10-15-08 - 2 O for NaBr the conjugate base of a strong...

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318M – Krische, Lecture 21: W - 10/15/08, Substitution/Elimination Today, we begin substitution and elimination. It is observed that nucleophiles (species that are electron rich) will react with aliphatic alkyl halides and other electrophiles (species that are electron deficient at carbon by virtue of a polarized carbon heteroatom bond) to give products of “substitution,” i.e. products whereby the halide has been “displaced” or substituted by the nucleophile. Such reactions are termed “nucleophilic substitutions.” The following reaction is an example of nucleophilic substitution: NaOH + CH 3 Br HOCH 3 + NaBr For this reaction, NaOH is the nucleophile and CH 3 Br is the electrophile. The thermodynamic driving force for this reaction is easily recognized. First, we are breaking a relatively weak C-Br bond (70 Kcal/mol) and forming a relatively strong C-O bond (91 Kcal/mol). Secondly, we are exchanging NaOH, the conjugate base of a weak acid (H
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Unformatted text preview: 2 O), for NaBr, the conjugate base of a strong acid (HBr). The reaction is exothermic. • Kinetics studies reveal that the reaction is overall second order: first order in NaOH and first order in CH 3 Br, i.e. the rate law is Rate = k[NaOH][CH 3 Br]. This means that both NaOH and CH 3 Br are present in the transition state of the rate-determining step. Also, second order kinetics suggests a concerted mechanism, i.e. bond formation and bond cleavage likely occur simultaneously. We refer to such nucleophilic substitutions that exhibit second order kinetics as S N 2 reactions (Substitution Nucleophilic 2 nd order). • What is the mechanism of the S N 2 reaction? There are two limiting possibilities. Path A – Backside attack, which would occur with inversion of stereochemistry at carbon and Path B – Frontal attack, which would occur with retention of stereochemistry at carbon. MO theory predicts backside attack....
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