The use of curved arrows is a powerful tool that illustrates even complex reactions.

a. Add curved arrows to show how carbocation A is converted to carbocation B. Label each new bond formed. Similar reactions have been used in elegant synthesis of steroids.

b. Draw the product by following the curved arrows. This reaction is an example of a [3,3] sigma tropic rearrangement, as we will learn in chapter 27.

Short Answer

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Answer

a.

Rearrangement of the carbocation A

The formation of the carbocation B

The new sigma bonds in carbocation B

b.

The products of the reaction b

Step by step solution

01

Step-by-Step SolutionStep 1: Synthesis of the organic compounds

The synthesis of the organic compounds is done by planning a certain group of reactions by using readily available starting materials. The synthesis happens by using more than one step at a time.

The proper selectivity of the reagents is essential for the synthesis of organic compounds.

Arrows are used to describe chemical reactions. The use of curved arrows demonstrates the movement of the electron pairs for understanding the mechanism of the reactions during organic synthesis.

02

The conversion of carbocation A to carbocation B.

The carbocation A is converted into carbocation B as shown in the following structure:Rearrangement of the carbocation A

The arrows indicate the movement of electrons in the above structure. In the following structure, the carbocation B is finally formed.The formation of the carbocation B

The new sigma bonds in carbocation B

b. The products of the given reaction are as follows:

The products of the reaction b

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Most popular questions from this chapter

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?

Consider the following reaction: CH4+Cl··CH3+HCl.

a. Use curved arrows to show the movement of electrons.

b. Calculate H°using the bond dissociation energies in Table 6.2.

c. Draw an energy diagram assuming that Ea=16kJ/mol.

d. What is Eafor the reverse reaction (·CH3+HClCH4+Cl·)?

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

The following is a concerted, bimolecular reaction:CH3+NaCNCH3CN+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?

Esterification is the reaction of a carboxylic acid (RCOOH) with an alcohol (R'OH) to form an ester (RCOOR') with a loss of water. Equation [1] is an example of an intermolecular esterification reaction. Equation [2] is an example of an intramolecular esterification reaction; that is, the carboxylic acid and alcohol are contained in the same starting material, forming a cyclic ester as the product. The equilibrium constants for both reactions are given. Explain why Keqis different for these two apparently similar reactions.

[1]

[2]

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