Draw the products of homolysis or heterolysis of each indicated bond. Use electronegativity differences to decide on the location of charges in the heterolysis reaction. Classify each carbon reactive intermediate as a radical, carbocation, or carbanion.

a.

b.

Short Answer

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Answer

a. A radical intermediate is generated.

b. A carbocation intermediate is generated.

Step by step solution

01

Step-by-Step SolutionStep 1: Homolysis and Heterolysis

The cleavage of a bond in which each atom involved in the bonding retains one electron is termed homolytic cleavage or homolysis.

The cleavage of a bond in which both the electrons involved in bonding are transferred to one of the atoms (the more electronegative atom) is termed heterolytic cleavage or heterolysis.

02

Carbon intermediates

The intermediate involving a carbon atom bearing a positive charge (indicating deficiency of electrons) are termed carbocations. The ones bearing a negative charge (indicating an excess of electrons) are termed carbanions.

The carbon species having an unshared electron over them are termed carbon radicals.

03

Products of the given bond cleavage

a.

Homolysis of the C-H bond

The homolytic cleavage of the bond between the carbon and the hydrogen atom generates a carbon radical as both the carbon and the hydrogen atom get one electron each.

b.Heterolysis of the C-O bond

In the given case, the shared pair of electrons are transferred to the electronegative oxygen atom, and hence a carbocation is generated.

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Most popular questions from this chapter

Consider the following two-step reaction:

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

b. How many bonds are broken and formed in Step [2]? Would you predict the H°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 H°overallis negative for this two-step reaction, draw an energy diagram illustrating all of the information in parts (a)–(d).

The conversion of (CH3)3Clto (CH3)2C=CH2can occur by either a one-step or a two-step mechanism, as shown in Equations [1] and [2].

a. What rate equation would be observed for the mechanism in Equation [1]?

b. What rate equation would be observed for the mechanism in Equation [2]?

c. What is the order of each rate equation (i.e., first, second, and so forth)?

d. How can these rate equations be used to show which mechanism is the right one for this reaction?

e. Assume Equation [1] represents an endothermic reaction and draw an energy diagram for the reaction. Label the axes, reactants, products, Ea, and H°. Draw the structure for the transition state.

f. Assume Equation [2] represents an endothermic reaction and that the product of the rate-determining step is higher in energy than the reactants or products. Draw an energy diagram for this two-step reaction. Label the axes, reactants and products for each step, and the Eaand H°for each step. Label H°overall. Draw the structure for both transition states.

For a reaction with Keq=0.8and Ea=80kJ/mol, decide which of the following statements is (are) true. Correct any false statement to make it true. Ignore entropy considerations. (a) The reaction is faster than a reaction with Keq=8and Ea=80kJ/mol. (b) The reaction is faster than a reaction with Keq=0.8and Ea=40kJ/mol. (c) G° for the reaction is a positive value. (d) The starting materials are lower in energy than the products of the reaction. (e) The reaction is exothermic.

Identify the catalyst in each equation.

a.

b.

Which value (if any) corresponds to a faster reaction: (a) Ea=40kJ/molor Ea=4kJ/mol; (b) a reaction temperature of 0°C or a reaction temperature of 25°C; (c) Keq=10or Keq=100; (d) role="math" localid="1648275832528" H=-10kJ/molor H=10kJ/mol ?

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