ORGANIC CHEMISTRY (LL) W/CONNECT
ORGANIC CHEMISTRY (LL) W/CONNECT
5th Edition
ISBN: 9781260692570
Author: SMITH
Publisher: MCG
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Textbook Question
Chapter 1, Problem 1.38P

Zingerone gives ginger its pungent taste.

Chapter 1, Problem 1.38P, Zingerone gives ginger its pungent taste. a.What is the molecular formula of zingerone? b.How many

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 ( [ 1 ] [ 4 ] ) ?

Expert Solution
Check Mark
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 C atoms, each gray ball represents H atoms, and each red ball represents O atoms.

Answer to Problem 1.38P

The molecular formula of zingerone is C11H14O3.

Explanation of Solution

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

ORGANIC CHEMISTRY (LL) W/CONNECT, Chapter 1, Problem 1.38P , additional homework tip  1

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 C atoms, each gray ball represents H atoms, and each red ball represents O atoms.

In the above model,

• There are three red balls. Thus, there are three O atoms.

• There are eleven black balls. Thus, there are eleven C atoms.

• There are fourteen grey balls. Thus, there are fourteen H atoms.

Hence, the molecular formula of zingerone is C11H14O3.

Conclusion

The molecular formula of zingerone is C11H14O3.

Expert Solution
Check Mark
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 C11H14O3. 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 (3×2=6) in zingerone.

Conclusion

There are total 6 lone pairs in zingerone.

Expert Solution
Check Mark
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 C, gray ball with H, and red ball with O. Omit the H atom on carbon, but not in the case of heteroatom.

Answer to Problem 1.38P

A skeletal structure of zingerone is,

ORGANIC CHEMISTRY (LL) W/CONNECT, Chapter 1, Problem 1.38P , additional homework tip  2

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 C, gray ball with H, red ball with O, and blue ball with N. Omit the H atom on carbon, but not in the case of heteroatom.

A skeletal structure of zingerone is shown in Figure 2.

ORGANIC CHEMISTRY (LL) W/CONNECT, Chapter 1, Problem 1.38P , additional homework tip  3

Figure 2

Conclusion

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

Expert Solution
Check Mark
Interpretation Introduction

(d)

Interpretation: The number of sp2 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 sp2 hybridized, and an atom that is surrounded with four groups is sp3 hybridized.

Answer to Problem 1.38P

There are seven sp2 hybridized carbon atoms in zingerone.

Explanation of Solution

The Lewis structure of zingerone is,

ORGANIC CHEMISTRY (LL) W/CONNECT, Chapter 1, Problem 1.38P , additional homework tip  4

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 sp2 hybridized, and an atom that is surrounded with four groups is sp3 hybridized.

The sp2 hybridized carbon atoms (surrounded by three groups) are shown in Figure 4.

ORGANIC CHEMISTRY (LL) W/CONNECT, Chapter 1, Problem 1.38P , additional homework tip  5

Figure 4

Thus, there are seven sp2 hybridized carbon atoms in zingerone.

Conclusion

There are seven sp2 hybridized carbon atoms in zingerone.

Expert Solution
Check Mark
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 sp2 hybridized, and an atom that is surrounded with four groups is sp3 hybridized.

Answer to Problem 1.38P

Bond [1] is formed by Csp2Csp3 hybridized orbitals. Bond [2] is formed by Csp3Csp3 hybridized orbitals. Bond [3] is formed by Csp3H1s orbitals. Bond [4] is formed by Csp2H1s hybridized orbitals.

Explanation of Solution

Bond [1] represents bonding between the carbon atom of benzene (C6H6) and the carbon atom of CH2 group. The carbon atom of benzene is sp2 hybridized and the carbon atom of CH2 group is sp3 hybridized.

Thus, [1] is formed by Csp2Csp3 hybridized orbitals.

Bond [2] represents CC bond in which both carbon atoms are sp3 hybridized. Thus, [2] is formed by Csp3Csp3 hybridized orbitals.

Bond [3] represents, CH bond. This bond is formed through sp3 hybridized orbital of carbon and 1s orbital of hydrogen.

Thus, [3] is formed by Csp3H1s orbitals.

Bond [4] represents bonding between the carbon atom of benzene (C6H6), and hydrogen. This bond is formed through sp2 hybridized orbital of carbon and 1s orbital of hydrogen.

Thus, bond [4] is formed by Csp2H1s 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.

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Chapter 1 Solutions

ORGANIC CHEMISTRY (LL) W/CONNECT

Ch. 1 - Prob. 1.11PCh. 1 - Prob. 1.12PCh. 1 - Prob. 1.13PCh. 1 - Draw a second resonance structure for each species...Ch. 1 - Prob. 1.15PCh. 1 - Prob. 1.16PCh. 1 - Prob. 1.17PCh. 1 - Prob. 1.18PCh. 1 - Using the principles of VSEPR theory, you can...Ch. 1 - Convert each condensed formula to a Lewis...Ch. 1 - Prob. 1.21PCh. 1 - Prob. 1.22PCh. 1 - Convert each skeletal structure to a complete...Ch. 1 - What is the molecular formula of quinine, the...Ch. 1 - Draw in all hydrogens and lone pairs on the...Ch. 1 - Prob. 1.26PCh. 1 - What orbitals are used to form each of the CC, and...Ch. 1 - What orbitals are used to form each bond in the...Ch. 1 - Determine the hybridization around the highlighted...Ch. 1 - The unmistakable odor of a freshly cut cucumber is...Ch. 1 - Prob. 1.31PCh. 1 - Rank the following atoms in order of increasing...Ch. 1 - Prob. 1.33PCh. 1 - Prob. 1.34PCh. 1 - Provide the following information about...Ch. 1 - Use the ball-and-stick model to answer each...Ch. 1 - Citric acid is responsible for the tartness of...Ch. 1 - Zingerone gives ginger its pungent taste. a.What...Ch. 1 - Assign formal charges to each carbon atom in the...Ch. 1 - Assign formal charges to each and atom in the...Ch. 1 - Prob. 1.41PCh. 1 - Prob. 1.42PCh. 1 - Prob. 1.43PCh. 1 - Draw all possible isomers for each molecular...Ch. 1 - 1.45 Draw Lewis structures for the nine isomers...Ch. 1 - Prob. 1.46PCh. 1 - Prob. 1.47PCh. 1 - Prob. 1.48PCh. 1 - Prob. 1.49PCh. 1 - Prob. 1.50PCh. 1 - Prob. 1.51PCh. 1 - Prob. 1.52PCh. 1 - Consider compounds A-D, which contain both a...Ch. 1 - Prob. 1.54PCh. 1 - Prob. 1.55PCh. 1 - 1.56 Consider the compounds and ions with curved...Ch. 1 - 1.57 Predict all bond angles in each...Ch. 1 - 1.58 Predict the geometry around each highlighted...Ch. 1 - Prob. 1.59PCh. 1 - Draw in all the carbon and hydrogen atoms in each...Ch. 1 - 1.61 Convert each molecule into a skeletal...Ch. 1 - Prob. 1.62PCh. 1 - Prob. 1.63PCh. 1 - Predict the hybridization and geometry around each...Ch. 1 - Prob. 1.65PCh. 1 - Ketene, , is an unusual organic molecule that has...Ch. 1 - Rank the following bonds in order of increasing...Ch. 1 - Prob. 1.68PCh. 1 - Two useful organic compounds that contain Cl atoms...Ch. 1 - Use the symbols + and to indicate the polarity of...Ch. 1 - Prob. 1.71PCh. 1 - Anacin is an over-the-counter pain reliever that...Ch. 1 - Answer the following questions about acetonitrile...Ch. 1 - Prob. 1.74PCh. 1 - 1.75 The principles of this chapter can be...Ch. 1 - a. What is the hybridization of each N atom in...Ch. 1 - 1.77 Stalevo is the trade name for a medication...Ch. 1 - 1.78 and are two highly reactive carbon...Ch. 1 - 1.79 The N atom in (acetamide) is hybridized,...Ch. 1 - Prob. 1.80PCh. 1 - Prob. 1.81PCh. 1 - Prob. 1.82PCh. 1 - Prob. 1.83PCh. 1 - Prob. 1.84PCh. 1 - Prob. 1.85P
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