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I read about the competition between nucleophilic substitution and nucleophilic elimination depending on temperature here. Though the webpage clearly says higher temperature favors elimination while lower temperature favors substitution, the explanation on the webpage is unclear.

Need a clear explanation regarding the phenomenon.

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  • $\begingroup$ @Zhe Good point. You see Mockingbird, elimination is not a "nucleophile" phenomenon, its rather a "base" phenomenon. Regarding your question, consider the fact that eliminations involve more bonds breaking, and hence a higher activational energy. Does that give you any idea why eliminations require higher temperatures? $\endgroup$ – Pritt Balagopal Jun 16 '17 at 14:31
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    $\begingroup$ An even better question is (selectivity aside) what reactions are not favored by heat? $\endgroup$ – Zhe Jun 16 '17 at 14:36
  • $\begingroup$ @Pritt But the hyperlinked page also says the reaction 'rate decreases' for substitution decreases as temperature increases. What about that? $\endgroup$ – Mockingbird Jun 16 '17 at 15:29
  • $\begingroup$ @Mockingbird I'm not very sure, but it could be due to the opposing backward reactions. $\endgroup$ – Pritt Balagopal Jun 16 '17 at 15:30
  • $\begingroup$ That's why I hyperlinked. Some points are explained not so clearly in the webpage. Please visit the page. $\endgroup$ – Mockingbird Jun 16 '17 at 15:32
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For the nucleophilic substitutions and eliminations, we can draw up these four generalised schemes:

$$\begin{align} &\mathrm{S_N1}{:} & \ce{R-X + Nu- &-> R+ + X- + Nu- -> R-Nu + X-}\\ &\mathrm{S_N2}{:} & \ce{R-X + Nu- &-> R-Nu + X-}\\ &\mathrm{E1}{:} & \ce{RCH2-CR2-X + B- &-> RCH2-CR2+ + X- + B- -> RCH=CR2 + X- + HB}\\ &\mathrm{E2}{:} & \ce{RCH2-CH2-X + B- &-> RCH=CH2 + X- + HB}\end{align}$$

Clearly, each substitution keeps the number of freely diffusing particles the same while each elimination increases it by one. Therefore, the entropic term of the eliminations is likely to be higher than that of the substitutions. The rest is given by the Gibbs free energy equation: $$\Delta G = \Delta H - T\Delta S$$

The entropic factor scales with temperature as Bryce already mentioned so this outweighs any other effects at high temperatures.

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When the question is asking about favorability, it is really asking about the magnitude of the Gibbs free energy. The more negative this value, the more favorable a reaction is. Gibbs depends on enthalpy and a temperature scaled entropy: $$\Delta G=\Delta H-T\Delta S$$

When we think of nucleophilic elimination, we know that a hydrogen atom is removed by a nucleophilic group resulting in our final product. Since a hydrogen is removed in elimination, while left intact during substitution, there are more final species formed during an elimination reaction. According to general chemistry principles, this means that there was a larger increase in entropy compared to the substitution. As temperature increases this larger entropy becomes more noticeable as it is now scaled by a larger $T$.

Therefore at low temperature, the enthalpy term dominates Gibbs and the substitution reaction prevails. However as $T$ grows, so does the entropy term and at high $T$ this overrides enthalpy, making the elimination reaction more favorable.

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Breaking bonds requires an input of energy. Elimination reactions require the breakage of more bonds than substitution reactions, and therefore require a higher input of energy. This indicates that the activation energy of elimination reactions is higher than that of substitution reactions. Increasing temperature will allow more molecules to reach the required activation energy threshold. Therefore, as temperatures is increased, elimination is favored.

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