Fujio and co-workers studied the reaction of pyridine with a...

Fujio and co-workers studied the reaction of pyridine with a wide range of 1 -arylethyl bromides in acetonitrile. By careful analysis of the kinetic data, they were able to dissect each reaction into a first-order and a second-order component, as shown in the table below. The first-order components were correlated by a Yukawa-Tsuno equation: $\log k / k_{o}=5.0\left(\sigma^{\circ}+1.15 \bar{\sigma}^{+}\right)$. The second-order component gave a curved plot, as shown in Figure 4.P18. Analyze the responses of the reaction to the aryl substituents in terms of transition state structures.

Key Concepts

Dual Reaction Mechanism Pathways

Some reactions proceed via multiple pathways, such as unimolecular (first-order) and bimolecular (second-order) mechanisms. Analyzing the sensitivity of each pathway to substituent effects provides insight into differences in transition state structures and the extent of charge development during bond-making or bond-breaking processes.

Electronic Substituent Effects

Substituent effects refer to the influence that electron-donating or -withdrawing groups exert on a reaction site. These effects can modulate the development of positive or negative charges in the transition state, either stabilizing or destabilizing it, thereby impacting the reaction kinetics.

Linear Free Energy Relationships (LFER)

LFERs, such as the Hammett and Yukawa–Tsuno relationships, are mathematical correlations that relate reaction rates or equilibria to electronic parameters of substituents. They help in interpreting how changes in substituent electronic properties influence the energy of the transition state and, consequently, the overall reaction rate.

Transition State Theory

This concept involves understanding the high?energy configuration along the pathway from reactants to products. It emphasizes that the structure and energy of the transition state dictate the rate of the reaction, with substituents potentially stabilizing or destabilizing partial charges or bonds in this fleeting state.

Transcript

00:01 We have a reaction between 2 -1 -633 -cbl with hydroxide iron, and then the hydroxide ion, the bone ion is going to be replaced by the hydroxide iron.

00:13 All right, so for this reaction is the first order.

00:16 We respect to our reactant over here.

00:21 And then in several experiments, the weight constant will determine at different temperature.

00:27 And then it's going to plot as a lateral lot of the weight constant.

00:30 Versus the 1 over the temperature.

00:33 And we had the slope equals to 1 .10 .10 power of 4, and then y introduces 43 .5.

00:39 So we have to find out the activation energy, the frequency factor a or ironized factor, and also the value of k at 25 degrees celsius.

00:50 All right, so before you move on to solve for this question, we have to understand why we have to plot a lecture lot of k versus 1 over t.

01:00 So from here, we can start with our arrhenius equation.

01:04 So we have a weight constant equals to the iranus factor of fencing factor time the lateral loomberg to the power of levert xea times rt.

01:13 This is the component.

01:16 If we take the lateral lot of both sides, so we will have lateral log a e to the relative to a relative to a rt, and then if you actually further to to simplify it or to do some rearrangement, we should be able to see that, we should be able to divide this equation, divide this equation.

01:50 So if i rearrange that properly, so we have ga over r, time 1 over t, and then plus lateral law of a, natural law of a.

02:03 If you can see that, you are reaching this form is essentially similar to y equals mx plus c.

02:12 So it's a general formula for a straight line.

02:16 So if we go back to our question, we plot a lateral of k versus 1 over t.

02:23 Essentially we have a letter of k s y, 1 over t x.

02:27 So we have this equation.

02:29 And if we can correspond to this loop, we can respond to this loop.

02:33 Will be left va overall, will be the sloop.

02:39 Lack of va overall will be equals to the sloop.

02:41 I will just add represents the sloop, and also the lateral a will be equals to the intercept.

02:47 Okay, so before we move on to look at the a, we are going to start with defining out the activation energy.

02:54 Okay, so we know the sloop is equals to 1 .1...

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