Potential-energy versus reaction-coordinate diagrams for the two propagation steps for monobromination of methane should be sketched. Concept introduction: The mechanism for monobromination comprises of three stages illustrated as follows: Step1: Initiation via homolytic cleavage of Br − Br bond: Br 2 initiates the reaction; undergoes homolysis and forms bromine radical atom. Energy for homolytic cleavage is provided by the heat or ultraviolet light. Step2: Propagation: In first of the propagation step bromine radical from step 1 abstracts hydrogen radical from methane as follows: In subsequent propagation step methyl radical abstracts Br · radical from another Br 2 , and liberates bromoethane and new bromine radical. Step3: Termination: Radicals generated in propagation steps get quenched upon combination with one another. Thus termination steps are essentially the radical− radical combination illustrated as follows:

Question

Want to see the full answer?

Answer 1

Question

Chapter 3, Problem 23P

Interpretation Introduction

Interpretation: Potential-energy versus reaction-coordinate diagrams for the two propagation steps for monobromination of methane should be sketched.

Concept introduction: The mechanism for monobromination comprises of three stages illustrated as follows:

Step1: Initiation via homolytic cleavage of Br−Br bond: Br2 initiates the reaction; undergoes homolysis and forms bromine radical atom. Energy for homolytic cleavage is provided by the heat or ultraviolet light.

Step2: Propagation: In first of the propagation step bromine radical from step 1 abstracts hydrogen radical from methane as follows:

In subsequent propagation step methyl radical abstracts Br· radical from another Br2 , and liberates bromoethane and new bromine radical.

Step3: Termination: Radicals generated in propagation steps get quenched upon combination with one another. Thus termination steps are essentially the radical− radical combination illustrated as follows:

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

(a 4 shows scanning electron microscope (SEM) images of extruded actions of packing bed for two capillary columns of different diameters, al 750 (bottom image) and b) 30-μm-i.d. Both columns are packed with the same stationary phase, spherical particles with 1-um diameter. A) When the columns were prepared, the figure shows that the column with the larger diameter has more packing irregularities. Explain this observation. B) Predict what affect this should have on band broadening and discuss your prediction using the van Deemter terms. C) Does this figure support your explanations in application question 33? Explain why or why not and make any changes in your answers in light of this figure. Figure 4 SEM images of sections of packed columns for a) 750 and b) 30-um-i.d. capillary columns.³

fcrip = ↓ bandwidth Il temp 32. What impact (increase, decrease, or no change) does each of the following conditions have on the individual components of the van Deemter equation and consequently, band broadening? Increase temperature Longer column Using a gas mobile phase instead of liquid Smaller particle stationary phase Multiple Paths Diffusion Mass Transfer

34. Figure 3 shows Van Deemter plots for a solute molecule using different column inner diameters (i.d.). A) Predict whether decreasing the column inner diameters increase or decrease bandwidth. B) Predict which van Deemter equation coefficient (A, B, or C) has the greatest effect on increasing or decreasing bandwidth as a function of i.d. and justify your answer. Figure 3 Van Deemter plots for hydroquinone using different column inner diameters (i.d. in μm). The data was obtained from liquid chromatography experiments using fused-silica capillary columns packed with 1.0-μm particles. 35 20 H(um) 큰 20 15 90 0+ 1500 100 75 550 01 02 594 05 μ(cm/sec) 30 15 10