Secondary: See Primary. -
Steroids (Section 27.6): Lipids whose structure is based on a tetracyclic carbon skeleton with three 6 -membered and one 5 -membered ring. Steroids occur in both plants and animals and have a variety of important hormonal functions.
Stork enamine reaction (Section 23.11): The conjugate addition of an enamine to an $\alpha, \beta$-unsaturated carbonyl compound, followed by hydrolysis to yield a 1,5-dicarbonyl product.
STR loci (Chapter 28 Chemistry Matters): Short tandem repeat sequences of noncoding DNA that are unique to every individual and allow DNA fingerprinting.
Straight-chain alkanes (Section 3.2): Alkanes whose carbon atoms are connected without branching.
Substitution reactions (Section 6.1): What occurs when two reactants exchange parts to give two new products. $\mathrm{S}_{\mathrm{N}} 1$ and $\mathrm{S}_{\mathrm{N}} 2$ reactions are examples.
Sulfides (Section 3.1, Section 18.7): A class of compounds that has two organic substituents bonded to the same sulfur atom, RSR'.
Sulfonation (Section 16.2): The substitution of a sulfonic acid group $\left(-\mathrm{SO}_{3} \mathrm{H}\right)$ onto an aromatic ring.
Sulfone (Section 18.7): A compound of the general structure $\mathrm{RSO}_{2} \mathrm{R}^{\prime}$.
Sulfonium ions (Section 18.7): A species containing a positively charged, trivalent sulfur atom, $\mathrm{R}_{3} \mathrm{~S}^{+}$.
Sulfoxide (Section 18.7): A compound of the general structure RSOR'.
Suprafacial (Section 30.5): A word used to describe the geometry of pericyclic reactions. Suprafacial reactions take place on the same side of the two ends of a $\pi$ electron system.
Suzuki-Miyaura reaction (Section 10.7): The palladiumcatalyzed coupling reaction of an aromatic or vinylic halide with an aromatic or vinylic boronic acid.


Secondary: See Primary. -
Suzuki-Miyaura reaction (Section 10.7): The palladiumcatalyzed coupling reaction of an aromatic or vinylic halide with an aromatic or vinylic boronic acid.
Symmetry-allowed, symmetry-disallowed (Section 30.1): A symmetry-allowed reaction is a pericyclic process that has a favorable orbital symmetry for reaction through a concerted pathway. A symmetry-disallowed reaction is one that does not have favorable orbital symmetry for reaction through a concerted pathway.
Symmetry plane (Section 5.2): A plane that bisects a molecule such that one half of the molecule is the mirror image of the other half. Molecules containing a plane of symmetry are achiral.
Syn periplanar (Section 11.8): Describing a stereochemical relationship in which two bonds on adjacent carbons lie in the same plane and are eclipsed.
Syn stereochemistry (Section 8.5): The opposite of anti. A syn addition reaction is one in which the two ends of the double bond react from the same side. A syn elimination is one in which the two groups leave from the same side of the molecule. Syndiotactic (Section 31.2): A chain-growth polymer in which the stereochemistry of the substituents alternates regularly on opposite sides of the backbone.


Secondary: See Primary. -
Tautomers (Section 9.4, Section 22.1): Isomers that interconvert spontaneously, usually with the change in position of a hydrogen.

Terpenoids (Chapter 8 Chemistry Matters, Section 27.5): Lipids that are formally derived by head-to-tail polymerization of isoprene units.


Tertiary: See Primary -
Tertiary structure (Section 26.9): The level of protein structure that involves the manner in which the entire protein chain is folded into a specific three-dimensional arrangement.

Thermodynamic control (Section 14.3): An equilibrium reaction that yields the lowest-energy, most stable product is said to be thermodynamically controlled.

Thermoplastics (Section 31.7): Polymers that have a high $T_{\mathrm{g}}$ and are hard at room temperature but become soft and viscous when heated.

Thermosetting resins (Section 31.7): Polymers that become highly cross-linked and solidify into a hard, insoluble mass when heated.

Thioesters (Chapter 21 Introduction): A class of compounds with the RCOSR' functional group.
Thiols (Section 3.1, Chapter 18 Introduction): A class of compounds containing the -SH functional group.
Thiolate ion (Section 18.7): The anion of a thiol, $\mathrm{RS}^{-}$.
TMS (Section 13.3): Tetramethylsilane; used as an NMR calibration standard.


Tertiary: See Primary -
TOF (Section 12.4): Time-of-flight mass spectrometry; a sensitive method of mass detection accurate to about 3 ppm .
Tollens' reagent (Section 25.6): A solution of $\mathrm{Ag}_{2} \mathrm{O}$ in aqueous ammonia; used to oxidize aldehydes to carboxylic acids.
Torsional strain (Section 3.6, Section 4.3): The strain in a molecule caused by electron repulsion between eclipsed bonds. Torsional strain is also called eclipsing strain.
Tosylate (Section 11.1): A p-toluenesulfonate ester; useful as a leaving group in nucleophilic substitution reactions.
Transamination (Section 29.9): The exchange of an amino group and a keto group between reactants.
Transcription (Section 28.4): The process by which the genetic information encoded in DNA is read and used to synthesize RNA in the nucleus of the cell. A small portion of doublestranded DNA uncoils, and complementary ribonucleotides line up in the correct sequence for RNA synthesis.
Transfer RNA (tRNA) (Section 28.4): A kind of RNA that transports amino acids to the ribosomes, where they are joined together to make proteins.
Transimination (Section 29.9): The exchange of an amino group and an imine group between reactants.

Transition state (Section 6.9): An activated complex between reactants, representing the highest energy point on a reaction curve. Transition states are unstable complexes that can't be isolated.

Translation (Section 28.5): The process by which the genetic information transcribed from DNA onto mRNA is read by tRNA
and used to direct protein synthesis.
Tree diagram (Section 13.8): A diagram used in NMR to sort out the complicated splitting patterns that can arise from multiple couplings.
Triacylglycerols (Section 27.1): Lipids, such as those found in animal fat and vegetable oil, that are a triester of glycerol with long-chain fatty acids.

Tricarboxylic acid cycle (Section 29.7): An alternative name for the citric acid cycle by which acetyl CoA is degraded to $\mathrm{CO}_{2}$.


Tertiary: See Primary -
Tricarboxylic acid cycle (Section 29.7): An alternative name for the citric acid cycle by which acetyl CoA is degraded to $\mathrm{CO}_{2}$.

Triple bonds (Section 1.8): A type of covalent bond formed by sharing three electron pairs between atoms.

Triplet (Section 13.6): A symmetrical three-line splitting pattern observed in the ${ }^{1} \mathrm{H}$ NMR spectrum when a proton has two equivalent neighbor protons.

Turnover number (Section 26.10): The number of substrate molecules acted on by an enzyme molecule per unit time.
Twist-boat conformation (Section 4.5): A conformation of cyclohexane that is somewhat more stable than a pure boat conformation.

Ultraviolet (UV) spectroscopy (Section 14.7): An optical spectroscopy employing ultraviolet irradiation. UV spectroscopy provides structural information about the extent of $\pi$ electron conjugation in organic molecules.
Unimolecular reaction (Section 11.4): A reaction that occurs by spontaneous transformation of the starting material without the intervention of other reactants. For example, the dissociation of a tertiary alkyl halide in the $\mathrm{S}_{\mathrm{N}} 1$ reaction is a unimolecular process.
Unsaturated (Section 7.2): A molecule that has one or more multiple bonds.
Upfield (Section 13.3): The right-hand portion of the NMR chart.

Urethane (Section 31.4): A functional group in which a carbonyl group is bonded to both an -OR and an $-\mathrm{NR}_{2}$.
Uronic acid (Section 25.6): A monocarboxylic acid formed by oxidizing the $-\mathrm{CH}_{2} \mathrm{OH}$ end of an aldose without affecting the - CHO end


Tertiary: See Primary -
Valence bond theory (Section 1.5): A bonding theory that describes a covalent bond as resulting from the overlap of two atomic orbitals.
Valence shell (Section 1.4): The outermost electron shell of an atom.
van der Waals forces (Section 2.12): Intermolecular forces that are responsible for holding molecules together in the liquid and solid states.
Vegetable oils (Section 27.1): Liquid triacylglycerols derived from a plant source.
Vicinal (Section 9.2): A term used to refer to a 1,2-disubstitution pattern. For example, 1,2-dibromoethane is a vicinal dibromide.
Vinyl group (Section 7.3): $\mathrm{A}_{2} \mathrm{C}=\mathrm{CH}$ - substituent.
Vinyl monomer (Section 8.10, Section 31.1): A substituted alkene monomer used to make a chain-growth polymer.


Tertiary: See Primary -
Vinylic (Section 9.3): A term that refers to a substituent at a double-bond carbon atom. For example, chloroethylene is a vinylic chloride, and enols are vinylic alcohols.
Vitamin (Section 26.10): A small organic molecule that must be obtained in the diet and is required in trace amounts for proper growth and function.
Vulcanization (Section 14.6): A technique for cross-linking and hardening a diene polymer by heating with a few percent by weight of sulfur.
Walden inversion (Section 11.1): The inversion of configuration at a chirality center that accompanies an $\mathrm{S}_{\mathrm{N}} 2$ reaction.
Wave equation (Section 1.2): A mathematical expression that defines the behavior of an electron in an atom.
Wave function (Section 1.2): A solution to the wave equation for defining the behavior of an electron in an atom. The square of the wave function defines the shape of an orbital.
Wavelength, $\lambda$ (Section 12.5): The length of a wave from peak to peak. The wavelength of electromagnetic radiation is inversely proportional to frequency and inversely proportional to energy.
Wavenumber, $\widetilde{\boldsymbol{v}}$ (Section 12.6): The reciprocal of the wavelength in centimeters.
Waxes (Section 27.1): A mixture of esters of long-chain carboxylic acids with long-chain alcohols.
Williamson ether synthesis (Section 18.2): A method for synthesizing ethers by $\mathrm{S}_{\mathrm{N}} 2$ reaction of an alkyl halide with an alkoxide ion.
Wittig reaction (Section 19.11): The reaction of a phosphorus
ylide with a ketone or aldehyde to yield an alkene.
Wohl degradation (Section 25.6): A method for shortening the chain of an aldose sugar by one carbon.
Wolff-Kishner reaction (Section 19.9): The conversion of an aldehyde or ketone into an alkane by reaction with hydrazine and base.


Tertiary: See Primary -
Wolff-Kishner reaction (Section 19.9): The conversion of an aldehyde or ketone into an alkane by reaction with hydrazine and base.
X-ray crystallography (Chapter 12 Chemistry Matters): A technique that uses $X$ rays to determine the structure of molecules.
Ylide (Section 19.11): A neutral species with adjacent + and charges, such as the phosphoranes used in Wittig reactions.
$\boldsymbol{Z}$ geometry (Section 7.5): A term used to describe the stereochemistry of a carbon-carbon double bond. The two groups on each carbon are ranked according to the Cahn-Ingold-Prelog sequence rules, and the two carbons are compared. If the higher ranked groups on each carbon are on the same side of the double bond, the bond has $Z$ geometry.
Zaitsev's rule (Section 11.7): A rule stating that E2 elimination reactions normally yield the more highly substituted alkene as major product.
Ziegler-Natta catalysts (Section 31.2): Catalysts of an alkylaluminum and a titanium compound used for preparing alkene polymers.
Zwitterion (Section 26.1): A neutral dipolar molecule in which the positive and negative charges are not adjacent. For example, amino acids exist as zwitterions, $\mathrm{H}_{3} \stackrel{+}{\mathrm{N}}-\mathrm{CHR}-\mathrm{CO}_{2}{ }^{-}$


APPENDIX D -
Periodic Table


Chapter 1 -
PROBLEM (a) $1 s^{2} 2 s^{2} 2 p^{4}$ (b) $1 s^{2} 2 s^{2} 2 p^{3}$ (c) $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{6} 3 p^{4}$
1-1
PROBLEM (a) 2 (b) 2 (c) 6
1-2
PROBLEM

PROBLEM
1-4

PROBLEM
(a) $\mathrm{CCl}_{4}$
(b) $\mathrm{AlH}_{3}$
(c) $\mathrm{CH}_{2} \mathrm{Cl}_{2}$
(d) $\mathrm{SiF}_{4}$
(e) $\mathrm{CH}_{3} \mathrm{NH}_{2}$

1-5
PROBLEM (a)

(c)

(d)

PROBLEM $\mathrm{C}_{2} \mathrm{H}_{7}$ has too many hydrogens for a compound with two carbons.
1-7
PROBLEM
1-8

All bond angles are near $109^{\circ}$.
PROBLEM
1-9

PROBLEM The $\mathrm{CH}_{3}$ carbon is $s p^{3}$; the double-bond carbons are $s p^{2}$; the $\mathrm{C}=\mathrm{C}-\mathrm{C}$ and $\mathrm{C}=\mathrm{C}-\mathrm{H}$ bond angles are
1-10 approximately $120^{\circ}$; other bond angles are near $109^{\circ}$.

PROBLEM All carbons are $s p^{2}$, and all bond angles are near $120^{\circ}$.
1-11

PROBLEM All carbons except $\mathrm{CH}_{3}$ are $s p^{2}$.
1-12

PROBLEM The $\mathrm{CH}_{3}$ carbon is $s p^{3}$; the triple-bond carbons are $s p$; the $\mathrm{C} \equiv \mathrm{C}-\mathrm{C}$ and $\mathrm{H}-\mathrm{C} \equiv \mathrm{C}$ bond angles are
1-13 approximately $180^{\circ}$.


Chapter 1 -
PROBLEM (a) O has 2 lone pairs and is $s p^{3}$-hybridized. (b) N has 1 lone pair and is $s p^{3}$-hybridized.
1-14 (c) P has 1 lone pair and is $s p^{3}$-hybridized. (d) S has 2 lone pairs and is $s p^{3}$-hybridized.

PROBLEM (a)
1-15

(b)

Estrone $-\mathrm{C}_{18} \mathrm{H}_{22} \mathrm{O}_{2}$

PROBLEM (a) There are numerous possibilities, such as: (b) There are numerous possibilities, such as:
1-16

(c) There are numerous possibilities, such as:

(d) There are numerous possibilities, such as:

PROBLEM
1-17


Chapter 2 -
PROBLEM (a) H (b) Br (c) Cl (d) C
2-1
PROBLEM
(c)
$\mathrm{H}_{2}{ }_{\mathrm{S}}^{\mathrm{N}}-\mathrm{S}_{\mathrm{H}}{ }^{\delta+}$
(d) $\mathrm{H}_{3} \mathrm{C}-\mathrm{SH}$

Carbon and sulfur have identical electronegativities.

PROBLEM $\mathrm{H}_{3} \mathrm{C}-\mathrm{OH}<\mathrm{H}_{3} \mathrm{C}-\mathrm{MgBr}<\mathrm{H}_{3} \mathrm{C}-\mathrm{Li}=\mathrm{H}_{3} \mathrm{C}-\mathrm{F}<\mathrm{H}_{3} \mathrm{C}-\mathrm{K}$
2-3
PROBLEM The nitrogen is electron-rich, and the carbon is electron-poor.
2-4

PROBLEM The two $\mathrm{C}-\mathrm{O}$ dipoles cancel because of the symmetry of the molecule:
2-5

PROBLEM (a)

(b)

2-6

(c)

(d)

PROBLEM (a) For carbon: $\mathrm{FC}=4-8 / 2-0=0$ For the middle nitrogen: $\mathrm{FC}=5-8 / 2-0=+1$ For the end 2-7 nitrogen: $\mathrm{FC}=5-4 / 2-4=-1$
(b) For nitrogen: $\mathrm{FC}=5-8 / 2-0=+1$ For oxygen: $\mathrm{FC}=6-2 / 2-6=-1$
(c) For nitrogen: $\mathrm{FC}=5-8 / 2-0=+1$ For the triply bonded carbon: $\mathrm{FC}=4-6 / 2-2=-1$

PROBLEM
2-8

PROBLEM The structures in (a) are resonance forms.
2-9
PROBLEM (a)
2-10

(b)

(c)

(d)

PROBLEM
2-11

PROBLEM Phenylalanine is stronger.
2-12
PROBLEM Water is a stronger acid.


Chapter 2 -
(b)

(c)

(d)

PROBLEM
2-11

PROBLEM Phenylalanine is stronger.
2-12
PROBLEM Water is a stronger acid.

2-13
PROBLEM Neither reaction will take place.
2-14
PROBLEM Reaction will take place.
2-15
PROBLEM $K_{\mathrm{a}}=4.9 \times 10^{-10}$
2-16
PROBLEM (a)
2-17

(b)

PROBLEM (a)
2-18

(b)

PROBLEM Vitamin C is water-soluble (hydrophilic); vitamin A is fat-soluble (hydrophilic).


Chapter 3 -
PROBLEM (a) Sulfide, carboxylic acid, amine (b) Aromatic ring, carboxylic acid
3-1 (c) Ether, alcohol, aromatic ring, amide, $\mathrm{C}=\mathrm{C}$ bond
PROBLEM (a) $\mathrm{CH}_{3} \mathrm{OH}$
3-2
(b)

(c)

(d) $\mathrm{CH}_{3} \mathrm{NH}_{2}$
(e)

(f)


PROBLEM -
3-3

PROBLEM
3-4
$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$

PROBLEM (a)
3-5

(b)

$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{N}$
(c) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{SSCH}_{2} \mathrm{CH}_{3}$
$\mathrm{CH}_{3} \mathrm{SSCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$

PROBLEM
(a) Two (b) Four
(c) Four

3-6
PROBLEM
3-7

PROBLEM (a)

(b)

(c)

PROBLEM Primary carbons have primary hydrogens, secondary carbons have secondary hydrogens, and 3-9 tertiary carbons have tertiary hydrogens.

PROBLEM (a)
3-10

(b)

(c)

PROBLEM (a) Pentane, 2-methylbutane, 2,2-dimethylpropane (b) 2,3-Dimethylpentane
3-11 (c) 2,4-Dimethylpentane (d) 2,2,5-Trimethylhexane
PROBLEM (a)

(b)

(c)

(d)

PROBLEM Pentyl, 1-methylbutyl, 1-ethylpropyl, 2-methylbutyl, 3-methylbutyl, 1,1-dimethylpropyl,
3-13 1,2-dimethylpropyl, 2,2-dimethylpropyl
PROBLEM
3-14

3,3,4,5-Tetramethylheptane

PROBLEM
3-15

PROBLEM (a)
3-16

(b)

(c)

(d)

PROBLEM
3-17

PROBLEM
3-18


PROBLEM -
PROBLEM
3-15

PROBLEM (a)
3-16

(b)

(c)

(d)

PROBLEM
3-17

PROBLEM
3-18

Chapter 4
PROBLEM (a) 1,4-Dimethylcyclohexane (b) 1-Methyl-3-propylcyclopentane (c) 3-Cyclobutylpentane
4-1 (d) 1-Bromo-4-ethylcyclodecane (e) 1-Isopropyl-2-methylcyclohexane
(f) 4-Bromo-1-tert-butyl-2-methylcycloheptane

PROBLEM (a)
4-2

(b)

(c)

(d)

PROBLEM 3-Ethyl-1,1-dimethylcyclopentane
4-3
PROBLEM (a) trans-1-Chloro-4-methylcyclohexane (b) cis-1-Ethyl-3-methylcycloheptane
4-4
PROBLEM (a)
4-5

(b)

(c)

PROBLEM The two hydroxyl groups are cis. The two side chains are trans.
4-6
PROBLEM (a) cis-1,2-Dimethylcyclopentane (b) cis-1-Bromo-3-methylcyclobutane
4-7
PROBLEM Six interactions; 21\% of strain
4-8
PROBLEM The cis isomer is less stable because the methyl groups nearly eclipse each other.
4-9
PROBLEM Ten eclipsing interactions; $40 \mathrm{~kJ} / \mathrm{mol} ; 35 \%$ is relieved.
4-10
PROBLEM Conformation (a) is more stable because the methyl groups are farther apart.
4-11
PROBLEM
4-12

PROBLEM
4-13


PROBLEM -
PROBLEM
4-13

PROBLEM Before the ring-flip, red and blue are equatorial and green is axial. After the ring-flip, red and blue
4-14 are axial and green is equatorial.
PROBLEM $4.2 \mathrm{~kJ} / \mathrm{mol}$
4-15
PROBLEM Cyano group points straight up.
4-16
PROBLEM Equatorial $=70 \%$; $a x i a l=30 \%$
4-17
PROBLEM (a) $2.0 \mathrm{~kJ} / \mathrm{mol}$ (axial Cl) (b) $11.4 \mathrm{~kJ} / \mathrm{mol}\left(\mathrm{axial} \mathrm{CH}_{3}\right)$ (c) $2.0 \mathrm{~kJ} / \mathrm{mol}$ (axial Br)
4-18 (d) $8.0 \mathrm{~kJ} / \mathrm{mol}\left(\right.$ axial $\left.\mathrm{CH}_{2} \mathrm{CH}_{3}\right)$
PROBLEM

1-Chloro-2,4-dimethylcyclohexane
(less stable chair form)
PROBLEM trans-Decalin is more stable because it has no 1,3-diaxial interactions.
4-20
PROBLEM Both ring-fusions are trans.
4-21
Chapter 5
PROBLEM Chiral: screw, shoe

5-1
PROBLEM (a)
5-2

(b)

(c)

PROBLEM
5-3

and

PROBLEM (a)
5-4

(b)

PROBLEM Levorotatory
5-5
PROBLEM $+16.1^{\circ}$
5-6
PROBLEM (a) -Br
(b) -Br (c) $-\mathrm{CH}_{2} \mathrm{CH}_{3}$
(d) -OH
(e) $-\mathrm{CH}_{2} \mathrm{OH}$ (f) $-\mathrm{CH}=\mathrm{O}$


PROBLEM -
5-7
PROBLEM (a) $-\mathrm{OH},-\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH},-\mathrm{CH}_{2} \mathrm{CH}_{3},-\mathrm{H}$ (b) $-\mathrm{OH},-\mathrm{CO}_{2} \mathrm{CH}_{3},-\mathrm{CO}_{2} \mathrm{H},-\mathrm{CH}_{2} \mathrm{OH}$
5-8 (c) $-\mathrm{NH}_{2},-\mathrm{CN},-\mathrm{CH}_{2} \mathrm{NHCH}_{3},-\mathrm{CH}_{2} \mathrm{NH}_{2}$ (d) $-\mathrm{SSCH}_{3},-\mathrm{SH},-\mathrm{CH}_{2} \mathrm{SCH}_{3},-\mathrm{CH}_{3}$
PROBLEM (a) $S$ (b) $R$ (c) $S$
5-9
PROBLEM (a) $S$ (b) $S$ (c) $R$
5-10
PROBLEM
5-11

PROBLEM $S$
5-12
PROBLEM Compound (a) is D-erythrose 4-phosphate, (d) is its enantiomer, and (b) and (c) are diastereomers.
5-13
PROBLEM Five chirality centers and $2^{5}=32$ stereoisomers
5-14
PROBLEM $S, S$
5-15
PROBLEM Compounds (a) and (d) are meso.
5-16
PROBLEM Compounds (a) and (c) have meso forms.
5-17
PROBLEM
5-18

PROBLEM Two diastereomeric salts: ( $R$ )-lactic acid plus ( $S$ )-1-phenylethylamine and ( $S$ )-lactic acid plus 5-20 (S)-1-phenylethylamine

PROBLEM (a) Constitutional isomers (b) Diastereomers
5-21
PROBLEM (a)
5-22

(b) pro- $\mathrm{C} \longrightarrow$ pro-S

PROBLEM (a)
5-23

(b)


PROBLEM -
(b) pro- $\mathrm{C} \longrightarrow$ pro-S

PROBLEM (a)
5-23

(b)

PROBLEM (S)-Lactate
5-24
PROBLEM The -OH adds to the Re face of C 2 , and -H adds to the Re face of C 3 . The overall addition has anti
5-25 stereochemistry.


Chapter 6 -
PROBLEM (a) Substitution (b) Elimination (c) Addition
6-1
PROBLEM (a) Carbon is electrophilic. (b) Sulfur is nucleophilic. (c) Nitrogens are nucleophilic.
6-2 (d) Oxygen is nucleophilic; carbon is electrophilic.
PROBLEM
6-3
PROBLEM Bromocyclohexane; chlorocyclohexane
6-4
PROBLEM
6-5

The mechanism is shown in Figure 6.4
PROBLEM (a)
6-6

(b)

(c)

PROBLEM
6-7

PROBLEM 1-Chloro-2-methylpentane, 2-chloro-2-methylpentane, 3-chloro-2-methylpentane, 6-8 2-chloro-4-methylpentane, 1-chloro-4-methylpentane
PROBLEM

Chapter 7
PROBLEM (a) 1 (b) 2 (c) 2
7-1
PROBLEM (a) 5 (b) 5 (c) 3 (d) 1 (e) 6 (f) 5
7-2
PROBLEM $\mathrm{C}_{16} \mathrm{H}_{13} \mathrm{ClN}_{2} \mathrm{O}$
7-3
PROBLEM (a) 3,4,4-Trimethyl-1-pentene (b) 3-Methyl-3-hexene (c) 4,7-Dimethyl-2,5-octadiene
$7-4$ (d) 6-Ethyl-7-methyl-4-nonene
7-4 (d) 6-Ethyl-7-methyl-4-nonene
PROBLEM (a)
7-5

(b)

(c)

(d)


Chapter 6 -
(b)

(c)

(d)

PROBLEM (a) 1,2-Dimethylcyclohexene
(b) 4,4-Dimethylcycloheptene (c) 3-Isopropylcyclopentene 7-6
PROBLEM (a) 2,5,5-Trimethylhex-2-ene (b) 2,3-Dimethylcyclohexa-1,3-diene
7-7
PROBLEM
7-8

PROBLEM Compounds (c), (e), and (f) have cis-trans isomers.
7-9
PROBLEM (a) cis-4,5-Dimethyl-2-hexene (b) trans-6-Methyl-3-heptene 7-10
PROBLEM
(a) $-\mathrm{CH}_{3}$
(b) -Cl
(c) $-\mathrm{CH}=\mathrm{CH}_{2}$
(d) $-\mathrm{OCH}_{3}$
(e) $-\mathrm{CH}=\mathrm{O}$ (f) $-\mathrm{CH}=\mathrm{O}$

7-11
PROBLEM (a) $-\mathrm{Cl},-\mathrm{OH},-\mathrm{CH}_{3},-\mathrm{H}$ (b) $-\mathrm{CH}_{2} \mathrm{OH},-\mathrm{CH}=\mathrm{CH}_{2},-\mathrm{CH}_{2} \mathrm{CH}_{3},-\mathrm{CH}_{3}$
7-12 (c) $-\mathrm{CO}_{2} \mathrm{H},-\mathrm{CH}_{2} \mathrm{OH},-\mathrm{C} \equiv \mathrm{N},-\mathrm{CH}_{2} \mathrm{NH}_{2}$ (d) $-\mathrm{CH}_{2} \mathrm{OCH}_{3},-\mathrm{C} \equiv \mathrm{N},-\mathrm{C} \equiv \mathrm{CH},-\mathrm{CH}_{2} \mathrm{CH}_{3}$
PROBLEM (a) $Z$ (b) $E$ (c) $Z$ (d) $E$
7-13
PROBLEM
7-14

z


Chapter 6 -
z

PROBLEM (a) 2-Methylpropene is more stable than 1-butene.
7-15 (b) trans-2-Hexene is more stable than cis-2-hexene.
(c) 1-Methylcyclohexene is more stable than 3-methylcyclohexene.

PROBLEM (a) Chlorocyclohexane (b) 2-Bromo-2-methylpentane (c) 4-Methyl-2-pentanol
7-16 (d) 1-Bromo-1-methylcyclohexane
PROBLEM (a) Cyclopentene (b) 1-Ethylcyclohexene or ethylidenecyclohexane (c) 3-Hexene
7-17 (d) Vinylcyclohexane (cyclohexylethylene)
PROBLEM (a)
7-18

(b)

PROBLEM In the conformation shown, only the methyl- group $\mathrm{C}-\mathrm{H}$ that is parallel to the carbocation $p$ orbital
7-19 can show hyperconjugation.
PROBLEM The second step is exergonic; the transition state resembles the carbocation.
7-20
PROBLEM
7-21


Chapter 8 -
PROBLEM 2-Methyl-2-butene and 2-methyl-1-butene
8-1
PROBLEM Five

8-2
PROBLEM trans-1,2-Dichloro-1,2-dimethylcyclohexane
8-3
PROBLEM
8-4

PROBLEM trans-2-Bromocyclopentanol
8-5
PROBLEM Markovnikov
8-6
PROBLEM (a) 2-Pentanol (b) 2-Methyl-2-pentanol
8-7
PROBLEM (a) Oxymercuration of 2-methyl-1-hexene or 2-methyl-2-hexene
8-8 (b) Oxymercuration of cyclohexylethylene or hydroboration of ethylidenecyclohexane
PROBLEM (a)

(b)

PROBLEM (a) 3-Methyl-1-butene (b) 2-Methyl-2-butene (c) Methylenecyclohexane
$\mathbf{8 - 1 0}$
PROBLEM

8-11

and

(b) 1,1-Dimethylcyclopentane (c) tert-Butylcyclohexane 8-12
PROBLEM
8-13
cis-2,3-Epoxybutane

PROBLEM (a) 1-Methylcyclohexene (b) 2-Methyl-2-pentene (c) 1,3-Butadiene 8-14

PROBLEM 8-15

PROBLEM (a) 2-Methylpropen
(b) 3-Hexene

8-16
PROBLEM (a)
8-17

(b)

PROBLEM (a) $\mathrm{H}_{2} \mathrm{C}=\mathrm{CHOCH}_{3}$
(b) $\mathrm{ClCH}=\mathrm{CHCl}$

8-18
PROBLEM
8-19


Chapter 8 -
8-18
PROBLEM
8-19

PROBLEM An optically inactive, non-50:50 mixture of two racemic pairs: $(2 R, 4 R)+(2 S, 4 S)$ and $(2 R, 4 S)+$
8-20 $(2 S, 4 R)$
PROBLEM Non-50:50 mixture of two racemic pairs: $(1 S, 3 R)+(1 R, 3 S)$ and $(1 S, 3 S)+(1 R, 3 R)$


Chapter 9 -
PROBLEM (a) 2,5-Dimethyl-3-hexyne (b) 3,3-Dimethyl-1-butyne (c) 3,3-Dimethyl-4-octyne
9-1 (d) 2,5,5-Trimethyl-3-heptyne (e) 2,4-Octadiene-6-yne
PROBLEM 1-Hexyne, 2-hexyne, 3-hexyne, 3-methyl-1-pentyne, 4-methyl-1-pentyne, 4-methyl-2-pentyne, 9-2 3,3-dimethyl-1-butyne
PROBLEM (a) 1,1,2,2-Tetrachloropentane (b) 1-Bromo-1-cyclopentylethylene
9-3 (c) 2-Bromo-2-heptene and 3-bromo-2-heptene
PROBLEM (a) 4-Octanone (b) 2-Methyl-4-octanone and 7-methyl-4-octanone 9-4
PROBLEM (a) 1-Pentyne (b) 2-Pentyne 9-5
PROBLEM (a) $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{C} \equiv \mathrm{CH}$ (b) 2,5-Dimethyl-3-hexyne 9-6
PROBLEM (a) Mercuric sulfate-catalyzed hydration of phenylacetylene
9-7 (b) Hydroboration/oxidation of cyclopentylacetylene
PROBLEM (a) Reduce 2-octyne with Li/ $\mathrm{NH}_{3}$. (b) Reduce 3-heptyne with $\mathrm{H}_{2}$ /Lindlar catalyst.
9-8 (c) Reduce 3-methyl-1-pentyne.
PROBLEM No: (a), (c), (d); yes: (b)
9-9


Chapter 9 -
9-8 (c) Reduce 3-methyl-1-pentyne.
PROBLEM No: (a), (c), (d); yes: (b)
9-9
PROBLEM (a) 1-Pentyne $+\mathrm{CH}_{3} \mathrm{I}$, or propyne $+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{I}$ (b) 3-Methyl-1-butyne $+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{I}$ 9-10 (c) Cyclohexylacetylene $+\mathrm{CH}_{3} \mathrm{I}$
PROBLEM
9-11
$\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{CH} \xrightarrow[\text { 2. } \mathrm{CH}_{3} \mathrm{I}]{\text { 1. } \mathrm{NaNH}_{2}} \quad \mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{CCH}_{3} \xrightarrow[\begin{array}{c}\text { Lindlar } \\ \text { cat. }\end{array}]{\mathrm{H}_{2}}$ cis $-\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCH}_{3}$
PROBLEM (a) $\mathrm{KMnO}_{4}, \mathrm{H}_{3} \mathrm{O}^{+}$(b) $\mathrm{H}_{2} /$ Lindlar (c) 1. $\mathrm{H}_{2} /$ Lindlar; 2. HBr
9-12 (d) 1. $\mathrm{H}_{2} /$ Lindlar; 2. $\mathrm{BH}_{3} ; 3 . \mathrm{NaOH}, \mathrm{H}_{2} \mathrm{O}_{2}$ (e) 1. $\mathrm{H}_{2} /$ Lindlar; $2 . \mathrm{Cl}_{2}$ (f) $\mathrm{O}_{3}$


Chapter 9 -
PROBLEM (a) 1. $\mathrm{HC} \equiv \mathrm{CH}+\mathrm{NaNH}_{2} ; 2 . \mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{6} \mathrm{CH}_{2} \mathrm{Br} ; 3.2 \mathrm{H}_{2} / \mathrm{Pd}$
9-13 (b) $\mathrm{HC} \equiv \mathrm{CH}+\mathrm{NaNH}_{2} ; 2 .\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH}_{2} \mathrm{CH}_{2} \mathrm{I} ; 3.2 \mathrm{H}_{2} / \mathrm{Pd}$
(c) 1. $\mathrm{HC} \equiv \mathrm{CH}+\mathrm{NaNH}_{2} ; 2 . \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{I} ; 3 . \mathrm{BH}_{3} ; 4 . \mathrm{H}_{2} \mathrm{O}_{2}$
(d) 1. $\mathrm{HC} \equiv \mathrm{CH}+\mathrm{NaNH}_{2} ; 2 . \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{I} ; 3 . \mathrm{HgSO}_{4}, \mathrm{H}_{3} \mathrm{O}^{+}$


Chapter 9 -
Chapter 10
PROBLEM (a) 1-Iodobutane (b) 1-Chloro-3-methylbutane (c) 1,5-Dibromo-2,2-dimethylpentane
10-1 (d) 1,3-Dichloro-3-methylbutane (e) 1-Chloro-3-ethyl-4-iodopentane
(f) 2-Bromo-5-chlorohexane

PROBLEM (a)
10-2

(b)

(c)

(d)

(e)

(f)

PROBLEM Chiral: 1-chloro-2-methylpentane, 3-chloro-2-methylpentane, 2-chloro-4-methylpentane Achiral:
10-3 2-chloro-2-methylpentane, 1-chloro-4-methylpentane

$\mathrm{ClCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}\left(\mathrm{CH}_{3}\right)_{2}$
PROBLEM 1-Chloro-2-methylbutane (29\%), 1-chloro-3-methylbutane (14\%), 2-chloro-2-methylbutane (24\%),
10-4 2-chloro-3-methylbutane (33\%)
PROBLEM
10-5


Chapter 9 -
PROBLEM The intermediate allylic radical reacts at the more accessible site and gives the more highly 10-6 substituted double bond.
PROBLEM (a) 3-Bromo-5-methylcycloheptene and 3-bromo-6-methylcycloheptene (b) Four products 10-7
PROBLEM (a) 2-Methyl-2-propanol +HCl (b) 4-Methyl-2-pentanol + $\mathrm{PBr}_{3}$
10-8 (c) 5-Methyl-1-pentanol $+\mathrm{PBr}_{3}$ (d) 3,3-Dimethyl-cyclopentanol + HF, pyridine
PROBLEM Both reactions occur
10-9
PROBLEM React Grignard reagent with $\mathrm{D}_{2} \mathrm{O}$.
10-10
PROBLEM (a) 1. NBS; 2. $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CuLi}$ (b) 1. Li; 2. $\mathrm{CuI} ; 3 . \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}$
10-11 (c) 1. $\mathrm{BH}_{3} ; 2 . \mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{NaOH} ; 3 . \mathrm{PBr}_{3} ; 4$. Li, then $\mathrm{CuI} ; 5 . \mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{4} \mathrm{Br}$
PROBLEM (a)
10-12


Chapter 9 -
(b)
$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2}<\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}<\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{N}$
PROBLEM (a) Reduction (b) Neither
10-13
Chapter 11
PROBLEM ( $R$ )-1-Methylpentyl acetate, $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{CH}\left(\mathrm{CH}_{3}\right) \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$
11-1
PROBLEM (S)-2-Butanol
11-2
PROBLEM
(S)-2-Bromo-4-methylpentane $\longrightarrow$

11-3

PROBLEM (a) 1-Iodobutane (b) 1-Butanol (c) 1-Hexyne (d) Butylammonium bromide
11-4
PROBLEM
(a) $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{~N}^{-}$
(b) $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}$
(c) $\mathrm{H}_{2} \mathrm{~S}$

11-5
PROBLEM $\mathrm{CH}_{3} \mathrm{OTos}>\mathrm{CH}_{3} \mathrm{Br}>\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCl}>\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCl}$
11-6
PROBLEM Similar to protic solvents
11-7
PROBLEM Racemic 1-ethyl-1-methylhexyl acetate

11-8
PROBLEM 90.1\% racemization, 9.9\% inversion
11-9
PROBLEM
11-10


Chapter 9 -
11-8
PROBLEM 90.1\% racemization, 9.9\% inversion
11-9
PROBLEM
11-10

PROBLEM $\mathrm{H}_{2} \mathrm{C}=\mathrm{CHCH}(\mathrm{Br}) \mathrm{CH}_{3}>\mathrm{CH}_{3} \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_{3}>\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Br}>\mathrm{H}_{2} \mathrm{C}=\mathrm{CHBr}$
11-11
PROBLEM The same allylic carbocation intermediate is formed.
11-12
PROBLEM (a) $\mathrm{S}_{\mathrm{N}} 1$ (b) $\mathrm{S}_{\mathrm{N}} 2$
11-13
PROBLEM
11-14

Linalyl diphosphate

PROBLEM (a) Major: 2-methyl-2-pentene; minor: 4-methyl-2-pentene
11-15 (b) Major: 2,3,5-trimethyl-2-hexene; minor: 2,3,5-trimethyl-3-hexene and 2-isopropyl-4-methyl-1-pentene
(c) Major: ethylidenecyclohexane; minor: cyclohexylethylene

PROBLEM
(a) 1-Bromo-3,6-dimethylheptane (b) 4-Bromo-1,2-dimethylcyclopentane


Chapter 9 -
PROBLEM
(a) 1-Bromo-3,6-dimethylheptane (b) 4-Bromo-1,2-dimethylcyclopentane

11-16
PROBLEM (Z)-1-Bromo-1,2-diphenylethylene
11-17
PROBLEM (Z)-3-Methyl-2-pentene
11-18
PROBLEM Cis isomer reacts faster because the bromine is axial.
11-19
PROBLEM (a) $\mathrm{S}_{\mathrm{N}} 2$ (b) E2 (c) $\mathrm{S}_{\mathrm{N}} 1$ (d) E1cB
11-20
Chapter 12
PROBLEM $\mathrm{C}_{19} \mathrm{H}_{28} \mathrm{O}_{2}$
12-1
PROBLEM (a) 2-Methyl-2-pentene (b) 2-Hexene
12-2
PROBLEM (a) 43,71 (b) 82 (c) 58 (d) 86
12-3
PROBLEM $102\left(\mathrm{M}^{+}\right)$, 84 (dehydration), 87 (alpha cleavage), 59 (alpha cleavage)
12-4
PROBLEM X-ray energy is higher; $\lambda=9.0 \times 10^{-6} \mathrm{~m}$ is higher in energy.


12-5 -
PROBLEM (a) $\left.2.4 \times 10^{6} \mathrm{~kJ} / \mathrm{mol}\right]$ (b) $4.0 \times 10^{4} \mathrm{~kJ} / \mathrm{mol}$ (c) $2.4 \times 10^{3} \mathrm{~kJ} / \mathrm{mol}$ (d) $2.8 \times 10^{2} \mathrm{~kJ} / \mathrm{mol}$
$12-6$ (e) $6.0 \mathrm{~kJ} / \mathrm{mol}$ (f) $4.0 \times 10^{-2} \mathrm{~kJ} / \mathrm{mol}$
PROBLEM (a) Ketone or aldehyde (b) Nitro compound (c) Carboxylic acid
12-7
PROBLEM (a) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}$ has an -OH absorption. (b) 1-Hexene has a double-bond absorption.
12-8 (c) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}$ has a very broad -OH absorption.
PROBLEM $1450-1600 \mathrm{~cm}^{-1}$ : aromatic ring; $2100 \mathrm{~cm}^{-1}: \mathrm{C} \equiv \mathrm{C} ; 3300 \mathrm{~cm}^{-1}: \mathrm{C} \equiv \mathrm{C}-\mathrm{H}$
12-9
PROBLEM (a) $1715,1640,1250 \mathrm{~cm}^{-1}$ (b) $1730,2100,3300 \mathrm{~cm}^{-1}$
12-10 (c) $1720,2500-3100,3400-3650 \mathrm{~cm}^{-1}$
PROBLEM 1690, 1650, $2230 \mathrm{~cm}^{-1}$
12-11


Chapter 13 -
PROBLEM $7.5 \times 10^{-5} \mathrm{~kJ} / \mathrm{mol}$ for ${ }^{19} \mathrm{~F} ; 8.0 \times 10^{-5} \mathrm{~kJ} / \mathrm{mol}$ for ${ }^{1} \mathrm{H}$
13-1
PROBLEM $1.2 \times 10^{-4} \mathrm{~kJ} / \mathrm{mol}$
13-2
PROBLEM The vinylic $\mathrm{C}-\mathrm{H}$ protons are nonequivalent.
$13-3$ b

PROBLEM (a) $7.27 \delta$ (b) $3.05 \delta$ (c) $3.46 \delta$ (d) $5.30 \delta$
13-4
PROBLEM (a) 420 Hz (b) $2.1 \delta$ (c) 1050 Hz
13-5
PROBLEM
(a) $1.43 \delta$ (b) $2.17 \delta$ (c) $7.37 \delta$ (d) $5.30 \delta$ (e) $9.70 \delta$ (f) $2.12 \delta$

13-6
PROBLEM There are seven kinds of protons labeled. The types and expected range of absorption of each
13-7 follow. a: ether, 3.5-4.5 $\delta$; b: aryl, 6.5-8.0 $\delta$; $\boldsymbol{c}$ : aryl, 6.5-8.0; $\boldsymbol{d}$ : vinylic, 4.5-6.5 $\delta$; e: vinylic, 4.5-6.5 $\delta$; $\boldsymbol{f}$ : alkyl (secondary), 1.2-1.6 $\delta$; $\boldsymbol{g}$ : alkyl (primary), 0.7-1.3 $\delta$.


Chapter 13 -
PROBLEM Two peaks; 3:2 ratio
13-8
PROBLEM (a) $-\mathrm{CHBr}_{2}$, quartet; $-\mathrm{CH}_{3}$, doublet (b) $\mathrm{CH}_{3} \mathrm{O}-$, singlet; $-\mathrm{OCH}_{2}-$, triplet; $-\mathrm{CH}_{2} \mathrm{Br}$, triplet
13-9 (c) $\mathrm{ClCH}_{2}-$, triplet; $-\mathrm{CH}_{2}{ }^{-}$, quintet
(d) $\mathrm{CH}_{3}{ }^{-}$, triplet; $-\mathrm{CH}_{2}{ }^{-}$, quartet; $-\mathrm{CH}-$, septet; $\left(\mathrm{CH}_{3}\right)_{2}$, doublet
(e) $\mathrm{CH}_{3}-$, triplet; $-\mathrm{CH}_{2}-$, quartet; - $\mathrm{CH}-$, septet; $\left(\mathrm{CH}_{3}\right)_{2}$, doublet
(f) $=\mathrm{CH}$, triplet, $-\mathrm{CH}_{2}-$, doublet, aromatic $\mathrm{C}-\mathrm{H}$, two multiplets


Chapter 13 -
PROBLEM (a) $\mathrm{CH}_{3} \mathrm{OCH}_{3}$ (b) $\mathrm{CH}_{3} \mathrm{CH}(\mathrm{Cl}) \mathrm{CH}_{3}$ (c) $\mathrm{ClCH}_{2} \mathrm{CH}_{2} \mathrm{OCH}_{2} \mathrm{CH}_{2} \mathrm{Cl}$
13-10 (d) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{CH}_{3}$ or $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$
PROBLEM $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OCH}_{2} \mathrm{CH}_{3}$
13-11
PROBLEM (a) Enantiotopic (b) Diastereotopic (c) Diastereotopic (d) Diastereotopic
13-12 (e) Diastereotopic (f) Homotopic
PROBLEM (a) 2 (b) 4 (c) 3 (d) 4 (e) 5 (f) 3

13-13
PROBLEM 4
13-14
PROBLEM $J_{1-2}=16 \mathrm{~Hz} ; J_{2-3}=8 \mathrm{~Hz}$
13-15


Chapter 13 -
13-13
PROBLEM 4
13-14
PROBLEM $J_{1-2}=16 \mathrm{~Hz} ; J_{2-3}=8 \mathrm{~Hz}$
13-15

un
PROBLEM 1-Chloro-1-methylcyclohexane has a singlet methyl absorption.
13-16
PROBLEM (a) 4 (b) 7 (c) 4 (d) 5 (e) 5 (f) 7
13-17
PROBLEM (a) 1,3-Dimethylcyclopentene (b) 2-Methylpentane (c) 1-Chloro-2-methylpropane
13-18
PROBLEM $-\mathrm{CH}_{3}, 9.3 \delta ;-\mathrm{CH}_{2}-27.6 \delta$; $\mathrm{C}=\mathrm{O}, 174.6 \delta ;-\mathrm{OCH}_{3}, 51.4 \delta$
13-19
PROBLEM
13-20

PROBLEM

PROBLEM
13-22


Chapter 13 -
PROBLEM

PROBLEM
13-22

PROBLEM A DEPT-90 spectrum would show two absorptions for the non-Markovnikov product ( $\mathrm{RCH}=\mathrm{CHBr}$ )
13-23 but no absorptions for the Markovnikov product $\left(\mathrm{RBrC}=\mathrm{CH}_{2}\right)$.
Chapter 14
PROBLEM Expected $\Delta H^{\circ}{ }_{\text {hydrog }}$ for allen is $-252 \mathrm{~kJ} / \mathrm{mol}$. Allen is less stable than a nonconjugated dene,
14-1 which is less stable than a conjugated dene.
PROBLEM 1-Chloro-2-pentene, 3-chloro-1-pentene, 4-chloro-2-pentene
14-2
PROBLEM 4-Chloro-2-pentene predominates in both.
14-3
PROBLEM 1,2 Addition: 6-bromo-1,6-dimethylcyclohexene 1,4 Addition: 3-bromo-1,2-dimethylcyclohexene 14-4
PROBLEM Interconversion occurs by $\mathrm{S}_{\mathrm{N}} 1$ dissociation to a common intermediate cation.
14-5
PROBLEM The double bond is more highly substituted.
14-6

PROBLEM

PROBLEM Good dienophiles: (a), (d)
14-8
PROBLEM Compound (a) is $s$-cis. Compound (c) can rotate to $s$-cis.
14-9
PROBLEM
14-10

PROBLEM
14-11

PROBLEM

PROBLEM $300-600 \mathrm{~kJ} / \mathrm{mol}$; UV energy is greater than IR or NMR energy.
14-13
PROBLEM $1.46 \times 10^{-5} \mathrm{M}$
14-14
PROBLEM All except (a) have UV absorptions.
14-15


Chapter 15 -
PROBLEM (a) Meta (b) Para (c) Ortho
15-1
PROBLEM (a) $m$-Bromochlorobenzene (b) (3-Methylbutyl)benzene (c) $p$-Bromoaniline
15-2 (d) 2,5-Dichlorotoluene (e) 1-Ethyl-2,4-dinitrobenzene (f) 1,2,3,5-Tetramethylbenzene
PROBLEM (a)

(b)

(c)

(d)

PROBLEM Pyridine has an aromatic sextet of electrons.
15-4

Pyridine

PROBLEM Cyclodecapentaene is not flat because of steric interactions.
15-5
PROBLEM All $\mathrm{C}-\mathrm{C}$ bonds are equivalent; one resonance line in both ${ }^{1} \mathrm{H}$ and ${ }^{13} \mathrm{C}$ NMR spectra.
15-6
PROBLEM The cyclooctatetraenyl dianion is aromatic (ten $\pi$ electrons) and flat.

| 15-7
 PROBLEM
 15-8
 15 | - |  |  |  |  |
| ---: | :---: | :---: | :---: | :---: | :---: |
|  |  | - | - | - | - |
|  | $\uparrow \downarrow$ |  | $\uparrow$ | $\uparrow$ | $\uparrow \downarrow$ |


PROBLEM -
15-9

Furan

PROBLEM The thiazolium ring has six $\pi$ electrons.
15-10

PROBLEM Yes, it's aromatic.
15-11

PROBLEM The three nitrogens in double bonds each contribute one; the remaining nitrogen contributes two. 15-12


Chapter 16 -
PROBLEM $o-$, $m$-, and $p$-Bromotoluene
16-1
PROBLEM
16-2

PROBLEM $o$-xylene: $2 ; p$-xylene: $1 ; m$-xylene: 3
16-3
PROBLEM $\mathrm{D}^{+}$does electrophilic substitutions on the ring.
16-4
PROBLEM No rearrangement: (a), (b), (e)
16-5
PROBLEM tert-Butylbenzene
16-6
PROBLEM (a) $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCOCl}$ (b) PhCOCl
16-7
PROBLEM (a) Phenol $>$ Toluene $>$ Benzene $>$ Nitrobenzene
16-8 (b) Phenol $>$ Benzene $>$ Chlorobenzene $>$ Benzoic acid
(c) Aniline $>$ Benzene $>$ Bromobenzene $>$ Benzaldehyde

PROBLEM (a) $o$ - and $p$-Bromonitrobenzene (b) $m$-Bromonitrobenzene (c) $o$ - and $p$-Chlorophenol
16-9 (d) $o$ - and $p$-Bromoaniline
PROBLEM Alkylbenzenes are more reactive than benzene itself, but acylbenzenes are less reactive. 16-10


Chapter 16 -
PROBLEM Toluene is more reactive; the trifluoromethyl group is electron-withdrawing. 16-11
PROBLEM The nitrogen electrons are donated to the nearby carbonyl group by resonance and are less 16-12 available to the ring.
PROBLEM The meta intermediate is most favored. 16-13
PROBLEM
(a) Ortho and para to $-\mathrm{OCH}_{3}$
(b) Ortho and para to $-\mathrm{NH}_{2}$
(c) Ortho and para to -Cl 16-14
PROBLEM (a) Reaction occurs ortho and para to the $-\mathrm{CH}_{3}$ group.
16-15 (b) Reaction occurs ortho and para to the $-\mathrm{OCH}_{3}$ group.
PROBLEM The phenol is deprotonated by KOH to give an anion that carries out a nucleophilic acyl substitution 16-16 reaction on the fluoronitrobenzene.
PROBLEM Only one benzyne intermediate can form from $p$-bromotoluene; two different benzyne 16-17 intermediates can form from $m$-bromotoluene.
PROBLEM (a) m-Nitrobenzoic acid (b) p-tert-Butylbenzoic acid 16-18
PROBLEM A benzyl radical is more stable than a primary alkyl radical by $52 \mathrm{~kJ} / \mathrm{mol}$ and is similar in stability 16-19 to an allyl radical.
PROBLEM 1. $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}, \mathrm{AlCl}_{3} ; 2$. NBS; 3. KOH , ethanol 16-20
PROBLEM 1. $\mathrm{PhCOCl}, \mathrm{AlCl}_{3} ; 2 . \mathrm{H}_{2} / \mathrm{Pd}$ 16-21


Chapter 16 -
PROBLEM 1. $\mathrm{PhCOCl}, \mathrm{AlCl}_{3} ; 2 . \mathrm{H}_{2} / \mathrm{Pd}$ 16-21
PROBLEM (a) 1. $\mathrm{HNO}_{3}, \mathrm{H}_{2} \mathrm{SO}_{4} ; 2 . \mathrm{Cl}_{2}, \mathrm{FeCl}_{3}$ (b) 1. $\mathrm{CH}_{3} \mathrm{COCl}, \mathrm{AlCl}_{3} ; 2 . \mathrm{Cl}_{2}, \mathrm{FeCl}_{3} ; 3 . \mathrm{H} 2 / \mathrm{Pd}$
16-22 (c) 1. $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COCl}, \mathrm{AlCl}_{3} ; 2 . \mathrm{Cl}_{2}, \mathrm{FeCl}_{3} ; 3 . \mathrm{H}_{2} / \mathrm{Pd} ; 4 . \mathrm{HNO}_{3}, \mathrm{H}_{2} \mathrm{SO}_{4}$
(d) 1. $\mathrm{CH}_{3} \mathrm{Cl}, \mathrm{AlCl}_{3} ; 2 . \mathrm{Br}_{2}, \mathrm{FeBr}_{3} ; 3 . \mathrm{SO}_{3}, \mathrm{H}_{2} \mathrm{SO}_{4}$


Chapter 16 -
PROBLEM (a) Friedel-Crafts acylation does not occur on a deactivated ring.
16-23 (b) Rearrangement occurs during Friedel-Crafts alkylation with primary halides; chlorination occurs ortho to the alkyl group.

Chapter 17
PROBLEM (a) 5-Methyl-2,4-hexanediol (b) 2-Methyl-4-phenyl-2-butanol (c) 4,4-Dimethylcyclohexanol
17-1 (d) trans-2-Bromocyclopentanol (e) 4-Bromo-3-methylphenol (f) 2-Cyclopenten-1-ol
PROBLEM
(a)

(b)

(c)

(d)

(e)

(f)

PROBLEM Hydrogen-bonding is more difficult in hindered alcohols.
17-3
PROBLEM (a) $\mathrm{HC} \equiv \mathrm{CH}<\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHOH}<\mathrm{CH}_{3} \mathrm{OH}<\left(\mathrm{CF}_{3}\right)_{2} \mathrm{CHOH}$
17-4 (b) $p$-Methylphenol < Phenol < p-(Trifluoromethyl)phenol
(c) Benzyl alcohol < Phenol < p-Hydroxybenzoic acid

PROBLEM The electron-withdrawing nitro group stabilizes an alkoxide ion, but the electron-donating
17-5 methoxyl group destabilizes the anion.
PROBLEM (a) 2-Methyl-3-pentanol (b) 2-Methyl-4-phenyl-2-butanol (c) meso-5,6-Decanediol 17-6
PROBLEM (a) $\mathrm{NaBH}_{4}$ (b) $\mathrm{LiAlH}_{4}$ (c) $\mathrm{LiAlH}_{4}$


Chapter 16 -
17-7
PROBLEM (a) Benzaldehyde or benzoic acid (or ester) (b) Acetophenone (c) Cyclohexanone
17-8 (d) 2-Methylpropanal or 2-methylpropanoic acid (or ester)
PROBLEM (a) 1-Methylcyclopentanol (b) 1,1-Diphenylethanol (c) 3-Methyl-3-hexanol
17-9
PROBLEM (a)
17-10

or

2-Methyl-2-propanol

(b)

1-Methylcyclohexanol
(c)

or

or

(d)

or

(e)

(f)


Chapter 16 -
or

2-Methyl-2-propanol

(b)

1-Methylcyclohexanol
(c)

or

or

(d)

or

(e)

(f)

PROBLEM Cyclohexanone $+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{MgBr}$ 17-11
PROBLEM 1. $p$-TosCl, pyridine; 2. NaCN 17-12
PROBLEM (a) 2-Methyl-2-pentene (b) 3-Methylcyclohexene (c) 1-Methylcyclohexene 17-13 (d) 2,3-Dimethyl-2-pentene (e) 2-Methyl-2-pentene
PROBLEM (a) 1-Phenylethanol (b) 2-Methyl-1-propanol (c) Cyclopentanol 17-14
PROBLEM (a) Hexanoic acid, hexanal (b) 2-Hexanone (c) Hexanoic acid, no reaction 17-15
PROBLEM $\mathrm{S}_{\mathrm{N}} 2$ reaction of $\mathrm{F}^{-}$on silicon with displacement of alkoxide ion. 17-16
PROBLEM Protonation of 2-methylpropene gives the tert-butyl cation, which carries out an electrophilic 17-17 aromatic substitution reaction.
PROBLEM Disappearance of -OH absorption; appearance of $\mathrm{C}=\mathrm{O}$ 17-18
PROBLEM (a) Singlet (b) Doublet (c) Triplet (d) Doublet (e) Doublet (f) Singlet 17-19


Chapter 18 -
PROBLEM (a) Diisopropyl ether (b) Cyclopentyl propyl ether
18-1 (c) $p$-Bromoanisole or 4-bromo-1-methoxybenzene
(d) 1-Methoxycyclohexene
(e) Ethyl isobutyl ether (f) Allyl vinyl ether

PROBLEM A mixture of diethyl ether, dipropyl ether, and ethyl propyl ether is formed in a $1: 1: 2$ ratio. 18-2
PROBLEM (a) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{O}^{-}+\mathrm{CH}_{3} \mathrm{Br}$ (b) $\mathrm{PhO}^{-}+\mathrm{CH}_{3} \mathrm{Br}$ (c) $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHO}^{-}+\mathrm{PhCH}_{2} \mathrm{Br}$
18-3 (d) $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH}_{2} \mathrm{O}^{-}+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Br}$
PROBLEM
18-4

PROBLEM (a) Either method (b) Williamson (c) Alkoxymercuration (d) Williamson 18-5
PROBLEM (a) Bromoethane $>$ 2-Bromopropane $>$ Bromobenzene
18-6 (b) Bromoethane $>$ Chloroethane $>$ 1-Iodopropene

PROBLEM (a)
18-7

(b)

PROBLEM Protonation of the oxygen atom, followed by E1 reaction
18-8
PROBLEM $\mathrm{Br}^{-}$and $\mathrm{I}^{-}$are better nucleophiles than $\mathrm{Cl}^{-}$.
18-9
PROBLEM
18-10

PROBLEM Epoxidation of cis-2-butene yields cis-2,3-epoxybutane, while epoxidation of trans-2-butene yields 18-11 trans-2,3-epoxybutane.
PROBLEM (a)

(b)


Chapter 18 -
(b)

18-12

PROBLEM (a) 1-Methylcyclohexene $+\mathrm{OsO}_{4}$; then $\mathrm{NaHSO}_{3}$
18-13 (b) 1-Methylcyclohexene + m-chloroperoxybenzoic acid, then $\mathrm{H}_{3} \mathrm{O}^{+}$
PROBLEM (a)

(b)

(c)

PROBLEM
18-15

PROBLEM (a) 2-Butanethiol (b) 2,2,6-Trimethyl-4-heptanethiol (c) 2-Cyclopentene-1-thiol
18-16 (d) Ethyl isopropyl sulfide (e) o-Di(methylthio)benzene (f) 3-(Ethylthio)cyclohexanone
PROBLEM (a) 1. $\mathrm{LiAlH}_{4} ; 2 . \mathrm{PBr}_{3} ; 3 .\left(\mathrm{H}_{2} \mathrm{~N}\right)_{2} \mathrm{C}=\mathrm{S} ; 4 . \mathrm{H}_{2} \mathrm{O}, \mathrm{NaOH}$
18-17 (b) 1. HBr ; 2. $\left(\mathrm{H}_{2} \mathrm{~N}\right)_{2} \mathrm{C}=\mathrm{S} ; 3 . \mathrm{H}_{2} \mathrm{O}, \mathrm{NaOH}$
PROBLEM 1,2-Epoxybutane
18-18
PROBLEM Acetyl chloride is more electrophilic than acetone.
18-69
PROBLEM
18-70

PROBLEM (a) Nucleophilic acyl substitution (b) Nucleophilic addition (c) Carbonyl condensation 18-71


Chapter 19 -
PROBLEM (a) 2-Methyl-3-pentanone (b) 3-Phenylpropanal (c) 2,6-Octanedione
19-1 (d) trans-2-Methylcyclohexanecarbaldehyde (e) 4-Hexenal
(f) cis-2,5-Dimethylcyclohexanone

PROBLEM (a)
19-2

(b)

(c)

(d)

(e)

(f)

PROBLEM (a) Dess-Martin periodinane (b) $1 . \mathrm{O}_{3} ; 2 \mathrm{Zn}$ (c) DIBAH
19-3 (d) 1. $\mathrm{BH}_{3}$, then $\mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{NaOH}$; 2. Dess-Martin periodinane
PROBLEM (a) $\mathrm{HgSO}_{4}, \mathrm{H}_{3} \mathrm{O}^{+}$(b) $1 . \mathrm{CH}_{3} \mathrm{COCl}, \mathrm{AlCl}_{3} ; 2 . \mathrm{Br}_{2}, \mathrm{FeBr}_{3}$ (c) 1. $\mathrm{Mg} ; 2 . \mathrm{CH}_{3} \mathrm{CHO} ; 3 . \mathrm{H}_{3} \mathrm{O}^{+} ; 4 . \mathrm{CrO}_{3}$
19-4 (d) 1. $\mathrm{BH}_{3} ; 2 . \mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{NaOH} ; 3 . \mathrm{CrO}_{3}$
PROBLEM
19-5

PROBLEM The electron-withdrawing nitro group in $p$-nitrobenzaldehyde polarizes the carbonyl group.
19-6
PROBLEM $\mathrm{CCl}_{3} \mathrm{CH}(\mathrm{OH})_{2}$
19-7
PROBLEM Labeled water adds reversibly to the carbonyl group.
19-8
PROBLEM The equilibrium is unfavorable for sterically hindered ketones.
19-9
PROBLEM
19-10

and


Chapter 19 -
and

PROBLEM The steps are the exact reverse of the forward reaction shown in Figure 19.7.
19-11
PROBLEM
19-12

PROBLEM (a) $\mathrm{H}_{2} / \mathrm{Pd}$ (b) $\mathrm{N}_{2} \mathrm{H}_{4}, \mathrm{KOH}$ (c) 1. $\mathrm{H}_{2} / \mathrm{Pd} ; 2 . \mathrm{N}_{2} \mathrm{H}_{4}, \mathrm{KOH}$
19-13
PROBLEM The mechanism is identical to that between a ketone and 2 equivalents of a monoalcohol, shown in
19-14 Figure 19.11.
PROBLEM
19-15

PROBLEM
(a) Cyclohexanone $+(\mathrm{Ph})_{3} \mathrm{P}=\mathrm{CHCH}_{3}$
(b) Cyclohexanecarbaldehyde $+(\mathrm{Ph})_{3} \mathrm{P}=\mathrm{CH}_{2}$

19-16
(e) Acetone $+(\mathrm{Ph})_{3} \mathrm{P}=\mathrm{CHCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$ (d) Acetone $+(\mathrm{Ph})_{3} \mathrm{P}=\mathrm{CHPh}$
(f) 2-Cyclohexenone $+(\mathrm{Ph})_{3} \mathrm{P}=\mathrm{CH}_{2}$

PROBLEM
19-17

$\beta$-Carotene
PROBLEM Intramolecular Cannizzaro reaction
19-18
PROBLEM Addition of the pro- $R$ hydrogen of NADH takes place on the Re face of pyruvate.
19-19

PROBLEM The -OH group adds to the Re face at C 2 , and -H adds to the Re face at C 3 , to yield $(2 R, 3 S)$-isocitrate.
19-20
PROBLEM
19-21


Chapter 19 -
PROBLEM (a) 3-Buten-2-one $+\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2}\right)_{2} \mathrm{CuLi}$ (b) 3-Methyl-2-cyclohexenone $+\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CuLi}$
19-22 (c) 4-tert-Butyl-2-cyclohexenone $+\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{2} \mathrm{CuLi}$ (d) Unsaturated ketone $+\left(\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}\right)_{2} \mathrm{CuLi}$
PROBLEM Look for appearance of either an alcohol or a saturated ketone in the product. 19-23
PROBLEM (a) $1715 \mathrm{~cm}^{-1}$ (b) $1685 \mathrm{~cm}^{-1}$ (c) $1750 \mathrm{~cm}^{-1}$ (d) $1705 \mathrm{~cm}^{-1}$ (e) $1715 \mathrm{~cm}^{-1}$ $\mathbf{1 9 - 2 4}$ (f) $1705 \mathrm{~cm}^{-1}$
PROBLEM (a) Different peaks due to McLafferty rearrangement
19-25 (b) Different peaks due to $\alpha$ cleavage and McLafferty rearrangement
(c) Different peaks due to McLafferty rearrangement

PROBLEM IR: $1750 \mathrm{~cm}^{-1}$; MS: 140, 84
19-26


Chapter 20 -
PROBLEM (a) 3-Methylbutanoic acid (b) 4-Bromopentanoic acid (c) 2-Ethylpentanoic acid
20-1 (d) cis-4-Hexenoic acid (e) 2,4-Dimethylpentanenitrile
(f) cis-1,3-Cyclopentanedicarboxylic acid

PROBLEM (a)
20-2

(c)

(d)

(e)

(f) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{CHCN}$


Chapter 20 -
PROBLEM Dissolve the mixture in ether, extract with aqueous NaOH , separate and acidify the aqueous layer,
20-3 and extract with ether.
PROBLEM 43\%
20-4
PROBLEM (a) $82 \%$ dissociation (b) $73 \%$ dissociation 20-5
PROBLEM Lactic acid is stronger because of the inductive effect of the -OH group.
20-6
PROBLEM The dianion is destabilized by repulsion between charges.
20-7
PROBLEM More reactive
20-8
PROBLEM (a) p-Methylbenzoic acid < Benzoic acid < $p$-Chlorobenzoic acid
20-9 (b) Acetic acid < Benzoic acid < $p$-Nitrobenzoic acid
PROBLEM (a) 1. $\mathrm{Mg} ; 2 . \mathrm{CO}_{2} ; 3 . \mathrm{H}_{3} \mathrm{O}^{+}$(b) 1. $\mathrm{Mg} ; 2 . \mathrm{CO}_{2} ; 3 . \mathrm{H}_{3} \mathrm{O}^{+}$or 1. $\mathrm{NaCN} ; 2 . \mathrm{H}_{3} \mathrm{O}^{+}$ 20-10
PROBLEM 1. NaCN; 2. $\mathrm{H}_{3} \mathrm{O}^{+}$; 3. $\mathrm{LiAlH}_{4}$ 20-11
PROBLEM 1. $\mathrm{PBr}_{3} ; 2 . \mathrm{NaCN} ; 3 . \mathrm{H}_{3} \mathrm{O}^{+} ; 4 . \mathrm{LiAlH}_{4}$
20-12


Chapter 20 -
20-12
PROBLEM (a) Propanenitrile $+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{MgBr}$, then $\mathrm{H}_{3} \mathrm{O}^{+}$(b) $p$-Chlorobenzonitrile $+\mathrm{CH}_{3} \mathrm{MgBr}$, then $\mathrm{H}_{3} \mathrm{O}^{+}$ 20-13
PROBLEM 1. NaCN ; 2. $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{MgBr}$, then $\mathrm{H}_{3} \mathrm{O}^{+}$ 20-14


Chapter 20 -
PROBLEM A carboxylic acid has a very broad -OH absorption at $2500-3300 \mathrm{~cm}^{-1}$.
20-15
PROBLEM 4-Hydroxycyclohexanone: $\mathrm{H}-\mathrm{C}-\mathrm{O}$ absorption near $4 \delta$ in the ${ }^{1} \mathrm{H}$ spectrum and $\mathrm{C}=\mathrm{O}$ absorption near 20-16 $210 \delta$ in the ${ }^{13} \mathrm{C}$ spectrum. Cyclopentanecarboxylic acid: $-\mathrm{CO}_{2} \mathrm{H}$ absorption near $12 \delta$ in the ${ }^{1} \mathrm{H}$ spectrum and $-\mathrm{CO}_{2} \mathrm{H}$ absorption near $170 \delta$ in the ${ }^{13} \mathrm{C}$ spectrum.

Chapter 21
PROBLEM (a) 4-Methylpentanoyl chloride (b) Cyclohexylacetamide (c) Isopropyl 2-methylpropanoate
21-1 (d) Benzoic anhydride
(e) Isopropyl cyclopentanecarboxylate
(f) Cyclopentyl 2-methylpropanoate (g) N -Methyl-4-pentenamide
(h) (R)-2-Hydroxypropanoyl phosphate (i) Ethyl 2,3-dimethyl-2-butenethioate

PROBLEM
(a) $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{C}_{6} \mathrm{H}_{5}$
(b)
$\mathrm{H}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CON}\left(\mathrm{CH}_{3}\right) \mathrm{CH}_{2} \mathrm{CH}_{3}$
(c) $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCH}_{2} \mathrm{CH}\left(\mathrm{CH}_{3}\right) \mathrm{COCl}$

21-2
(d)

(e)

(f)

(g)

(h)


Chapter 20 -
21-2
(d)

(e)

(f)

(g)

(h)

PROBLEM
21-3

PROBLEM (a) Acetyl chloride $>$ Methyl acetate $>$ Acetamide
21-4 (b) Hexafluoroisopropyl acetate $>2,2,2$-Trichloroethyl acetate $>$ Ethyl acetate
PROBLEM
(a) $\mathrm{CH}_{3} \mathrm{CO}_{2}^{-} \mathrm{Na}^{+}$
(b) $\mathrm{CH}_{3} \mathrm{CONH}_{2}$
(c) $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{CH}_{3}+\mathrm{CH}_{3} \mathrm{CO}_{2}^{-} \mathrm{Na}^{+}$
(d) $\mathrm{CH}_{3} \mathrm{CONHCH}_{3}$

21-5
PROBLEM

PROBLEM (a) Acetic acid + 1-butanol (b) Butanoic acid + methanol
21-7 (c) Cyclopentanecarboxylic acid + isopropyl alcohol
PROBLEM
21-8


Chapter 20 -
PROBLEM (a) Propanoyl chloride + methanol (b) Acetyl chloride + ethanol
21-9 (c) Benzoyl chloride + ethanol
PROBLEM Benzoyl chloride + cyclohexanol
21-10
PROBLEM This is a typical nucleophilic acyl substitution reaction, with morpholine as the nucleophile and
21-11 chloride as the leaving group.
PROBLEM (a) Propanoyl chloride + methylamine (b) Benzoyl chloride + diethylamine
21-12 (c) Propanoyl chloride + ammonia
PROBLEM (a) Benzoyl chloride $+\left[\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}\right]_{2} \mathrm{CuLi}$, or 2-methylpropanoyl chloride $+\mathrm{Ph}_{2} \mathrm{CuLi}$
21-13 (b) 2-Propenoyl chloride $+\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2}\right)_{2} \mathrm{CuLi}$, or butanoyl chloride $+\left(\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}\right)_{2} \mathrm{CuLi}$
PROBLEM This is a typical nucleophilic acyl substitution reaction, with $p$-hydroxyaniline as the nucleophile


Chapter 20 -
21-14 and acetate ion as the leaving group.
PROBLEM Monomethyl ester of benzene-1,2-dicarboxylic acid
21-15
PROBLEM Reaction of a carboxylic acid with an alkoxide ion gives the carboxylate ion.
21-16
PROBLEM LiAlH 4 gives $\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}$; DIBAH gives $\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CHO}$.
21-17
PROBLEM (a) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}\left(\mathrm{CH}_{3}\right) \mathrm{CH}_{2} \mathrm{OH}+\mathrm{CH}_{3} \mathrm{OH}$ (b) $\mathrm{PhOH}+\mathrm{PhCH}_{2} \mathrm{OH}$
21-18
PROBLEM (a) Ethyl benzoate $+2 \mathrm{CH}_{3} \mathrm{MgBr}$ (b) Ethyl acetate +2 PhMgBr
21-19 (c) Ethyl pentanoate $+2 \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{MgBr}$
PROBLEM (a) $\mathrm{H}_{2} \mathrm{O}, \mathrm{NaOH}$ (b) Benzoic acid $+\mathrm{LiAlH}_{4}$ (c) $\mathrm{LiAlH}_{4}$
21-20
PROBLEM 1. Mg; 2. $\mathrm{CO}_{2}$, then $\mathrm{H}_{3} \mathrm{O}^{+}$; 3. $\mathrm{SOCl}_{2}$; 4. $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}$; 5. $\mathrm{LiAlH}_{4}$
21-21
PROBLEM
21-22

$\downarrow$

PROBLEM (a)
21-23

(b)

(c)

PROBLEM
21-24


Chapter 20 -
$\downarrow$

PROBLEM (a)
21-23

(b)

(c)

PROBLEM
21-24

PROBLEM (a) Ester (b) Acid chloride (c) Carboxylic acid (d) Aliphatic ketone or cyclohexanone
21-25
PROBLEM (a) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$ and other possibilities (b) $\mathrm{CH}_{3} \mathrm{CON}\left(\mathrm{CH}_{3}\right)_{2}$
21-26 (c) $\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCOCl}$ or $\mathrm{H}_{2} \mathrm{C}=\mathrm{C}\left(\mathrm{CH}_{3}\right) \mathrm{COCl}$

Chapter 22
PROBLEM (a)
22-1

(b)

(c)

(d)

(e)

(f)

PROBLEM
(a) 4
(b) 3
(c) 3
(d) 2
(e) $4 \quad$ (f) 5

22-2
PROBLEM
22-3

PROBLEM Acid-catalyzed formation of an enol is followed by deuteronation of the enol double bond and
22-4 dedeuteronation of oxygen.
PROBLEM 1. $\mathrm{Br}_{2}$; 2. Pyridine, heat
22-5
PROBLEM The intermediate $\alpha$-bromo acid bromide undergoes a nucleophilic acyl substitution reaction with
22-6 methanol to give an $\alpha$-bromo ester.
PROBLEM (a) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHO}$ (b) $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCOCH}_{3}$
(c) $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}$ (d) $\mathrm{PhCONH}_{2}$ (e) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CN}$


Chapter 20 -
22-7 (f) $\mathrm{CH}_{3} \mathrm{CON}\left(\mathrm{CH}_{3}\right)_{2}$
PROBLEM
22-8 ${ }^{-}: \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{N}: \longleftrightarrow \mathrm{H}_{2} \mathrm{C}=\mathrm{C}=\ddot{\mathrm{N}}:^{-}$
PROBLEM Acid is regenerated, but base is used stoichiometrically.
22-9
PROBLEM (a) 1. $\mathrm{Na}^{+-} \mathrm{OEt}$; 2. $\mathrm{PhCH}_{2} \mathrm{Br}$; 3. $\mathrm{H}_{3} \mathrm{O}^{+}$
22-10 (b) 1. $\mathrm{Na}^{+-}{ }^{-} \mathrm{OEt}$; 2. $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}$; 3. $\mathrm{Na}^{+{ }^{-}} \mathrm{OEt}$; 4. $\mathrm{CH}_{3} \mathrm{Br} ;$ 5. $\mathrm{H}_{3} \mathrm{O}^{+}$
(c) 1. $\mathrm{Na}^{+-} \mathrm{OEt}$; 2. $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCH}_{2} \mathrm{Br} ; 3 . \mathrm{H}_{3} \mathrm{O}^{+}$


Chapter 20 -
PROBLEM Malonic ester has only two acidic hydrogens to be replaced. 22-11
PROBLEM 1. $\mathrm{Na}^{+-}$OEt; 2. $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCH}_{2} \mathrm{Br}$; 3. $\mathrm{Na}^{+-}{ }^{-} \mathrm{OEt}$; 4. $\mathrm{CH}_{3} \mathrm{Br} ; 5 . \mathrm{H}_{3} \mathrm{O}^{+}$ 22-12
PROBLEM (a) $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCH}_{2} \mathrm{Br}$ (b) $\mathrm{PhCH}_{2} \mathrm{CH}_{2} \mathrm{Br}$ 22-13
PROBLEM None can be prepared.
22-14
PROBLEM 1.2 $\mathrm{Na}^{+-} \mathrm{OEt}$; 2. $\mathrm{BrCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}$; 3. $\mathrm{H}_{3} \mathrm{O}^{+}$
22-15
PROBLEM (a) Alkylate phenylacetone with $\mathrm{CH}_{3} \mathrm{I}$ (b) Alkylate pentanenitrile with $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{I}$
22-16 (c) Alkylate cyclohexanone with $\mathrm{H}_{2} \mathrm{C}=\mathrm{CHCH}_{2} \mathrm{Br}$ (d) Alkylate cyclohexanone with excess $\mathrm{CH}_{3} \mathrm{I}$
(e) Alkylate $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COCH}_{2} \mathrm{CH}_{3}$ with $\mathrm{CH}_{3} \mathrm{I}$ (f) Alkylate methyl 3-methylbutanoate with $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{I}$

Chapter 23


Chapter 20 -
Chapter 23

PROBLEM (a)
23-1

(b)

(c)

PROBLEM The reverse reaction is the exact opposite of the forward reaction shown in Figure 23.2.

23-2

PROBLEM (a)
23-3

(b)

(c)

PROBLEM
23-4

PROBLEM (a) Not an aldol product (b) 3-Pentanone
23-5
PROBLEM 1. NaOH; 2. $\mathrm{LiAlH}_{4} ; 3 . \mathrm{H}_{2} / \mathrm{Pd}$
23-6
PROBLEM
23-7

PROBLEM
(a) $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}+\mathrm{CH}_{3} \mathrm{COCH}_{3}$
(b) Not easily prepared
(c) Not easily prepared

23-8
PROBLEM The $\mathrm{CH}_{2}$ position between the two carbonyl groups is so acidic that it is completely deprotonated to
23-9 give a stable enolate ion.
PROBLEM
23-10

PROBLEM (a)
23-11

(b)

(c)

PROBLEM The cleavage reaction is the exact reverse of the forward reaction.
23-12
PROBLEM
23-13

PROBLEM
23-14

PROBLEM
23-15

PROBLEM (a)
23-16

(b) $\left(\mathrm{CH}_{3} \mathrm{CO}\right)_{2} \mathrm{CHCH}_{2} \mathrm{CH}_{2} \mathrm{CN}$
(c)

PROBLEM (a)
23-17

(b)

PROBLEM $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COCH}=\mathrm{CH}_{2}+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NO}_{2}$ 23-18
PROBLEM (a)

(b)

(c)

PROBLEM (a) Cyclopentanone enamine + propenenitrile
23-20 (b) Cyclohexanone enamine + methyl propenoate
PROBLEM
23-21


Chapter 20 -
(b)

(c)

PROBLEM (a) Cyclopentanone enamine + propenenitrile
23-20 (b) Cyclohexanone enamine + methyl propenoate
PROBLEM
23-21

PROBLEM 2,5,5-Trimethyl-1,3-cyclohexanedione + 1-penten-3-one
23-22


Chapter 24 -
PROBLEM (a) $N$-Methylethylamine (b) Tricyclohexylamine (c) $N$-Ethyl- $N$-methylcyclohexylamine
24-1 (d) $N$-Methylpyrrolidine
(e) Diisopropylamine (f) 1,3-Butanediamine

PROBLEM
(a) $\left[\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}\right]_{3} \mathrm{~N}$
(b) $\left(\mathrm{H}_{2} \mathrm{C}=\mathrm{CHCH}_{2}\right)_{3} \mathrm{~N}$
(c)

(d)

(e)

(f)

PROBLEM (a)

(b)

(c)

(d)

PROBLEM (a) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2}$
(b) NaOH
(c) $\mathrm{CH}_{3} \mathrm{NHCH}_{3}$

24-4
PROBLEM Propylamine is stronger; benzylamine $\mathrm{p} K_{\mathrm{b}}=4.67$; propylamine $\mathrm{p} K_{\mathrm{b}}=3.29$
24-5
PROBLEM (a) $p$-Nitroaniline $<p$-Aminobenzaldehyde $<p$-Bromoaniline
24-6 (b) $p$-Aminoacetophenone $<p$-Chloroaniline $<p$-Methylaniline
(c) $p$-(Trifluoromethyl)aniline $<p$-(Fluoromethyl)aniline $<p$-Methylaniline


Chapter 24 -
PROBLEM Pyrimidine is essentially 100\% neutral (unprotonated).
24-7
PROBLEM (a) Propanenitrile or propanamide (b) $N$-Propylpropanamide (c) Benzonitrile or benzamide
24-8 (d) $N$-Phenylacetamide
PROBLEM The reaction takes place by two nucleophilic acyl substitution reactions.
24-9
PROBLEM
24-10

PROBLEM (a) Ethylamine + acetone, or isopropylamine + acetaldehyde (b) Aniline + acetaldehyde
24-11 (c) Cyclopentylamine + formaldehyde, or methylamine + cyclopentanone
PROBLEM
24-12

PROBLEM (a) 4,4-Dimethylpentanamide or 4,4-dimethylpentanoyl azide
24-13 (b) $p$-Methylbenzamide or $p$-methylbenzoyl azide
PROBLEM (a) 3-Octene and 4-octene (b) Cyclohexene (c) 3-Heptene (d) Ethylene and cyclohexene 24-14
PROBLEM $\mathrm{H}_{2} \mathrm{C}=\mathrm{CHCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{~N}\left(\mathrm{CH}_{3}\right)_{2}$


24-15 -
PROBLEM 1. $\mathrm{HNO}_{3}, \mathrm{H}_{2} \mathrm{SO}_{4} ; 2 . \mathrm{H}_{2} / \mathrm{PtO}_{2}$; 3. $\left(\mathrm{CH}_{3} \mathrm{CO}\right)_{2} \mathrm{O} ; 4 . \mathrm{HOSO}_{2} \mathrm{Cl}$; 5. aminothiazole; 6. $\mathrm{H}_{2} \mathrm{O}, \mathrm{NaOH}$ 24-16
PROBLEM (a) 1. $\mathrm{HNO}_{3}, \mathrm{H}_{2} \mathrm{SO}_{4} ; 2 . \mathrm{H}_{2} / \mathrm{PtO}_{2} ; 3.2 \mathrm{CH}_{3} \mathrm{Br}$
24-17 (b) 1. $\mathrm{HNO}_{3}, \mathrm{H}_{2} \mathrm{SO}_{4} ; 2 . \mathrm{H}_{2} / \mathrm{PtO}_{2}$; 3. $\left(\mathrm{CH}_{3} \mathrm{CO}\right)_{2} \mathrm{O} ; 4 . \mathrm{Cl}_{2} ; 5 . \mathrm{H} 2 \mathrm{O}, \mathrm{NaOH}$
(c) 1. $\mathrm{HNO}_{3}, \mathrm{H}_{2} \mathrm{SO}_{4} ; 2 . \mathrm{Cl}_{2}, \mathrm{FeCl}_{3} ; 3 . \mathrm{SnCl}_{2}$
(d) 1. $\mathrm{HNO}_{3}, \mathrm{H}_{2} \mathrm{SO}_{4} ; 2 . \mathrm{H}_{2} / \mathrm{PtO}_{2}$; 3. $\left(\mathrm{CH}_{3} \mathrm{CO}\right)_{2} \mathrm{O} ; 4.2 \mathrm{CH}_{3} \mathrm{Cl}, \mathrm{AlCl}_{3} ; 5 . \mathrm{H}_{2} \mathrm{O}, \mathrm{NaOH}$


24-15 -
PROBLEM (a) 1. $\mathrm{CH}_{3} \mathrm{Cl}, \mathrm{AlCl}_{3} ; 2 . \mathrm{HNO}_{3}, \mathrm{H}_{2} \mathrm{SO}_{4} ; 3 . \mathrm{SnCl}_{2} ; 4 . \mathrm{NaNO}_{2}, \mathrm{H}_{2} \mathrm{SO}_{4} ; 5 . \mathrm{CuBr} ; 6 . \mathrm{KMnO}_{4}, \mathrm{H}_{2} \mathrm{O}$
24-18 (b) 1. $\mathrm{HNO}_{3}, \mathrm{H}_{2} \mathrm{SO}_{4} ; 2 . \mathrm{Br}_{2}, \mathrm{FeBr}_{3} ; 3 . \mathrm{SnCl}_{2}, \mathrm{H}_{3} \mathrm{O}^{+} ; 4 . \mathrm{NaNO}_{2}, \mathrm{H}_{2} \mathrm{SO}_{4} ; 5 . \mathrm{CuCN} ; 6 . \mathrm{H}_{3} \mathrm{O}^{+}$
(c) 1. $\mathrm{HNO}_{3}, \mathrm{H}_{2} \mathrm{SO}_{4} ; 2 . \mathrm{Cl}_{2}, \mathrm{FeCl}_{3} ; 3 . \mathrm{SnCl}_{2} ; 4 . \mathrm{NaNO}_{2}, \mathrm{H}_{2} \mathrm{SO}_{4} ; 5 . \mathrm{CuBr}$
(d) 1. $\mathrm{CH}_{3} \mathrm{Cl}, \mathrm{AlCl}_{3} ; 2 . \mathrm{HNO}_{3}, \mathrm{H}_{2} \mathrm{SO}_{4} ; 3 . \mathrm{SnCl}_{2} ; 4 . \mathrm{NaNO}_{2}, \mathrm{H}_{2} \mathrm{SO}_{4} ; 5 . \mathrm{CuCN} ; 6 . \mathrm{H}_{3} \mathrm{O}^{+}$


24-15 -
(e) 1. $\mathrm{HNO}_{3}, \mathrm{H}_{2} \mathrm{SO}_{4} ; 2 . \mathrm{H}_{2} / \mathrm{PtO}_{2}$; 3. $\left(\mathrm{CH}_{3} \mathrm{CO}\right)_{2} \mathrm{O}$; $4.2 \mathrm{Br}_{2}$; 5. $\mathrm{H}_{2} \mathrm{O}, \mathrm{NaOH} ; 6 . \mathrm{NaNO}_{2}, \mathrm{H}_{2} \mathrm{SO}_{4}$; 7. CuBr


24-15 -
PROBLEM 1. $\mathrm{HNO}_{3}, \mathrm{H}_{2} \mathrm{SO}_{4} ; 2 . \mathrm{SnCl}_{2}$; 3a. 2 equiv. $\mathrm{CH}_{3} \mathrm{I} ; 3 \mathrm{~b} . \mathrm{NaNO}_{2}, \mathrm{H}_{2} \mathrm{SO}_{4} ; 4$. product of $3 \mathrm{a}+$ product of 3 b 24-19
PROBLEM
24-20

PROBLEM 4.1\% protonated
24-21
PROBLEM Attack at C2:
24-22

Unfavorable

PROBLEM The side-chain nitrogen is more basic than the ring nitrogen.

24-23
PROBLEM Reaction at C 2 is disfavored because the aromaticity of the benzene ring is lost. 24-24

PROBLEM $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCOCH}_{3} \rightarrow\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH}\left(\mathrm{NH}_{2}\right) \mathrm{CH}_{3}$ 24-25

Chapter 25
PROBLEM (a) Aldotetrose (b) Ketopentose (c) Ketohexose (d) Aldopentose
25-1
PROBLEM (a) $S$ (b) $R$ (c) $S$
25-2
PROBLEM A, B, and C are the same.
25-3
PROBLEM
25-4
PROBLEM
25-5

PROBLEM (a) L-Erythrose; $2 S, 3 S$ (b) D-Xylose; $2 R, 3 S, 4 R$ (c) D-Xylulose; $3 S, 4 R$
25-6
PROBLEM
25-7

PROBLEM (a)
25-8

(c)

PROBLEM 16 D and 16 L aldoheptoses
25-9
PROBLEM
25-10

PROBLEM
25-11

PROBLEM
25-12

$\alpha$-D-Fructofuranose


PROBLEM -
25-13

$\beta$-D-Galactopyranose

$\beta$-D-Mannopyranose

PROBLEM
25-14
PROBLEM $\alpha$-D-Allopyranose
25-15
PROBLEM
25-16

PROBLEM D-Galactitol has a plane of symmetry and is a meso compound, whereas D-glucitol is chiral.
25-17
PROBLEM The -CHO end of L-gulose corresponds to the $-\mathrm{CH}_{2} \mathrm{OH}$ end of D -glucose after reduction.
25-18
PROBLEM D-Allaric acid has a symmetry plane and is a meso compound, but D-glucaric acid is chiral.
25-19
PROBLEM D-Allose and D-galactose yield meso aldaric acids; the other six D-hexoses yield optically active
25-20 aldaric acids.
PROBLEM D-Allose + D-altrose
25-21
PROBLEM L-Xylose
25-22
PROBLEM D-Xylose and D-lyxose
25-23


PROBLEM -
25-24

PROBLEM (a) The hemiacetal ring is reduced. (b) The hemiacetal ring is oxidized.
25-25 (c) All hydroxyl groups are acetylated.
Chapter 26
PROBLEM Aromatic: Phe, Tyr, Trp, His; sulfur-containing: Cys, Met; alcohols: Ser, Thr; hydrocarbon side
26-1 chains: Ala, Ile, Leu, Val, Phe
PROBLEM The sulfur atom in the $-\mathrm{CH}_{2} \mathrm{SH}$ group of cysteine makes the side chain higher in ranking than the
26-2 $-\mathrm{CO}_{2} \mathrm{H}$ group.
PROBLEM
26-3

L-Threonine

Diastereomers of L-threonine

PROBLEM Net positive at $\mathrm{pH}=5.3$; net negative at $\mathrm{pH}=7.3$
26-4
PROBLEM (a) Start with 3-phenylpropanoic acid: 1. $\mathrm{Br}_{2}, \mathrm{PBr}_{3} ; 2 . \mathrm{NH}_{3}$
26-5 (b) Start with 3-methylbutanoic acid: 1. $\mathrm{Br}_{2}, \mathrm{PBr}_{3} ; 2 . \mathrm{NH}_{3}$
PROBLEM
26-6
(a) $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCH}_{2} \mathrm{Br}$
(b)

(c)

(d) $\mathrm{CH}_{3} \mathrm{SCH}_{2} \mathrm{CH}_{2} \mathrm{Br}$

PROBLEM
26-7

PROBLEM Val-Tyr-Gly (VYG), Tyr-Gly-Val (YGV), Gly-Val-Tyr (GVY), Val-Gly-Tyr (VGY), Tyr-Val-Gly (YVG), Gly-
26-8 Tyr-Val (GYV)
PROBLEM
26-9

PROBLEM
26-10

PROBLEM
26-11


PROBLEM -
PROBLEM
26-10

PROBLEM
26-11

PROBLEM Trypsin: Asp-Arg + Val-Tyr-Ile-His-Pro-Phe
26-12 Chymotrypsin: Asp-Arg-Val-Tyr + Ile-His-Pro-Phe
PROBLEM Methionine
26-13
PROBLEM
26-14

PROBLEM (a) Arg-Pro-Leu-Gly-Ile-Val (b) Val-Met-Trp-Asp-Val-Leu (VMWNVL)
26-15
PROBLEM This is a typical nucleophilic acyl substitution reaction, with the amine of the amino acid as the
26-16 nucleophile and tert-butyl carbonate as the leaving group. The tert-butyl carbonate then loses $\mathrm{CO}_{2}$ and gives tert-butoxide, which is protonated.
PROBLEM (1) Protect the amino group of leucine.
26-17 (2) Protect the carboxylic acid group of alanine.
(3) Couple the protected amino acids with DCC.
(4) Remove the leucine protecting group.
(5) Remove the alanine protecting group.

PROBLEM (a) Lyase (b) Hydrolase (c) Oxidoreductase
26-18


Chapter 27 -
PROBLEM $\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{18} \mathrm{CO}_{2} \mathrm{CH}_{2}\left(\mathrm{CH}_{2}\right)_{30} \mathrm{CH}_{3}$
27-1
PROBLEM Glyceryl tripalmitate is higher melting.
27-2
PROBLEM $\left[\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{7} \mathrm{CH}=\mathrm{CH}\left(\mathrm{CH}_{2}\right)_{7} \mathrm{CO}_{2}^{-}\right] 2 \mathrm{Mg}^{2+}$
27-3
PROBLEM Glyceryl dioleate monopalmitate $\rightarrow$ glycerol +2 sodium oleate + sodium palmitate
27-4
PROBLEM
27-5

PROBLEM The pro- $S$ hydrogen is cis to the $-\mathrm{CH}_{3}$ group; the pro- $R$ hydrogen is trans.
27-6

PROBLEM (a)
27-7

(b)

$\gamma$-Bisabolene

PROBLEM (a)
27-8

(b)

PROBLEM
27-9

PROBLEM Three methyl groups are removed, the side-chain double bond is reduced, and the double bond in $\mathbf{2 7 - 1 0}$ the $B$ ring is migrated.

Chapter 28

PROBLEM
28-2

PROBLEM (5') ACGGATTAGCC (3')
28-3


PROBLEM -
28-4

PROBLEM (3') CUAAUGGCAU (5')
28-5
PROBLEM (5') ACTCTGCGAA (3')
28-6
PROBLEM (a) GCU, GCC, GCA, GCG (b) UUU, UUC (c) UUA, UUG, CUU, CUC, CUA, CUG (d) UAU, UAC
28-7
PROBLEM
28-8
PROBLEM Leu-Met-Ala-Trp-Pro-Stop
28-9
PROBLEM (5') TTA-GGG-CCA-AGC-CAT-AAG (3')
28-10
PROBLEM The cleavage is an $\mathrm{S}_{\mathrm{N}} 1$ reaction that occurs by protonation of the oxygen atom followed by loss of
28-11 the stable triarylmethyl carbocation.
PROBLEM
28-12


Chapter 29 -
PROBLEM $\mathrm{HOCH}_{2} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{OH}+\mathrm{ATP} \rightarrow \mathrm{HOCH}_{2} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{OPO}_{3}{ }^{2-}+\mathrm{ADP}$
29-1
PROBLEM Caprylyl CoA $\rightarrow$ Hexanoyl CoA $\rightarrow$ Butyryl CoA $\rightarrow 2$ Acetyl CoA 29-2
PROBLEM (a) 8 acetyl CoA; 7 passages (b) 10 acetyl CoA; 9 passages
29-3
PROBLEM The dehydration is an E1cB reaction.
29-4
PROBLEM At C2, C4, C6, C8, and so forth
29-5
PROBLEM The Si face
29-6
PROBLEM Steps 7 and 10
29-7
PROBLEM Steps 1, 3: Phosphate transfers; steps 2, 5, 8: isomerizations; step 4: retro-aldol reaction; step 29-8 5: oxidation and nucleophilic acyl substitution; steps 7, 10: phosphate transfers; step 9: E1cB dehydration
PROBLEM C 1 and C 6 of glucose become $-\mathrm{CH}_{3}$ groups; C 3 and C 4 become $\mathrm{CO}_{2}$. 29-9
PROBLEM Citrate and isocitrate 29-10
PROBLEM E1cB elimination of water, followed by conjugate addition 29-11
PROBLEM pro- $R$; anti geometry


Chapter 29 -
PROBLEM The reaction occurs by two sequential nucleophilic acyl substitutions, the first by a cysteine residue
29-13 in the enzyme, with phosphate as leaving group, and the second by hydride donation from NADH, with the cysteine residue as leaving group.
PROBLEM Initial imine formation between PMP and $\alpha$-ketoglutarate is followed by double-bond 29-14 rearrangement to an isomeric imine and hydrolysis.
PROBLEM $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCH}_{2} \mathrm{COCO}_{2}{ }^{-}$
29-15
PROBLEM Asparagine
29-16
Chapter 30
PROBLEM Ethylene: $\psi_{1}$ is the HOMO and $\psi_{2}{ }^{*}$ is the LUMO in the ground state; $\psi_{2}{ }^{*}$ is the HOMO and there is no
30-1 LUMO in the excited state. 1,3-Butadiene: $\psi_{2}$ is the HOMO and $\psi_{3}{ }^{*}$ is the LUMO in the ground state; $\psi_{3}{ }^{*}$ is the HOMO and $\psi_{4}{ }^{*}$ is the LUMO in the excited state.
PROBLEM Disrotatory: cis-5,6-dimethyl-1,3-cyclohexadiene;
30-2 conrotatory: trans-5,6-dimethyl-1,3-cyclohexadiene. Disrotatory closure occurs.
PROBLEM The more stable of two allowed products is formed.
30-3
PROBLEM trans-5,6-Dimethyl-1,3-cyclohexadiene; cis-5,6-dimethyl-1,3-cyclohexadiene 30-4
PROBLEM cis-3,6-Dimethylcyclohexene; trans-3,6-dimethylcyclohexene 30-5
PROBLEM A [6 + 4] suprafacial cycloaddition 30-6
PROBLEM An antarafacial [1,7] sigmatropic rearrangement
30-7
PROBLEM
30-8


Chapter 29 -
PROBLEM A series of [1,5] hydrogen shifts occur. 30-9
PROBLEM o-(1-Methylallyl)phenol
30-10
PROBLEM Claisen rearrangement is followed by a Cope rearrangement. 30-11
PROBLEM (a) Conrotatory (b) Disrotatory (c) Suprafacial (d) Antarafacial (e) Suprafacial 30-12


Chapter 31 -
PROBLEM $\mathrm{H}_{2} \mathrm{C}=\mathrm{CHCO}_{2} \mathrm{CH}_{3}<\mathrm{H}_{2} \mathrm{C}=\mathrm{CHCl}<\mathrm{H}_{2} \mathrm{C}=\mathrm{CHCH}_{3}<\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}-\mathrm{C}_{6} \mathrm{H}_{5}$ 31-1
PROBLEM $\mathrm{H}_{2} \mathrm{C}=\mathrm{CHCH}_{3}<\mathrm{H}_{2} \mathrm{C}=\mathrm{CHC}_{6} \mathrm{H}_{5}<\mathrm{H}_{2} \mathrm{C}=\mathrm{CHC} \equiv \mathrm{N}$ 31-2
PROBLEM The intermediate is a resonance-stabilized benzylic carbanion, $\mathrm{Ph}-\overline{\ddot{\mathrm{C}}} \mathrm{HR}$ 31-3
PROBLEM The polymer has no chirality centers. 31-4
PROBLEM The polymers are racemic and have no optical rotation. 31-5


PROBLEM
 31-6
 -
PROBLEM
31-7
Polybutadiene chain

Polystyrene chain
PROBLEM
31-8


PROBLEM -
31-9

PROBLEM Vestenamer: ADMET polymerization of 1,9-decadiene or ROMP of cyclooctene; Norsorex: ROMP of 31-10 norbornene.


PROBLEM -
31-12

are a class of saturated hydrocarbons with the general formula $\mathrm{C}_{n} \mathrm{H}_{2 n+2}$. They contain no functional groups, are relatively inert, and can be either straight-chain (normal) or branched. Alkanes are named by a series of IUPAC rules of nomenclature. Compounds that have the same chemical formula but different


1. CHAPTER 1 Essential Ideas -
Figure 1.1 Chemical substances and processes are essential for our existence, providing sustenance, keeping us clean and healthy, fabricating electronic devices, enabling transportation, and much more. (credit "left": modification of work by "vxla"/Flickr; credit "left middle": modification of work by "the Italian voice"/Flickr; credit "right middle": modification of work by Jason Trim; credit "right": modification of work by "gosheshe"/Flickr)


2. CHAPTER OUTLINE - 2.1. Chemistry in Context
1.2 Phases and Classification of Matter
1.3 Physical and Chemical Properties


2. CHAPTER OUTLINE - 2.2. Measurements
1.5 Measurement Uncertainty, Accuracy, and Precision
1.6 Mathematical Treatment of Measurement Results

INTRODUCTION Your alarm goes off and, after hitting "snooze" once or twice, you pry yourself out of bed. You make a cup of coffee to help you get going, and then you shower, get dressed, eat breakfast, and check your phone for messages. On your way to school, you stop to fill your car's gas tank, almost making you late for the first day of chemistry class. As you find a seat in the classroom, you read the question projected on the screen: "Welcome to class! Why should we study chemistry?"

Do you have an answer? You may be studying chemistry because it fulfills an academic requirement, but if you consider your daily activities, you might find chemistry interesting for other reasons. Most everything you do and encounter during your day involves chemistry. Making coffee, cooking eggs, and toasting bread involve chemistry. The products you use-like soap and shampoo, the fabrics you wear, the electronics that keep you connected to your world, the gasoline that propels your car-all of these and more involve chemical substances and processes. Whether you are aware or not, chemistry is part of your everyday world. In this course, you will learn many of the essential principles underlying the chemistry of modern-day life.


3. LEARNING OBJECTIVES -
By the end of this section, you will be able to:

- Outline the historical development of chemistry
- Provide examples of the importance of chemistry in everyday life
- Describe the scientific method
- Differentiate among hypotheses, theories, and laws
- Provide examples illustrating macroscopic, microscopic, and symbolic domains

Throughout human history, people have tried to convert matter into more useful forms. Our Stone Age
ancestors chipped pieces of flint into useful tools and carved wood into statues and toys. These endeavors involved changing the shape of a substance without changing the substance itself. But as our knowledge increased, humans began to change the composition of the substances as well-clay was converted into pottery, hides were cured to make garments, copper ores were transformed into copper tools and weapons, and grain was made into bread.

Humans began to practice chemistry when they learned to control fire and use it to cook, make pottery, and smelt metals. Subsequently, they began to separate and use specific components of matter. A variety of drugs such as aloe, myrrh, and opium were isolated from plants. Dyes, such as indigo and Tyrian purple, were extracted from plant and animal matter. Metals were combined to form alloys-for example, copper and tin were mixed together to make bronze-and more elaborate smelting techniques produced iron. Alkalis were extracted from ashes, and soaps were prepared by combining these alkalis with fats. Alcohol was produced by fermentation and purified by distillation.

Attempts to understand the behavior of matter extend back for more than 2500 years. As early as the sixth century BC, Greek philosophers discussed a system in which water was the basis of all things. You may have heard of the Greek postulate that matter consists of four elements: earth, air, fire, and water. Subsequently, an amalgamation of chemical technologies and philosophical speculations was spread from Egypt, China, and the eastern Mediterranean by alchemists, who endeavored to transform "base metals" such as lead into "noble metals" like gold, and to create elixirs to cure disease and extend life (Figure 1.2).


3. LEARNING OBJECTIVES -
FIGURE 1.2 (a) This portrayal shows an alchemist's workshop circa 1580. Although alchemy made some useful contributions to how to manipulate matter, it was not scientific by modern standards. (b) While the equipment used by Alma Levant Hayden in this 1952 picture might not seem as sleek as you might find in a lab today, her approach was highly methodical and carefully recorded. A department head at the FDA, Hayden is most famous for exposing an aggressively marketed anti-cancer drug as nothing more than an unhelpful solution of common substances. (credit a: Chemical Heritage Foundation; b: NIH History Office)

From alchemy came the historical progressions that led to modern chemistry: the isolation of drugs from natural sources, such as plants and animals. But while many of the substances extracted or processed from those natural sources were critical in the treatment of diseases, many were scarce. For example, progesterone, which is critical to women's health, became available as a medicine in 1935, but its animal sources produced extremely small quantities, limiting its availability and increasing its expense. Likewise, in the 1940s, cortisone came into use to treat arthritis and other disorders and injuries, but it took a 36-step process to synthesize. Chemist Percy Lavon Julian turned to a more plentiful source: soybeans. Previously, Julian had developed a lab to isolate soy protein, which was used in firefighting among other applications. He focused on using the soy sterols-substances mostly used in plant membranes-and was able to quickly produce progesterone and later testosterone and other hormones. He later developed a process to do the same for cortisone, and laid the groundwork for modern drug design. Since soybeans and similar plant sources were extremely plentiful, the drugs soon became widely available, saving many lives.


4. Chemistry: The Central Science -
Chemistry is sometimes referred to as "the central science" due to its interconnectedness with a vast array of other STEM disciplines (STEM stands for areas of study in the science, technology, engineering, and math fields). Chemistry and the language of chemists play vital roles in biology, medicine, materials science, forensics, environmental science, and many other fields (Figure 1.3). The basic principles of physics are essential for understanding many aspects of chemistry, and there is extensive overlap between many subdisciplines within the two fields, such as chemical physics and nuclear chemistry. Mathematics, computer science, and information theory provide important tools that help us calculate, interpret, describe, and generally make sense of the chemical world. Biology and chemistry converge in biochemistry, which is crucial to understanding the many complex factors and processes that keep living organisms (such as us) alive. Chemical engineering, materials science, and nanotechnology combine chemical principles and empirical findings to produce useful substances, ranging from gasoline to fabrics to electronics. Agriculture, food science, veterinary science, and brewing and wine making help provide sustenance in the form of food and drink to the world's population. Medicine, pharmacology, biotechnology, and botany identify and produce substances that help keep us healthy. Environmental science, geology, oceanography, and atmospheric science incorporate many chemical ideas to help us better understand and protect our physical world. Chemical ideas are used to help understand the universe in astronomy and cosmology.

FIGURE 1.3 Knowledge of chemistry is central to understanding a wide range of scientific disciplines. This diagram shows just some of the interrelationships between chemistry and other fields.


4. Chemistry: The Central Science -
FIGURE 1.3 Knowledge of chemistry is central to understanding a wide range of scientific disciplines. This diagram shows just some of the interrelationships between chemistry and other fields.

What are some changes in matter that are essential to daily life? Digesting and assimilating food, synthesizing polymers that are used to make clothing, containers, cookware, and credit cards, and refining crude oil into gasoline and other products are just a few examples. As you proceed through this course, you will discover many different examples of changes in the composition and structure of matter, how to classify these changes and how they occurred, their causes, the changes in energy that accompany them, and the principles and laws involved. As you learn about these things, you will be learning chemistry, the study of the composition, properties, and interactions of matter. The practice of chemistry is not limited to chemistry books or laboratories: It happens whenever someone is involved in changes in matter or in conditions that may lead to such changes.


5. The Scientific Method -
Chemistry is a science based on observation and experimentation. Doing chemistry involves attempting to answer questions and explain observations in terms of the laws and theories of chemistry, using procedures that are accepted by the scientific community. There is no single route to answering a question or explaining an observation, but there is an aspect common to every approach: Each uses knowledge based on experiments that can be reproduced to verify the results. Some routes involve a hypothesis, a tentative explanation of
observations that acts as a guide for gathering and checking information. A hypothesis is tested by experimentation, calculation, and/or comparison with the experiments of others and then refined as needed.

Some hypotheses are attempts to explain the behavior that is summarized in laws. The laws of science summarize a vast number of experimental observations, and describe or predict some facet of the natural world. If such a hypothesis turns out to be capable of explaining a large body of experimental data, it can reach the status of a theory. Scientific theories are well-substantiated, comprehensive, testable explanations of particular aspects of nature. Theories are accepted because they provide satisfactory explanations, but they can be modified if new data become available. The path of discovery that leads from question and observation to law or hypothesis to theory, combined with experimental verification of the hypothesis and any necessary modification of the theory, is called the scientific method (Figure 1.4).

FIGURE 1.4 The scientific method follows a process similar to the one shown in this diagram. All the key components are shown, in roughly the right order. Scientific progress is seldom neat and clean: It requires open inquiry and the reworking of questions and ideas in response to findings.


6. The Domains of Chemistry -
Chemists study and describe the behavior of matter and energy in three different domains: macroscopic, microscopic, and symbolic. These domains provide different ways of considering and describing chemical behavior.

Macro is a Greek word that means "large." The macroscopic domain is familiar to us: It is the realm of everyday things that are large enough to be sensed directly by human sight or touch. In daily life, this includes the food you eat and the breeze you feel on your face. The macroscopic domain includes everyday and laboratory chemistry, where we observe and measure physical and chemical properties such as density, solubility, and flammability.

Micro comes from Greek and means "small." The microscopic domain of chemistry is often visited in the imagination. Some aspects of the microscopic domain are visible through standard optical microscopes, for example, many biological cells. More sophisticated instruments are capable of imaging even smaller entities such as molecules and atoms (see Figure 1.5 (b)).

However, most of the subjects in the microscopic domain of chemistry are too small to be seen even with the most advanced microscopes and may only be pictured in the mind. Other components of the microscopic
domain include ions and electrons, protons and neutrons, and chemical bonds, each of which is far too small to see.

The symbolic domain contains the specialized language used to represent components of the macroscopic and microscopic domains. Chemical symbols (such as those used in the periodic table), chemical formulas, and chemical equations are part of the symbolic domain, as are graphs, drawings, and calculations. These symbols play an important role in chemistry because they help interpret the behavior of the macroscopic domain in terms of the components of the microscopic domain. One of the challenges for students learning chemistry is recognizing that the same symbols can represent different things in the macroscopic and microscopic domains, and one of the features that makes chemistry fascinating is the use of a domain that must be imagined to explain behavior in a domain that can be observed.


6. The Domains of Chemistry -
A helpful way to understand the three domains is via the essential and ubiquitous substance of water. That water is a liquid at moderate temperatures, will freeze to form a solid at lower temperatures, and boil to form a gas at higher temperatures (Figure 1.5) are macroscopic observations. But some properties of water fall into the microscopic domain-what cannot be observed with the naked eye. The description of water as comprising two hydrogen atoms and one oxygen atom, and the explanation of freezing and boiling in terms of attractions between these molecules, is within the microscopic arena. The formula $\mathrm{H}_{2} \mathrm{O}$, which can describe water at either the macroscopic or microscopic levels, is an example of the symbolic domain. The abbreviations ( $g$ ) for gas, ( $S$ ) for solid, and ( $I$ ) for liquid are also symbolic.

FIGURE 1.5 (a) Moisture in the air, icebergs, and the ocean represent water in the macroscopic domain. (b) At the molecular level (microscopic domain), gas molecules are far apart and disorganized, solid water molecules are close together and organized, and liquid molecules are close together and disorganized. (c) The formula $\mathrm{H}_{2} \mathrm{O}$ symbolizes water, and (g), (s), and ( $)$ symbolize its phases. Note that clouds are actually comprised of either very small liquid water droplets or solid water crystals; gaseous water in our atmosphere is not visible to the naked eye, although it may be sensed as humidity. (credit a: modification of work by "Gorkaazk"/Wikimedia Commons)


7. LEARNING OBJECTIVES -
By the end of this section, you will be able to:

- Describe the basic properties of each physical state of matter: solid, liquid, and gas
- Distinguish between mass and weight
- Apply the law of conservation of matter
- Classify matter as an element, compound, homogeneous mixture, or heterogeneous mixture with regard to its physical state and composition
- Define and give examples of atoms and molecules

Matter is defined as anything that occupies space and has mass, and it is all around us. Solids and liquids are more obviously matter: We can see that they take up space, and their weight tells us that they have mass. Gases are also matter; if gases did not take up space, a balloon would not inflate (increase its volume) when filled with gas.

Solids, liquids, and gases are the three states of matter commonly found on earth (Figure 1.6). A solid is rigid and possesses a definite shape. A liquid flows and takes the shape of its container, except that it forms a flat or slightly curved upper surface when acted upon by gravity. (In zero gravity, liquids assume a spherical shape.) Both liquid and solid samples have volumes that are very nearly independent of pressure. A gas takes both the shape and volume of its container.

FIGURE 1.6 The three most common states or phases of matter are solid, liquid, and gas.
A fourth state of matter, plasma, occurs naturally in the interiors of stars. A plasma is a gaseous state of matter that contains appreciable numbers of electrically charged particles (Figure 1.7). The presence of these charged particles imparts unique properties to plasmas that justify their classification as a state of matter distinct from gases. In addition to stars, plasmas are found in some other high-temperature environments (both natural and man-made), such as lightning strikes, certain television screens, and specialized analytical instruments used to detect trace amounts of metals.

FIGURE 1.7 A plasma torch can be used to cut metal. (credit: "Hypertherm"/Wikimedia Commons)


8. LINK TO LEARNING -
In a tiny cell in a plasma television, the plasma emits ultraviolet light, which in turn causes the display at that location to appear a specific color. The composite of these tiny dots of color makes up the image that you see. Watch this video (http://openstax.org/1/16plasma) to learn more about plasma and the places you encounter it.

Some samples of matter appear to have properties of solids, liquids, and/or gases at the same time. This can occur when the sample is composed of many small pieces. For example, we can pour sand as if it were a liquid because it is composed of many small grains of solid sand. Matter can also have properties of more than one state when it is a mixture, such as with clouds. Clouds appear to behave somewhat like gases, but they are actually mixtures of air (gas) and tiny particles of water (liquid or solid).

The mass of an object is a measure of the amount of matter in it. One way to measure an object's mass is to measure the force it takes to accelerate the object. It takes much more force to accelerate a car than a bicycle because the car has much more mass. A more common way to determine the mass of an object is to use a balance to compare its mass with a standard mass.

Although weight is related to mass, it is not the same thing. Weight refers to the force that gravity exerts on an object. This force is directly proportional to the mass of the object. The weight of an object changes as the force of gravity changes, but its mass does not. An astronaut's mass does not change just because she goes to the moon. But her weight on the moon is only one-sixth her earth-bound weight because the moon's gravity is only one-sixth that of the earth's. She may feel "weightless" during her trip when she experiences negligible external forces (gravitational or any other), although she is, of course, never "massless."


8. LINK TO LEARNING -
The law of conservation of matter summarizes many scientific observations about matter: It states that there is no detectable change in the total quantity of matter present when matter converts from one type to another (a chemical change) or changes among solid, liquid, or gaseous states (a physical change). Brewing beer and the operation of batteries provide examples of the conservation of matter (Figure 1.8). During the brewing of beer, the ingredients (water, yeast, grains, malt, hops, and sugar) are converted into beer (water, alcohol, carbonation, and flavoring substances) with no actual loss of substance. This is most clearly seen during the bottling process, when glucose turns into ethanol and carbon dioxide, and the total mass of the substances does not change. This can also be seen in a lead-acid car battery: The original substances (lead, lead oxide, and sulfuric acid), which are capable of producing electricity, are changed into other substances (lead sulfate and water) that do not produce electricity, with no change in the actual amount of matter.

FIGURE 1.8 (a) The mass of beer precursor materials is the same as the mass of beer produced: Sugar has become alcohol and carbon dioxide. (b) The mass of the lead, lead oxide, and sulfuric acid consumed by the production of electricity is exactly equal to the mass of lead sulfate and water that is formed.

Although this conservation law holds true for all conversions of matter, convincing examples are few and far between because, outside of the controlled conditions in a laboratory, we seldom collect all of the material that is produced during a particular conversion. For example, when you eat, digest, and assimilate food, all of the matter in the original food is preserved. But because some of the matter is incorporated into your body, and much is excreted as various types of waste, it is challenging to verify by measurement.


9. Classifying Matter -
Matter can be classified into several categories. Two broad categories are mixtures and pure substances. A pure substance has a constant composition. All specimens of a pure substance have exactly the same makeup and properties. Any sample of sucrose (table sugar) consists of $42.1 \%$ carbon, $6.5 \%$ hydrogen, and $51.4 \%$ oxygen by mass. Any sample of sucrose also has the same physical properties, such as melting point, color, and sweetness, regardless of the source from which it is isolated.

Pure substances may be divided into two classes: elements and compounds. Pure substances that cannot be broken down into simpler substances by chemical changes are called elements. Iron, silver, gold, aluminum, sulfur, oxygen, and copper are familiar examples of the more than 100 known elements, of which about 90 occur naturally on the earth, and two dozen or so have been created in laboratories.

Pure substances that are comprised of two or more elements are called compounds. Compounds may be broken down by chemical changes to yield either elements or other compounds, or both. Mercury(II) oxide, an orange, crystalline solid, can be broken down by heat into the elements mercury and oxygen (Figure 1.9). When heated in the absence of air, the compound sucrose is broken down into the element carbon and the compound water. (The initial stage of this process, when the sugar is turning brown, is known as caramelization-this is what imparts the characteristic sweet and nutty flavor to caramel apples, caramelized onions, and caramel). Silver(I) chloride is a white solid that can be broken down into its elements, silver and chlorine, by absorption of light. This property is the basis for the use of this compound in photographic films and photochromic eyeglasses (those with lenses that darken when exposed to light).

FIGURE 1.9 (a) The compound mercury(II) oxide, (b) when heated, (c) decomposes into silvery droplets of liquid mercury and invisible oxygen gas. (credit: modification of work by Paul Flowers)


10. LINK TO LEARNING -
Many compounds break down when heated. This site (http://openstax.org/l/16mercury) shows the breakdown of mercury oxide, HgO. You can also view an example of the photochemical decomposition of silver chloride (http://openstax.org/l/16silvchloride) (AgCl), the basis of early photography.

The properties of combined elements are different from those in the free, or uncombined, state. For example, white crystalline sugar (sucrose) is a compound resulting from the chemical combination of the element carbon, which is a black solid in one of its uncombined forms, and the two elements hydrogen and oxygen, which are colorless gases when uncombined. Free sodium, an element that is a soft, shiny, metallic solid, and free chlorine, an element that is a yellow-green gas, combine to form sodium chloride (table salt), a compound that is a white, crystalline solid.

A mixture is composed of two or more types of matter that can be present in varying amounts and can be separated by physical changes, such as evaporation (you will learn more about this later). A mixture with a composition that varies from point to point is called a heterogeneous mixture. Italian dressing is an example of a heterogeneous mixture (Figure 1.10). Its composition can vary because it may be prepared from varying amounts of oil, vinegar, and herbs. It is not the same from point to point throughout the mixture-one drop may be mostly vinegar, whereas a different drop may be mostly oil or herbs because the oil and vinegar separate and the herbs settle. Other examples of heterogeneous mixtures are chocolate chip cookies (we can see the separate bits of chocolate, nuts, and cookie dough) and granite (we can see the quartz, mica, feldspar, and more).


10. LINK TO LEARNING -
A homogeneous mixture, also called a solution, exhibits a uniform composition and appears visually the same throughout. An example of a solution is a sports drink, consisting of water, sugar, coloring, flavoring, and electrolytes mixed together uniformly (Figure 1.10). Each drop of a sports drink tastes the same because each drop contains the same amounts of water, sugar, and other components. Note that the composition of a sports drink can vary-it could be made with somewhat more or less sugar, flavoring, or other components, and still be a sports drink. Other examples of homogeneous mixtures include air, maple syrup, gasoline, and a solution of salt in water.

FIGURE 1.10 (a) Oil and vinegar salad dressing is a heterogeneous mixture because its composition is not uniform throughout. (b) A commercial sports drink is a homogeneous mixture because its composition is uniform throughout. (credit a "left": modification of work by John Mayer; credit a "right": modification of work by Umberto Salvagnin; credit b "left: modification of work by Jeff Bedford)

Although there are just over 100 elements, tens of millions of chemical compounds result from different combinations of these elements. Each compound has a specific composition and possesses definite chemical and physical properties that distinguish it from all other compounds. And, of course, there are innumerable ways to combine elements and compounds to form different mixtures. A summary of how to distinguish between the various major classifications of matter is shown in (Figure 1.11).

FIGURE 1.11 Depending on its properties, a given substance can be classified as a homogeneous mixture, a heterogeneous mixture, a compound, or an element.

Eleven elements make up about $99 \%$ of the earth's crust and atmosphere (Table 1.1). Oxygen constitutes nearly one-half and silicon about one-quarter of the total quantity of these elements. A majority of elements on earth are found in chemical combinations with other elements; about one-quarter of the elements are also found in the free state.

Elemental Composition of Earth


10. LINK TO LEARNING -
Elemental Composition of Earth

| Element | Symbol |  | Percent Mass | Element | Symbol |  | Percent Mass |
| :--- | :--- | :--- | :--- | :--- | :--- | :---: | :---: |
| oxygen | O | 49.20 | chlorine | Cl | 0.19 |  |  |
| silicon | Si | 25.67 | phosphorus | P | 0.11 |  |  |
| aluminum | Al | 7.50 | manganese | Mn | 0.09 |  |  |
| calcium | Ca | 3.39 | carbon | C | 0.08 |  |  |
| sodium | Na | 2.63 | sulfur | S | 0.06 |  |  |
| potassium | K | 2.40 | barium | Ba | 0.04 |  |  |
| magnesium | Mg | 1.93 | fluorine | F | 0.03 |  |  |
| hydrogen | H | 0.87 | strontium | Sr | 0.02 |  |  |
| titanium | Ti | 0.58 | all others | - | 0.47 |  |  |

TABLE 1.1


11. Atoms and Molecules -
An atom is the smallest particle of an element that has the properties of that element and can enter into a chemical combination. Consider the element gold, for example. Imagine cutting a gold nugget in half, then cutting one of the halves in half, and repeating this process until a piece of gold remained that was so small that it could not be cut in half (regardless of how tiny your knife may be). This minimally sized piece of gold is an atom (from the Greek atomos, meaning "indivisible") (Figure 1.12). This atom would no longer be gold if it were divided any further.

FIGURE 1.12 (a) This photograph shows a gold nugget. (b) A scanning-tunneling microscope (STM) can generate views of the surfaces of solids, such as this image of a gold crystal. Each sphere represents one gold atom. (credit a: modification of work by United States Geological Survey; credit b: modification of work by "Erwinrossen"/Wikimedia Commons)

The first suggestion that matter is composed of atoms is attributed to the Greek philosophers Leucippus and Democritus, who developed their ideas in the 5th century BCE. However, it was not until the early nineteenth century that John Dalton (1766-1844), a British schoolteacher with a keen interest in science, supported this hypothesis with quantitative measurements. Since that time, repeated experiments have confirmed many aspects of this hypothesis, and it has become one of the central theories of chemistry. Other aspects of Dalton's atomic theory are still used but with minor revisions (details of Dalton's theory are provided in the chapter on atoms and molecules).