ORGANIC CHEMISTRY SAPLING ACCESS + ETEX
ORGANIC CHEMISTRY SAPLING ACCESS + ETEX
6th Edition
ISBN: 9781319306977
Author: LOUDON
Publisher: INTER MAC
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Chapter 7, Problem 7.42AP
Interpretation Introduction

(a)

Interpretation:

The structure of all products formed in the given reaction is to be shown.

Concept Introduction:

The compounds which have same molecular formula but different connectivity of atoms are known as constitutional isomers. Chiral compounds are those compounds which contain an asymmetric carbon atom. Chiral molecules are optically active molecules. Stereocentre can be an atom, bond, or any point in molecule at which interchange of two groups form a stereoisomer.

Answer:

(1) The structures of all products formed in the given reactions are shown below.

ORGANIC CHEMISTRY SAPLING ACCESS + ETEX, Chapter 7, Problem 7.42AP , additional homework tip  1

(2) The structures of all products formed in the given reactions are shown below.

ORGANIC CHEMISTRY SAPLING ACCESS + ETEX, Chapter 7, Problem 7.42AP , additional homework tip  2

(3) The structures of all products formed in the given reactions are shown below.

ORGANIC CHEMISTRY SAPLING ACCESS + ETEX, Chapter 7, Problem 7.42AP , additional homework tip  3

(4) The structures of all products formed in the given reactions are shown below.

ORGANIC CHEMISTRY SAPLING ACCESS + ETEX, Chapter 7, Problem 7.42AP , additional homework tip  4

(5) The structures of all products formed in the given reactions are shown below.

ORGANIC CHEMISTRY SAPLING ACCESS + ETEX, Chapter 7, Problem 7.42AP , additional homework tip  5

(6) The structures of all products formed in the given reactions are shown below.

ORGANIC CHEMISTRY SAPLING ACCESS + ETEX, Chapter 7, Problem 7.42AP , additional homework tip  6

Explanation:

(1) The alkene undergoes hydroboration-oxidation reaction to form alcohol compounds. The borane gets added to the double bond. Then it will undergo oxidation reaction to form the alcohol compound. The products formed in the given reaction are shown below.

ORGANIC CHEMISTRY SAPLING ACCESS + ETEX, Chapter 7, Problem 7.42AP , additional homework tip  7

Figure 1

(2) The electrophilic addition of hydrogen bromide takes place on the carbon-carbon double bond of the alkene. The addition follows markonikov’s rule, the hydrogen atom goes to the less substituted carbon atom. The products formed in the given reaction are shown below.

ORGANIC CHEMISTRY SAPLING ACCESS + ETEX, Chapter 7, Problem 7.42AP , additional homework tip  8

Figure 2

(3) The addition reaction of bromine on the alkene takes place. The addition of bromine on alkene is anti addition. The cyclic bromonium ion intermediate is formed in the reaction. The product formed in the given reaction is shown below.

ORGANIC CHEMISTRY SAPLING ACCESS + ETEX, Chapter 7, Problem 7.42AP , additional homework tip  9

Figure 3

(4) The addition reaction of bromine on the alkene takes place. The addition of bromine on alkene is anti addition. The cyclic bromonium ion intermediate is formed in the reaction. The product formed in the given reaction is shown below.

ORGANIC CHEMISTRY SAPLING ACCESS + ETEX, Chapter 7, Problem 7.42AP , additional homework tip  10

Figure 4

(5) In the reduction reaction of alkene in presence of palladium metal the hydrogen gets adsorbed on the metal surface. Alkene reduction to form alkane takes place forming syn addition product. The products formed in the given reaction are shown below.

ORGANIC CHEMISTRY SAPLING ACCESS + ETEX, Chapter 7, Problem 7.42AP , additional homework tip  11

Figure 5

(6) In the reduction reaction of alkene in presence of palladium metal, D2 gets adsorbed on the metal surface. Alkene reduction to alkane takes place forming syn addition product. The products formed in the given reaction are shown below.

ORGANIC CHEMISTRY SAPLING ACCESS + ETEX, Chapter 7, Problem 7.42AP , additional homework tip  12

Figure 6

Conclusion:

The products formed in the given reactions are shown in Figure 1, Figure 2, Figure 3, Figure 4, Figure 5 and Figure 6.

Expert Solution
Check Mark

Answer to Problem 7.42AP

(1) The structures of all products formed in the given reactions are shown below.

ORGANIC CHEMISTRY SAPLING ACCESS + ETEX, Chapter 7, Problem 7.42AP , additional homework tip  13

(2) The structures of all products formed in the given reactions are shown below.

ORGANIC CHEMISTRY SAPLING ACCESS + ETEX, Chapter 7, Problem 7.42AP , additional homework tip  14

(3) The structures of all products formed in the given reactions are shown below.

ORGANIC CHEMISTRY SAPLING ACCESS + ETEX, Chapter 7, Problem 7.42AP , additional homework tip  15

(4) The structures of all products formed in the given reactions are shown below.

ORGANIC CHEMISTRY SAPLING ACCESS + ETEX, Chapter 7, Problem 7.42AP , additional homework tip  16

(5) The structures of all products formed in the given reactions are shown below.

ORGANIC CHEMISTRY SAPLING ACCESS + ETEX, Chapter 7, Problem 7.42AP , additional homework tip  17

(6) The structures of all products formed in the given reactions are shown below.

ORGANIC CHEMISTRY SAPLING ACCESS + ETEX, Chapter 7, Problem 7.42AP , additional homework tip  18

Explanation of Solution

(1) The alkene undergoes hydroboration-oxidation reaction to form alcohol compounds. The borane gets added to the double bond. Then it will undergo oxidation reaction to form the alcohol compound. The products formed in the given reaction are shown below.

ORGANIC CHEMISTRY SAPLING ACCESS + ETEX, Chapter 7, Problem 7.42AP , additional homework tip  19

Figure 1

(2) The electrophilic addition of hydrogen bromide takes place on the carbon-carbon double bond of the alkene. The addition follows markonikov’s rule, the hydrogen atom goes to the less substituted carbon atom. The products formed in the given reaction are shown below.

ORGANIC CHEMISTRY SAPLING ACCESS + ETEX, Chapter 7, Problem 7.42AP , additional homework tip  20

Figure 2

(3) The addition reaction of bromine on the alkene takes place. The addition of bromine on alkene is anti addition. The cyclic bromonium ion intermediate is formed in the reaction. The product formed in the given reaction is shown below.

ORGANIC CHEMISTRY SAPLING ACCESS + ETEX, Chapter 7, Problem 7.42AP , additional homework tip  21

Figure 3

(4) The addition reaction of bromine on the alkene takes place. The addition of bromine on alkene is anti addition. The cyclic bromonium ion intermediate is formed in the reaction. The product formed in the given reaction is shown below.

ORGANIC CHEMISTRY SAPLING ACCESS + ETEX, Chapter 7, Problem 7.42AP , additional homework tip  22

Figure 4

(5) In the reduction reaction of alkene in presence of palladium metal the hydrogen gets adsorbed on the metal surface. Alkene reduction to form alkane takes place forming syn addition product. The products formed in the given reaction are shown below.

ORGANIC CHEMISTRY SAPLING ACCESS + ETEX, Chapter 7, Problem 7.42AP , additional homework tip  23

Figure 5

(6) In the reduction reaction of alkene in presence of palladium metal, D2 gets adsorbed on the metal surface. Alkene reduction to alkane takes place forming syn addition product. The products formed in the given reaction are shown below.

ORGANIC CHEMISTRY SAPLING ACCESS + ETEX, Chapter 7, Problem 7.42AP , additional homework tip  24

Figure 6

Conclusion

The products formed in the given reactions are shown in Figure 1, Figure 2, Figure 3, Figure 4, Figure 5 and Figure 6.

Interpretation Introduction

(b)

Interpretation:

The stereochemical relationship between products formed is to be stated.

Concept Introduction:

Stereoisomers which are non-superimposable and not mirror image are known as diastereomer. The compound must contain two or more than two stereocentre. Diastereomer are non identical stereoisomers. The pair of stereoisomer which are mirror image of each other are known as enantiomers. Enantiomers are non-congruent mirror images. If the molecules are placed on top of each other they will not give same molecule.

Expert Solution
Check Mark

Answer to Problem 7.42AP

(1) The products formed are diastereomers.

(2) The products formed are enantiomers.

(3) The products formed are enantiomers.

(4) The two pairs of products formed are diastereomers.

(5) Only one product is formed.

(6) Only one product is formed.

Explanation of Solution

(1) The products formed are non-congruent and not mirror image of each other. Hence they are diastereomers.

(2) The products formed are non-congruent mirror image of each other. Therefore, they are enantiomers.

(3) The products formed are non-congruent mirror image of each other. Therefore, they are enantiomers.

(4) The products formed are non-congruent and not mirror image of each other. Hence they are diastereomers.

(5) Only one product is formed.

(6) Only one product is formed.

Conclusion

The stereochemical relationship between products formed in reaction (1) and (4) are diastereomers and reaction (2) and (3) are enantiomers.

Interpretation Introduction

(c)

Interpretation:

The products are formed in identical or different amount is to be stated.

Concept Introduction:

The electrophilic addition reaction on the alkene can takes place from any side of alkene. The electrophile can attack the carbon double bond from above or below the plane. The probability of attack from both sides is equal.

Expert Solution
Check Mark

Answer to Problem 7.42AP

(1) The products are formed in equal amount.

(2) The products are formed in equal amount.

(3) The products are formed in equal amount.

(4) The products are formed in equal amount.

(5) Only one product is formed.

(6) Only one product is formed.

Explanation of Solution

(1) The alkene is a planar molecule. The reaction on the carbon-carbon double bond of alkene can takes place from above or below the plane. Therefore, products are formed in equal amount.

(2) The alkene is a planar molecule. The reaction on the carbon-carbon double bond of alkene can takes place from above or below the plane. Therefore, products are formed in equal amount.

(3) The alkene is a planar molecule. The reaction on the carbon-carbon double bond of alkene can takes place from above or below the plane. Therefore, products are formed in equal amount.

(4) The alkene is a planar molecule. The reaction on the carbon-carbon double bond of alkene can takes place from above or below the plane. Therefore, products are formed in equal amount.

(5) Only one product is formed.

(6) Only one product is formed.

Conclusion

The products are formed in equal amount in reaction (1), (2), (3), (4) and only one product is formed in reaction (5) and (6).

Interpretation Introduction

(d)

Interpretation:

The products which have different physical properties is to be stated.

Concept Introduction:

Stereoisomers which are non-superimposable and not mirror image are known as diastereomer. The compound must contain two or more than two stereocentre. Diastereomer are non identical stereoisomers. The pair of stereoisomer which are mirror image of each other are known as enantiomers. Enantiomers are non-congruent mirror images. Diastereomers shows different physical properties and enantiomers show same physical properties.

Expert Solution
Check Mark

Answer to Problem 7.42AP

(1) The products formed in the given reaction will have different physical properties.

(2) The products formed in the given reaction will have identical physical properties.

(3) The products formed in the given reaction will have identical physical properties.

(4) The products formed in the given reaction will have different physical properties.

(5) Only one product is formed.

(6) Only one product is formed.

Explanation of Solution

(1) The products formed in the given reaction are diastereomer. Diastereomers show different physical properties. Therefore, the products formed in the given reaction will have different physical properties.

(2) The products formed in the given reaction are enantiomers. Enantiomers show identical physical properties. Therefore, the products formed in the given reaction will have identical physical properties.

(3) The products formed in the given reaction are enantiomers. Enantiomers show identical physical properties. Therefore, the products formed in the given reaction will have identical physical properties.

(4) The products formed in the given reaction are diastereomer. Diastereomers show different physical properties. Therefore, the products formed in the given reaction will have different physical properties.

(5) Only one product is formed.

(6) Only one product is formed.

Conclusion

The products formed in the reaction (1) and reaction (4) will have different physical properties. The products formed in the reaction (2) and reaction (3) will have identical physical properties. In reaction (5) and (6) only one product is formed.

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Give the MAJOR product(s) for the following reactions. For example, if BOTHelimination and substitution would occur, write both products. Or, if two carboncontaining groups would be formed as a result of a molecule splitting, write bothorganic molecules. Also, as indicated, describe the stereochemistry of theproducts. If no stereochemistry exists in the final product, but if astereochemistry answer box is included for that problem, then write NOSTEREOCHEMISTRY, NONE or N/A in the corresponding box. At the end, youwill be asked to label which problems yielded products that were examples ofhemiacetals/ acetals, hemiketals/ketals, and Schiff bases. There is onehemiacetal/acetal or hemi-ketal/ketal, and one Schiff base.
Give the MAJOR product(s) for the following reactions. For example, if BOTHelimination and substitution would occur, write both products. Or, if two carboncontaining groups would be formed as a result of a molecule splitting, write bothorganic molecules. Also, as indicated, describe the stereochemistry of theproducts. If no stereochemistry exists in the final product, but if astereochemistry answer box is included for that problem, then write NOSTEREOCHEMISTRY, NONE or N/A in the corresponding box. At the end, youwill be asked to label which problems yielded products that were examples ofhemiacetals/ acetals, hemiketals/ketals, and Schiff bases. There is onehemiacetal/acetal or hemi-ketal/ketal, and one Schiff base. Do them in structural format given and show you work step by step
Give the MAJOR product(s) for the following reactions. For example, if BOTHelimination and substitution would occur, write both products. Or, if two carboncontaining groups would be formed as a result of a molecule splitting, write bothorganic molecules. Also, as indicated, describe the stereochemistry of theproducts. If no stereochemistry exists in the final product, but if astereochemistry answer box is included for that problem, then write NOSTEREOCHEMISTRY, NONE or N/A in the corresponding box. At the end, youwill be asked to label which problems yielded products that were examples ofhemiacetals/ acetals, hemiketals/ketals, and Schiff bases. There is onehemiacetal/acetal or hemi-ketal/ketal, and one Schiff base. Do them in the structural format they are in

Chapter 7 Solutions

ORGANIC CHEMISTRY SAPLING ACCESS + ETEX

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