does.
