Chemistry, Student Solutions Manual Chapter 10
Chapter 10 Organic Chemistry
Solutions to Problems in Chapter 10
10.1
10.3 (a) 2-Methylpentane
(b) 2,3-Dimethylbutane
(c) 2-Methylpentane
(d) 2,4-Dimethylpentane
(e) 3-Methylpentane
10.5
10.7 (a) o-Bromonitrobenzene
(b) p-Dicyclohexylbenzene
(c) m-Dinitrobenzene
10.9
10.11
(a) 1-Chloropropane (or n-propyl chloride)
(b) 2-Chlorobutane
(c) 2-Chloro-3,3-dimethylbutane
(d) 2-Chloro-3-methylbutane
(e) 3-Chloro-3-methylheptane
10.13 (a) Secondary 2°
(b) Primary 1°
(c) Secondary 2°
(d) Tertiary 3°
(e) Secondary 2°
10.15
10.17 (a) Tertiary 3°
(b) Primary 1°
(c) Primary 1°
(d) Secondary 2°
(e) Tertiary 3°
10.19 (a) Cyclohexanone
(b) Butanal
(c) Ethyl butanoate
(d) Butanoic acid
(e) Butanamide
10.21 (a) No E/Z isomerization
(b) No E/Z isomerization
(c)
(d) No E/Z isomerization
(e) No E/Z isomerization
10.23 (a) Achiral
(b) Achiral
(c) Chiral
(d) Chiral
(e) Chiral
10.25
Amoxicillin
10.27 Good leaving groups are conjugate bases of strong acids (like Cl– and Br–); that is, weak bases. Poor leaving groups are conjugate bases of weak acids (like OH–); that is, strong bases.
(a) Br– = good leaving group
(b) CH3–O– = poor leaving group
(c) NH2– = poor leaving group
(d) Cl– = good leaving group
(e) CH3COO– (acetate anion) = good leaving group
10.29
(a)
(b)
(c)
(d)
(e)
10.31
10.33
10.35
10.37
(a)
(b)
(c)
(d)
(e)
10.39
10.41
10.43
(a)
(b)
(c)
(d)
(e)
10.45 Markovnikov’s rule states that in the addition of HX to an alkene, the hydrogen atom adds to the carbon atom of the double bond that already has the greater number of hydrogen atoms. An example is
10.47
10.49
(a)
(b)
(c)
(d)
(e)
10.51
(a)
(b)
(c)
(d)
(e)
10.53 The Pauling electronegativity number for carbon is 2.5 and for hydrogen is 2.1. This small difference (0.4) means that the two electrons in the C–H bond are essentially equally shared between the two atoms. The result is a covalent bond that is essentially non-polar.
10.55 Hexane is an essentially non-polar molecule with relatively weak dispersion forces as the main intermolecular forces. Water is a highly polar molecule with the comparatively stronger hydrogen bonding as the main intermolecular force. Non-polar hexane has no groups that can act as hydrogen-bond donors or acceptors. As a result, the intermolecular forces holding water molecules together (hydrogen bonds) are much stronger than any forces between water molecules and hexane molecules, thus the liquids are immiscible.
10.57 In CCl4, even thought there is an electronegative chlorine (χ = 3.0) covalently bonded to a less electronegative carbon (χ = 2.5) the four dipoles cancel each other out and the net result is a non-polar molecule. In CH2Cl2, there are two non-polar bonds (C–H) and two polar (C–Cl) bonds. These dipoles do not cancel each other out and the result is a polar molecule.
10.59 The carbonyl group of acetone would have a dipole, as shown below. The + end of the dipole is centred on the carbon atom, whereas the – end is centred on the oxygen atom. Attack by a nucleophile (such as the OH– ion) would occur at the positive end of the dipole, as shown below.
10.61 There are two factors that make acetic acid more acidic than ethanol. First is the presence of a second electronegative oxygen atom. This helps make the negative charge that exists on the acetate anion more favourable overall. Second is the fact that the negative charge on the acetate anion can be delocalized over two oxygen atoms through a process known as resonance, which helps to stabilize the negative charge. There is neither an extra electronegative atom nor resonance stabilization to help make the ethoxide anion more stable.
10.63 The configuration changes only because the atomic number for sulphur is higher than the atomic number for oxygen. This makes the CH2–SH substituent a higher priority than the COOH substituent on cystine. On serine, the COOH substituent has a higher priority (because of two oxygen atoms) than the CH2–OH group. This change in priority results in a change in the configuration of the two different chirality centres.
10.65 The two stereoisomers (2R,3R)-2,3-dibromopentane and (2S,3S)-2,3-dibromopentane are enantiomers of each other. The two stereoisomers (2R,3S)-2,3-dibromopentane and (2S,3R)-2,3-dibromopentane are also enantiomers of each other. The (2S,3S)-2,3-dibromopentane and the (2R,3R)-2,3-dibromopentane stereoisomers are diastereomers of the (2R,3S)-2,3-dibromopentane and the (2S,3R)-2,3-dibromopentane. This is illustrated below.
10.67 In the absence of acid, the leaving group on tert-butanol would be a hydroxide ion, which is the conjugate base of a weak acid (H2O). Hydroxide ion is, therefore, a very poor leaving group. By adding acid, the first step in the mechanism is to protonate the oxygen atom of the alcohol. In this case then, the leaving group is a molecule of water, the conjugate base of a strong acid (H3O+). This makes water a very good leaving group.
10.69 In the addition of HCl to a double bond, the first step is protonation of the double bond and the formation of a carbocation on one of the carbon atoms. The presence of the highly electronegative fluorine atoms strongly disfavours formation of this carbocation.
10.71 The critical step in the SN1 reaction is the formation of the carbocation intermediate. This carbocation is formally an sp2-hybridized carbon atom that is planar, with 120° bond angles and an empty p orbital. The incoming nucleophile has no preference as to which side of the carbocation to attack to form the new covalent bond. In this example, the result is an equimolar mixture of the (S)-1-bromo-1-phenylethane, the result of attack from the “top” of the carbocation, and (R)-1-bromo-1-phenylethane from attack at the “bottom” of the carbocation.
10.73 The reactions show that with a small base (i.e., NaOCH3), substitution is the result via an SN2 mechanism. The larger base (NaO-tert—Bu) results in only the elimination product being formed. The conclusion is that elimination is favoured by larger bases.
10.75 The tert-butyl chloride reacts faster, because under SN1 conditions the rate-determining step is formation of the carbocation. Tertiary carbocations are more stable than primary carbocations and this is demonstrated by this example. It is much easier to form a tertiary than a primary carbocation. If you picture each C–H bond on the adjacent methyl groups as a “cloud” of electrons (and this is a pretty accurate picture), then each bond is able to donate a little of this cloud into the empty p orbital on the carbocation through a process called hyperconjugation. Branched alkyl groups have a stronger electron donation effect than straight-chain alkyl groups. It is this slight donation of each of the CH bonds in the alkyl branches that accounts for the extra stability of the tertiary carbocation.
10.77 The fact that more energy is released when Z-2-butene is hydrogenated must mean that there was more energy stored in this double bond than in E-2-butene. This leads to the conclusion that E-alkenes are more stable (i.e., lower in energy) than Z-alkenes. The reason for this is that in Z-2-butene, the two methyl groups that are on the same side of the double bond can interact with each other, raising the energy of the molecule. In E-2-butene, this unfavourable interaction is removed as the two methyl groups are as far apart as possible, lowering the overall energy of the system.
10.79 The cycloheptatrienyl cation is an aromatic system, the same as benzene. This is demonstrated by the fact that all of the bond lengths are of equal length and the molecule is planar owing to the presence of the sp2-hybridized positively charged carbon and the sp2-hybridized carbons of the double bonds.
10.81 The molecular orbital for the π bond would be formed by combination of the two atomic p orbitals. The “in-phase” overlap would then result in a region of greatest probability for finding the two electrons in the bonding molecular orbital between the two carbon atoms. This overlap would exist in a region above and below the plane of the carbon–carbon sigma bond, preventing the rotation about this bond. Any rotation out of the plane would result in loss of the favourable atomic orbital overlap that generates the bonding molecular orbital.
10.83 The enantiomer of D-glucose is L-glucose and is the complete mirror image of the molecule, as shown below. All of the chiral centres in D-glucose have the (S) configuration, whereas all of the chiral centres in L-glucose have the (R) configuration.
10.85 In general, an E2 reaction can take place on any substrate with a hydrogen atom on a carbon atom adjacent to the leaving group. For the E2 reaction, it makes little difference if the leaving group is on a primary, secondary, or tertiary carbon atom. However, the SN2 mechanism is favoured by primary and some secondary but never by a tertiary substrate. Moreover, the size of the base has an effect. Large, strong, and bulky bases, such as tert-butoxide, favour the elimination pathway. Smaller bases that are derived from weaker acids favour the substitution pathway.
10.87 The electronegativity numbers for hydrogen (2.1) and boron (2.0) mean that the hydrogen atom is more electronegative than the boron atom and that the electrons in the B–H bond are polarized toward the hydrogen atom. This means that in an electrophilic addition, the mechanism occurs as shown below, where the boron actually adds first to the double bond. The second step is the attack of an H– (a hydride ion) on the carbocation. This mechanism agrees with the definition of Markovnikov’s rule in Problem 10.86, since the most stable carbocation is still the intermediate in the reaction.
10.89In this reaction the relative energy difference between starting material and product is the same as in Problem 10.88, and there will be a net release of 63 kJ/mol. The major and significant difference is that this reaction proceeds with the initial formation of a carbocation intermediate. This intermediate will be higher in energy than either the starting material or the product but will have a measurable existence (unlike a transition state). Therefore, the diagram has to take this into account as shown below.
10.91 This trend is explainable based on the electronegativity of chlorine (3.0) compared with hydrogen (2.1). The electronegative chlorine atom will help to stabilize the negative charge that forms on the conjugate base of each of these acids. Each chlorine induces a dipole in the C–Cl bond that is then transferred through the other C–C bonds to the oxygen atom that carries the negative charge. The more chlorine atoms on the acid, the stronger this inductive effect and the more stable the anion, which results in a stronger base. This is shown below.
10.93 Note: There in fact may be more than one possible synthesis of each product from the given starting material; the solutions suggested below may not be the only possible legitimate pathway.
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