Chapter 1, Problem 1.38P

Zingerone gives ginger its pungent taste.

a. What is the molecular formula of zingerone?

b. How many lone pairs are present?

c. Draw a skeletal structure.

d. How many $s{p}^2$ hybridized carbon atoms are present?

e. What orbitals are used to form each indicated bond $\left(\left[1\right]-\left[4\right]\right)$?

Interpretation Introduction

(a)

Interpretation: The molecular formula of zingerone is to be stated.

Concept introduction: In ball-and-stick model, each colored ball represents a specific atom and each stick represents a bond. In this model, each black ball represents $\text{C}$ atoms, each gray ball represents $\text{H}$ atoms, and each red ball represents $\text{O}$ atoms.

Answer to Problem 1.38P

The molecular formula of zingerone is ${\text{C}}_{\text{11}}{\text{H}}_{\text{14}}{\text{O}}_{\text{3}}$.

Explanation of Solution

The given ball-and-stick model of zingerone is,

Figure 1

In ball-and-stick model, each colored ball represents a specific atom and each stick represents a bond. In this model, each black ball represents $\text{C}$ atoms, each gray ball represents $\text{H}$ atoms, and each red ball represents $\text{O}$ atoms.

In the above model,

• There are three red balls. Thus, there are three $\text{O}$ atoms.

• There are eleven black balls. Thus, there are eleven $\text{C}$ atoms.

• There are fourteen grey balls. Thus, there are fourteen $\text{H}$ atoms.

Hence, the molecular formula of zingerone is ${\text{C}}_{\text{11}}{\text{H}}_{\text{14}}{\text{O}}_{\text{3}}$.

Conclusion

The molecular formula of zingerone is ${\text{C}}_{\text{11}}{\text{H}}_{\text{14}}{\text{O}}_{\text{3}}$.

Interpretation Introduction

(b)

Interpretation: The number of lone pairs in zingerone is to be stated.

Concept introduction: In a compound or molecule, the lone pairs represent number of unshared electrons on atom. An atom may or may not have unshared electrons. For example, carbon and hydrogen atoms have no lone pair but each oxygen atom has two lone pairs.

Answer to Problem 1.38P

There are total $6$ lone pairs in zingerone.

Explanation of Solution

The molecular formula of citric acid is ${\text{C}}_{\text{11}}{\text{H}}_{\text{14}}{\text{O}}_{\text{3}}$. Carbon and hydrogen atoms have no lone pair in citric acid, but each oxygen atom has two lone pairs. There are three oxygen atoms in zingerone. Thus, there are total $6$ lone pairs $\left(3\times 2=6\right)$ in zingerone.

Conclusion

There are total $6$ lone pairs in zingerone.

Interpretation Introduction

(c)

Interpretation: A skeletal structure of zingerone is to be drawn.

Concept introduction: A ball-and-stick model is converted into skeletal structure by replacing black ball with $\text{C}$, gray ball with $\text{H}$, and red ball with $\text{O}$. Omit the $\text{H}$ atom on carbon, but not in the case of heteroatom.

Answer to Problem 1.38P

A skeletal structure of zingerone is,

Explanation of Solution

In ball-and-stick model each colored ball represents a specific atom and each stick represents a bond. A ball-and-stick model is converted into skeletal structure by replacing black ball with $\text{C}$, gray ball with $\text{H}$, red ball with $\text{O}$, and blue ball with $\text{N}$. Omit the $\text{H}$ atom on carbon, but not in the case of heteroatom.

A skeletal structure of zingerone is shown in Figure 2.

Figure 2

Conclusion

In ball-and-stick model each colored ball represents a specific atom and each stick represents a bond.

Interpretation Introduction

(d)

Interpretation: The number of $s{p}^2$ hybridized carbon in zingerone is to be stated.

Concept introduction: According to the rule of hybridization, an atom that is surrounded with two groups is $sp$ hybridized, an atom that is surrounded with three groups is $s{p}^2$ hybridized, and an atom that is surrounded with four groups is $s{p}^3$ hybridized.

Answer to Problem 1.38P

There are seven $s{p}^2$ hybridized carbon atoms in zingerone.

Explanation of Solution

The Lewis structure of zingerone is,

Figure 3

According to the rules of hybridization, an atom that is surrounded with two groups is $sp$ hybridized, an atom that is surrounded with three groups is $s{p}^2$ hybridized, and an atom that is surrounded with four groups is $s{p}^3$ hybridized.

The $s{p}^2$ hybridized carbon atoms (surrounded by three groups) are shown in Figure 4.

Figure 4

Thus, there are seven $s{p}^2$ hybridized carbon atoms in zingerone.

Conclusion

There are seven $s{p}^2$ hybridized carbon atoms in zingerone.

Interpretation Introduction

(e)

Interpretation: The orbitals that are used to form each indicated bond is to be stated.

Concept introduction: According to the rule of hybridization, an atom that is surrounded with two groups is $sp$ hybridized, an atom that is surrounded with three groups is $s{p}^2$ hybridized, and an atom that is surrounded with four groups is $s{p}^3$ hybridized.

Answer to Problem 1.38P

Bond $\left[1\right]$ is formed by ${\text{C}}_{s{p}^2}-{\text{C}}_{s{p}^3}$ hybridized orbitals. Bond $\left[2\right]$ is formed by ${\text{C}}_{s{p}^3}-{\text{C}}_{s{p}^3}$ hybridized orbitals. Bond $\left[3\right]$ is formed by ${\text{C}}_{s{p}^3}-{\text{H}}_{1s}$ orbitals. Bond $\left[4\right]$ is formed by ${\text{C}}_{s{p}^2}-{\text{H}}_{1s}$ hybridized orbitals.

Explanation of Solution

Bond [1] represents bonding between the carbon atom of benzene $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{6}}\right)$ and the carbon atom of $-{\text{CH}}_{\text{2}}-$ group. The carbon atom of benzene is $s{p}^2$ hybridized and the carbon atom of $-{\text{CH}}_{\text{2}}-$ group is $s{p}^3$ hybridized.

Thus, $\left[1\right]$ is formed by ${\text{C}}_{s{p}^2}-{\text{C}}_{s{p}^3}$ hybridized orbitals.

Bond $\left[2\right]$ represents $\text{C}-\text{C}$ bond in which both carbon atoms are $s{p}^3$ hybridized. Thus, $\left[2\right]$ is formed by ${\text{C}}_{s{p}^3}-{\text{C}}_{s{p}^3}$ hybridized orbitals.

Bond $\left[3\right]$ represents, $\text{C}-\text{H}$ bond. This bond is formed through $s{p}^3$ hybridized orbital of carbon and $1s$ orbital of hydrogen.

Thus, $\left[3\right]$ is formed by ${\text{C}}_{s{p}^3}-{\text{H}}_{1s}$ orbitals.

Bond $\left[4\right]$ represents bonding between the carbon atom of benzene $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{6}}\right)$, and hydrogen. This bond is formed through $s{p}^2$ hybridized orbital of carbon and $1s$ orbital of hydrogen.

Thus, bond $\left[4\right]$ is formed by ${\text{C}}_{s{p}^2}-{\text{H}}_{1s}$ orbitals.

Conclusion

The number of surrounded group around any atom predicts the hybridization of that atom, which is further helpful in predicting the orbitals involved in the bond formation.