Chapter 3, Problem 3.54P

Interpretation Introduction

(a)

Interpretation:

The hybridization of each non-hydrogen atom in Octocrylene is to be determined.

Concept introduction:

A hybrid atomic orbital is a cross between two or more pure AOs from the valence shell of a single atom. Typically $\text{2s and 2p}$ orbitals are involved in hybridization, resulting in ${\text{sp, sp}}^{\text{2}}{\text{or sp}}^{\text{3}}$ hybridization. The electron geometry for an atom indicates its hybridization. According to the VSEPR theory, if an atom is surrounded by two electron groups, either two bonds or one bond and one lone pair, then the geometry of such an atom is linear and thus, it must be $\text{sp}$ hybridized. If an atom is surrounded by three electron groups, either three bonds or two bond and one lone pair, then the geometry of such an atom is trigonal planar and thus, it must be ${\text{sp}}^2$ hybridized. If an atom is surrounded by four electron groups, either four bonds or three bond and one lone pair, then the geometry of such an atom is tetrahedral and thus, it must be ${\text{sp}}^3$ hybridized.

Answer to Problem 3.54P

Atoms shown in blue color are ${\text{sp}}^2$ hybridized. Atoms shown in red are $\text{sp}$ hybridized, and atoms shown in green color are ${\text{sp}}^3$ hybridized.

Explanation of Solution

The given structure for Octocrylene is:

The above molecule has two benzene rings. Each carbon atom in the benzene ring has three electron groups: two single bonds and one double bond. Thus, the geometry for all these carbon atoms must be trigonal planar and hybridization must be ${\text{sp}}^2$. Apart from these carbon atoms, the two carbon atoms forming the double bond and a carbon atom forming a double bond with oxygen atom, also have three electron groups. Thus, the hybridization for these three carbon atoms is also ${\text{sp}}^2$. The double bonded oxygen in the structure also possess two lone pairs of electrons along with one double bond. Thus, it has three electron groups and it is also ${\text{sp}}^2$ hybridized. Thus, there are a total of 15 ${\text{sp}}^2$ atoms in the above structure.

The triple bonded carbon atom has two electron groups: one single bond and one triple bond. Thus, the hybridization of this carbon atom must be $\text{sp}$. Similarly, the triply bonded nitrogen atom also has two electron groups: one triple bond and one lone pair. Thus, the hybridization of this carbon atom must be $\text{sp}$. Both the atoms with $\text{sp}$ hybridization are shown in red:

The single bonded oxygen atom and each carbon atom in the alkyl chain has four electrons groups. Thus, its geometry must be tetrahedral and hybridization must be ${\text{sp}}^3$. These atoms are shown in green color:

Conclusion

Hybridization of all non-hydrogen atoms is shown in three figures as above.

Interpretation Introduction

(b)

Interpretation:

All the atoms bonded to the acyclic $\text{C=C}$ that are required to be in the same plane are to be circled.

Concept introduction:

Free rotation can occur about single bonds but not about double bonds. According to the VSEPR theory, if an atom has three electron groups: three bonds, or two bonds and one pair of electrons then, the atom is said to be ${\text{sp}}^2$ hybridized and all the atoms that are directly bonded to this ${\text{sp}}^2$ hybridized are in one plane. Thus, when two atoms are connected by a double bond, those atoms and any atoms to which they are directly bonded prefer to lie in the same plane.

Answer to Problem 3.54P

Explanation of Solution

The given structure for Octocrylene is:

In the above structure, the $\text{C=C}$ double bond located outside the two rings is the acyclic $\text{C=C}$ double bond. When two atoms are connected by a double bond, those atoms and any atoms to which they are directly bonded prefer to lie in the same plane. All the atoms bonded to the acyclic $\text{C=C}$ that are required to be in the same plane are shown as:

Conclusion

All the atoms bonded to the acyclic $\text{C=C}$ that are required to be in the same plane are circled and are shown in the figure above.

Interpretation Introduction

(c)

Interpretation:

It is to be determined if there are two unique configurations possible about the acyclic $\text{C=C}$ double bond that are required to be in the same plane.

Concept introduction:

No free rotation takes place about the double bonds. Hence, in such compounds, molecules that differ by the exchange of two groups on one of the doubly bonded atoms are said to have different configurations. It is a cis configuration if the two non-hydrogen substituents are on present the same side of the double bond, and it is trans if they are on the opposite sides.

Answer to Problem 3.54P

It is not possible to have two unique configurations about the acyclic $\text{C=C}$ double bond that are required to be in the same plane. Because the two groups attached to one of the doubly bonded carbon atoms are identical.

Explanation of Solution

The given structure for Octocrylene is:

In the structure above, the acyclic $\text{C=C}$ double bond is circled. For the first doubly bonded carbon atom on the left, the two groups that are present on the same side of the double bond are the two phenyl rings. Since there are two same groups attached at one of the doubly bonded carbon atoms, even if these two groups are exchanged, it is not possible to have a cis or a trans configuration.

Thus, it is not possible to have two unique configurations about the acyclic $\text{C=C}$ double bond that are required to be in the same plane.

Conclusion

For a $\text{C=C}$ double bond having two identical non-hydrogen atoms on one side cannot have unique configurations.

Interpretation Introduction

(d)

Interpretation:

Out of the two indicated $\text{C}-\text{C}$ single bonds, the one which is expected to be shorter is to be determined and the reason for it is to be explained.

Concept introduction:

The bond distance decrease and bond energies increase as the hybridization of the atom goes changes ${\text{sp}}^{\text{3}}$ to ${\text{sp}}^2$ to $\text{sp}$. Thus, an atom with an $\text{sp}$ hybridization is going to have a shorter and stronger bond. However, an atom with a ${\text{sp}}^3$ hybridization is going to have a longer and weaker bond. Thus, as the percentage of s character increases, the bond strength increases and bond distance decreases.

Answer to Problem 3.54P

The $\text{C}-\text{C}$ single bond between a doubly bonded and a triply bonded carbon atom labeled as ‘a’ is expected to be shorter than the $\text{C}-\text{C}$ single bond present between two doubly bonded carbon atoms labeled as ‘b’

Explanation of Solution

The given structure for Octocrylene is:

In the structure above, the two $\text{C}-\text{C}$ single bonds are indicated by ‘a’ and ‘b’.

The two carbon atoms involved in ‘a’ type of are $\text{sp}$ and ${\text{sp}}^2$ hybridised.

The two carbon atoms involved in ‘b’ type of bond are both ${\text{sp}}^2$ hybridized. The $\text{C}-\text{C}$ single bond in ‘a’ possesses more s character than ‘b’ type of bonds. Hence, the $\text{C}-\text{C}$ single bond indicated by ‘a’ is expected to be shorter and stronger than the $\text{C}-\text{C}$ single bond indicated by ‘b’. Thus, the $\text{C}-\text{C}$ single bond between a doubly bonded and a triply bonded carbon atom labeled as ‘a’ is expected to be shorter than the $\text{C}-\text{C}$ single bond present between two doubly bonded carbon atoms.

Conclusion

The bond distance decrease and bond energies increase as the hybridization of the atom goes changes ${\text{sp}}^{\text{3}}$ to ${\text{sp}}^2$ to $\text{sp}$.