Exam (elaborations) TEST BANK FOR McMurry's Organic chemistry 8th Edition By John McMurry and Susan (Study Guide and Student Solutions Manual)
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Organic Chemistry
EIGHTH EDITION
John McMurry
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Contents
Solutions to Problems
Chapter 1 Structure and Bonding 1
Chapter 2 Polar Covalent Bonds; Acids and Bases 20
Review Unit 1 38
Chapter 3 Organic Compounds: Alkanes and Their Stereochemistry 41
Chapter 4 Organic Compounds Cycloalkanes and Their Stereochemistry 64
Chapter 5 Stereochemistry 88
Review Unit 2 112
Chapter 6 An Overview of Organic Reactions 116
Chapter 7 Alkenes: Structure and Reactivity 132
Chapter 8 Alkenes: Reactions and Synthesis 158
Review Unit 3 186
Chapter 9 Alkynes: An Introduction to Organic Synthesis 190
Chapter 10 Organohalides 213
Chapter 11 Reactions of Alkyl Halides: Nucleophilic Substitutions and Eliminations 233
Review Unit 4 264
Chapter 12 Structure Determination: Mass Spectrometry and Infrared Spectroscopy 268
Chapter 13 Structure Determination: Nuclear Magnetic Resonance Spectroscopy 289
Review Unit 5 316
Chapter 14 Conjugated Dienes and Ultraviolet Spectroscopy 319
Chapter 15 Benzene and Aromaticity 342
Chapter 16 Chemistry of Benzene: Electrophilic Aromatic Substitution 361
Review Unit 6 400
Chapter 17 Alcohols and Phenols 404
Chapter 18 Ethers and Epoxides; Thiols and Sulfides 440
Review Unit 7 469
Carbonyl Preview 472
Chapter 19 Aldehydes and Ketones: Nucleophilic Addition Reactions 474
Chapter 20 Carboxylic Acids and Nitriles 518
Chapter 21 Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions 544
Review Unit 8 584
Chapter 22 Carbonyl Alpha-Substitution Reactions 588
Chapter 23 Carbonyl Condensation Reactions 616
Chapter 24 Amines and Heterocycles 654
Review Unit 9 698
Chapter 25 Biomolecules: Carbohydrates 701
Chapter 26 Biomolecules: Amino Acids, Peptides, and Proteins 733
Review Unit 10 762
Chapter 27 Biomolecules: Lipids 765
Chapter 28 Biomolecules: Nucleic Acids 790
Chapter 29 The Organic Chemistry of Metabolic Pathways 807
Review Unit 11 832
Chapter 30 Orbitals and Organic Chemistry: Pericyclic Reactions 836
Chapter 31 Synthetic Polymers 857
Review Unit 12 874
Appendices
Functional-Group Synthesis 877
Functional-Group Reactions 882
Reagents in Organic Chemistry 886
Name Reactions in Organic Chemistry 893
Abbreviations 901
Infrared Absorption Frequencies 904
Proton NMR Chemical Shifts 907
Nobel Prize Winners in Chemistry 908
Answers to Review-Unit Questions 917
iii
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Chapter 1 – Structure and Bonding
Chapter Outline
I. Atomic Structure (Sections 1.1–1.3).
A. Introduction to atomic structure (Section 1.1).
1. An atom consists of a dense, positively charged nucleus surrounded by negatively
charged electrons.
a. The nucleus is made up of positively charged protons and uncharged neutrons.
b. The nucleus contains most of the mass of the atom.
c. Electrons move about the nucleus at a distance of about 2 x 10–10 m (200 pm).
2. The atomic number (Z) gives the number of protons in the nucleus.
3. The mass number (A) gives the total number of protons and neutrons.
4. All atoms of a given element have the same value of Z.
a. Atoms of a given element can have different values of A.
b. Atoms of the same element with different values of A are called isotopes.
B. Orbitals (Section 1.2).
1. The distribution of electrons in an atom can be described by a wave equation.
a. The solution to a wave equation is an orbital, represented by Ψ.
b. Ψ
2 predicts the volume of space in which an electron is likely to be found.
2. There are four different kinds of orbitals (s, p, d, f).
a. The s orbitals are spherical.
b. The p orbitals are dumbbell-shaped.
c. Four of the five d orbitals are cloverleaf-shaped.
3. An atom's electrons are organized into electron shells.
a. The shells differ in the numbers and kinds of orbitals they contain.
b. Electrons in different orbitals have different energies.
c. Each orbital can hold up to a maximum of two electrons.
4. The two lowest-energy electrons are in the 1s orbital.
a. The 2s orbital is the next higher in energy.
b. The next three orbitals are 2px, 2py and 2pz, which have the same energy.
i. Each p orbital has a region of zero density, called a node.
c. The lobes of a p orbital have opposite algebraic signs.
C. Electron Configuration (Section 1.3).
1. The ground-state electron configuration of an atom is a listing of the orbitals
occupied by the electrons of the atom in the lowest energy configuration.
2. Rules for predicting the ground-state electron configuration of an atom:
a. Orbitals with the lowest energy levels are filled first.
i. The order of filling is 1s, 2s, 2p, 3s, 3p, 4s, 3d.
b. Only two electrons can occupy each orbital, and they must be of opposite spin.
c. If two or more orbitals have the same energy, one electron occupies each until
all are half-full (Hund's rule). Only then does a second electron occupy one of
the orbitals.
i. All of the electrons in half-filled shells have the same spin.
II. Chemical Bonding Theory (Sections 1.4–1.5).
A. Development of chemical bonding theory (Section 1.4).
1. Kekulé and Couper proposed that carbon has four "affinity units"; carbon is
tetravalent.
2. Kekulé suggested that carbon can form rings and chains.
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2 Chapter 1
3. Van't Hoff and Le Bel proposed that the 4 atoms to which carbon forms bonds sit
at the corners of a regular tetrahedron.
4. In a drawing of a tetrahedral carbon, a wedged line represents a bond pointing
toward the viewer, a dashed line points behind the plane of the page, and a solid
line lies in the plane of the page..
B. Covalent bonds.
1. Atoms bond together because the resulting compound is more stable than the
individual atoms.
a. Atoms tend to achieve the electron configuration of the nearest noble gas.
b. Atoms in groups 1A, 2A and 7A either lose electrons or gain electrons to form
ionic compounds.
c. Atoms in the middle of the periodic table share electrons by forming covalent
bonds.
d. The neutral collection of atoms held together by covalent bonds is a molecule.
2. Covalent bonds can be represented two ways.
a. In electron-dot structures, bonds are represented as pairs of dots.
b. In line-bond structures, bonds are represented as lines drawn between two
bonded atoms.
3. The number of covalent bonds formed by an atom depends on the number of
electrons it has and on the number it needs to achieve an octet.
4. Valence electrons not used for bonding are called lone-pair (nonbonding) electrons.
a. Lone-pair electrons are often represented as dots.
C. Valence bond theory (Section 1.5).
1. Covalent bonds are formed by the overlap of two atomic orbitals, each of which
contains one electron. The two electrons have opposite spins.
2. Bonds formed by the head-on overlap of two atomic orbitals are cylindrically
symmetrical and are called σ bonds.
3. Bond strength is the measure of the amount of energy needed to break a bond.
4. Bond length is the optimum distance between nuclei.
5. Every bond has a characteristic bond length and bond strength.
III. Hybridization (Sections 1.6–1.10).
A. sp3 Orbitals (Sections 1.6, 1.7).
1. Structure of methane (Section 1.6).
a. When carbon forms 4 bonds with hydrogen, one 2s orbital and three 2p orbitals
combine to form four equivalent atomic orbitals (sp3 hybrid orbitals).
b. These orbitals are tetrahedrally oriented.
c. Because these orbitals are unsymmetrical, they can form stronger bonds than
unhybridized orbitals can.
d. These bonds have a specific geometry and a bond angle of 109.5°.
2. Structure of ethane (Section 1.7).
a. Ethane has the same type of hybridization as occurs in methane.
b. The C–C bond is formed by overlap of two sp3 orbitals.
c. Bond lengths, strengths and angles are very close to those of methane.
B. sp2 Orbitals (Section 1.8).
1. If one carbon 2s orbital combines with two carbon 2p orbitals, three hybrid sp2
orbitals are formed, and one p orbital remains unchanged.
2. The three sp2 orbitals lie in a plane at angles of 120°, and the unhybridized p orbital
is perpendicular to them.
3. Two different types of bonds form between two carbons.
a. A σ bond forms from the overlap of two sp2 orbitals.
b. A π bond forms by sideways overlap of two p orbitals.
c. This combination is known as a carbon–carbon double bond.
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Structure and Bonding 3
4. Ethylene is composed of a carbon–carbon double bond and four σ bonds formed
between the remaining four sp2 orbitals of carbon and the 1s orbitals of hydrogen.
a. The double bond of ethylene is both shorter and stronger than the C–C bond of
ethane.
C. sp Orbitals (Section 1.10).
1. If one carbon 2s orbital combines with one carbon 2p orbital, two hybrid sp orbitals
are formed, and two p orbitals are unchanged.
2. The two sp orbitals are 180° apart, and the two p orbitals are perpendicular to them
and to each other.
3. Two different types of bonds form.
a. A σ bond forms from the overlap of two sp orbitals.
b. Two π bonds form by sideways overlap of four unhybridized p orbitals.
c. This combination is known as a carbon–carbon triple bond.
4. Acetylene is composed of a carbon–carbon triple bond and two σ bonds formed
between the remaining two sp orbitals of carbon and the 1s orbitals of hydrogen.
a. The triple bond of acetylene is the strongest carbon–carbon bond.
D. Hybridization of nitrogen and oxygen (Section 1.10).
1. Covalent bonds between other elements can be described by using hybrid orbitals.
2. Both the nitrogen atom in ammonia and the oxygen atom in water form sp3 hybrid
orbitals.
a. The lone-pair electrons in these compounds occupy sp3 orbitals.
3. The bond angles between hydrogen and the central atom is often less than 109°
because the lone-pair electrons take up more room than the σ bond.
4. Because of their positions in the third row, phosphorus and sulfur can form more
than the typical number of covalent bonds.
IV. Molecular orbital theory (Section 1.11).
A. Molecular orbitals arise from a mathematical combination of atomic orbitals and belong
to the entire molecule.
1. Two 1s orbitals can combine in two different ways.
a. The additive combination is a bonding MO and is lower in energy than the two
hydrogen 1s atomic orbitals.
b. The subtractive combination is an antibonding MO and is higher in energy than
the two hydrogen 1s atomic orbitals.
2. Two p orbitals in ethylene can combine to form two π MOs.
a. The bonding MO has no node; the antibonding MO has one node.
3. A node is a region between nuclei where electrons aren't found.
a. If a node occurs between two nuclei, the nuclei repel each other.
V. Chemical structures (Section 1.12).
A. Drawing chemical structures.
1. Condensed structures don't show C–H bonds and don't show the bonds between
CH3, CH2 and CH units.
2. Skeletal structures are simpler still.
a. Carbon atoms aren't usually shown.
b. Hydrogen atoms bonded to carbon aren't usually shown.
c. Other atoms (O, N, Cl, etc.) are shown.
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4 Chapter 1
Solutions to Problems
1 . 1 (a) To find the ground-state electron configuration of an element, first locate its atomic
number. For oxygen, the atomic number is 8; oxygen thus has 8 protons and 8 electrons.
Next, assign the electrons to the proper energy levels, starting with the lowest level. Fill
each level completely before assigning electrons to a higher energy level.
Notice that the 2p electrons are in different orbitals. According to Hund's rule, we must
place one electron into each orbital of the same energy level until all orbitals are half-filled.
2p
2s
1s
Oxygen
Remember that only two electrons can occupy the same orbital, and that they must be of
opposite spin.
A different way to represent the ground-state electron configuration is to simply write
down the occupied orbitals and to indicate the number of electrons in each orbital. For
example, the electron configuration for oxygen is 1s2 2s2 2p4.
(b) Nitrogen, with an atomic number of 7, has 7 electrons. Assigning these to energy
levels:
Nitrogen 2p
2s
1s
The more concise way to represent ground-state electron configuration for nitrogen:
1s2 2s2 2p3
(c) Sulfur has 16 electrons.
1s2 2s2 2p6 3s2 3p4
Sulfur 2p
2s
1s
3s
3p
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Structure and Bonding 5
1 . 2 The elements of the periodic table are organized into groups that are based on the number of
outer-shell electrons each element has. For example, an element in group 1A has one outershell
electron, and an element in group 5A has five outer-shell electrons. To find the
number of outer-shell electrons for a given element, use the periodic table to locate its
group.
(a) Magnesium (group 2A) has two electrons in its outermost shell.
(b) Cobalt is a transition metal, which has two electrons in the 4s subshell, plus seven
electrons in its 3d subshell.
(c) Selenium (group 6A) has six electrons in its outermost shell.
1 . 3 A solid line represents a bond lying in the plane of the page, a wedged bond represents a
bond pointing out of the plane of the page toward the viewer, and a dashed bond represents
a bond pointing behind the plane of the page.
C Chloroform
H
Cl
Cl
Cl
1 . 4
C C Ethane
H
H
H
H
H
H
1 . 5 Identify the group of the central element to predict the number of covalent bonds the
element can form.
(a) Carbon (Group 4A) has four electrons in its valence shell and forms four bonds to
achieve the noble-gas configuration of neon. A likely formula is CCl4.
Element Group Likely Formula
(b) Al 3A AlH3
(c) C 4A CH2Cl2
(d) Si 4A SiF4
(e) N 5A CH3NH2
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6 Chapter 1
1 . 6 Start by drawing the electron-dot structure of the molecule.
(1) Determine the number of valence, or outer-shell electrons for each atom in the
molecule. For chloroform, we know that carbon has four valence electrons, hydrogen
has one valence electron, and each chlorine has seven valence electrons.
C
H
Cl
total valence electrons
4 x 1 = 4
1 x 1 = 1
7 x 3 = 21
26
.
.
.
.
.
: .
. .
. .
(2) Next, use two electrons for each single bond.
Cl C
H. .
:. .: Cl
Cl
(3) Finally, use the remaining electrons to achieve an noble gas configuration for all atoms.
For a line-bond structure, replace the electron dots between two atoms with a line.
Cl C
H. .
:. .: Cl
Cl
. . . .
: . . . . :
: . . :
(a) CHCl3 C Cl
Cl
Cl
H
(b) H2S H S
8 valence electrons H
(c) CH3NH2
14 valence electrons
H C
H
H
N
H [Show Less]