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Solutions Manual - Organic Chemistry, 6th Edition (Smith, 2020), Chapter 1-29 + Spectroscopy A+B+C | All Chapters
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Solutions Manual - Organic Chemistry, 6th Edition (Smith, 2020), Chapter 1-29 + Spectroscopy A+B+C | All Chapters

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SOLUTIONS MANUAL
Organic Chemistry


Janice Gorzynski Smith
6th Edition

,Table of Contents

Chapter 01 Structure and Bonding 1
Chapter 02 Acids and Bases 35
Chapter 03 Introduction to Organic Molecules and Functional Groups 63
Chapter 04 Alkanes 87
Chapter 05 Stereochemistry 121
Chapter 06 Understanding Organic Reactions 149
Chapter 07 Alkyl Halides and Nucleophilic Substitution 171
Chapter 08 Alkyl Halides and Elimination Reactions 203
Chapter 09 Alcohols, Ethers, and Related Compounds 233
Chapter 10 Alkenes 267
Chapter 11 Alkynes 293
Chapter 12 Oxidation and Reduction 317
Chapter 13 Radical Reactions 347
Chapter 14 Conjugation, Resonance, and Dienes 373
Chapter 15 Benzene and Aromatic Compounds 401
Chapter 16 Reactions of Aromatic Compounds 425
Chapter 17 Introduction to Carbonyl Chemistry; Organometallic Reagents; Oxidation and
Reduction 463
Chapter 18 Aldehydes and Ketones—Nucleophilic Addition 493
Chapter 19 Carboxylic Acids and Nitriles 527
Chapter 20 Carboxylic Acids and Their Derivatives—Nucleophilic Acyl Substitution 555
Chapter 21 Substitution Reactions of Carbonyl Compounds at the 585
Chapter 22 Carbonyl Condensation Reactions 613
Chapter 23 Amines 643
Chapter 24 Carbon–Carbon Bond-Forming Reactions in Organic Synthesis 673
Chapter 25 Pericyclic Reactions 697
Chapter 26 Carbohydrates 721
Chapter 27 Amino Acids and Proteins 753
Chapter 28 Synthetic Polymers 785
Chapter 29 Lipids 811
Spectroscopy A Mass Spectrometry 827
Spectroscopy B Infrared Spectroscopy 837
Spectroscopy C Nuclear Magnetic Resonance Spectroscopy 851

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Solutions Manual - Organic Chemistry, 6th Edition (Smith, 2020)
Structure and Bonding 1–1

Chapter 1: Structure and Bonding

Chapter Review

Important facts

· The general rule of bonding: Atoms strive to attain a complete outer shell of valence electrons
(Section 1.2). H “wants” 2 electrons. Second-row elements “want” 8 electrons.




· Formal charge (FC) is the difference between the number of valence electrons on an atom and the
number of electrons it “owns” (Section 1.3C). See Sample Problem 1.3 for a stepwise example.




· Curved arrow notation shows the movement of an electron pair. The tail of the arrow always
begins at an electron pair, either in a bond or a lone pair. The head points to where the electron pair
“moves” (Section 1.6).




· Electrostatic potential plots are color-coded maps of electron density, indicating electron rich (red)
and electron deficient (blue) regions (Section 1.12).




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Solutions Manual - Organic Chemistry, 6th Edition (Smith, 2020)
Chapter 1–2

The importance of Lewis structures (Sections 1.3–1.5)

A properly drawn Lewis structure shows the number of bonds and lone pairs present around each atom
in a molecule. In a valid Lewis structure, each H has two electrons, and each second-row element has
no more than eight. This is the first step needed to determine many properties of a molecule.




Resonance (Section 1.6)

The basic principles:
· Resonance occurs when a compound cannot be represented by a single Lewis structure.
· Two resonance structures differ only in the position of nonbonded electrons and p bonds.
· The resonance hybrid is the only accurate representation for a resonance-stabilized compound. A
hybrid is more stable than any single resonance structure because electron density is delocalized.




The difference between resonance structures and isomers:
· Two isomers differ in the arrangement of both atoms and electrons.
· Resonance structures differ only in the arrangement of electrons.




Geometry and hybridization

The number of groups around an atom determines both its geometry (Section 1.7) and hybridization
(Section 1.9).

Number of Geometry Bond angle (°) Hybridization Examples
groups
2 linear 180 sp HCºCH
2
3 trigonal planar 120 sp CH2=CH2
4 tetrahedral 109.5 sp3 CH4, NH3, H2O




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Solutions Manual - Organic Chemistry, 6th Edition (Smith, 2020)
Structure and Bonding 1–3

Drawing organic molecules (Section 1.8)

· Shorthand methods are used to abbreviate the structure of organic molecules.




· A carbon bonded to four atoms is tetrahedral in shape. The best way to represent a tetrahedron is to
draw two bonds in the plane, one in front, and one behind.




Bond length

· Bond length decreases as you go from left to right across a row and increases down a column of the
periodic table (Section 1.7A).




· Bond length decreases as the number of electrons between two nuclei increases (Section 1.11A).




· Bond length increases as the percent s-character decreases (Section 1.11B).




· Bond length and bond strength are inversely related. Shorter bonds are stronger bonds
(Section 1.11).




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Solutions Manual - Organic Chemistry, 6th Edition (Smith, 2020)
Chapter 1–4

· Sigma (s) bonds are generally stronger than p bonds (Section 1.10).




Electronegativity and polarity (Sections 1.12, 1.13)

· Electronegativity increases from left to right across a row and decreases down a column of the
periodic table.
· A polar bond results when two atoms with different electronegativities are bonded together.
Whenever C or H is bonded to N, O, or any halogen, the bond is polar.
· A polar molecule has either one polar bond, or two or more bond dipoles that reinforce.

Drawing Lewis structures: A shortcut

Chapter 1 devotes a great deal of time to drawing valid Lewis structures. For molecules with many
bonds, it may take quite awhile to find acceptable Lewis structures by using trial-and-error to place
electrons. Fortunately, a shortcut can be used to figure out how many bonds are present in a molecule.

Shortcut on drawing Lewis structures—Determining the number of bonds:
[1] Count up the number of valence electrons.
[2] Calculate how many electrons are needed if there are no bonds between atoms and every atom
has a filled shell of valence electrons; that is, hydrogen gets two electrons, and second-row
elements get eight.
[3] Subtract the number obtained in Step [1] from the sum obtained in Step [2]. This difference
tells how many electrons must be shared to give every H two electrons and every second-row
element eight. Because there are two electrons per bond, dividing this difference by two tells
how many bonds are needed.

To draw the Lewis structure:
[1] Arrange the atoms as usual.
[2] Count up the number of valence electrons.
[3] Use the shortcut to determine how many bonds are present.
[4] Draw in the two-electron bonds to all the H’s first. Then, draw the remaining bonds between
other atoms, making sure that no second-row element gets more than eight electrons and that you
use the total number of bonds determined previously.
[5] Finally, place unshared electron pairs on all atoms that do not have an octet of electrons, and
calculate formal charge. You should have now used all the valence electrons determined in the
second step.




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Solutions Manual - Organic Chemistry, 6th Edition (Smith, 2020)
Structure and Bonding 1–5

Example: Draw all valid Lewis structures for CH3NCO using the shortcut procedure.

[1] Arrange the atoms.
· In this case the arrangement of atoms is implied by the way the structure is
drawn.


[2] Count up the number of valence electrons.




[3] Use the shortcut to figure out how many bonds are needed.
· Number of electrons needed if there were no bonds:




· Number of electrons that must be shared:




· Every bond requires two electrons, so 16/2 = 8 bonds are needed.

[4] Draw all possible Lewis structures.
· Draw the bonds to the H’s first (three bonds). Then add five more bonds. Arrange them between
the C’s, N, and O, making sure that no atom gets more than eight electrons. There are three
possible arrangements of bonds; that is, there are three resonance structures.
· Add additional electron pairs to give each atom an octet and check that all 22 electrons are used.




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Solutions Manual - Organic Chemistry, 6th Edition (Smith, 2020)
Chapter 1–6

· Calculate the formal charge on each atom.




· You can evaluate the Lewis structures you have drawn. The middle structure is the best
resonance structure, because it has no charged atoms.

Note: This method works for compounds that contain second-row elements in which every element gets
an octet of electrons. It does NOT necessarily work for compounds with an atom that does not have an
octet (such as BF3), or compounds that have elements located in the third row and later in the periodic
table.

Practice Test on Chapter Review

1.a. Which compound(s) contain a labeled atom with a +1 formal charge? All lone pairs of electrons
have been drawn in.




4. Both (1) and (2) have labeled atoms with a +1 charge.
5. Compounds (1), (2), and (3) all contain labeled atoms with a +1 formal charge.

b. Which of the following compounds is a valid resonance structure for A?




4. Both (1) and (2) are valid resonance structures for A.
5. Cations (1), (2), and (3) are all valid resonance structures for A.
2
c. Which species contains a labeled carbon atom that is sp hybridized?




4. Both (1) and (2) contain labeled sp2 hybridized atoms.
5. Species (1), (2), and (3) all contain labeled sp2 hybridized carbon atoms.




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Solutions Manual - Organic Chemistry, 6th Edition (Smith, 2020)
Structure and Bonding 1–7

d. Which of the following compounds has a net dipole?
1. CH3CH2NHCH2CH3 4. Compounds (1) and (2) both have net dipoles.
2. CH3CH2CH2OH 5. Compounds (1), (2), and (3) all have net dipoles.
3. FCH2CH2CH2F

2. Rank the labeled bonds in order of increasing bond length. Label the shortest bond as 1, the longest
bond as 4, and the bonds of intermediate length as 2 and 3.




3. Answer the following questions about compounds A–D.




a. What is the hybridization of the labeled atom in A?
b. What is the molecular shape around the labeled atom in B?
c. In what type of orbital does the lone pair in C reside?
d. What orbitals are used to form bond [1] in D?
e. Which orbitals are used to form the carbon–oxygen double bond [2] in D?

4. Draw an acceptable Lewis structure for CH3NO3. Assume that the atoms are arranged as drawn.




5. Follow the curved arrows and draw the product with all the needed charges and lone pairs.




Answers to Practice Test

1. a. 4 2. A – 1 3. a. sp3 4. One possibility: 5.
b. 1 B–4 b. trigonal planar
c. 1 C–2 c. sp2
d. 5 D–3 d. Csp3–Csp2
e. Csp–Osp2,
Cp–Op




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Solutions Manual - Organic Chemistry, 6th Edition (Smith, 2020)
Chapter 1–8

Answers to Problems

1.1 Two isotopes differ in the number of neutrons. Two isotopes have the same number of protons and
electrons, group number, and number of valence electrons. The mass number is the number of
protons and neutrons. The atomic number is the number of protons and is the same for all
isotopes.
Nitrogen-14 Nitrogen-13
a. number of protons = atomic number for N = 7 7 7
b. number of neutrons = mass number – atomic number 7 6
c. number of electrons = number of protons 7 7
d. group number 5A 5A
e. number of valence electrons 5 5

1.2 Ionic bonds form when an element on the far left side of the periodic table transfers an electron to
an element on the far right side of the periodic table. Covalent bonds result when two atoms
share electrons.




1.3 Atoms with one, two, three, or four valence electrons form one, two, three, or four bonds,
respectively. Atoms with five or more valence electrons form [8 – (number of valence electrons)]
bonds.
a. O 8 - 6 valence e- = 2 bonds c. Br 8 - 7 valence e- = 1 bond
b. Al 3 valence e- = 3 bonds d. Si 4 valence e- = 4 bonds

1.4 [1] Arrange the atoms with the H’s on the periphery.
[2] Count the valence electrons.
[3] Arrange the electrons around the atoms. Give the H’s 2 electrons first, and then fill the octets
of the other atoms.
[4] Assign formal charges (Section 1.3C).




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