Q1(d). 3-Chlorobut-1-ene and hydroxide: allylic substitution
For the reaction given below, write the structures of major and minor products, and discuss the mechanism of their formation:
CH₃-CHCl-CH=CH₂ → (OH)
Answer in brief
Major: but-3-en-2-ol (attack at C-3); minor: (E)-but-2-en-1-ol (allylic rearrangement). Both come from the same allylic cation (SN1 and SN1'); the unrearranged alcohol predominates because the leaving chloride stays close to C-3 in the ion pair.
Key points
- In aqueous hydroxide the C–Cl bond ionises (SN1, rate-determining step) to the resonance-stabilised allylic cation; the positive charge is shared by C-1 and C-3 (SN1' when the attack is at C-1).
- Water or OH⁻ attacks C-3 (the carbon that carried Cl) to give but-3-en-2-ol, CH₃CH(OH)CH=CH₂, and C-1 to give but-2-en-1-ol (crotyl alcohol), CH₃CH=CHCH₂OH, mainly E.
- The cation is formed first as an intimate ion pair: Cl⁻ still sits near C-3 and polarises the cation, so C-3 is attacked more often. The product is therefore richer in the unrearranged alcohol (product spread): with 0.8 M…
- With a higher concentration of OH⁻ a bimolecular path adds to this: direct SN2 at C-3 (unrearranged alcohol) and SN2' attack at C-1 with shift of the double bond (rearranged alcohol).
The full model answer
Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.
Introduction. 3-Chlorobut-1-ene is a secondary allylic chloride. Hydroxide can replace Cl at the carbon that carries it, or at the far end of the allyl system with a shift of the double bond (allylic rearrangement). Two isomeric alcohols result.
Scheme: Attack at C-3 gives the secondary alcohol; attack at C-1 with shift of the double bond gives crotyl alcohol.
Mechanism: one allylic cation, two ends
- In aqueous hydroxide the C–Cl bond ionises (SN1, rate-determining step) to the resonance-stabilised allylic cation; the positive charge is shared by C-1 and C-3 (SN1' when the attack is at C-1).
- Water or OH⁻ attacks C-3 (the carbon that carried Cl) to give but-3-en-2-ol, CH₃CH(OH)CH=CH₂, and C-1 to give but-2-en-1-ol (crotyl alcohol), CH₃CH=CHCH₂OH, mainly E.
- The cation is formed first as an intimate ion pair: Cl⁻ still sits near C-3 and polarises the cation, so C-3 is attacked more often. The product is therefore richer in the unrearranged alcohol (product spread): with 0.8 M aqueous NaOH at 25°C 3-chlorobut-1-ene gives 62 % but-3-en-2-ol and 38 % but-2-en-1-ol, whereas 1-chlorobut-2-ene gives 40 % but-3-en-2-ol and 60 % but-2-en-1-ol.
Scheme: Ionisation gives one delocalised cation, which water traps at either end.
- With a higher concentration of OH⁻ a bimolecular path adds to this: direct SN2 at C-3 (unrearranged alcohol) and SN2' attack at C-1 with shift of the double bond (rearranged alcohol).
- A small amount of buta-1,3-diene forms by elimination (hydroxide as a base).
Conclusion. Major: but-3-en-2-ol (attack at C-3); minor: (E)-but-2-en-1-ol (allylic rearrangement). Both come from the same allylic cation (SN1 and SN1'); the unrearranged alcohol predominates because the leaving chloride stays close to C-3 in the ion pair.
Application. Allylic rearrangement explains why 3-chlorobut-1-ene and 1-chlorobut-2-ene give the same pair of alcohols on hydrolysis; it is used in making allylic alcohols for perfumery (geraniol and linalool interconvert in the same way).
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All parts of Question 1
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