Chapter 3, Problem 3.38P

Interpretation Introduction

(a)

Interpretation:

The hybridization of each nonhydrogen atom in norethynodrel is to be determined.

Concept introduction:

Answer to Problem 3.38P

Atom	Hybridization
${\text{O}}_{\text{1}}$, ${\text{O}}_{\text{12}}$	${\text{sp}}^{\text{2}}$
${\text{C}}_{\text{2}}$, ${\text{C}}_{\text{4}}$, ${\text{C}}_{\text{20}}$	${\text{sp}}^{\text{2}}$
${\text{C}}_{\text{3}}$, ${\text{C}}_{\text{5}}$, ${\text{C}}_{\text{6}}$, ${\text{C}}_{\text{7}}$, ${\text{C}}_{\text{8}}$, ${\text{C}}_{\text{9}}$, ${\text{C}}_{\text{10}}$, ${\text{C}}_{\text{11}}$, ${\text{C}}_{\text{15}}$, ${\text{C}}_{\text{16}}$, ${\text{C}}_{\text{17}}$, ${\text{C}}_{\text{18}}$, ${\text{C}}_{\text{19}}$, ${\text{C}}_{\text{21}}$, ${\text{C}}_{\text{22}}$	${\text{sp}}^{\text{3}}$
${\text{C}}_{\text{13}}$, ${\text{C}}_{\text{14}}$	sp

Explanation of Solution

The structure of Norethynodrel is as below:

The nonhydrogen atoms are highlighted with red color and numbered.

According to VSEPR theory, ${\text{O}}_{\text{1}}$ and ${\text{O}}_{\text{12}}$ atom is surrounded by three electron groups — two lone pairs and a double bond. So, the electron geometry of ${\text{O}}_{\text{1}}$ and ${\text{O}}_{\text{12}}$ is trigonal planar. Therefore, the atom ${\text{O}}_{\text{1}}$ and ${\text{O}}_{\text{12}}$ is ${\text{sp}}^{\text{2}}$ hybridized.

According to VSEPR theory, ${\text{C}}_{\text{2}}$, ${\text{C}}_{\text{4}}$, and ${\text{C}}_{\text{20}}$ atom is surrounded by three electron groups — two single bonds and a double bond. So, the electron geometry of ${\text{C}}_{\text{2}}$, ${\text{C}}_{\text{4}}$, and ${\text{C}}_{\text{20}}$ is trigonal planar. Therefore, the atom ${\text{C}}_{\text{2}}$, ${\text{C}}_{\text{4}}$, and ${\text{C}}_{\text{20}}$ is ${\text{sp}}^{\text{2}}$ hybridized.

According to VSEPR theory, the ${\text{C}}_{\text{3}}$, ${\text{C}}_{\text{5}}$, ${\text{C}}_{\text{6}}$, ${\text{C}}_{\text{7}}$, ${\text{C}}_{\text{8}}$, ${\text{C}}_{\text{9}}$, ${\text{C}}_{\text{10}}$, ${\text{C}}_{\text{11}}$, ${\text{C}}_{\text{15}}$, ${\text{C}}_{\text{16}}$, ${\text{C}}_{\text{17}}$, ${\text{C}}_{\text{18}}$, ${\text{C}}_{\text{19}}$, ${\text{C}}_{\text{21}}$ and ${\text{C}}_{\text{22}}$ atom is surrounded by four electron groups — four single bonds. So, the electron geometry of ${\text{C}}_{\text{3}}$, ${\text{C}}_{\text{5}}$, ${\text{C}}_{\text{6}}$, ${\text{C}}_{\text{7}}$, ${\text{C}}_{\text{8}}$, ${\text{C}}_{\text{9}}$, ${\text{C}}_{\text{10}}$, ${\text{C}}_{\text{11}}$, ${\text{C}}_{\text{15}}$, ${\text{C}}_{\text{16}}$, ${\text{C}}_{\text{17}}$, ${\text{C}}_{\text{18}}$, ${\text{C}}_{\text{19}}$, ${\text{C}}_{\text{21}}$ and ${\text{C}}_{\text{22}}$ is tetrahedral. Therefore, the atom ${\text{C}}_{\text{3}}$, ${\text{C}}_{\text{5}}$, ${\text{C}}_{\text{6}}$, ${\text{C}}_{\text{7}}$, ${\text{C}}_{\text{8}}$, ${\text{C}}_{\text{9}}$, ${\text{C}}_{\text{10}}$, ${\text{C}}_{\text{11}}$, ${\text{C}}_{\text{15}}$, ${\text{C}}_{\text{16}}$, ${\text{C}}_{\text{17}}$, ${\text{C}}_{\text{18}}$, ${\text{C}}_{\text{19}}$, ${\text{C}}_{\text{21}}$ and ${\text{C}}_{\text{22}}$ is ${\text{sp}}^{\text{3}}$ hybridized.

According to VSEPR theory, ${\text{C}}_{\text{13}}$ and ${\text{C}}_{\text{14}}$ atom is surrounded by two electron groups — a single bond and a triple bond. So, the electron geometry of ${\text{C}}_{\text{13}}$ and ${\text{C}}_{\text{14}}$ is linear. Therefore, the atom ${\text{C}}_{\text{13}}$ and ${\text{C}}_{\text{14}}$ is sp hybridized.

Conclusion

According to VSEPR theory, geometry of an atom has been determined, and from the geometry of the atom, its hybridization has been determined.

Interpretation Introduction

(b)

Interpretation:

The total number of $\sigma$ bonds and $\pi$ bonds in norethynodrel is to be determined.

Concept introduction:

Answer to Problem 3.38P

There are total fifty-one $\sigma$ bonds and four $\pi$ bonds in one norethynodrel molecule.

Explanation of Solution

The structure of Norethynodrel is as below:

There is one C=C triple bond in norethynodrel molecule. Each triple bond is composed of two $\pi$ bonds and one $\sigma$ bond, for a total of two $\pi$ bonds and one $\sigma$ bond. There is one C=C double bond in norethynodrel molecule. Each double bond is composed of one $\pi$ bond and one $\sigma$ bond, for a total of one $\pi$ bond and one $\sigma$ bond. There is one C=O double bond in norethynodrel molecule. Each double bond is composed of one $\pi$ bond and one $\sigma$ bond, for a total of one $\pi$ bond and one $\sigma$ bond. There are twenty-one C-C single bonds, twenty-five C-H single bonds, one C-O single bond, and one O-H single bond. Each C-C, C-H, C-O, and O-H bond is a $\sigma$ bond, giving a total forty-eight $\sigma$ bonds. Therefore, total fifty-one $\sigma$ bonds and four $\pi$ bonds are in one norethynodrel molecule.

Conclusion

From the $\sigma$ bonds and $\pi$ bonds, the total number $\sigma$ bonds and $\pi$ bonds has been determined.