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The following two questions prompted me to ask this question.

  1. Determine the order of the following compounds according to reactivity in Williamson Ether Synthesis.
    a. $\ce{CH3CH2CH2Br}$
    b. $\ce{CH2=CHCH2Cl}$

  2. Which of these represent the ether($\ce{CH2=CHCH2OC3H7}$) on forming peroxide in air
    a. $\ce{CH2=CHCH2OCH(OOH)CH2CH3}$
    b. $\ce{CH2=CH-CH(OOH)OC3H7}$

In both the case, I think the question is the relative ease of nucleophilic attack on carbon with $\alpha$-unsaturation. In the former, the bromine is a better leaving group. But, the presence of $\alpha$-unsaturation in the chloro compound, changes the order to b>a. In the latter, the attack of oxygen depends on the most stable electrophilic centre.

My question:-

Is $\alpha$-unsaturation favourable for nucleophilic attack? Relevant factors would be the increased electrophilicity of the carbon attached to the $sp^2$ hybridised ethylinic carbon and the repulsion of the negative nucleophile by the pi electrons.

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    $\begingroup$ As a note, the peroxide forming reaction in #2 is a radical reaction substituting H for OOH, so arguments involving nucleophilicity and electrophilicity are not explicitly relevant since these concepts are defined for ionic reactions. $\endgroup$
    – Ben Norris
    Jan 16, 2014 at 16:14

1 Answer 1

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As Ben Norris said, the aspects of nucleophilic and electrophilic attacks are not relevant in the second case.

Talking about the compound given in question 1(b), the first step of nucleophilic attack is opposed by the conditions you mentioned, i.e., of the electron withdrawal effect of the double bond and replusion of nucleophile by $\pi$ electron cloud.

But talking in terms of yield of the product, it is necessary to consider two things:

  1. Chlorine is a good leaving group.
  2. The stability of the carbocation intermediate formed by $SN_1$ mechanism (steric factors don't favour $SN_2$). The carbocation formed will be very stable due to conjugation with the double bond. Hence, the yield of the product is not much low.

While solving the above question, we should look at the stability of carbocation intermediate formed,which is a more dominating effect. Though Bromine is a better leaving group, and there is decrease in tendency for nucleophilic attack due to $\alpha $-unsaturation; both factors which support higher nucleophilic attack for the first compound, stabilisation of carbocation by resonance (conjugation) is highly stabilising, and this shadows the opposing factors.

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  • $\begingroup$ Thank you for the answer.Your arguments are logically satisfying. But can you provide some reference to support the claims. $\endgroup$
    – stochastic13
    Jan 27, 2014 at 5:00
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    $\begingroup$ Stability of compounds/intermediates is compared using Internal energy(IE) of the compound and activation energy(AE) of the reaction.Any system tries to minimise IE,to become stable.High AE makes it difficult for a reaction to proceed,so such an intermediate isn't stable.In the 2nd compound there's a delocalised pi electron cloud,due to resonance.Resonance is known to decrease the IE of a compound significantly.Stable reaction intermediates imply low AE.Couldn't give you direct references,you can find the data regarding AE and resonance energy of compounds in detailed textbooks and Wikipedia. $\endgroup$
    – scienceauror
    Jan 27, 2014 at 11:06
  • $\begingroup$ Whenever we compare stability, it's all about thermodynamics.That's why detail organic chemistry textbooks mention activation energies,yield of each possible product,all energies supplied and evolved during a reaction;while explaining a reaction mechanism. $\endgroup$
    – scienceauror
    Jan 27, 2014 at 11:06

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