solutions for chapter 5 end-of-chapter problemsquantum.bu.edu/.../text/answers/chapter05.pdf ·...

January 2005 ACS Chemistry Chapter 5 suggested solutions 1

Solutions for Chapter 5 End-of-Chapter Problems

Problem 5.1. This is the structure of 2-methyl-1-propanol, C4H10O.

H C C C O H

H

H

H

C

H

H

H

H H

We are asked to identify which, if any, of these other structures is also a representation of 2-methyl-1-propanol.

(i)

H C C C H

H

H

O

C

H

H

H

H H

H

(ii)O C C C H

H

H

H

H

C

H

H

H

H H

(iii)

H

C

C

C

H

HH

C

O HH

H

H H

H

Structure (i) is not a representation of 2-methyl-1-propanol. Note that the –OH group is attached to the second carbon atom of the three-carbon chain, not the first. You can check this answer by making models of 2-methyl-1-propanol and of structure (i). You will find that you cannot superimpose the two models. Structure (i) is 2-methyl-2-propanol. [NOTE: Compounds that are not identical always have different names.] Structure (ii) is not a representation of 2-methyl-1-propanol. Although the –OH group is correctly attached to the first carbon atom, the four carbons are attached in a chain, not branched as in 2-methyl-1-propanol. The fact that the last –CH3 group is written above the line of the other three carbon atoms is just a different paper and pencil representation and this –CH3 group is not attached to the second carbon atom. You can check this by making models of 2-methyl-1-propanol and of structure (i). You will find that you cannot superimpose the two models. Structure (ii) is 1-butanol. Structure (iii) is a representation of 2-methyl-1-propanol. Note that the –OH group is attached to the first carbon atom in a three carbon chain, and the –CH3 group is attached to the second carbon atom in this chain. You can check this by making models of 2-methyl-1-propanol and of structure (iii). You will find that you can superimpose the two models.

Problem 5.2. Since the molecular formulas of the green and reddish-purple compounds the student made are the same, Co(H2NCH2CH2NH2)2Cl3, the compounds must be isomers.

Structure of Molecules Chapter 5

2 ACS Chemistry Chapter 5 suggested solution

[NOTE: In Chapter 6, metal ion complexes are introduced. The compounds here are metal ion complexes with each of the 1,2-diaminoethane (ethylenediamine) molecules bonded at two of the six octahedral positions around a central Co(III) ion and chlorides bonded at the other two positions. The green compound has the chlorides bonded on opposite sides (trans) of the metal ion as far from one another as possible in this structure. In the reddish-purple compound the two chlorides are bonded at adjacent positions (cis).]

Problem 5.3. Both 1-pentanol and 2-pentanol have the same molecular formula, C5H12O. (a) Draw the structure of each alcohol. (b) Predict which of these two alcohols has the higher boiling point. Explain the reason for your prediction. (c) Draw all alcohols with the formula C5H12O. Hint: there are eight total including 1-pentanol and 2-pentanol. (a) The structures of 1-pentanol and 2-pentanol (molecular formulas, C5H12O) are:

H C C C C C O H

H

H

H

H

H

H

H

H

H

H

H C C C C C H

H

H

H

H

H

H

H

O

H

H

H 1-pentanol 2-pentanol* (b) Based on a comparison with 1-butanol and 2-butanol in Table 5.1, the prediction (and fact) is that the boiling point of the straight-chain alcohol, 1-pentanol, is likely to be greater than that of the branched alcohol, 2-pentanol. Although hydrogen-bonding attractions among molecules of both alcohols is about the same, dispersion forces between molecules of straight-chain alcohols are greater than the dispersion forces between branched alcohols. This is because the branched-chain molecules are more compact and have less surface area to interact with their neighbors. More energy is required to change the straight-chain molecules from the liquid to the gas phase, which is reflected in higher boiling points for straight-chain alcohols compared to branched alcohols of the same molecular mass. (c) The structures of the other six C5H12O alcohols are:

H C C C C O

H

H

H

H H

H

H

HC

HHH

H C C C C O

H

H

C

H

H

H

H

H

H

HHH

H C C C O

H

H

H

H

HC

HHH

CHHH

2-methyl-1-butanol* 3-methyl-1-butanol 2,2-dimethyl-1-propanol

H C C C C H

H

H

H

H

C

O

H

H

H

HHH

H C C C C H

H

H H

H

O

H

H

H

CHHH

H C C C C C H

H

H

H

H

H

O

H H

H

H

H

2-methyl-2-butanol 3-methyl-2-butanol* 3-pentanol

Chapter 5 Structure of Molecules

ACS Chemistry FROG 3

[NOTE: Three of the compound names (2-pentanol, 3-pentanol, and 3-methyl-2-butanol) are marked with an asterisk to denote that these structures actually exist in two stereoisomeric forms as mirror image or optical isomers that are introduced in Section 5.9. In each of these molecules, there is a carbon atom to which four different groups/atoms are bonded. For example, in 2-pentanol, the four groups are –H, –OH, –CH3, and –CH2CH2CH3. What are the four groups in the other two cases?]

Problem 5.4. When comparing isomeric branched and straight-chain alkanes, the branched alkanes invariably have lower boiling points than their straight-chain isomeric cousins, because the branched structures are more compact and have a smaller surface area for dispersion attractions between the molecules. Thus the branched isomers require less energy to break free and vaporize, so can do so at a lower temperature. A similar effect applies to the isomeric variants of other functional groups (see the solution to Problem 5.3). Therefore, we predict that 2-propanol should boil at a lower temperature than 1-propanol. This prediction is borne out by experiment: 2-propanol boils at 82 °C and 1-propanol at 97 °C.

Problem 5.5. (a) One way to represent the propane molecule in three dimensions is to focus on the central carbon atom and try to represent the geometry around this atom, as in structure (i). Another, somewhat more complicated representation is shown in structure (ii), where the shape around each carbon is represented. The symbolism for these representations is explained in Section 5.8: lines represent bonds in the plane of the paper and wedges represent bonds coming forward out of the paper (full wedge) or backward out of the paper (dashed wedge).

(i)

H

CH

CH3

CH3 (ii)

CCC

HH

HH

H HH H

(b) The shape (geometry) around the central carbon atom is tetrahedral (a tetrahedron, regular triangular prism, is formed by connecting the four atoms directly bonded to the central carbon atom), as shown most easily in structure (i) in part (a). (c) There are no other isomers of this structure. At least four carbons are necessary before more than one linkage of carbon atoms is possible for an alkane.

Problem 5.6. (a) Each bond line shown in a Lewis structure represents a pair of bonding electrons, so the 16 bond lines in this Lewis structure of pentane represent 32 valence electrons in the molecule:

H C C C C C H

H

H

H

H

H

H

H

H

H

H

(b) In a Lewis structure, the atomic symbols represent the atomic core of the element. A carbon atomic core has two electrons (and a nucleus with six protons and six neutrons) and a hydrogen atomic core has no electrons (it’s just a nucleus with a single proton). There are five carbon



atomic cores represented in this Lewis structure of pentane, so there are a total of 10 core electrons implicitly represented or a total of 32 + 10 = 42 total electrons represented.

Problem 5.7. Lewis structures of the three pentane isomers are shown in Worked Example 5.8 and photographs of their molecular models in Figure 5.2. Because hydrocarbons are nonpolar, the only intermolecular forces among these molecules are dispersion forces. The molecular volume of the three pentane isomers must be almost the same because the volume of each consists of five carbon atoms and twelve hydrogen atoms. However, their surface areas are not the same. The most nearly linear isomer (pentane) has the greatest surface area and therefore has the most intermolecular attractions due to dispersion forces. This in turn leads to the highest boiling point, 36.1 C, because more energy is required for molecules to break free and enter the gas phase. 2,2-dimethylpropane has the most nearly spherical shape and has the smallest surface area of the three pentane isomers. Therefore, it will have the lowest boiling point, 9.5 C. (Recall from geometry that the sphere is the geometric solid that minimizes the surface area for any given volume.) In general, branching leads to more compact structures and correspondingly lower boiling points.

Problem 5.8. All three isomeric ethers with the formula C4H10O have polar carbon-oxygen bonds, but only the circled isomer is branched.

H3C

H2C

CH2

OCH3

H3C

H2C

O

H2C

CH3 H3CCH

OCH3

CH3

As we have found in the solutions for Problems 5.3, 5.4, and 5.7, branched isomers with similar polar interactions have lower boiling points than their straight-chain counterparts, because the dispersion forces between the branched molecules are smaller than between the more linear structures. The boiling points of these three ethers are, respectively, 38-9 C, 34.5 C, and 32.5 C. The affect of this small amount of branching is not large, but is in the predicted direction.

Problem 5.9. These are the structures of the four isomeric amines with the formula C3H9N:

H3C

H2C

CH2

NH2 H3C

H2C

NH

CH3 H3CCH

CH3

NH2

H3CN

CH3

CH3

A B C D Two of the amines (A and B) are linear and two (C and D) are branched to the same extent. In previous problems, we have argued that branching lowers the boiling points of isomers, because the dispersion forces are smaller between the more compact molecular structures. However, this argument is only valid for isomers whose polar interactions, if any, are roughly the same. For these amines, branching is not the most important factor influencing the boiling points. Compounds A, B, and C are all capable of forming intermolecular hydrogen bonds, whereas



compound D is not. Since, for rather small molecules like these, hydrogen bonding is a stronger intermolecular force than dispersion, compound D (lacking hydrogen bonds) has the lowest boiling point.

Problem 5.10. Valence electrons are generally responsible for chemical bonding, because these electrons are in the shell farthest from the nucleus and, are therefore, the ones that come in contact with the valence electrons of another atom. It is interactions of these valence electrons with the two atomic cores that can result in the formation of a chemical bond.

Problem 5.11. Atomic orbitals describe the energy and spatial distribution of electrons attracted to a single atomic nucleus (and interacting with the other electrons also attracted to this nucleus). Molecular orbitals describe the energy and spatial distribution of valence electrons simultaneously attracted to two (or more) atomic cores. [In one model of molecular orbital formation, molecular orbitals are imagined to be formed from combinations atomic orbitals.]

Problem 5.12. Explain the difference(s) between a bonding and a nonbonding molecular orbital. Bonding molecular orbitals describe the valence electrons simultaneously attracted to two (or more) atomic cores and responsible for chemical bonds between (among) them. Nonbonding molecular orbitals describe valence electrons in molecules that remain associated with a single atomic core and are not involved in bonding.

Problem 5.13. (a) In the Lewis structure for bonding, a hydrogen molecule is represented as H:H or H–H. We can imagine that two hydrogen atoms combine to form a hydrogen molecule like this:

H• + •H H–H

In this molecule, each hydrogen atomic core is associated with a bonding electron pair (the same electron pair), which is just what the Lewis model suggests for all hydrogen atomic cores bonded in molecules. (b) The MO model for bonding in the hydrogen molecule is essentially that shown in Figure 5.5(b). That figure is supposed to represent the probability distribution for a single electron simultaneously attracted to two hydrogen atomic cores (protons), but it can also represent the probability distribution for a pair of electrons simultaneously attracted to the two hydrogen atomic cores. The resulting electron distribution is an electron-pair sigma (�) bonding orbital. The molecule is stable (relative to the separate atoms) because the negative potential energy of attraction between the electrons and two nuclei is enough to overcome the positive potential energy of mutual repulsion between the nuclei and between the electrons as well as the positive kinetic energy of the electron waves.

Problem 5.14. (a) If we imagine that two helium atoms were to combine to form a helium molecule, He2, the Lewis model would not be able to describe its bonding. The bonding between two first period elements, as in H2, is described as an electron pair shared between the two nuclei. [See the solution for Problem 5.13(a).] Such a bond between two helium atomic cores, He:He, would



accommodate two electrons, but there are two more furnished by the two He atoms and there is no place in the Lewis model to accommodate them. This leads to the conclusion that two helium atoms do not form a molecule more stable than the separated atoms. Another way to look at this is to consider that, in the Lewis model, first period elements can share or have only two valence electrons and the He atom already has two valence electrons, so there is no drive for two of these atoms to bond. (b) If we apply the MO model to bonding in the imagined He2, we would first form a sigma bonding orbital like that depicted in Figure 5.5(b). This orbital could accommodate two of the four electrons that would be present in He2. At this point in the text, we have no indication what to do with the remaining two electrons. By analogy with atomic structure, we might think about accommodating these electrons in a higher energy orbital sort of analogous to another electron “shell” (or wave) farther from the nuclei. Such an orbital would be unlikely to stabilize the He2 molecule, since it would place electrons farther from the nuclei than they would be in their atomic orbitals in He, which would require an unfavorable energy change. Furthermore, such an orbital would not place electron density between the two 2+ nuclei that are repelling one another, so would not help to offset this repulsion. Thus, the MO model leads to the conclusion that formation of He2 is likely to be an energetically unfavorable process that will not occur.

Problem 5.15. Molecules exist, even though there is always a repulsive force between their positively charged nuclei (atomic cores), because the contributions of the potential energy from simultaneous attractions of electrons to two (or more) nuclei makes the total energy of the molecule lower than the combined energies of these electrons in separated atoms.

Problem 5.16. The number of valence electrons in a molecule is equal to the sum of the number of valence electrons in the atoms that are bonded to form the molecule. This is also the number of electrons that must be represented in a Lewis structure that shows both bonding and nonbonding electrons. For elements in the representative families (groups) that are designated with a Roman numeral (in the periodic table on the inside front cover of the textbook), the number of valence electrons in the elemental atom is the same as the group number. Thus: (a) Na (group I) has 1 valence electron. (b) CO2 has 16 (= 4 + 6 + 6) valence electrons. (c) NH3 has 8 (=5 + 1 + 1 + 1) valence electrons. (d) Se (group VI) has 6 valence electrons.



Problem 5.17. As the two nuclei move toward each other, relative energies for this two-nuclei-one-electron molecular orbital system are shown here (and in Figure 5.4) with the energy curves labeled 1-4. Each curve provides us information about what is happening in the system. (a) Curve 1 shows that the kinetic energy of the one-electron molecular orbital system increases as the nuclei get closer together and, hence, the bonding electron wave becomes smaller and higher in energy. (b) Curve 2 is the reference baseline for energy and does not change as the two nuclei approach each other. (c) Curve 3 shows that, as the two nuclei approach each other, the total energy of the one-electron molecular orbital system falls to a minimum and then increases rapidly as the nuclei are brought closer together and the electron wave is constrained even further. (d) Curve 4 shows that, as the two nuclei approach each other, the potential energy of the system decreases because the distances between the electron and the nuclei get smaller and the potential energy of attraction between the electron and the two nuclei grows.

Problem 5.18.

(a) The Lewis structure for the methide anion, CH3–, is

C HHH .

(b) The sigma orbitals, three bonding and one nonbonding, in the methide anion are tetrahedrally arranged around the carbon (just as the four sigma bonding orbitals are in methane, CH4). (c) The geometry (the arrangement of the nuclei) of the methide anion is a trigonal pyramid

with carbon at the apex and the hydrogen atomic cores forming the base, C

HHH

. (d) The arrangement of sigma orbitals and the geometry of the methide anion are not the same, because the nonbonding electron pair is not included in the geometric description. The structure of the methide anion is much like the structure of ammonia, Figure 5.10.

Problem 5.19.

(a) The Lewis structure for the amide anion, NH2–, is

N HH

(b) The sigma orbitals, two bonding and two nonbonding, in the amide anion are tetrahedrally arranged around the nitrogen (just as the three sigma bonding and one sigma nonbonding orbitals are in ammonia, NH3).



(c) The geometry of the amide anion is bent,

N

HH.

(d) The arrangement of sigma orbitals and the geometry of the amide anion are not the same, because the nonbonding electron pairs are not included in the geometric description. The structure of the amide ion is much like the structure of water, Figure 5.10.

Problem 5.20. (a) The Lewis structure for IF5 has 42 valence electrons, seven valence electrons from each of the six group VII elements in the compound. (b) A molecule of IF5 has a total of 98 electrons, including all of the core electrons. This total is obtained by adding the atomic numbers of all the constituent atoms (5·9 + 53 = 98). The number of electrons in each atom is equal to the number of protons, which is the atomic number of the atom. (c) A Lewis structure for the 42 valence electrons in IF5 can be represented like this:

IF F

F FF

We don’t usually try to show geometries in Lewis structures, but, in this case, a roughly octahedral geometry of five sigma bonding electron pairs and one sigma nonbonding electron pair on the central iodine atomic core is indicated, in order to help visualize the molecular geometry discussed in part (d). (d) The geometry of the fluorine atoms in the IF5 molecule is square pyramidal (a pyramid with a square base), as represented in the structure shown in part (c). This is different from the molecular shape of PF5, in which the geometry of the fluorine atoms is trigonal bipyramidal, as shown in Figure 5.11(b). There are only 40 valence electrons in PF5 and, therefore, only five sigma orbitals (all bonding) around the central phosphorus atomic core. The most efficient way to pack the five sigma electron pairs around the central atom is in the trigonal bipyramidal arrangement. For six sigma electron pairs, an octahedral arrangement is most efficient, which leads to the structure shown in part (c).

Problem 5.21. Lewis structures, including all non bonding pairs of electrons, are shown for five molecules or ions. Also shown are 3-d structures to explain the geometries described in Problem 5.22. (a) I3

– has 22 (= 3·7 + 1) valence electrons (4 � and 18 �n). The central iodine is surrounded by five pairs of electrons that will adopt a trigonal bipyramidal arrangement with the two other iodine atomic cores 180° apart at the opposite ends of the structure. For clarity, only the electron pairs on the central iodine are shown in the 3-d structure on the right.



I I I I I I

(b) BF4– has 32 (= 3 + 4·7 +1) valence electrons total (8 � and 24 �n). Four pairs of bonding

electrons surround the in a tetrahedral arrangement.

F

BF

FF

F

BF

FF

(c) SF4 has 34 (= 6 + 4·7) valence electrons total (8 � and 26 �n). The sulfur has five pairs of electrons that will adopt a trigonal bipyramidal arrangement about this central atom. The nonbonding pair is located at one of the sites on the triangular base shared by the two pyramids. This arrangement keeps the nonbonding electron pair orbital on S, which occupies more space (because it is attracted by only one atomic core), as far from the fluorine nonbonding electrons as possible in this geometry. The sigma bonding orbitals define the geometry of the fluorine atomic cores, which looks like a “see-saw” with the S as the fulcrum (pivot point).

SF F

F F

F S F

F F

F S F

F F

(d) XeF4 has 36 (= 8 + 4·7) valence electrons total (8 � and 28 �n). The xenon has six pairs of sigma electrons that will adopt an octahedral arrangement with the two nonbonding pairs of electrons 180° apart at the opposite sides of the structure. This arrangement keeps the nonbonding electron pair orbitals, which occupy more space [see part (c)] as far apart as possible in this geometry The sigma bonding orbitals and the fluorine atomic cores are in a square planar arrangement with respect to the xenon.

XeF F

F

F

F F

F

F

Xe



(e) PF6– has 48 (= 5 + 6·7 + 1) valence electrons total (12 � and 36 �n). Six pairs of bonding

electrons surround the phosphorus in an octahedral arrangement.

P

F

F

F

F

F

F

F

F

F

F

F

P

F

Problem 5.22. The reasoning for the geometry of each molecule or ion in Problem 5.21, is given in the solution to Problem 5.21 and explains these observations: (a) I3

– is linear. (b) BF4

– is tetrahedral.

(c) SF4 is shaped like a “see-saw”, , with S at the fulcrum. (d) XeF4 is square planar, with the Xe at the center of the square. (e) PF6

– is octahedral with the P in the center of the octahedron.

Problem 5.23. The second period elements, carbon, nitrogen, and oxygen are capable of making double bonds. Lithium is not included, because its compounds are almost all ionic (the Li+ cation). Beryllium compounds are also ionic (the Be2+ cation), although showing some covalent character as well, Boron compounds are covalent, but usually “electron deficient” (see the solution to Problem 5.58) with fewer than eight bonding or nonbonding electron pairs surrounding the boron atomic core. The low atomic core charge, 3+, means that attraction for electrons is weaker than in the elements further along in the period, so the lower number of electrons makes sense. The second period elements that form double bonds have eight bonding or nonbonding electron pairs surrounding the atomic core in essentially all of their compounds. Fluorine does not form double bonds because there is no place to accommodate more than one bonding electron pair on the atomic core that already has seven electrons.

Problem 5.24. Hydrogen and fluorine never make double bonds. The hydrogen atomic core (proton) can only accommodate one pair of bonding electrons (one sigma bonding orbital). Fluorine does not form double bonds because there is no place to accommodate more than one bonding electron pair on the atomic core that already has seven electrons. (All the higher period halogens can accommodate more than eight valence electrons and form many doubly-bonded molecules and ions, especially with oxygen.) No compounds of the lighter noble gases have been made and it is unlikely that either He or Ne would form double bonds (first and second period elements with filled electron shells); Xe bonds to O via double bonds. The alkali metals and alkaline earth metals form almost exclusively ionic compounds; the concept of multiple bonds in ionic compounds makes no sense because the cations are not bonded to any particular anion and vice versa. For the rest of the elements (including the representative elements and the transition



metals), double bonds are either common or have been found for virtually every element in some compound.

Problem 5.25. The Lewis structure and the molecular geometry for each carbon atom are shown here for five molecules, each of which has a multiple bond. (a) Propene, C3H6, has one carbon-carbon double bond:

CC

CH

H

H

H

HHtetrahedral

trigonal planar

trigonal planar

(b) Propyne, C3H6, has one carbon-carbon triple bond:

C C CH

HH

H

tetrahedrallinear

linear

(c) Allene, C3H4, has two carbon-carbon double bonds:

C C C

H

HHH

trigonal planar

linear It is important to make a molecular model of this molecule, in order to see how the CH2 groups at the ends of the molecule are oriented at right angles to one another. It is as though the Hs were directly bonded to the central carbon atomic core, instead of to carbons doubly bonded to the central carbon atomic core. (d) Ethanal, C2H4O, has one carbon-oxygen double bond:

CC

O

H

H

HHtetrahedral

trigonal planar



(e) Acetonitrile, C2H3N, has one carbon-nitrogen triple bond:

C C NH

HH

tetrahedral

linear

Problem 5.26. Both sigma (�) and a pi (�) orbitals are molecular orbitals that describe the interaction of electrons waves with two (or more) atomic cores (or a single atomic core bonded to others). Sigma orbitals are cylindrically symmetric about the axis connecting two atomic cores and concentrate electron density between the positively charged cores. Pi bonds concentrate electron density off-axis on opposite sides of a plane that contains the axis connecting the two sigma bonded atomic cores. Because sigma bonds concentrate electron density between the atomic cores, they are typically stronger than pi bonds.

Problem 5.27. There is a two-electron sigma bonding orbital between every pair of bonded atom cores in a molecule. A sigma orbital has the greatest electron density along the bond axis, forming a bond with the lowest possible total energy for the two nuclei system. A pi-bonding orbital is not needed if sigma bonding (and nonbonding) orbitals successfully account for all of the valence electrons.

Problem 5.28. If two carbon atoms are bonded by a sigma bonding orbital, a second bond between those two atoms is a pi orbital rather than another sigma bonding orbital because another orbital that put two more electrons in the same space as the first two would be energetically very unfavorable (as well as possibly disobeying the Pauli exclusion principle). The pi bonding orbital has its greatest electron density away from the bond axis, not directly between the bonded atom cores, so the pi electrons are more-or-less out of the way of the sigma electrons.

Problem 5.29. Molecules with multiple bonds are generally more reactive than similar compounds without multiple bonds, because the pi bonding orbitals have a considerable amount of electron density away from the bond axis and these are more readily available for interactions with other molecules than are sigma bonding electrons.

Problem 5.30. Lewis structures (or partial structures highlighting multiple bonds) are shown for several molecules and ions. Each one that contains a � bond is noted with an indication whether it is a delocalized or localized � orbital. (a) Ethene, H2CCH2, has a double bond, H2C=CH2, so it has a localized � orbital. (b) Methanal, H2CO, has a double bond, H2C=O, so it has a localized � orbital. (c) The nitrate ion, NO3

–, has these three equivalent Lewis structures, each containing a double bond, so its � orbital is delocalized:



NOO

ONO

O

ON

O

OO

(d) Ammonia, NH3, has no multiple bonds, so it has no � orbital. (e) Carbon dioxide, CO2, has two double bonds, O=C=O, so it has two � orbitals. Although a single Lewis structure that obeys all the rules and accounts for all the electrons can be written for this molecule, the � orbitals are delocalized. See the solution for Problem 5.33 to see why. (f) Ethanoic acid (acetic acid), CH3C(O)OH, has a double bond, so it has a localized � orbital.

CH3CO

OH

(g) Methane, CH4, has no multiple bonds, so it has no � orbital. (h) Sulfur dioxide, SO2, has these two equivalent Lewis structures, each containing a double bond, so its � orbital is delocalized:

SOO

SO O

(i) Water, H2O, has no multiple bonds, so it has no � orbital.

Problem 5.31. (a) The hydrogen carbonate anion, HOCO2

–, has 24 valence electrons and the ozone molecule, O3, has 18 valence electrons. (b) The Lewis structures for the hydrogen carbonate anion and the ozone molecule are:

C OO

O

H

O

OO

(c) The central C atom in HOCO2– is surrounded by three sigma bonding orbitals and the central

O atom in O3 has two sigma bonding orbitals (and one sigma nonbonding orbital). (d) The O atoms around the C in HOCO2

– are arranged in a plane with O–C–O angles of about 120 ; the geometry about the C atom is trigonal planar. Ozone, O3, is a bent molecule; the three sigma orbitals are arranged roughly trigonally (but the angles between the orbitals are not all equal). (e) There is one pi bonding orbital around each central atom in these species. (f) Each species has one delocalized pi orbital. In the HOCO2

–, the pi orbital is delocalized over the central C atom and the two O atoms not bonded to an H atom. In ozone, the pi orbital is delocalized over all three atoms. In each case, a second Lewis structure, identical in energy to the ones above, can be written:

C OO

O

HC O

O

O

H

O

OOO

O O



Whenever two or more Lewis structures with the same energy can be written for a species, neither is an accurate representation; the pi orbitals involved are delocalized. (g) The bond orders between the C atom and the two O atoms not bonded to an H atom in HOCO2

– are each 1.5, one-and-one-half bonds. The same is true for the bonds between the central O atom and the other O atoms in ozone. In both cases, the sigma bond between the atoms is a full bond and the delocalized pi bond contributes one half a bond to each bond, for a 1.5 bond. Although these species are quite different in many ways, some aspects of their bonding (delocalization and 1.5 order bonds) are very much alike. Sketches of the delocalized � orbital would look something like the one shown in Figure 5.23(a), but over only three atom centers instead of four. [NOTE: Sometimes the delocalized � orbital is represented as shown here for ozone:

OOO

The problem is that the dashed curve representing the pair of electrons in the delocalized � orbital may be confusing and is too easily misinterpreted, especially when one is just beginning to learn to count electrons in Lewis molecular structures.]

Problem 5.32. (a) The Lewis structure for CO2 is:

C OO

There does not appear to be any other way to arrange the electrons that is equivalent to this one, so delocalized pi orbitals are not predicted by the Lewis structure. (b) The Lewis structure appears to have the two pairs of pi electrons localized in two-center bonds between the carbon and an oxygen atom. This is not consistent with the delocalized pi orbitals in Figure 5.24 where the pi electron pairs spread over all three atoms. (c) The model shown is one way to arrange the pi orbitals that are perpendicular to one another. The second picture here represents another way.

In these two models, the atoms have not switched places, but the localized pi orbitals are shown on different pairs of atoms. There are two energy-equivalent ways to arrange the pi orbitals. (d) Since there are two energy-equivalent ways to arrange the pi orbitals (which are not represented by the two-dimensional Lewis structure), delocalization of the pi electrons occurs and lowers the total energy of the molecule. Carbon dioxide is more stable (lower energy) than



localized pi orbitals would predict. The delocalization implied by having two equivalent ways to arrange the pi-bond paddles is the same as that shown in Figure 5.24 and the explanation for the C-O bond lengths that are shorter than for a C-O bond with a localized pi bond, is given in the answer to Check This 5.39(b).

Problem 5.33. For the nitrate anion, NO3

–, the nitrite anion, NO2–, and the nitronium cation, NO2

+, this table gives: (a) the total number of valence electrons; (b) the Lewis structure; (c) the number of � bonding orbitals around the central atom; (d) the predicted geometry; (e) the number of � bonding orbitals around the central atom; (f) the number of localized and delocalized � bonding orbitals; and (g) the bond order for nitrogen-to-oxygen.

nitrate anion nitrite anion nitronium cation (a) 1(5) + 3(6) + 1 = 24 e– 1(5) + 2(6) + 1 = 18 e– 1(5) + 2(6) – 1 = 16 e– (b)

NO

OO

NO O

N OO

(c) 3 � 2 � 2 � (d) trigonal planar bent linear (e) 1 � 1 � 2 � (f) 0 loc �, 1 deloc � 0 loc �, 1 deloc � 0 loc �, 2 deloc � (g) 11/3 11/2 2

(f) For the nitrate anion, there are two more Lewis structures (double bond to one of the other two carbons) that are identical in energy to the one in the table and for the nitrite anion there is one more such Lewis structure. Our criterion for delocalized � orbitals is that there be more than one suitable Lewis structure of the same energy for the species. Both anions fulfill this criterion, so their � orbital is delocalized. For the nitronium ion, there is no other Lewis structure we can write that is different from that one shown. However, models like those pictured in the solution for Problem 5.34(c) show that there are two pi orbitals delocalized over all three atoms and perpendicular to one another. (g) Each of the bonds in the nitrate and nitrite anions consists of a � bonding orbital and partial contribution from a delocalized � bonding orbital. In the nitrate anion, the delocalization is over the three N–O bonds, so the � contribution to each bond is 1/3 and the overall bond order for each bond is 1 + 1/3 = 11/3. In the nitrite anion, the delocalization is over the two N–O bonds, so the � contribution to each bond is 1/2 and the overall bond order for each bond is 1 + 1/2 = 11/2. In the nitronium cation, each of the bonds consists of a � bonding orbital and partial contributions from two delocalized � bonding orbitals. The delocalization is over the two N–O bonds, so the contribution of each � orbital to each bond is 1/2 and the overall bond order for each bond is 1 + 1/2 + 1/2 = 2. These bonds are both double bonds (as indicated by the Lewis structure, but not localized, that structure implies). (h) Bond lengths should go in the inverse order of bond order: the higher the bond order, the stronger the bond and the closer the atomic centers can be held to one another. Thus the bond



orders from shortest to longest bond lengths would be: 2 < 11/2 < 11/3. Nitronium would have the shortest N-O bonds and nitrate the longest.

Problem 5.34. (a) The energies of each of these three structures would be identical, if the structures existed separately, because each has two single N–O bonds and one double N=O bond. The double bond is just to a different one of the oxygen atomic cores in each structure.

N

O

OON

O

OON

O

O O

(b) Since there are three identical structures with identical energies, each would be expected to contribute exactly one-third to an intermediate structure based on equal contributions from each individual structure. (c) If we focus on one of the oxygen atoms, say the one at the bottom, we see that it is singly bonded in two structures and doubly bonded in the third. Thus, the sum of its bonding in the three structures is 4, but each contributes only a third, so its net is 4/3 or 11/3. , just as we found in Section 5.7 and the solution to Problem 5.33.

Problem 5.35. The electrons in the diffuse cloud representing the delocalized electrons in Figure 5.27 are relatively free to move about anywhere within the metal crystal lattice. If an extra electron enters the sea of electrons from a negative electrical terminal, another can immediately leave the electron sea at the positive terminal, in order to maintain the electrical neutrality of the crystal. The electron that enters does not have to be the one that leaves, which makes the flow of electrons through the crystal relatively easy and rapid. That is, the metal is a good conductor of electricity.

Problem 5.36. Dislocations in a metal crystal lattice weaken the lattice because the atomic centers are no longer symmetrically surrounded by other atom centers and the delocalized molecular orbitals are disrupted. A discontinuity in the bonding occurs at dislocations and, if they grow by further mechanical manipulation of the metal, the overall bonding becomes so weakened that the piece of metal can break along the discontinuities, which is why you can often break a piece of wire or thin strip of metal by bending it back and forth repeatedly.

Problem 5.37. [NOTE: This problem is probably out of place here and should go into Section 5.8, since part (c) involves optical isomers (or it can).]

The number of isomers possible for any particular molecular formula depends on the geometry about the atoms that make up the molecule. This is one of the ways chemists of the last century figured out that carbon (and other atoms) assumed a tetrahedral and not a square planar geometry. See for yourself! (a) For methyl bromide, CH3Br, only one isomer is possible for either square planar or tetrahedral geometry about the carbon atomic core. Two possible ways of drawing each



structure on paper are shown here. In both cases, the structures are identical and can be superimposed on one another by appropriate rotation without breaking any connections. You may need to make models to convince yourself of this for the 3-d tetrahedral structure.

C HBr

HH

CH

BrH

H

C

Br

HH

HC

H

BrH

H

square planar tetrahedral (b) For dichloromethane, CH2Cl2, two isomers are possible for square planar geometry. These structures cannot be superimposed without breaking and remaking connections in one of the structures.

C HCl

HCl

CH

ClCl

H

For tetrahedral geometry, only one isomer is possible. The two structures shown here can be superimposed on one another by appropriate rotation without breaking and remaking any connections. You may need to make models to convince yourself of this.

C

Cl

HCl

HC

H

ClH

Cl

(c) For propylene glycol, CH3CH(OH)CH2OH, three isomers are possible that cannot be superimposed without breaking and remaking connections in two of the structures. For ease in drawing and interpreting the structures, the –CH2OH group is represented as –R. The three structures can be characterized by saying that the –CH3, –R, and –H groups, respectively, are across the diagonal of the square from the –OH group.

C ROH

HCH3

CH

CH3

OHRC CH3

OHH

R For tetrahedral geometry, two isomers are possible, but these differ in a subtle way; they are mirror images of one another (see Section 5.9 and the discussion of optical isomers). Make models to convince yourself that these two structures cannot be superimposed on one another without breaking and remaking connections in one of the structures.

C

OH

RH3CH

C

OH

RCH3

H

mirror



Problem 5.38. The condensed representations and molecular formulas corresponding to skeletal structures are given for several molecules. Recall that each vertex or line end (which does not have an atomic core indicated) is understood to represent a carbon atomic core. Because carbon cores will always form four bonds in stable organic molecules, the number of hydrogen atomic cores bonded to any carbon atomic core you choose to examine is simply the mathematical difference between the number four and the number of bonds already specified at that carbon core.

(a) aspirin

O

O OH

O

HCHC

CH

CHC

CC

O

O OH

CCH3

O

C9H8O4

(b) nicotine N

N

HCHC

NCHC

HC CH N

CH2

CH2H2C

CH3 C10H14N2

(c) vitamin C

O OHO

HO

HO OH C C

CO

CH OCHCH2

HO

HO

HO OH C6H8O6

(d) cholesterol HO

H2CCH

CH2

CC

H2C

CH

CH2

CHCH CH

C

H2C

H2C

CH2

CH2CH

HO

CH3

CH3CH

H3CH2C

CH2

H2C

CHCH3

CH3

C27H46O

Note that the condensed structures begin to look very crowded and hard to interpret when the number of atomic cores represented grows. For large molecules, skeletal structures are often a good deal easier to interpret than other representations.

Problem 5.39. These pairs of structures are identified as conformers (i.e., the same molecule whose forms differ only in that there have been rotations about single bonds) and those that are actually different molecules (not interconvertible by any combination of rotations about single bonds or motions of the molecule as a whole). We have tried to show how to make the comparisons we have noted for each pair. [NOTE: It is extremely difficult to answer questions about molecular structure without the aid of molecular models. If you build and manipulate molecular models a lot, the two-dimensional drawings will eventually take on three-dimensional significance. Even the most experienced chemists still resort to physical models when investigating new or unusual structures. The answers to questions such as this one can usually be arrived at confidently if you have built the models.]



(a) These structures are conformers, that is, the same compound (not isomers of any kind). The curved arrows show how to rotate the right-hand structure in the plane of the paper and then rotate the methyl groups on the ends of the “arms” to bring them into the same geometry as the original left-hand structure.

(b) These structures are conformers, that is, the same compound (not isomers of any kind). The curved arrows show how to rotate the right-hand structure in the plane of the paper and then rotate the methyl group on one end of an “arm” to bring it into the same geometry as the original left-hand structure.

(c) These structures are mirror images of one another, that is, they are optical isomers. The curved arrow shows how to rotate the right-hand structure by 180 about an axis that coincides with the C–O bond to show that it is the mirror image of the original left-hand structure.

OH

H3C HCH2CH3

OH

H3C CH2CH3H

OH

CH3H

H3CH2C

(d) These structures are identical to one another, that is, the same compound. The curved arrow shows how to rotate the right-hand structure by 120 about an axis that coincides with the C–CH3 bond to show that it is identical to the original left-hand structure.

CH2CH3

H3C OHH

OH

H3C HCH2CH3

OH

H3C HCH2CH3

(e) These structures are identical to one another, that is, the same compound. The curved arrow shows how to rotate the right-hand structure by 180 about an axis down the center of the ring to show that it is identical to the original left-hand structure.

Br

Br

Br

Br

Br

Br (f) These structures are mirror images of one another, that is, they are optical isomers. The curved arrow shows how to rotate the right-hand structure in the plane of the paper to show that it is the mirror image of the original left-hand structure.

Br

Br Br

Br Br

Br



Problem 5.40. A 3-d structure of a conformation that allows all the carbon atomic cores in these molecules to lie in a single plane (if possible) is shown. (a) A structure for butane, CH3CH2CH2CH3, that allows all the carbon cores to lie in a plane is:

CC

CC

H H

H HHH

HH

HH

(b) There is no structure for 2-methylpropane, CH(CH3)3, that allows all the carbon cores to lie in a plane:

CC

CH

H H

H HHH

H

(c) A structure for butadiene, CH2CHCHCH2, that allows all the carbon cores to lie in a plane is:

CC

CCH

H

H

H H

H

(d) A structure for 1,2-propanediene, CH2CCH2, that allows all the carbon cores to lie in a plane is:

C CCH

HH

H

Problem 5.41. A 3-d structure of a conformation that allows all the carbon AND all the hydrogen atomic cores in these molecules to lie in a single plane (if possible) is shown in the solution to Problem 5.40, where the 3-d structures were drawn to put as many atomic cores as possible in the plane of the paper. The bonds connecting these cores are represented as solid straight lines in the drawings. (a) It is impossible for all the atomic cores in butane, CH3CH2CH2CH3, to lie in a plane. The molecular bonds can be rotated so that four carbon cores and two of the hydrogen cores can lie in a plane. (b) It is impossible for all the atomic cores in 2-methylpropane, CH(CH3)3, to lie in a plane. The molecular bonds can be rotated so that three carbon cores and two of the hydrogen cores can lie in a plane. (c) It is possible for all the atomic cores in 1,3-butadiene, CH2CHCHCH2, to lie in a plane. (d) It is impossible for all the atomic cores in 1,2-propanediene, CH2CCH2, to lie in a plane. There is only one possible molecular structure for this molecule and it has three carbon cores and two of the hydrogen cores at one end of the molecule lying in a plane.

Problem 5.42. A 3-d structure of a conformation that allows all the atomic centers in these molecules to lie in a single plane (if possible) is shown. The bonds connecting these cores are represented as solid straight lines in the drawings.



(a) A conformation that puts four of the six atomic cores in methanol, CH3OH, in a plane is possible, but two of the hydrogen cores bonded to the carbon core must lie outside (front and back) of the plane:

CO

H

HH

H

(b) A conformation that puts all four of the atomic cores in hydrogen peroxide, H2O2, in a plane is possible:

OO

HH

(c) All three atomic cores in a molecule with only three atomic cores. like hydrogen cyanide, HCN, must lie in a plane (or on a line), because three points define a plane.

C NH (d) A conformation that puts all five of the atomic cores in nitric acid, (HO)NO2, in a plane is possible:

NO

OO

H

(e) A conformation that puts all four of the atomic cores in nitrous acid, (HO)NO, in a plane is possible:

NO

OH

Problem 5.43. Each of these pairs of structures is identified as identical, optical isomers, and other types of stereoisomers.

(a) These structures are mirror images of one another, that is, they are optical isomers. The curved arrow shows how to rotate the right-hand structure in the plane of the paper to show that it is the mirror image of the original left-hand structure.

Br

Br Br

Br Br

Br (b) These are another type of isomer. They can’t be identical, because one structure has both bromine atomic cores on the same side (face) of the ring (cis) and the other has them on opposite sides (trans). Nor, for the same reason, can they be mirror images of one another, that is, they cannot be optical isomers.

Br

Br Br

Br



(c) These structures are conformers, that is, the same compound (not isomers of any kind). The curved arrows show how to rotate the right-hand structure about the axis of the molecule, then in the plane of the paper, and finally rotate the right end of the structure to bring it into the same geometry as the original left-hand structure.

(d) These are another type of isomer. They can’t be identical, because one structure has both groups on the same side of the double bond (cis) and the other has them on opposite sides (trans). Nor, for the same reason, can they be mirror images of one another, that is, they cannot be optical isomers.

OH OH (e) These structures are mirror images of one another, that is, they are optical isomers. The mirror shows how the structures reflect into one another.

OH OH

mirror (f) These structures are identical (not isomers of any kind). The curved arrow shows how to rotate the right-hand structure about the axis shown to bring it into the same geometry as the original left-hand structure. Rotation of the molecular structure brings the –OH group from in back of the plane of the paper to in front of the plane.

OH OH OH

(g) These structures are mirror images of one another, that is, they are optical isomers. They are redrawn from the structure in the textbook to show all four groups attached to the second carbon from the right and then the arrows show how 120 rotation of the end of the right-hand structure followed by rotation of the entire structure by 180 makes the right-hand structure the mirror image of the left-hand structure.

CH3H3C

OH

OH

OHH

OH

CH3H

CH3

HHO

HO Hmirror



(h) These structures are mirror images of one another, that is, they are optical isomers. They are redrawn from the structure in the textbook to emphasize the 5-carbon chain that lies in the plane of the paper. Then each structure is rotated 90 with the edge-on view of the plane of the 5-carbon chain represented by the thick bar. The direction that the chlorine atomic centers project from the plane is shown and the mirror shows that the structures are mirror images of one another. The edge of the plane that is facing toward you is different from the edge at the other end of the structure, so the mirror images shown cannot be simply flipped around to make them superimposable (that is, identical).

Cl Cl

Cl ClCl

Cl

Cl

Cl

Cl

Cl

Cl

Clmirror

(i) These are another type of isomer. They can’t be identical, because the structure on the left has both chlorines on the same side of the plane defined by the 5-carbon chain [see part (h)] and the right-hand structure has them on opposite sides of the plane. Nor, for the same reason, can they be mirror images of one another, that is, they cannot be optical isomers.

Cl Cl

Cl Cl

(j) These structures are mirror images of one another, that is, they are optical isomers. They are redrawn and rotated as in part (h) to show their mirror image relationship.

Cl Cl

Cl ClCl

Cl

Cl

Cl

Cl Cl

mirrorCl Cl



Problem 5.44. (a) This table compares four properties of structural and stereoisomers:

Stereoisomers Property Structural

Isomers Optical Isomers Other molar mass same same same boiling point different same different density different same different bond connectivity different same same

(b) For one of these data columns, all the entries are “the same.” By definition, two compounds cannot be isomers if all of their properties are the same. What is the property that allows us to distinguish the isomers in this column?

(b) For optical isomers, all entries in the table are “the same.” By definition, two compounds cannot be isomers if all of their properties are the same. Optical rotation (measured by a polarimeter) distinguishes the “right” and “left” handed optical isomers from one another. Whatever the extent of rotation of plane polarized light by one isomer, the other rotates plane polarized light an equal amount, but in the opposite direction. One rotates clockwise; the other counterclockwise. In chemical reactions with other molecules, especially those that are also optically active, optical isomers often react differently to give products that may also be optical isomers or another kind of stereoisomeric pair.

Problem 5.45. There are six non-cyclic isomers and five cyclic isomers of the formula C5H10: 1-pentene trans-2-pentene

cis-2-pentene 2-methyl-1-butene

3-methyl-1-butene 2-methyl-2-butene

cyclopentane methylcyclobutane

1,1-dimethylcyclopropane

trans-1,2-dimethycyclopropane

cis-1,2-dimethycyclopropane There are eleven C5H10 isomers compared to six C4H8 isomers (Investigate This 5.55). As the number of carbon atoms increases, the number of possible isomers increases. This makes sense because there are more ways for the carbon chain to branch and rings to form as well as more possible places for multiple bonds to be located in molecules like these.



Problem 5.46. This is a model-building problem that is designed to help you see how flexible (or rigid) molecular structures may be. (a) A model of cyclopentane is not a planar pentagonal ring, but rather is slightly puckered. The interior angles of a regular pentagon are 108 , but the tetrahedral C–C–C bonding angle at each cyclopentane carbon center is a little larger than 109 . This is not a large difference, but it does cause the ring not to be planar. If you choose any three adjacent carbon atomic centers to define a plane, one of the remaining two will be a bit above this plane and the other a bit below. It is relatively easy to flex the ring a bit and change the puckering. The tetrahedral bond angles about each carbon center are easy to accommodate and the structure is easy to build. (b) A model of cyclobutane is harder to build than cyclopentane, because the interior angles of a square are 90 . In order to accommodate these angles, you have to bend the connectors between the atom centers in your model. Again, the structure is not planar, but slightly puckered, as this relieves some of the strain of reducing the bonding angle from about 109 to something close to 90 . This ring is not nearly as flexible as the five-membered ring in part (a). (c) A model of cyclopropane is much harder to build than cyclopentane, because the interior angles of a triangle are only 60 . In order to accommodate these angles, you have to bend the connectors between the atom centers in your model a large amount. Since three points define a plane, this structure (the three carbon centers) must be planar and rigid with very strained bonds holding the ring together. The strain on the ring bonds should make cyclopropane a good deal more reactive than cyclopentane in reactions that break the ring and permit the bonding angles to attain the unstrained 109 sigma bonding value or the larger angles associated with molecules that also have pi bonding. [Heating cyclopropane to temperatures above 200-300 C causes relatively rapid isomerization of the molecules to propene, the straight-chain isomer.]

Problem 5.47. This is another model-building problem that is designed to help you see the relationships among the parts of molecules and to help you develop an “inner eye” for structure. (a) There is only one structure for cyclopentanol. No matter how you build separate structures, they can all be superimposed on one another by slightly flexing the model to make the puckering the same in each structure. (b) There is more than one structure for 1,2-cyclopentanediol (“di” = two). Both –OH groups can be on the same face of the “plane” of the ring (cis) or they can be on opposite faces (trans).

OHOH

OH

OH

HO

HO

mirror The trans-isomer is a mixture of mirror image isomers, optical isomers [an enantiomeric pair]. Thus there are three isomers of the diol. This may be very hard to see without building models. The easiest way to see mirror image isomers is to build both, show that they are in fact mirror images, and then show that they are not superimposable. Alternatively, build a model, put it in front of a mirror, and build a model of the image you see in the mirror.



Problem 5.48. (a) Functional groups (from Table 5.4) that contain oxygen are alcohol, ether, aldehyde, ketone, carboxylic acid, ester, and amide. Note that this is all the groups shown, except amine. (b) Functional groups that contain nitrogen are amine and amide. (c) Functional groups that are most likely to ionize, that is, react with water to form ions, are carboxylic acid (transfer of a proton to water to produce a hydronium ion and a carboxylate anion) and amine (accept a proton from water to form an ammonium-like cation and hydroxide anion).

Problem 5.49. (a) To see how the number of Hs in an alkane is described as a function of the number of Cs consider the straight-chain alkanes with several Cs. These can all be written as a string of Cs with two Hs on each C except the first and last, which have three Hs. Thus, the number of Hs is twice the number of Cs plus two more to account for the ends. The formula for these alkanes is CnH2n+2. All isomers of an alkane, including the straight chain isomer, have the same molecular formula, so our alkane formula applies to branched isomers as well. (b) To see how the number of Hs in an alkene is described as a function of the number of Cs consider the relationship of an alkene to an alkane with the same number of Cs. The difference is that the alkene has a double bond between two of the Cs and, as a consequence, has two fewer Hs than the corresponding alkane. Thus, we look at the alkane formula and take away two Hs to get the general alkene formula, CnH2n. (c) To see how the number of Hs in an alkyne is described as a function of the number of Cs we use the same logic as in part (b). Consider the relationship of an alkyne to an alkene with the same number of Cs. The difference is that the alkyne has a triple bond between the two Cs connected by a double bond in the alkene. As a consequence, an alkyne has two fewer Hs than the corresponding alkene. Thus, we look at the alkene formula and take away two Hs to get the general alkene formula, CnH2n–2.

Problem 5.50. (a) An alkane is an acyclic (not cyclic) hydrocarbon--compound consisting of only hydrogen and carbon--in which every carbon is connected to four other atoms. These connections are entirely by means of single (sigma) bonds. Closely related are cycloalkanes, which have one or more rings in their structures. An alkene is a hydrocarbon that has at least one double bond as part of its structure. The doubly bonded carbons are connected to only three other atoms because one of those connections is made with four electrons, two in a sigma bond and two in a pi bond. A cycloalkene, as you might expect, has at least one ring and one double bond in a ring.

alkane: 2-methylpentane alkene: 2-methylpentene

(b) An alcohol has an oxygen atom linked by single (sigma) bonds to a hydrogen atom and a carbon atom; two pairs of sigma nonbonding electrons complete its octet. An equivalent description of an alcohol focuses on the carbon atom. In this definition, the key feature is a carbon atom singly bonded to an –OH group (also known as a hydroxyl group) and having all



three of its other bonds to either carbon or hydrogen. An aldehyde has an oxygen atom linked by a double bond to a carbon atom. As with alcohols, two pairs of sigma nonbonding electrons complete the octet on oxygen. The C=O double bond is known as a carbonyl group and is found as a substructure in many functional groups. In an aldehyde the carbonyl carbon is linked by single bonds to a hydrogen and a carbon.

O OH

alcohol: 4-methyl-1-pentanol aldehyde: 4-methylpentanal

(c) An ether has an oxygen atom linked by single (sigma) bonds to two carbon atoms; two pairs of sigma nonbonding electrons complete the oxygen atom octet. An ester is characterized by a carbonyl group, C=O, linked by single bonds to a carbon atom and a second oxygen also bound to carbon with its other single bond (also known as an alkoxy group).

O OO

ether: ethyl isopropyl ether ester: ethyl ethanoate (ethyl acetate)

(d) An amine has a nitrogen atom linked by three single bonds to either carbon or hydrogen atoms. An amide has a nitrogen atom linked by a single bond to a carbonyl carbon. The other two single bonds may connect the nitrogen to either carbon or hydrogen atoms. The definitions of amines and amides sound surprisingly similar, yet their physical and chemical properties are dramatically different.

NH2 NH2

O

amine: ethylamine ester: ethanamide Ethylamine (aminoethane) is a gas at room temperature that is miscible with water to form a basic solution (like ammonia with pH about 9). Ethanamide (acetamide) is a solid at room temperature that dissolves in water (not miscible) to form a neutral solution (pH the same as the water).

(e) The carbonyl group was defined in part (b). A carboxyl group consists of a carbonyl carbon singly bonded to a carbon or hydrogen and also singly bonded to an oxygen that is also bonded to a carbon (forming an ester) or hydrogen (forming a carboxylic acid). In each of these structures, R is either an H atom center or a carbon atom center.

R R

O

R O

OR

carbonyl group carboxyl group

Problem 5.51. (a) This is the structural formula for methanal (formaldehyde), H2CO, with the bond dipole shown as a dashed arrow (head pointed in the direction of the negative end of the dipole). All



the atom centers in this polar molecule lie in a plane because there are three sigma bonds on the carbon atom center.

C OH

H (b) These are structural formulas for aspartic acid, HO(O)CCH(NH2)CH2C(O)OH, one of the amino acids from which our proteins are constructed. As many of the atom centers as possible are shown in the plane of the paper. The structure on the right shows most of the bond dipoles (dotted arrows) in the molecule; the N–H bond dipoles (negative end at N) are omitted to avoid crowding. Aspartic acid is a polar molecule. The principal stereochemistry in this molecule is the arrangement around the central carbon center (to which the amino and carboxylic acid group are attached), There are four different groups bonded to this carbon center, so aspartic acid exists in mirror image forms (optical isomers). The structural formula here represents the isomer found in your proteins.

CCC

H NC

O

OH

H HO

OH

HH

CCC

H NC

O

OH

H HO

OH

HH

(c) These are structural formulas for 1,2-diaminoethane (ethylenediamine), H2NCH2CH2NH2. As many of the atom centers as possible are shown in the plane of the paper. The C–N bond dipoles are shown in the structure on the right, but the N–H bond dipoles (negative end at N) are omitted to avoid crowding. This is a polar molecule that is often used to form complexes with metal ions (see Chapter 6, Section 6.6).

C CNN

H HH

H

H

H

H

H

H2C CH2

H2N NH2

Problem 5.52. (a) In addition to the eight (or eleven, if you count optical isomers as two structures) C5H12O isomers that are the alcohols whose structures you drew in Problem 5.3, there are six more isomers that are ethers:

O OO

O O O

methyl n-butyl ether ethyl propyl ether methyl sec-butyl ether

iso-butyl methyl ether ethyl isopropyl ether methyl tert-butyl ether

(b) The criterion we have used to predict the lowest boiling isomer in a set of isomers with similar intermolecular attractions (their polarity in the case of these ethers) is branching. The



most highly branched isomer is the most compact, which means dispersion forces among molecules are lower and it takes less energy for a molecule to break free of the liquid, so the boiling point is lower. The most highly branched of this set of ethers is methyl tert-butyl ether (MTBE), so we predict that it will have the lowest boiling point among these isomers.

Problem 5.53. (a) This diagram shows how thymine and adenine can form two hydrogen bonds with the bases in the same relationship to the DNA deoxyribose-phosphate backbone as shown for the cytosine-guanine pair in the problem statement. That is, the attachment sites are located on the same edge of an imaginary rectangular plane on which the bases lie.

NN

O

O

attachedto deoxyribosein DNA chain

Thymine

NN

N

N


N

Adenine

HHH3C

H

You might also imagine that the thymine and adenine bases could be flipped relative to one another and the other doubly-bonded oxygen on thymine used in hydrogen bonding.


Adenine

NN

N

N

NHH

NN

O

O


Thymine

H3C

H

The problem with this pairing is that it puts the attachment sites on opposite edges of the imaginary rectangular plane in such a position that the regular helical structure of the backbone would be interrupted. Figure 1.33 in the textbook tries to represent the structure of an “unwound” DNA helix showing the bonding of the base pairs to one another and to the backbone. This figure shows the hydrogen bonding pair we first wrote. (b) Thymine and guanine do not have polar groups strategically arranged to facilitate multiple hydrogen bonding. You can see in this representation that there is only one possible hydrogen bonding possibility between these two bases and some spatial interference between the Hs that face one another.

NN

N

N

attachedto deoxyribosein DNA chainN

HH

H

O

Guanine

NN

O

O


Thymine

H3C

H



(c) In the process of DNA replication required when cells divide to form two daughter cells, the DNA of the mother cell must be reproduced as faithfully as possible in each daughter cell. The DNA double helix comes apart in this process and the cell synthesizes the complement of each half of the original helix by pairing each base in the new strand with its complement in the original. The specificity of base pairing due to hydrogen bonding is essential for the cellular machinery to “recognize” which base to add to the growing new strand. Although there are errors in replicating DNA, considerable repetition is built into the DNA double helix, helping to reduce the number of errors during the replication process.

Problem 5.54. (a) The activity in the Web Companion, Chapter 5, Section 5.11, page 3, and the one you did in Check This 5.72 both involve a molecular structure with four different groups bonded tetrahedrally to a central atom. There are two possible arrangements of the four groups (the two optical isomers). Only one of the isomers has the four groups arranged so that three of them can interact with three complementary groups on another molecule. (b) Two views of the model of aspartame are shown here. One is stretched out to show the parts a bit better and the second is curled around, as it will have to be to interact with the sweet receptor:

(c) To represent the negative, positive, and neutral (hydrophobic) sites on the sweet receptor, we have used sheets of colored paper, red, blue, and light green, respectively. The molecular structure is shown here with its negative carboxylate group, -C(O)O–, interacting with the positive (blue) site, the positive ammonium-like group, -NH3

+, interacting with the negative (red) site, and the uncharged phenyl ring, -C6H5, interacting with the neutral (light green) site. The hydrogen atom at the carbon whose stereochemistry is critical is marked with an arrow. Note that this hydrogen atom is pointing out toward you. This isomer fits the “sweetness triangle” and will taste sweet.



(d) The aspartame model with the -H and -NH3+ switched is shown here interacting with our

sweet receptor. The hydrogen atom on the critical carbon is again marked and shown in the same position, pointing out at you, as in the picture in part (c). Now the -C(O)O– and -NH3

+ cannot properly interact with sweet receptor and this isomer will not taste sweet.

(e) Exchange at the other stereocenter in the molecule (after the first is restored to its original form) gives a third isomer shown here. Again, the arrow identifies the hydrogen at the critical center. The hydrogen is pointing out toward you and both of the polar groups are interacting with their oppositely charged counterparts on the sweet receptor, as in part (c). The other end of the molecule can be rotated so that the phenyl ring can interact with the neutral site on the sweet receptor. This isomer also ought to taste sweet.

The difference between the results in part (d) and part (e) has to do with where in the molecule the structural change is made. If it is made at the critical stereocenter (as in (d)), then the molecule will not “fit” the receptor. There is no way for the molecule to maintain the same geometric relationship to the receptor (shown by the position of the hydrogen atom with respect to the receptor) and also have the appropriate interactions of the charged groups with the receptor. When the change is made at the other stereocenter, the structure can rotate to put the phenyl ring in the correct geometry to interact favorably with the sweet receptor while the overall molecule maintains its geometric relationship to the receptor.

Problem 5.55. Oxygen atoms are smaller than carbon atoms because of higher nuclear charge acting on electrons in same shell. This same factor is present in the sigma and pi molecular orbitals in oxygen vs. carbon. The orbital volumes are smaller in oxygen (increased attraction can balance greater kinetic energy) and the nuclei are closer together for the same bond order. This factor helps to explain why the double bond in O2 and a triple bond between two carbons have almost the same bond length, 121 and 120 pm, respectively. The double bond length between two



carbons is about 134 pm, which provides another measure of how much the increased nuclear charge of the oxygen core decreases the double bond length in O2.

Problem 5.56. (a) Molecular fluorine, F2, has 14 valence electrons. (b) Assume that the molecular orbital diagram shown in Figure 5.42 for molecular oxygen is also applicable to molecular fluorine. Missing from the diagram are two nonbonding sigma orbitals that accommodate four of the 14 valence electrons in F2, but do not affect the energetics of bonding between the two atomic cores. There are 10 additional valence electrons to be accommodated and they can be assigned to the molecular orbitals like this.

(c) Since the equal numbers of electrons in the � (bonding) and �* (antibonding) molecular orbitals effectively cancel out, the molecular orbital model predicts a bond order of 1 (from the two electrons in the sigma bonding orbital). (d) The Lewis model also predicts a bond order of 1.

Problem 5.57. (a) Two Lewis structures for the cyanate ion, OCN–, are:

(i) C NO

(ii) C NO

In each structure, we recognize atoms that do not have their usual complement of bonding orbitals. In structure (i), the N atom has only two bonding orbitals, but we usually see the N atom with three bonding orbitals, as in NH3. The N atom here is like the N atom in the amide ion, NH2

–, so we can think of the N atom in structure (i) carrying the overall negative charge of the ion. In structure (ii), the O atom has only one bonding orbital, so it is like the O atom in the hydroxide ion, OH–, and we can think of the O atom in structure (ii) carrying the overall negative charge of the ion. (b) In both structures (i) and (ii), the central C atomic core is sigma bonded to the O and N atomic cores and it has no non-bonding electron pairs. The geometry of the ion with two sigma bonds to the carbon atomic center is linear (Figure 5.21), just like the structure of carbon dioxide (Figure 5.24) and hydrogen cyanide (Figure 5.22 and Table 5.3). (c) Structure (i) will have two delocalized pi orbitals like carbon dioxide. These are shown for carbon dioxide in Figure 5.24 and further represented in the solution to Problem 5.32. The two pi orbitals in structure (ii) are more localized between the C and N atoms. (d) Two Lewis structures for the isocyanate ion, ONC–, are:

(iii) N CO

(iv) N CO



Again, we recognize atoms in both structures that do not have their usual complement of bonding orbitals. In structure (iii), the C atom has only two bonding orbitals, but we usually see the C atom with four bonding orbitals, as in CH4. The C atom in structure (iii) is like the C atom in the methylide ion, CH2

2–, so we can think of this C atom as carrying a –2 charge. The N atom has four bonding orbitals, as in NH4

+, so we can think of the N atom in both structures as carrying a +1 charge. In structure (iii), the –2 and +1 charges sum to an overall –1 charge, as they must to give the correct overall charge on the ion. In structure (iv), the O atom is like the one in structure (ii) and can be considered to carry a –1 charge. The C atom in structure (iv) is like the C atom in the methide ion, CH3

–, so we can think of this C atom as carrying a –1 charge. The sum of the charges in structure (iv) is (–1) + (+1) + (–1) = –1, as it must be to give the correct overall charge on the ion. The sigma-bonding framework for these two structures is, again, like those in carbon dioxide and hydrogen cyanide and the isocyanate ion is linear. The pi electron distributions in structures (iii) and (iv) are like those in structures (i) and (ii), respectively, so we expect delocalized pi orbitals in structure (iii). (e) The cyanate ion is likely to be more stable than the isocyanate ion. Two of the atoms in the cyanate ion (either Lewis structure) are in a bonding environment similar to that they would experience in stable, neutral molecules. The third atom is in a bonding environment similar to what it would experience as an ion. At least one atom in any Lewis structure for an ion has to be in a bonding environment like that of an ion, since the structure carries an overall charge. In the isocyanate ion Lewis structure (iii), two of the atoms are in bonding environments like that of ions and one of these is a multiply charged ion. In the isocyanate ion Lewis structure (iv), all three atoms are in bonding environments like that of ions. Lewis structures with bonding environments that are most like stable, neutral species are more stable. (In Chapter 6, we will introduce the concept of formal charge and these same kinds of arguments can then be made in terms of formal charge.) Lewis structure (i) for cyanate ion is further stabilized by the delocalization of the pi orbitals. Experimentally, we observe that isocyanates usually readily isomerize to cyanates.

Problem 5.58. (c) Under certain conditions BeF2 molecules can be observed in the gas phase. Write the Lewis structure for BeF2. What geometry do you predict for BeF2? What figure in the chapter helps you explain your choice of geometry? Explain. Hint: This compound does not obey the octet rule. Answer to 5.58: (a) The gaseous compound, BF3, is likely to be a molecular compound covalently bonded, since the molecules move about independently of one another as a gas. The solid, crystalline compound, BeF2, is likely to be ionic (or at least have a large degree of electron transfer from the beryllium atom to the highly electronegative fluorine atoms). Perhaps it can be characterized as, (Be2+)(F–)2.

(b) The Lewis structure for BF3 is: B

FF

F

. Since there are three sigma bonding orbitals around the central B atom, the geometry of the B and F atoms should be trigonal planar as in Figure 5.18 and Table 5.3.



(c) The Lewis structure is: Be FF

. Since there are two sigma bonding orbitals around the central Be atom, the geometry of the Be and F atoms should be linear as in Figure 5.21 and Table 5.3.

Problem 5.59. Lewis structures for methanoic acid, HC(O)OH, and the methanoate anion, HC(O)O– are

CH

O

O

H

CH

O

O and CH

O

O

There is a single Lewis structure for the acid with C-O single and double bond lengths 134 and 120 pm, respectively. There are two equivalent structures for the anion, so the pi electrons in the anion are delocalized over three atom centers, O-C-O. The electrons are shared between two C-O bonds and each has a bond order of 11/2. Thus, we would predict a bond length intermediate between a single and double C-O bond, about 127 pm [= (134 + 120)/2]. The experimental value is 126 pm.

solutions for chapter 5 end-of-chapter problemsquantum.bu.edu/.../text/answers/chapter05.pdf ·...

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