Account for the following observations when (S)- CH3 CH2 CH2 CHID is heated in acetone solution

Account for the following observations when (S)- CH3 CH2 CH2...

Account for the following observations when (S)- $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CHID}$ is heated in acetone solution with NaI: $(a)$ The enantiomer is racemized. (b) If radioactive ${ }^* 1^{-}$is present in excess, the rate of racemization is twice the rate at which the radioactive ${ }^* 1^{-}$is incorporated into the compound. (a) Since enantiomers have identical energy, reaction proceeds in both directions until a racemic equilibrium mixture is formed. (b) Each radioactive $* \mathrm{I}^{-}$incorporated into the compound forms one molecule of enantiomer. Now one unreacted molecule and one molecule of its enantiomer, resulting from reaction with ${ }^* \mathrm{I}^{-}$, form a racemic modification. Since two molecules are racemized when one ${ }^* 1^{-}$reacts, the rate of racemization will be twice that at which ${ }^* \mathrm{I}^{-}$ reacts.

Account for the following observations when (S)- $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CHID}$ is heated in acetone solution with NaI: $(a)$ The enantiomer is racemized. (b) If radioactive ${ }^* 1^{-}$is present in excess, the rate of racemization is twice the rate at which the radioactive ${ }^* 1^{-}$is incorporated into the compound.
(a) Since enantiomers have identical energy, reaction proceeds in both directions until a racemic equilibrium mixture is formed.
(b) Each radioactive $* \mathrm{I}^{-}$incorporated into the compound forms one molecule of enantiomer. Now one unreacted molecule and one molecule of its enantiomer, resulting from reaction with ${ }^* \mathrm{I}^{-}$, form a racemic modification. Since two molecules are racemized when one ${ }^* 1^{-}$reacts, the rate of racemization will be twice that at which ${ }^* \mathrm{I}^{-}$ reacts.

Key Concepts

Reaction Kinetics and Isotopic Labeling

The observation that the rate of racemization is twice the rate of incorporation of a labeled nucleophile highlights the kinetic details of the reaction mechanism. In this context, the use of a radioactive isotope allows for tracking the specific pathway of substitution. The fact that each incorporation event leads to the formation of one enantiomer while also affecting another molecule in the racemization process results in a doubling of the apparent rate of racemization relative to the rate of labeled nucleophile incorporation. This insight is crucial in understanding the stoichiometry and detailed steps of the reaction mechanism.

Reversible Reaction Equilibrium

In a reaction where the reactants and products are in equilibrium and the energy differences between the enantiomers are negligible, both forms are formed in equal amounts. This reversible nature of the reaction explains why even if one enantiomer is initially favored, the process will eventually lead to a racemic mixture as both directions of the reaction occur at the same rate once equilibrium is reached.

Racemization

Racemization refers to the process in which an optically active compound is converted into a mixture of equal amounts of two enantiomers. This occurs because the energy barrier between the two forms is low enough that the reaction allows for interconversion, leading to a racemic mixture at equilibrium. In many substitution reactions, such as an SN2 reaction, the intermediate or transition state can lose the memory of the original configuration, allowing nucleophilic attack to occur from either side equally.

Nucleophilic Substitution (SN2) Mechanism

The SN2 reaction mechanism involves a nucleophile attacking the substrate from the side opposite to the leaving group, resulting in a backside attack and inversion of configuration. However, when the reaction is reversible, as in the case with halide exchange, both the forward and reverse reactions can occur, leading to an equilibrium between opposite configurations. This mechanism is key to understanding how a single stereocenter can ultimately yield a racemic mixture.

Transcript

00:01 To determine the order of the reaction with respect to each reactant, we need to find two experiments where the reactant we're interested in has a changing concentration, and the other reactants do not.

00:16 So if we look at experiments one and two, where only br is being doubled, the other two concentrations are staying constant, we see an increase in the rate from about six, times 10 to the negative 5 to 12 times 10 to negative 5, suggesting when the concentration is doubled, the rate is doubled.

00:42 This is indicative of, no, hold on a second.

00:46 No, the rate stays constant.

00:49 5 .7 times 10 to the negative 5 .5.

00:52 So because it stays constant, this is representative of 0th order.

00:56 Now let's look at experiments 1 and 3.

00:59 When we look at experiments 1 and 3, the changing constant, concentration is that of h3o plus, where the br2 stays at 0 .05 and the acetone stays at 0 .3.

01:14 So when we only double the concentration of h3o plus, here is where we see about a doubling of the rate, about 6, going to 12.

01:27 And this is representative of first order.

01:30 If what we do to the concentration, we do to the rate, we double the concentration, we double the rate, we double the rate, we half the concentration, we half the rate.

01:37 This is representative of first order.

01:40 So it's first order with respect to h3o plus, zero order with respect to br2.

01:45 The last one's going to be a little bit more complicated.

01:48 We don't see a simple doubling of the concentration of acetone.

01:53 So let's look at experiments 1 and 4.

01:56 In experiments 1 and 4, the concentrations of br2 and h30 plus are all 0 .05.

02:05 The only changing concentration is acetone going from 0 .3 to 0 .04.

02:12 So what we can do is take a ratio of two rate laws.

02:19 For experiment one, the rate is 5 .7 times 10 to negative 5...