The Chemistry of Life: Organic and Biological Chemistry

25 The Chemistry of Life: Organic and Biological Chemistry Visualizing Concepts 25.1 Analyze/Plan. Given structural formulas, specify which molecul...
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The Chemistry of Life: Organic and Biological Chemistry

Visualizing Concepts 25.1

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.

25.3

Analyze/plan. Given 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. The molecule that experiences hydrogen bonding will have the highest boiling point. Solve. Only , and bonds fit the strict definition of hydrogen bonding. Molecule (b), an alcohol, forms hydrogen bonds with like molecules; it has the highest boiling point. [Molecules (a) and (d) have dipole-dipole and dispersion forces. The greater molar mass of (d) probably means it has stronger dispersion forces and a slightly higher boiling point than molecule (a). Molecule (c) has only dispersion forces and the lowest boiling point. The probable order of strength of forces and boiling points is: (b) > (d) > (a) > (c).]

25.5

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. (a)

Molecule (i) is a disaccharide composed of galactose (left) and glucose (right); it can be hydrolyzed to form a solution containing glucose. Since it is the only sugar molecule depicted, it was not necessary to know the exact structure of glucose to answer the question.

(b)

Amino acids form zwitterions, so the choice is molecule (iv).

(c)

Molecule (iii) is an organic base present in DNA (again, the only possible choice).

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Solutions to Exercises

(d)

Molecule (v) because alcohols react with carboxylic acids to form esthers.

(e)

Molecule (ii), because it has a long hydrocarbon chain and an ester functional group.

Introduction to Organic Compounds; Hydrocarbons 25.7

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. C1 has triagonal 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 tetrahedral electron domain geometry, 109° bond 3 angles and sp hybridization.

25.9

Ammonia, NH3, contains no carbon, so it is not, strictly speaking, considered an organic molecule. Carbon monoxide contains a carbon atom that does not form four bonds, so it certainly is not a typical organic molecule.

25.11

(a)

A straight-chain hydrocarbon has all carbon atoms connected in a continuous chain; no carbon atom is bound to more than two other carbon atoms. A branched-chain hydrocarbon has a branch; at least one carbon atom is bound to three or more carbon atoms.

(b)

An alkane is a complete molecule composed of carbon and hydrogen in which all bonds are single (sigma) bonds. An alkyl group is a substituent formed by removing a hydrogen atom from an alkane.

(c)

Alkanes are said to be saturated because they contain only single bonds. Multiple bonds that enable addition of H 2 or other substances are absent. The bonding capacity of each carbon atom is fulfilled with single bonds to C or H.

(d)

Ethylene (or ethene), CH2=CH2, is unsaturated.

25.13

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

CH 3 CH 2 CH 2 CH 2 CH 3 , C 5 H 1 2

(b)

(c) (d)

C 5 H 8 saturated: (a), (b); unsaturated: (c), (d)

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Solutions to Exercises

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 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 are: CH2=CH−C≡C−C≡CH

CH≡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. 25.17

Analyze/Plan. Follow the logic in Sample Exercise 25.3. Solve.

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707

Solutions to Exercises

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sp

3

(b)

sp

2

(a)

25.21

Analyze/Plan. Follow the rules for naming alkanes given in Section 25.3 and illustrated in Sample Exercise 25.1. Solve. 2-methylhexane

(b)

4-ethyl-2,4-dimethyldecane

sp

2

25.19

(a)

(c)

Solutions to Exercises

(d)

sp

(c)

(d)

(e)

25.23

25.25

Analyze/Plan. Follow the logic in Sample Exercises 25.1 and 25.4. (a)

2,3-dimethylheptane

(b)

cis-6-methyl-3-octene

(c)

para-dibromobenzene

(d)

4,4-dimethyl-1-hexyne

(e)

methylcyclobutane

Solve.

Each doubly bound carbon atom in an alkene has two unique sites for substitution. These sites cannot be interconverted because rotation about the double bond is restricted; geometric isomerism results. In an alkane, carbon forms only single bonds, so the three remaining sites are interchangeable by rotation about the single bond. Although there is also restricted rotation around the triple bond of an alkyne, there is only one additional bonding site on a triply bound carbon, so no isomerism results.

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25 Organic and Biological Chemistry 25.27

Solutions to Exercises

Analyze/Plan. In order for geometrical isomerism to be possible, the molecule must be an alkene with two different groups bound to each of the alkene C atoms. Solve.

(a)

(b)

25.29

(c)

no, not an alkene

(d)

no, not an alkene

Assuming that each component retains its effective octane number in the mixture (and this isn’t always the case), we obtain: octane number = 0.35(0) + 0.65(100) = 65.

Reactions of Hydrocarbons 25.31

(a)

An addition reaction is the addition of some reagent to the two atoms that form a multiple bond. In a substitution reaction, one atom or group of atoms replaces (substitutes for) another atom or group of atoms. In an addition reaction, two atoms and a multiple bond on the target molecule are altered; in a substitution reaction, the environment of one atom in the target molecule changes. Alkenes typically undergo addition, while aromatic hydrocarbons usually undergo substitution.

(b)

Plan. Consider the general form of addition across a double bond. The π bond is broken and one new substituent (in this case two Br atoms) adds to each of the C atoms involved in the π bond. Solve. CH3CH2CH=CH−CH3 + Br2

→ CH3CH2CH(Br)CH(Br)CH3

2-pentene

25.33

2,3-dibromopentane

(c)

Plan. Consider the general form of a substitution reaction. Two Cl atoms will replace two of the H atoms on the benzene ring. The term para means the Cl atoms will be opposite each other across the benzene ring in the product. Solve.

(a)

Plan. Consider the structures of cyclopropane, cyclopentane, and cyclohexane. Solve. The small 60° angles in the cyclopropane ring cause strain that provides a driving force for reactions that result in ring opening. There is no comparable strain in the five- or six-membered rings.

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Solutions to Exercises

Plan. First form an alkyl halide: C 2 H 4 (g) + HBr(g) → CH 3 CH 2 Br(l); then carry out a Friedel-Crafts reaction. Solve.

25.35

Not necessarily. That the rate laws are both first order in both reactants and second order overall indicates that the activated complex in the rate-determining step in each mechanism is bimolecular and contains one molecule of each reactant. This is usually an indication that the mechanisms are the same, but it does not rule out the possibility of different fast steps, or a different order of elementary steps.

25.37

Analyze/Plan. Both combustion reactions produce CO 2 and H 2 O: C 3 H 6 (g) + 9/2 O 2 (g) → 3CO 2 (g) + 3H 2 O(l) C 5 H 1 0(g) + 15/2 O 2 (g) → 5CO 2 (g) + 5H 2 O(l) Thus, we can calculate the ΔH c omb / CH 2 group for each compound.

Solve.

ΔH c omb/CH 2 group for cyclopropane is greater because C 3 H 6 contains a strained ring. When combustion occurs, the strain is relieved and the stored energy is released during the reaction.

Functional Groups and Chirality 25.39

Analyze/Plan. Match the structural features of various functional groups shown in Table 25.4 to the molecular structures in this exercise. Solve. (a)

−OH, alcohol

(b)

(c)

O , ether

(d)

(e) (f)

25.41

−NH− , amine;

C=C , alkene

, ketone;

C=C , alkene

( CHO), aldehyde C≡C ( CC ), alkyne;

COOH, carboxylic acid

Analyze/Plan. Given the name of a molecule, write the structural formula of an isomer molecular formula of the given molecule, draw the structural formula of a molecule with the same formula that contains the specified functional group. Solve. (a)

The formula of acetone is C 3 H 6 O. An aldehyde contains the group An aldehyde that is an isomer of acetone is propionaldehyde (or propanal):

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(b)

25.43

25.45

Solutions to Exercises

The formula of 1-propanol is C 3 H 8 O. An ether contains the group ether that is an isomer of 1-propanol is ethylmethyl ether:

An

Analyze/Plan. From the hydrocarbon name, deduce the number of C atoms in the acid; one carbon atom is in the carboxyl group. Solve. (a)

meth = 1 C atom

(b)

pent = 5 C atoms

(c)

dec = 10 C atoms in backbone

Analyze/Plan. In a condensation reaction between an alcohol and a carboxylic acid, the alcohol loses its hydrogen atom and the acid loses its group. The alkyl group from the acid is attached to the carbonyl group and the alkyl group from alcohol is attached to the ether oxygen of the ester. The name of the ester is the alkyl group from the alcohol plus the alkyl group from the acid plus the suffix -oate. Solve.

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Solutions to Exercises

25.47

Analyze/Plan. Follow the logic in Sample Exercise 25.6.

25.49

Yes, we expect acetic acid to be a strongly hydrogen-bonded substance. The carboxyl group has both and groups that participate in hydrogen bonding. The boiling point of acetic acid is higher than that of water (118°C vs. 100°C), indicating that hydrogen-bonding in acetic acid is even stronger than that of water (Figure 11.10). The melting points, 16.7°C for acetic acid and 0°C for water, show a similar trend.

25.51

Analyze/Plan. Follow the logic in Sample Exercise 25.2, incorporating functional group information from Table 25.4. Solve. CH3CH(OH)CH2OH (a) CH3CH2CH2CH(OH)CH3 (b)

(c)

(e)

25.53

Solve.

(d)

CH3OCH2CH3

Analyze/Plan. Review the rules for naming alkanes and haloalkanes; draw the structures. That is, draw the carbon chain indicated by the root name, place substituents, fill remaining positions with H atoms. Each C atom attached to four different groups is chiral. Solve. ∗

chiral C atoms

C2 is obviously attached to four different groups. C3 is chiral because the substituents on C2 render the C1-C2 group different than the C4-C5 group.

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Proteins 25.55

(a)

An α-amino acid contains an NH 2 group attached to the carbon that is bound to the carbon of the carboxylic acid function.

(b)

In forming a protein, amino acids undergo a condensation reaction between the amino group and carboxylic acid:

(c)

25.57

The bond that links amino acids in proteins is called the peptide bond.

Analyze/Plan. Two dipeptides are possible. Either peptide can have the terminal carboxyl group or the terminal amino group. Solve.

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25 Organic and Biological Chemistry 25.59

Solutions to Exercises

Analyze/Plan. Follow the logic in Sample Exercise 25.7.

Solve.

(a)

25.61

(b)

Three tripeptides are possible: Gly-Gly-His, GGH; Gly-His-Gly, GHG; His-GlyGly, HGG

(a)

The primary structure of a protein refers to the sequence of amino acids in the chain. Along any particular section of the protein chain the configuration may be helical, may be an open chain, or arranged in some other way. This is called the secondary structure. The overall shape of the protein molecule is determined by the way the segments of the protein chain fold together, or pack. The interactions which determine the overall shape are referred to as the tertiary structure.

(b)

X-ray crystallography is the primary and preferred technique for determining protein structure.

Carbohydrates and Lipids 25.63

25.65

(a)

Carbohydrates, or sugars, are composed of carbon, hydrogen, and oxygen. From a chemical viewpoint, they are polyhydroxyaldehydes or ketones. Carbohydrates are primarily derived from plants and are a major food source for animals.

(b)

A monosaccharide is a simple sugar molecule that cannot be decomposed into smaller sugar molecules by (acid) hydrolysis.

(c)

A disaccharide is a carbohydrate composed of two simple sugar units. Hydrolysis breaks the disaccharides into two monosaccharides.

(d)

A polysaccharide is a polymer composed of many simple sugar units.

(a)

In the linear form of mannose, the aldehydic carbon is C1. Carbon atoms 2, 3, 4, and 5 are chiral because they each carry four different groups. Carbon 6 is not chiral because it contains two H atoms.

(b)

Both the α (left) and β (right) forms are possible.

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Solutions to Exercises

25.67

The empirical formula of cellulose is C 6 H 1 0O 5 . As in glycogen, the six-membered ring form of glucose forms the monomer unit that is the basis of the polymer cellulose. In cellulose, glucose monomer units are joined by β linkages.

25.69

The term lipid refers to the broad class of naturally-occurring molecules that are soluble in nonpolar solvents. Two important subgroups are fats and fatty acids. Structurally, fatty acids are carboxylic acids with a hydrocarbon chain of more than four carbon atoms (typically 16-20 carbon atoms). Fats are esters formed by condensation of an alcohol and a fatty acid. In animals, the alcohol is the triol glycerol; three fatty acid molecules condense with one glycerol molecule to form a large, nonpolar molecule. For both fatty acids and fats it is the long hydrocarbon chains that define the polarity, solubility and other physical properties of the molecules. Phospholipids are glycerol esters formed from one phosphoric acid (RPO(OH)2 and two fatty acid (RCOOH) molecules. At body pH, the phosphate group is depronated and has a negative charge. It is the juxtaposition of two long, nonpolar hydrocarbon chains with the charged phosphate “head” that causes phospholipids to form bilayers in water. The nonpolar chains do not readily mix with polar water. They do interact with the nonpolar chains of other phospholipids molecules on the inside of the bilayer. The charged phosphate heads interact with polar water molecules on the outsides of the bilayer.

Nucleic Acids 25.71

Dispersion forces increase as molecular size (and molar mass) increases. The larger purines (2 rings vs. 1 ring for pyrimidines) have larger dispersion forces.

25.73

Analyze/Plan. Consider the structures of the organic bases in Section 25.10. The first base in the sequence is attached to the sugar with the free phosphate group in the 5′ position. The last base is attached to the sugar with a free −OH group in the 3′ position. Solve. The DNA sequence is 5′−TACG−3′.

25.75

Analyze/Plan. Recall that there is complimentary base pairing in nucleic acids because of hydrogen bond geometry. The DNA pairs are A−T and G−C. (The RNA pairs are A−U and G−C.) Solve. From the single strand sequence, formulate the complimentary strand. Note that 3′ of the complimentary strand aligns with 5′ of the parent strand. 5′−GCATTGGC−3′ 3′−CGTAACCG−5′

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Solutions to Exercises

Additional Exercises

25.77

Structures with the group attached to an alkene carbon atom are not included. These molecules are called “vinyl alcohols” and are not the major form at equilibrium. 25.80

Cyclopentene does not show cis-trans isomerism because the existence of the ring demands that the bonds be cis to one another. 25.83

(Structures with the group attached to an alkene carbon atom are not included. These molecules are called “vinyl alcohols” and are not the major form at equilibrium.) 25.85

(a)

(b)

(c)

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Solutions to Exercises

(d)

25.91

Glu-Cys-Gly is the only possible order. Glutamic acid has two carboxyl groups that can form a peptide bond with cysteine, so there are two possible structures.

Integrative Exercises 25.97

Determine the empirical formula, molar mass, and thus molecular formula of the compound. Confirm with physical data.

The empirical formula is C 4 H 8 O. Using Equation 10.11 (MM = molar mass):

The formula weight of C 4 H 8 O is 72, so the molecular formula is also C 4 H 8 O. Since the compound has a carbonyl group and cannot be oxidized to an acid, the only possibility is 2-butanone.

The boiling point of 2-butanone is 79.6°C, confirming the identification.

717