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Use curved arrows to show the movement of electrons in each equation.

a.

b.

Short Answer

Expert verified

Answer

a.

b.

Step by step solution

01

Step-by-Step SolutionStep 1: Carbocation and carbanion

Carbon atoms having a positive charge (lack of an electron) are called carbocations and the ones having excess electrons that are indicated using a negative charge over the carbon atom are called carbanions.

02

Carbon radical

Neutral carbon atoms that have an unpaired electron over them are termed carbon radicals.

03

Movement of electrons

a.

Movement of electrons in reaction a

Both the species consist of carbon radicals.

Each donates one electron each, resulting in a bond.

b.

Movement of electrons in reaction b

Electrons move from negatively charged species (species-rich in electrons) to positively charged species (electron-deficient).

Here, the bromine atom is electron-rich and it donates both the electrons to the carbocation resulting in a bond.

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Most popular questions from this chapter

For each rate equation, what effect does the indicated concentration change have on the overall rate of the reaction?

[1] rate=k[CH3CH2Br][-OH]

  1. tripling the concentration of CH3CH2Bronly
  2. tripling the concentration of – OH only
  3. tripling the concentration of both CH3CH2Br and – OH

[2]role="math" localid="1648280223497" rate=k[(CH3)3COH]

  1. doubling the concentration of (CH3)3COH
  2. increasing the concentration of (CH3)3COH by a factor of 10

As we learned in Chapter 4, propane (CH3CH2CH3)has both 1°and 2°hydrogens.

  1. Draw the carbon radical formed by homolysis of each type of C-H bond.
  2. Use the values in Table 6.2 to determine which C-H bond is stronger.
  3. Explain how this information can be used to determine the relative stability of the two radicals formed. Which radical formed from propane is more stable?

The following is a concerted, bimolecular reaction:CH3+NaCN→CH3CN+NaBr.

a. What is the rate equation for this reaction?

b. What happens to the rate of the reaction if[CH3Br] is doubled?

c. What happens to the rate of the reaction if [NaCN] is halved?

d. What happens to the rate of the reaction if [CH3Br] and [NaCN] are both increased by a factor of five?

Draw the structure for the transition state in each reaction.

a.

b.

As we will learn in Section 15.12, many antioxidants—compounds that prevent unwanted radical oxidation reactions from occurring—are phenols, compounds that contain an OH group bonded directly to a benzene ring.

  1. Explain why homolysis of the O-H bond in phenol requires considerably less energy than homolysis of the O-H bond in ethanol (362 kJ/mol vs 438 kJ/mol).
  2. Why is the C-O bond in phenol shorter than C-O bond in ethanol?
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