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SolutionsSN2-SN1 Competition

1. Order of reactivity

Give a reasoned order of SN1/SN2 reactivity for the following compounds towards: 1) NaSMe/DMF; 2) dilute H3O+.

Seven chlorides: (S)-3-chlorocyclohexene (1), benzyl chloride (2), 1,1-dichloro-2-methylpropene (3), 1-(chloromethyl)-3-methylcyclopentane (4), (R)-cis-1-chloro-3-methylcyclopentane (5), tert-butyl chloride (6) and chlorobenzene (7)

NaSMe/DMF

This is a very good nucleophile in a polar aprotic solvent, which favours the SN2 reaction if the structure of the substrate allows it.

  1. Compounds 3 and 7 do not react at all because the halogens are bonded to sp2 carbons.
  2. Compounds 1, 2 and 6 will react mainly by SN1, because the carbocations they form are less unstable (allylic, benzylic and tertiary, respectively).
  3. Compound 1 has a stereocentre of S configuration at the reacting carbon. The amount of R isomer produced in excess over the racemic mixture tells us how much of the reaction has gone through SN2, i.e. with inversion.
  4. Compound 4 is a primary halide but branched at the adjacent carbon, so SN2 will be difficult. SN1 will also be difficult, because the carbocation is primary.
  5. Compound 5 is secondary and has a stereocentre of R configuration at the reacting carbon. The amount of S isomer produced in excess over the racemic mixture tells us how much of the reaction has gone through SN2.

Probable order of overall reactivity (SN1 + SN2): 1, 2 and 6 > 5 > 4 >>> 3 and 7.

Dilute H3O+

The nucleophile is poor (H2O) and the solvent is polar and protic: SN2 is hindered and SN1 is favoured.

  1. Compounds 3 and 7 do not react, because the halogens are bonded to sp2 carbons.
  2. Compounds 1, 2 and 6 will react by SN1 (allylic, benzylic and tertiary carbocations), favoured by the polar protic medium.
  3. Compound 1 (S) will give the racemic alcohol, exclusively by SN1.
  4. In compound 4, primary and branched at the adjacent carbon, SN2 is impossible with such a poor nucleophile in a protic medium, and SN1 is difficult because the carbocation would be primary.
  5. Compound 5 (R) will give the racemic alcohol, exclusively by SN1.

Probable order of reactivity, now only by SN1: 1, 2 and 6 > 5 > 4 >>> 3 and 7.

2. SN1 or SN2?

Predict the main mechanism of each reaction:

  1. (CH3)3CBr in methanol.
  2. CH3CH2Br with NaCN in DMSO.
  3. (R)-2-bromobutane with NaN3 in DMF.
  4. Benzyl bromide in water.
  1. SN1: tertiary halide, weak nucleophile and protic solvent.
  2. SN2: primary halide and good nucleophile in an aprotic solvent.
  3. SN2, with inversion: (S)-2-azidobutane is obtained.
  4. SN1: the benzylic carbocation is stabilized by resonance.

3. Benzylic halides

Why does benzyl bromide react quickly by both SN1 and SN2?

It is an unhindered primary halide, which favours SN2, and at the same time it gives a benzylic carbocation stabilized by resonance with the ring, which favours SN1. The mechanism will depend on the nucleophile and the solvent.

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