Question

Consider the following two-step reaction:

a. How many bonds are broken and formed in Step [1]? Would you predict $\mathbf{∆}\mathbf{H}\mathbf{°}$of Step [1] to be positive or negative?

b. How many bonds are broken and formed in Step [2]? Would you predict the $\mathbf{∆}\mathbf{H}\mathbf{°}$of Step [2] to be positive or negative?

c. Which step is rate-determining?

d. Draw the structure for the transition state in both steps of the mechanism.

e. If ${\mathbf{∆}\mathbf{H}\mathbf{°}}_{\mathbf{overall}}$is negative for this two-step reaction, draw an energy diagram illustrating all of the information in parts (a)–(d).

Short answer

Answer

1. Two bonds are breaking, and only 1 bond is forming. must be positive.
2. Only one bond is formed, and no bond is broken. must be negative.
3. Step [1]must be the rate-determining step.
4. 
   

Transition State in Step [1]

Transition State in Step [2]

e.

Energy Diagram

Step-by-step solution

Step-by-Step SolutionStep 1: Rate-determining Step

The slowest step in a chemical reaction is the rate-determining step.

For example, if step [1] is the slowest in a reaction, all the other steps must wait for step[1] to get completed so that the reaction proceeds forward.

Enthalpy Change  

$∆\mathrm{H}°$is positive when energy is given to break bonds. $∆\mathrm{H}°$is negative when energy is released during the formation of two bonds.

Explanation

1. In step [1], one $\mathrm{\pi }-\mathrm{bond}$and one H-Cl are broken, and one C-H bond is formed. Since two bonds are breaking and only 1 bond is forming, the $∆{\mathrm{H}}^{°}$must be positive.
2. In step [2], only one C-Cl bond is formed, and no bond is broken. Hence, $∆{\mathrm{H}}^{°}$must be negative.
3. Since step [1] involves the breaking of two bonds and the formation of one new bond, it must be the rate-determining step.
4. Transition state in both the steps are:Transition State in Step [1]

Transition State in Step [2]

e. Transition State in Step [2]

Energy Diagram