ORGANIC CHEMISTRY-OWL V2 ACCESS
ORGANIC CHEMISTRY-OWL V2 ACCESS
8th Edition
ISBN: 9781305582422
Author: Brown
Publisher: CENGAGE L
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Chapter 6, Problem 6.15P

(a)

Interpretation Introduction

Interpretation:

The structure for isomeric carbocation intermediate formed on treatment of given alkene in presence of given acid has to be drawn, labelled and identified that whether the isomeric carbocation forms readily or not.

Concept Introduction:

Leaving group: it is the fragment that leaves the substrate in a chemical reaction with a pair of electrons.

Nucleophile: donates pair of electrons to positively charged substrate resulting in the formation of chemical bond.

Carbocation: carbon ion that bears a positive charge on it.

Carbocation stability order:

ORGANIC CHEMISTRY-OWL V2 ACCESS, Chapter 6, Problem 6.15P , additional homework tip  1

Resonance stabilization: Due to the delocalization of electrons within the molecule the overall energy becomes lower and makes that molecule more stable.

(a)

Expert Solution
Check Mark

Explanation of Solution

ORGANIC CHEMISTRY-OWL V2 ACCESS, Chapter 6, Problem 6.15P , additional homework tip  2

  1. 1. There are two possibilities present in attack of proton from acid to the given alkene.
  2. 2. First is H+ from HCl attacks the carbon with one hydrogen (secondary carbon) in C=C and results to form tertiary carbocation.
  3. 3. Second is H+ from HCl attacks the carbon with 0 hydrogen (tertiary carbon) in C=C and results to form secondary carbocation.
  4. 4. Of these the tertiary carbocation forms readily since it is more stable than the secondary carbocation.

(b)

Interpretation Introduction

Interpretation:

The structure for isomeric carbocation intermediate formed on treatment of given alkene in presence of given acid has to be drawn, labelled and identified that whether the isomeric carbocation forms readily or not.

Concept Introduction:

Leaving group: it is the fragment that leaves the substrate in a chemical reaction with a pair of electrons.

Nucleophile: donates pair of electrons to positively charged substrate resulting in the formation of chemical bond.

Carbocation: carbon ion that bears a positive charge on it.

Carbocation stability order:

ORGANIC CHEMISTRY-OWL V2 ACCESS, Chapter 6, Problem 6.15P , additional homework tip  3

Resonance stabilization: Due to the delocalization of electrons within the molecule the overall energy becomes lower and makes that molecule more stable.

(b)

Expert Solution
Check Mark

Explanation of Solution

ORGANIC CHEMISTRY-OWL V2 ACCESS, Chapter 6, Problem 6.15P , additional homework tip  4

  1. 1. There are two possibilities present in attack of proton from acid to the given alkene.
  2. 2. First is H+ from HCl attacks the carbon with one hydrogen and one methyl group (secondary carbon) in C=C and results to form secondary carbocation.
  3. 3. Second is H+ from HCl attacks the carbon with 1 hydrogen and one ethyl group (secondary carbon) in C=C and results to form another secondary carbocation.
  4. 4. Both carbocations formed are secondary hence both have equal rates.

(c)

Interpretation Introduction

Interpretation:

The structure for isomeric carbocation intermediate formed on treatment of given alkene in presence of given acid has to be drawn, labelled and identified that whether the isomeric carbocation forms readily or not.

Concept Introduction:

Leaving group: it is the fragment that leaves the substrate in a chemical reaction with a pair of electrons.

Nucleophile: donates pair of electrons to positively charged substrate resulting in the formation of chemical bond.

Carbocation: carbon ion that bears a positive charge on it.

Carbocation stability order:

ORGANIC CHEMISTRY-OWL V2 ACCESS, Chapter 6, Problem 6.15P , additional homework tip  5

Resonance stabilization: Due to the delocalization of electrons within the molecule the overall energy becomes lower and makes that molecule more stable.

(c)

Expert Solution
Check Mark

Explanation of Solution

ORGANIC CHEMISTRY-OWL V2 ACCESS, Chapter 6, Problem 6.15P , additional homework tip  6

  1. 1. There are two possibilities present in attack of proton from acid to the given alkene.
  2. 2. First is H+ from HCl attacks the carbon that does not have methyl bonded with it (secondary carbon) in C=C and results to form tertiary carbocation.
  3. 3. Second is H+ from HCl attacks the carbon with methyl substitution (tertiary carbon) in C=C and results to form secondary carbocation.
  4. 4. Of these the tertiary carbocation forms readily since it is more stable than the secondary carbocation.

(d)

Interpretation Introduction

Interpretation:

The structure for isomeric carbocation intermediate formed on treatment of given alkene in presence of given acid has to be drawn, labelled and identified that whether the isomeric carbocation forms readily or not.

Concept Introduction:

Leaving group: it is the fragment that leaves the substrate in a chemical reaction with a pair of electrons.

Nucleophile: donates pair of electrons to positively charged substrate resulting in the formation of chemical bond.

Carbocation: carbon ion that bears a positive charge on it.

Carbocation stability order:

ORGANIC CHEMISTRY-OWL V2 ACCESS, Chapter 6, Problem 6.15P , additional homework tip  7

Resonance stabilization: Due to the delocalization of electrons within the molecule the overall energy becomes lower and makes that molecule more stable.

(d)

Expert Solution
Check Mark

Explanation of Solution

ORGANIC CHEMISTRY-OWL V2 ACCESS, Chapter 6, Problem 6.15P , additional homework tip  8

  1. 1. There are two possibilities present in attack of proton from acid to the given alkene.
  2. 2. First is H+ from HCl attacks the carbon that has 0 hydrogens with it (tertiary carbon) in C=C and results to form primary carbocation.
  3. 3. Second is H+ from HCl attacks the carbon with 2 hydrogens (primary carbon) in C=C and results to form tertiary carbocation.
  4. 4. Of these the tertiary carbocation forms readily since it is more stable than the primary carbocation.

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Chapter 6 Solutions

ORGANIC CHEMISTRY-OWL V2 ACCESS

Ch. 6.5 - Prob. 6.11PCh. 6.5 - Prob. 6.12PCh. 6.5 - What alkene with the molecular formula C6H12, when...Ch. 6 - Prob. 6.15PCh. 6 - Prob. 6.16PCh. 6 - Predict the organic product(s) of the reaction of...Ch. 6 - Prob. 6.18PCh. 6 - Prob. 6.20PCh. 6 - Draw a structural formula for an alkene with the...Ch. 6 - Account for the fact that addition of HCl to...Ch. 6 - Account for the fact that treating propenoic acid...Ch. 6 - Draw a structural formula for the alkene with the...Ch. 6 - Draw the alternative chair conformations for the...Ch. 6 - Draw a structural formula for the cycloalkene with...Ch. 6 - Reaction of this bicycloalkene with bromine in...Ch. 6 - Terpin, prepared commercially by the...Ch. 6 - Propose a mechanism for this reaction and account...Ch. 6 - Treating 2-methylpropene with methanol in the...Ch. 6 - When 2-pentene is treated with Cl2 in methanol,...Ch. 6 - Treating cyclohexene with HBr in the presence of...Ch. 6 - Propose a mechanism for this reaction. 1-Pentane...Ch. 6 - Treating 4-penten-1-ol with bromine in water forms...Ch. 6 - Prob. 6.35PCh. 6 - Prob. 6.36PCh. 6 - Reaction of -pinene with borane followed by...Ch. 6 - Write structural formulas for the major organic...Ch. 6 - Draw the structural formula of the alkene that...Ch. 6 - Consider the following reaction. (a) Draw a...Ch. 6 - Prob. 6.42PCh. 6 - Prob. 6.43PCh. 6 - Show how to convert ethylene to these compounds....Ch. 6 - Show how to convert cyclopentene into these...Ch. 6 - Prob. 6.46PCh. 6 - Describe the stereochemistry of the bromohydrin...Ch. 6 - Prob. 6.49PCh. 6 - Treating 1,3-butadiene with 1 mole of HBr gives a...Ch. 6 - In this chapter, we studied the mechanism of the...Ch. 6 - As we have seen in this chapter, carbon-carbon...Ch. 6 - Prob. 6.53PCh. 6 - Prob. 6.54P
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