Solutions Manual For

Solutions Manual For Chemistry The Central Science 15 th Edition By Theodore Brown, Eugene LeMay, Bruce Bursten, Catherine Murphy, Patrick Woodward, Matthew Stoltzfus (All Chapters 1-24, 100% Original Verified, A+ Grade)

All Chapters Arranged Reverse:

24-1 This is The Original Solutions Manual For 15 th Edition, All other Files in The Market are Fake/Old/Wrong Edition. 1 / 4

24 The Chemistry of Life: Organic and Biological Chemistry

Visualizing Concepts 24.1 Analyze/Plan. Follow the logic in Sample Exercise 2.15 to name each compound. Decide which structures are the same compound. Solve.

• 2,2,4-trimethylpentane
• 3-ethyl-2-methylpentane
• 2,3,4-trimethylpentane
• 2,3,4-trimethylpentane

Structures (c) and (d) are the same molecule.

24.2 Analyze/Plan. Given structural formulas, specify which molecules are unsaturated. Consider the definition of unsaturated and apply it to the molecules in the exercise. Solve.Unsaturated molecules contain one or more multiple bonds. Saturated molecules contain only single bonds. Molecules (c) and (d) are unsaturated.

24.3 Analyze/Plan. Given structural formulas, decide which molecule will undergo addition. Consider which functional groups are present in the molecules and which are most susceptible to addition. Solve.

• Molecule (iii), an alkene, will readily undergo addition. Addition reactions are characteristic of alkenes. [Molecule (i) will not typically undergo addition, because its delocalized electron

cloud is too difficult to disrupt. Molecules (b) and (d) contain carbonyl groups (actually carboxylic acid groups) that do not typically undergo addition, except under special conditions.]

• Molecule (i) is an aromatic hydrocarbon.
• Molecule (i) most readily undergoes a substitution reaction.

24.4 Analyze/Plan. Given condensed structural formulas, predict which molecule will have the highest boiling point. Boiling point is determined by strength of intermolecular forces; for neutral molecules with similar molar masses, the strongest intermolecular force is hydrogen bonding.We are also asked to identify various functional groups, given condensed structural formulas. Refer to Table 24.2 for information on functional groups. Solve.

• Molecule (ii), an alcohol, forms hydrogen bonds with like molecules; it has the highest boiling point.
• Molecule (iv) is most oxidized; it has the most oxygen atoms.
• None of the compounds is an ether. Ethers are characterized by an oxygen atom bound to two carbon atoms. Compound (iv) is an ester, which has an ether linkage.
• Compound (iv) is an ester.
• None of the compounds is a ketone. Ketones are characterized by a carbonyl group bonded to two alkyl groups. Compound (i) is an aldehyde; the carbonyl group is bonded to one

alkyl group and one hydrogen atom.

24.5 Analyze. Given structural formulas, name the compound, identify the functional groups present, and determine if the molecule is chiral. Plan. Apply principles of organic nomenclature in Section 2.9, as well as 24.2–24.4. Identify functional groups using Table 24.2. Assess chirality by looking for one or more carbon atoms that are bonded to four different groups. Solve.

• Valine. This molecule is an amino acid shown in the zwitterion form. Use Figure 24.16 to name the amino acid. In the neutral amino acid, there is a carboxlyic acid group and an amine

group. (In the zwitterion, these become a carboxyl group and an ammonium group.) The amino acid is chiral. The C atom to which the −NH3+ group is bound is a chiral center, because it is bound to four different groups. All amino acids except glycine are chiral.

• 3-chlorobenzoic acid. The functional groups are a carboxylic acid group and an (aryl) halide. The molecule is not chiral; no C atom is bound to four different groups.
• 2-pentene. The molecule contains an alkene functional group and is not chiral.
• Propane. The molecule is an alkane, there are no functional groups, and it is not chiral.

24.6 Analyze/Plan. Given ball-and-stick models, select the molecule that fits the description given. From the models, decide the type of molecule or functional group represented.Solve. Molecule (i) is a sugar, (ii) is an ester with a long hydrocarbon chain, (iii) is an organic base and a component of nucleic acids, (iv) is an amino acid, and (v) is an alcohol.

• Molecule (i) is a disaccharide composed of galactose (left) and glucose (right); it can be hydrolyzed to form a solution containing glucose. Because it is the only sugar molecule

depicted, it was not necessary to know the exact structure of glucose to answer the question.

• Amino acids form zwitterions, so the choice is molecule (iv).
• Molecule (iii) is an organic base present in DNA (again, the only possible choice).
• Molecule (v), because alcohols react with carboxylic acids to form esters.
• Molecule (ii), because it has a long hydrocarbon chain and an ester functional group.

Introduction to Organic Compounds; Hydrocarbons (Sections 24.1 and 24.2) 24.7

• False. Butane is an alkane; it contains carbon atoms that are sp3 hybridized.
• False. Cyclohexane is a saturated hydrocarbon, whereas benzene is aromatic.
• True.
• False. Olefin is another name for alkene.

24.8

• False. Hexane contains six carbon atoms and pentane contains only five. 2 / 4

• True. For molecules with similar structures, the larger the molecule, the stronger their dispersion forces and the higher the boiling point.
• True. The carbon atoms involved in an alkyne group are sp hybridized.
• False. There is only one way to arrange the three carbon atoms in propane.

24.9 Analyze/Plan. Given a condensed structural formula, determine the bond angles and hybridization about each carbon atom in the molecule. Visualize the number of electron domains about each carbon. State the bond angle and hybridization based on electron domain geometry. Solve.Extended Description C1 has trigonal planar electron domain geometry, 120° bond angles, and sp2 hybridization. C3 and C4 have linear electron domain geometry, 180° bond angles, and sp hybridization. C2 and C5 both have tetra-hedral electron domain geometry, 109° bond angles, and sp3 hybridization.

24.10 Extended Description

• C2, C5, and C6 have sp3 hybridization (4 e- domains around C).
• C7 has sp hybridization (2 e- domains around C).
• C1, C3, and C4 have sp2 hybridization (3 e- domains around C).

24.11 Analyze/Plan. For each molecule, count the number of carbon atoms in the root name and in the substituent group(s). Solve.

• One. The root “meth” indicates one carbon atom. Methane is the simplest alkane.
• Ten. The root “dec” indicates 10 carbon atoms.
• Seven. The root “hex” indicates 6 carbon atoms, plus 1 in the “methyl” substituent.
• Five. Although this is a common name (not an IUPAC name), the root “pent” still indicates 5 total carbon atoms.
• Two. Again a common name, acetylene is the simplest two-carbon alkyne. The IUPAC name is ethyne.

24.12 True. Note that we are comparing enthalpies for only single bonds in the molecule.

24.13

• True.
• True.
• False. Alkenes contain carbon-carbon double bonds.
• False. Alkynes contain carbon-carbon triple bonds.
• True.
• False. Cyclohexane is a saturated hydrocarbon; benzene is aromatic.
• True.

24.14 All the classifications listed are hydrocarbons; they contain only the elements hydrogen and carbon.

• Alkanes are hydrocarbons that contain only single bonds.
• Cycloalkanes contain at least one ring of three or more carbon atoms joined by single bonds. Because it is a type of alkane, all bonds in a cycloalkane are single bonds.
• Alkenes contain at least one C ═ C double bond.
• Alkynes contain at least one C≡C triple bond.
• A saturated hydrocarbon contains only single bonds. Alkanes and cycloalkanes fit this definition.
• An aromatic hydrocarbon contains one or more planar, six-membered rings of carbon atoms with delocalized π-bonding throughout the ring.

24.15 Analyze/Plan. Follow the rules for naming alkanes given in Section 2.9 and illustrated in Sample Exercises 2.15 and 24.2. Solve.

• 2,3-dimethylheptane

2.

3.Extended Description

• 2,2,5-trimethylhexane
• 3-ethylheptane

24.16 3 / 4

1.Extended Description 2.

3.

• 2,4-dimethylhexane
• methylcyclobutane

Alkenes, Alkynes, and Aromatic Hydrocarbons (Section 24.3) 24.17

• C4H6 is an unsaturated hydrocarbon. The maximum number of hydrogen atoms for 4 C atoms in a saturated alkane is 2×4+2=10. C4H6 does not contain the maximum possible

hydrogen atoms and is unsaturated.

• Yes, all alkynes are unsaturated. The presence of a triple bond means that the alkyne carbon atoms are not bound to the maximum possible number of hydrogen atoms.

24.18

• The molecule CH3CH ═ CH2 is unsaturated because it contains a double bond. It is possible to add more hydrogen to the molecule.
• The formula CH3CH2CH ═ CH3 has too many H atoms bound to the right-most C atom; the formula as it stands implies 5 bonds to this atom. A correct formula is

CH3CH2CH ═ CH2, with 2 H atoms on the right-most C atom.

24.19 Analyze/Plan. Consider the definition of the stated classification and apply it to a compound containing five C atoms. Solve.

1. CH3CH2CH2CH2CH3,C5H12

2.

, C5H10

Extended Description

3. CH2 ═ CHCH2CH2CH3, C5H10

4. HC≡CCH2CH2CH3, C5H8

24.20 cyclic alkane, , C6H12 Extended Description cyclic alkene, , C6H10 Extended Description alkyne, CH3−CH2−C≡C−CH2−CH3,C6H10 aromatic hydrocarbon, , C6H6 Extended Description 24.21 Analyze/Plan. We are given the class of compounds “enediyne.” Based on organic nomenclature, determine the structural features of an enediyne. Construct a molecule with 6 C atoms in a row that has these features. Solve.The term “enediyne” contains the suffixes -ene and -yne. The suffix -ene is used to name alkenes, molecules with one double bond. The suffix -yne is used to name alkynes, molecules with one triple bond. An enediyne then features one double and two triple bonds. Possible arrangements of these bonds involving 6 C atoms in a row are:

CH2 ═ CH ─ C≡C ─ C≡CHCH≡C ─ CH ═ CH ─ C≡CH

Check. The formula of a saturated alkane is CnH2n+2. For each double bond subtract 2 H atoms, for each triple bond subtract 4 H atoms. A saturated 6 C alkane has 14 H atoms. For the enediyne, subtract 2+4+4=10 H atoms. The molecular formula is C6H4. That is the molecular formula of each structure above.

24.22 CnH2n-2 24.23 Analyze/Plan. Follow the logic in Sample Exercise 24.1.Solve. There are many correct structures that are alkenes or alkynes and have the molecular formula C6H10. The molecule will have two points of unsaturation. Molecules can include various arrangements of one alkyne group, two alkene groups, or one cyclic mono-alkene. A few of the possibilities are shown below.Extended Description Extended Description 24.24

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