How does each of the following changes affect the rate of an...

How does each of the following changes affect the rate of an E2 reaction? a. tripling [RX] b. halving [B:] c. changing the solvent from $\mathrm{CH}_{3} \mathrm{OH}$ to DMSO d. changing the leaving group from $I^{-}$ to $B r^{-}$ e. changing the base from ${ }^{-} \mathrm{OH}$ to $\mathrm{H}_{2} \mathrm{O}$ f. changing the alkyl halide from $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Br}$ to $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHBr}$

How does each of the following changes affect the rate of an E2 reaction?
a. tripling [RX]
b. halving [B:]
c. changing the solvent from $\mathrm{CH}_{3} \mathrm{OH}$ to DMSO
d. changing the leaving group from $I^{-}$ to $B r^{-}$
e. changing the base from ${ }^{-} \mathrm{OH}$ to $\mathrm{H}_{2} \mathrm{O}$
f. changing the alkyl halide from $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Br}$ to $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHBr}$

Key Concepts

Steric Effects and Substrate Structure

The structure of the substrate affects the rate of an E2 reaction through steric hindrance and the proper alignment of the leaving group and ?-hydrogen (anti-periplanar orientation). More substituted substrates might suffer from steric hindrance, making it more difficult for the base to access the hydrogen that needs to be removed, thus potentially reducing the reaction rate.

Base Strength

The strength of the base plays a vital role in the E2 mechanism. Stronger bases more effectively abstract beta-hydrogens, thereby increasing the rate of the elimination process. A change in the base that reduces its strength typically results in a slower reaction.

Leaving Group Ability

The ease with which a leaving group departs from the substrate is crucial in E2 reactions. A good leaving group stabilizes the negative charge upon leaving, facilitating the conversion to the alkene, so differences in leaving group stability can directly correlate with differences in reaction rate.

Rate Law for E2 Reactions

In an E2 reaction, the overall rate is given by the product of the concentrations of the substrate and the base. This means any change in the concentration of either reactant will proportionately affect the reaction rate.

E2 Reaction Mechanism

The E2 mechanism is a concerted, one?step process where deprotonation and leaving group departure occur simultaneously. This mechanism is generally characterized as second order, meaning the rate is dependent on both the substrate and the base concentrations.

Solvent Effects

The nature of the solvent can significantly impact E2 reactions. Polar aprotic solvents generally enhance the reactivity of the nucleophile or base by not strongly solvating anions, thus increasing the reaction rate, whereas polar protic solvents may slow it down due to better solvation of the base.

Transcript

00:01 All right, so we're going to look at a couple different scenarios and how they would affect the rate of reaction of an e2 reaction.

00:07 So our first step is going to be what would happen if we triple the concentration of our alkaliobromide.

00:17 So real quick, i'm going to go ahead and throw up the, i'm going to go ahead and draw up the rate law for a e2 reaction, which is a second order reaction.

00:30 So based on the rate law, our rate is going to be directly related to the concentration of the alkaliobromy.

00:36 So if we triple our concentration of alkaliobromine, we're going to go ahead and triple our rate.

00:41 Similarly, if we reduce the concentration of our base by half, the resulting effect on our rate is that it will be one half of its original.

00:57 All right, and so for our third scenario, we're moving from a methanol solvent, shown you here in black to a dimethyl sulfoxide solvent shown here in blue.

01:11 And so for this one, we need to know that for e2 reactions, polar apodic solvents are going to be stronger or increase our rate.

01:20 So both of these compounds are going to be slightly polar based on our highly electronegative oxygen.

01:26 However, the methanol is protic.

01:29 It has a hydrogen bonded to this oxygen, which is going to slightly interfere with our vision.

01:34 Since dimethyl sulfoxide is not protic, it is a protic, it is actually going to be a, it's going to increase our rate of reaction because it's not only just polar, it's also a protic.

01:49 Our fourth scenario is moving from an iodine leaving group to a bromine leaving group, which i'll go ahead and draw those up...

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