It is to be determined in which of the two solvents shown, Y or Z, will the reaction from Problem 9.26 proceed faster by the E1 mechanism. Concept introduction: An E1 reaction starts off the same way as an S N 1 reaction. The leaving group breaks off, producing a planar sp 2 hybridized carbocation in the first, rate-determining step. In the second step, a base extracts a proton from an adjacent carbon atom, with the C − H bond pair moving to the carbocation to form a π bond between the two carbon atoms. The leaving group departs with the pair of electrons from its bond with the carbon. Therefore, the leaving group is negatively charged. It can be strongly solvated by a protic solvent, driving the rate-determining step in the forward direction. An aprotic solvent will not effectively solvate the departing leaving group and, therefore, does not have much effect on the rate. Solvation of the base, often a negatively charged species, doesn’t noticeably affect the rate since it is not involved in the rate-determining step. A protic solvent will, therefore, favor an E1 reaction for the same reason for which it favors an S N 1 reaction.

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Chapter 9, Problem 9.27P

Interpretation Introduction

Interpretation:

It is to be determined in which of the two solvents shown, Y or Z, will the reaction from Problem 9.26 proceed faster by the E1 mechanism.

Concept introduction:

An E1 reaction starts off the same way as an SN1 reaction. The leaving group breaks off, producing a planar sp2 hybridized carbocation in the first, rate-determining step. In the second step, a base extracts a proton from an adjacent carbon atom, with the C−H bond pair moving to the carbocation to form a π bond between the two carbon atoms. The leaving group departs with the pair of electrons from its bond with the carbon. Therefore, the leaving group is negatively charged. It can be strongly solvated by a protic solvent, driving the rate-determining step in the forward direction. An aprotic solvent will not effectively solvate the departing leaving group and, therefore, does not have much effect on the rate. Solvation of the base, often a negatively charged species, doesn’t noticeably affect the rate since it is not involved in the rate-determining step.

A protic solvent will, therefore, favor an E1 reaction for the same reason for which it favors an SN1 reaction.

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