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Chapter 28, Problem 28.39P
Interpretation Introduction

(a)

Interpretation: The given compound is to be classified as identical to A or its enantiomer.

Concept introduction: The stereochemistry of the compound is determined by prioritizing the groups attached to its stereogenic center. The groups are prioritized on the basis of atomic number of their atoms. The group that contain atom with higher atomic number is given higher priority. Complete the circle in decreasing order of priority from 123. The configuration of the compound is R and S when circle rotates in clockwise and anticlockwise direction.

Expert Solution
Check Mark

Answer to Problem 28.39P

The given compound is classified as identical to A because the configuration of A is same to the configuration of the given compound i.e. R.

Explanation of Solution

The structure of compound A is,

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 28, Problem 28.39P , additional homework tip  1

Figure 1

In the compound A, CHO, OH, CH2CH3 and H substituents are attached to carbon atom. The atomic number of C, O and H is 6, 8 and 1 respectively. Thus, first priority is given to OH substituent. There are two substituents i.e. CHO, CH2CH3 connected to the stereogenic centre through the carbon atom. Now, the priority is given on the basis of atomic number of second atom which is attached to the carbon atom. The carbon of CHO is connected to the oxygen atom and the atomic number of oxygen is 8 while the another carbon atom of CH2CH3 is connected to the carbon atom and the atomic number of carbon is 6. Thus, the second priority is given to CHO substituent then third priority to CH2CH3 substituent. Complete the circle in decreasing order of priority from 123 is shown below.

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 28, Problem 28.39P , additional homework tip  2

Figure 2

The circle rotates in the anticlockwise direction and its configuration will be S but the lowest priority group should be away from the viewer i.e. lies on vertical lineas shown below.

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 28, Problem 28.39P , additional homework tip  3

Figure 3

Thus, the configuration will reverse. Therefore, the compound A is labeled as R.

The given compound is,

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 28, Problem 28.39P , additional homework tip  4

Figure 4

In the given compound, CHO, OH, CH2CH3 and H substituents are attached to carbon atom. The atomic number of C, O and H is 6, 8 and 1 respectively. Thus, first priority is given to OH substituent. There are two substituents i.e. CHO, CH2CH3 are connected to the stereogenic centre through carbon atom. Now, the priority is given on the basis of atomic number of second atom which is attached to the carbon atom. The carbon of CHO is to connected the oxygen atom and the atomic number of oxygen is 8 while the another carbon atom of CH2CH3 is connected to the carbon atom and the atomic number of carbon is 6. Thus, the second priority is given to CHO substituent then third priority to CH2CH3 substituent. Complete the circle in decreasing order of priority from 123 is shown below.

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 28, Problem 28.39P , additional homework tip  5

Figure 5

The circle rotates in the clockwise direction. Thus, the given compound is labeled as R.

Hence, the given compound is classified as identical to A because the configuration of A is same to the configuration of the given compound i.e. R.

Conclusion

The given compound is classified as identical to A because the configuration of A is same to the configuration of the given compound i.e. R.

Interpretation Introduction

(b)

Interpretation: The given compound is to be classified as identical to A or its enantiomer.

Concept introduction: The stereochemistry of the compound is determined by prioritizing the groups attached to its stereogenic center. The groups are prioritized on the basis of atomic number of their atoms. The group that contain atom with higher atomic number is given higher priority. Complete the circle in decreasing order of priority from 123. The configuration of the compound is R and S when circle rotates in clockwise and anticlockwise direction.

Expert Solution
Check Mark

Answer to Problem 28.39P

The given compound is classified as identical to A because the configuration of A is same to the configuration of the given compound i.e. R.

Explanation of Solution

The compound A is labeled as R.

The given compound is,

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 28, Problem 28.39P , additional homework tip  6

Figure 6

In the given compound, CHO, OH, CH2CH3 and H substituents are attached to carbon atom. The atomic number of C, O and H is 6, 8 and 1 respectively. Thus, first priority is given to OH substituent. There are two substituents i.e. CHO, CH2CH3 are connected to the stereogenic centre through carbon atom. Now, the priority is given on the basis of atomic number of second atom which is attached to the carbon atom. The carbon of CHO is connected to the oxygen atom and the atomic number of oxygen is 8 while the another carbon atom of CH2CH3 is connected to the carbon atom and the atomic number of carbon is 6. Thus, the second priority is given to CHO substituent then third priority to CH2CH3 substituent. Complete the circle in decreasing order of priority from 123 is shown below.

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 28, Problem 28.39P , additional homework tip  7

Figure 7

The circle rotates in the anticlockwise direction hence its configuration will be S but the lowest priority group should be away from the viewer as shown below.

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 28, Problem 28.39P , additional homework tip  8

Figure 8

Thus, the configuration will reverse. Therefore, the compound is labeled as R.

Hence, the given compound is classified as identical to A because the configuration of A is same to the configuration of the given compound i.e. R.

Conclusion

The given compound is classified as identical to A because the configuration of A is same to the configuration of the given compound i.e. R.

Interpretation Introduction

(c)

Interpretation: The given compound is to be classified as identical to A or its enantiomer.

Concept introduction: The stereochemistry of the compound is determined by prioritizing the groups attached to its stereogenic center. The groups are prioritized on the basis of atomic number of their atoms. The group that contain atom with higher atomic number is given higher priority. Complete the circle in decreasing order of priority from 123. The configuration of the compound is R and S when circle rotates in clockwise and anticlockwise direction.

Expert Solution
Check Mark

Answer to Problem 28.39P

The given compound is classified as enantiomer to A because the configuration of A is different from the configuration of the given compound.

Explanation of Solution

The compound A is labeled as R.

The given compound is,

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 28, Problem 28.39P , additional homework tip  9

Figure 9

In the given compound, CHO, OH, CH2CH3 and H substituents are attached to carbon atom. The atomic number of C, O and H is 6, 8 and 1 respectively. Thus, first priority is given to OH substituent. There are two substituents i.e. CHO, CH2CH3 are connected to the stereogenic centre through carbon atom. Now, the priority is given on the basis of atomic number of second atom which is attached to the carbon atom. The carbon of CHO is connected to the oxygen atom and the atomic number of oxygen is 8 while the another carbon atom of CH2CH3 is connected to the carbon atom and the atomic number of carbon is 6. Thus, the second priority is given to CHO substituent then third priority to CH2CH3 substituent. Complete the circle in decreasing order of priority from 123 is shown below.

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 28, Problem 28.39P , additional homework tip  10

Figure 10

The circle rotates in the clockwise direction hence its configuration will be R but the lowest priority group should be away from the viewer as shown below.

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 28, Problem 28.39P , additional homework tip  11

Figure 11

Thus, the configuration will reverse. Therefore, the compound is labeled as S.

The given compound and A are enantiomer as shown below.

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 28, Problem 28.39P , additional homework tip  12

Figure 12

Hence, the given compound is classified as enantiomer to A because the configuration of A is different from the configuration of the given compound.

Conclusion

The given compound is classified as enantiomer to A because the configuration of A is different from the configuration of the given compound.

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Chapter 28 Solutions

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card

Ch. 28 - Prob. 28.11PCh. 28 - Prob. 28.12PCh. 28 - Prob. 28.13PCh. 28 - Prob. 28.14PCh. 28 - Prob. 28.15PCh. 28 - Prob. 28.16PCh. 28 - Prob. 28.17PCh. 28 - Draw a stepwise mechanism for the following...Ch. 28 - Prob. 28.19PCh. 28 - Prob. 28.20PCh. 28 - Prob. 28.21PCh. 28 - Prob. 28.22PCh. 28 - Draw the products formed when D-arabinose is...Ch. 28 - Prob. 28.24PCh. 28 - Prob. 28.25PCh. 28 - Prob. 28.26PCh. 28 - Prob. 28.27PCh. 28 - Prob. 28.28PCh. 28 - Prob. 28.29PCh. 28 - Prob. 28.30PCh. 28 - Prob. 28.31PCh. 28 - Prob. 28.32PCh. 28 - Prob. 28.33PCh. 28 - Prob. 28.34PCh. 28 - Problem-28.35 Draw the structures of the...Ch. 28 - Prob. 28.36PCh. 28 - 28.37 Convert each ball-and-stick model to a...Ch. 28 - Prob. 28.38PCh. 28 - Prob. 28.39PCh. 28 - 28.40 Convert each compound to a Fischer...Ch. 28 - Prob. 28.41PCh. 28 - Prob. 28.42PCh. 28 - 28.43 Draw a Haworth projection for each compound...Ch. 28 - Prob. 28.44PCh. 28 - 28.45 Draw both pyranose anomers of each...Ch. 28 - Prob. 28.46PCh. 28 - Prob. 28.47PCh. 28 - Prob. 28.48PCh. 28 - Prob. 28.49PCh. 28 - 28.50 Draw the products formed when D-altrose is...Ch. 28 - Prob. 28.51PCh. 28 - Prob. 28.52PCh. 28 - Prob. 28.53PCh. 28 - Prob. 28.54PCh. 28 - Prob. 28.55PCh. 28 - Prob. 28.56PCh. 28 - Prob. 28.57PCh. 28 - 28.58 Draw a stepwise mechanism for the following...Ch. 28 - Prob. 28.59PCh. 28 - Prob. 28.60PCh. 28 - Prob. 28.61PCh. 28 - Prob. 28.62PCh. 28 - Prob. 28.63PCh. 28 - Prob. 28.64PCh. 28 - Prob. 28.65PCh. 28 - Prob. 28.66PCh. 28 - Prob. 28.67PCh. 28 - Prob. 28.68PCh. 28 - Prob. 28.69PCh. 28 - Prob. 28.70PCh. 28 - 28.71 Draw a stepwise mechanism for the following...
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