5.1-5.2 How do different alcohols react in a Breathalyzer test? This will be addressed in Week 12 lecture both in group work and in PowerPoint slides. Review the experiment results (PowerPoint Week 12) before attempting to answer the questions.

5.3 How do the properties of the butyl alcohol isomers differ?

(a) Some similarities: melting points of sec-butyl and iso-butyl alcohols, most of the alcohols have similar water solubilities. The major differences are all for tert-butyl alcohol. It has a much higher melting point, a lower boiling point, and is miscible with water.

(b) Most of the properties of tert-butyl alcohol are fairly close to those of ethanol; the melting point is an exception. This seems odd since the two structures are so different.

(c) A positive Breathalyzer test would occur for all the alcohols except tert-butyl alcohol. This åfitsπ in the sense that tert-butyl alcohol shows many exceptions to the patterns of the other alcohols.

Note: To melt a compound requires breaking some of its intermolecular forces so that some molecules can slip past one another (so the liquid can flow). To boil a compound requires that all intermolecular forces be overcome so that molecules can independently enter the gas phase. The temperatures at which melting and boiling occur depend on the strengths of intermolecular forces, and these are highly influenced by molecular structure. So, to understand all the information in Table 5.1 you must understand differences in molecular structure.

5.4 Comparison of the C₅H₁₂ and C₄H₁₀O isomers

(a) The properties of tert-butyl alcohol do not match the reactivity, melting point, boiling point, and solubility trends seen in the other C₄H₁₀O isomers. Something similar is seen with the C₅H₁₂ isomers. One of the isomers (2,2-dimethyl propane) has a higher than expected melting point and a lower than expected boiling point compared to the other two isomers.

(b) Many of the two sets of isomers include å2-methylπ something in the name. The isomer in each set whose properties are most out of line with the others (tert-butyl and 2,2-dimethylpropane) both have å2,2π in the name (this indicates a lot of åbranchingπ in the carbon chain).

(c) The fact that the two sets of isomers have similar molar masses tells you that they should have similar London forces. The generally higher boiling points of the C₄H₁₀O isomers can be explained by the strong hydrogen bonding between molecules. We observed this behavior for alcohols in earlier chapters.

5.5 Lewis structures of the C₂H₇N isomers

(a) There are 2(4) + 7(1) + 5 = 20 valence e- in C₂H₇N. Here are the Lewis structures for ethyl amine (left) and dimethyl amine (right):

(b) Build models of the molecules to make sure you understand this! Both have bent or zigzag structures, but one is more symmetrical.

5.7 Lewis structures and molecular models for C₃H₉N isomers

There are 3(4) + 9(1) +5 = 26 valence e- in C₃H₉N. The four possible structures are

5.9 More on the C₅H₁₂ isomers

(a) Do this exercise with your model kit. You will find that the three structures can be superimposed (you may have to rotate some bonds to see this), so they must be the same molecule.

(b) (aπ) is n-pentane since it has a straight 5 C chain, (bπ) is 2-methyl butane since it has a 4 C chain with a methyl group at the 2^nd C, and (eπ) is 2,2-dimethyl propane since it has a 3 C chain with two methyl groups on the center C. Always look for the longest C chain to get the årootπ name and then add the branch names.

5.10 What are the Lewis structures and molecular models for the C₄H₁₀O isomeric alcohols?

(a) ≠(c) These are problems in Week 12 group work. The C₄H₁₀O isomeric alcohols are the ones discussed earlier in Chapter 5 in the Breathalyzer reaction.

(d) Retinol and retinal are very similar except that retinol has an alcohol group at one end but retinal has the 0=C-H group (aldehyde group) instead. We saw with the isomeric C₄H₁₀O alcohols that a positive Breathalyzer test occurs only if the C with the alcohol group also has a H bound to it. Since retinol has such a C, we expect it to give a + Breathalyzer test.

5.11 Boiling points of the C₄H₈O alcohols

There is a typographical error here since Table 5.1 refers to the C₄H₁₀O alcohols. All the isomers can hydrogen bond and all will have similar London dispersion forces (since they are similar molar mass). However, straighter chain molecules can pack together better than ones that are more branched, so straighter chain isomers will have stronger IMFs. (Build models to see the differences in packing.) 2-methyl-2-propanol (tert-butyl alcohol) is much more branched than the other isomers, so it will have weaker IMFs and should have a lower boiling point.

5.12 How do the PEs and KE vary with the proton-proton distance?

(a) PE is given by Coulombπs Law: E of the interaction is proportional to q₁q₂/r². For oppositely charged particles (proton and e-), the E of interaction is -. It will become a larger magnitude - number as proton A and proton B move closer together since this will mean that the distance between proton B and the e- drops (r gets smaller).

(b) As the A to B distance drops, the distance between proton A and the e- also drops, and the E of the interaction becomes a larger magnitude number.

(c) As the A to B distance drops, the E of the interaction (which is always + since both charges are +) becomes a larger + number.

(d) The kinetic energy of a spherical e- wave is inversely proportional to the square of the wave åradius.π (As an e- is constrained to be closer to the nucleus, its kinetic energy increases ≠ see p. 271.) So, whatever keeps the e- closer to the nucleus will increase the kinetic energy. As the A to B distance drops, the e- wave radius drops, and the kinetic energy of the e- increases.

5.13 How do the PEs and KE vary with the proton-proton distance?

The energy and internuclear distance scales are arbitrary, so use any scale you want. Simply add the value of E for the red KE line to the value for the green PE line at any fixed value of internuclear distance. You will find that the sum always equals the value on the black total energy curve.

5.14 What are the relative sizes of atomic and molecular orbitals?

(a) The H₂⁺ molecular orbital has the higher kinetic energy. The e- in this orbital is constrained to a smaller volume (because it is attracted to two nuclei), so it has the higher KE.

(b) Most of the e- density in the MO is concentrated between the nuclei, so it is more stable than if it were in an atomic orbital with only one nucleius.

5.15 What is the symmetry of a nonbonding sigma orbital?

There is no bond axis to rotate around. Instead, imagine a line from the atomic core through the center of the nonbonding orbitalπs electron density. If you rotate around this line, the nonbonding orbital looks the same no matter how far you rotate it. So, the orbital has spherical symmetry.

5.16 What are the geometries of second period hydrides?

(a) We have seen these Lewis structures before:

(b) There are four e- pairs around each of the central atoms in these molecules (some are bonding e-, some are non-bonding lone pairs). The arrangement of these e- pairs is tetrahedral, something that is obvious if you make the models.

(c) The molecular geometries are found by the looking at the arrangement of atoms around the central atom. They are, respectively, tetrahedral, trigonal pyramidal, and bent.

(d) The electron pair and molecular geometries are the same for methane. For the other two molecules, the geometries are not the same because of the presence of lone pairs. See Figure 5.10 for sketches that explain the difference.

5.17 What are the geometries of second period hydrides?

(a) The Lewis structure of HF is

and the orbital arrangement is

The four sigma orbitals (one bonding and three non-bonding) are arranged tetrahedrally to accommodate the four e- pairs at the fluorine atom. The molecular geometry (shape of the molecule) is linear. The bond lengths for CH₄, NH₃, H₂O are 109, 101, and 94 pm (see p. 297). This decrease in bond length happens because the nuclear charge (atomic core charge) increases from +4 to +5 to +6 for this series of molecules. HF (with a nuclear charge of +7) would be expected to follow the trend, so the H-F bond length is probably a little lower than 90pm. You can predict all these shapes if you keep in mind the VSEPR geometry rules you learned in Chapter 1.

(b) The three molecules have different shapes despite the fact that they all have tetrahedral e- pair geometry. The differences are due entirely to the different numbers of lone pairs.

5.19 The structure of SF₆.

The number of valence e- is 6 + 6(7) = 48 and the Lewis structure is shown below at left. A molecular model is shown below at right.

5.20 How are all six octahedral positions equivalent?

(a) You can try the demonstration with pieces of your model kit. No matter how you rotate the model, all terminal atom positions look equivalent.

(b) The arrangements should be the same.

5.21 Sigma orbitals around third- and higher-period atoms in molecules.

The Lewis structures are shown below. You can see that the central atom has 5, 6 or 7 pairs of e-, respectively.

The number of fluorine atoms that can bond to a central atom is limited, in large part, by the size of the central atom. While iodine is large enough to accommodate 7 fluorine atoms, chlorine and bromine are much too small for so many fluorine atoms to fit (e- clouds of the fluorine atoms will repel each other so much as to make ClF₇ and BrF₇ unstable).

5.22-5.26 Do carbon-hydrogen compounds react with permanganate?

These will be discussed in Week 13 lecture.

5.26d Vitamin A has many C=C bonds, so it is expected to react with permanganate.

5.27 What is the geometry of double bonded molecules?

(a) This will be discussed in Week 13 lecture. The molecular geometry at the central atoms is trigonal planar.

(b) The bond angles in the real molecules are actually ~120∞, but because the models were made from tetrahedral model kit centers, the angles in the models are closer to 109∞C. (We will learn to make models using kit pieces that give the correct 120∞ geometry later in the text.)

(c) The C-C bond length in ethane is a little longer than the C=C bond in ethane. You can see this in the two models if you use the curved connectors to represent the double bond and a white connector for the C-C single bond. There are more bonding e- in C=C vs. C-C, and they draw the two nuclei closer together.

(d) A C-O bond is shorter than a C-C bond because of the smaller size of O vs. C (bond length depends on atom size).

5.28 Other bond angles in ethane and methanal

(a) The H-C-C bond angle in ethane is found by considering the information inf Figure 5.14. The H-C-H angle is 117∞, and the rest of the 360∞ space around the C is 360-117 = 243∞. The H-C-C bonds divide this space in two parts, so the H-C-C bond angle is 243∞/2 = 121.5∞. The H-C-O bond angle is found using the same approach: (360-116∞)/2 = 122∞.

(b) If we use the curved bond connectors and black (C) and red (O) atom centers, the H-C-H bond angles are tetrahedral (109.5^o). The H-C-C bond angle in ethane is found the same way as in (a): (360-109.5)/2 = ~125 ^o. The H-C-O bond angle in methanal is found in the same way and is also ~125 ^o. So, the model kit comes close but really doesnπt represent the true bond angles.

5.30 Lewis structure for ethyne (acetylene) molecule

(a) Number of valence e- = 2(1) + 2(4) = 10 e-. To have an octet of e- around each C requires multiple bonds:

(b) C-C is 154 pm, C=C is 133 pm, C=C is 120 pm. The trend seems to be more bonds = shorter bond distances. The more bonding e-, the closer together the nuclei are drawn.

5.32-5.33 Sigma-pi representation for methanal

Figure 5.19 shows a C=C pi bond. To illustrate the sigma and pi MOs in methanal (H₂C=O), one of the C atom centers must be replaced with a red O center (without any H atoms). The sigma orbitals of C and O are shown in pink in the diagram below. (The two orbitals to the left of the O atom åcontainπ the two non-bonding pairs.) The sigma orbitals of H are shown in yellow. The pi orbitals on C and O are shown in blue. The pi orbitals are perpendicular to the sigma bonding framework.

5.34 The sigma-pi representation for ethyne

(a) The N atom center in HCN would be replaced with another C atom center that is also bound to two H atoms. There would still be two pi bonds perpendicular to each other. In the diagram below, one of the pi bonds is light blue, the other is purple. The sigma bond between C-C is shown in yellow. The sigma bonds between C-H are shown as the interaction of yellow orbitals from C to green orbitals from H.

(b) The triple bond is shorter than a C=C bond because the extra pair of pi electrons helps to draw the two C atoms closer together.

5.36 Lewis structure for the nitrate ion

The nitrate ion, NO₃^-, has 5 + 3(6) + 1 = 24 e-. Since we are told all the O atoms are equivalent, none of the three Lewis structures shown below (where one of the N-O bonds is a double bond but the other two are single bonds) is entirely satisfactory.

5.37 Do Lewis structures give correct geometries for carbonate and nitrate?

The Lewis structures have three groups of e- around the central atoms, so trigonal planar e- pair geometry is predicted.

5.38 What are the bonding and structure of the carbon dioxide molecule?

(a) There are 2(6) + 4 = 16 valence e- in the molecule:

(b) From the Lewis structure, we see that there are 2 e- groups at the central C atom. These will be linearly arranged. Each of the 2 e- groups involves a double bond, so each involves a sigma bond and a pi bond. So, there are 2 sigma bonds linearly arranged at the central C.

(c) Make the model with your kit! You will see that the model is linear.

5.39 Bond order in carbon dioxide

(a) The bond order for the carbon-oxygen bond is 2. The bond order is just the sum of the sigma and pi bonds, and the Lewis structure shows a double bond between C and O.

(b) C-O is 140 pm and C=O is 120 pm. In carbon dioxide, there are C=O bonds, so the observed bond length of 115 pm makes sense. Itπs close to the 120 pm of an ordinary C=0 bond. The smaller than åordinaryπ length is due to delocalization of the pi bonding e- (see the bottom of p. 313, Figure 5.24 for a picture of what these delocalized orbitals look like).

5.40 Structure of ozone

(a) The molecule has 3(6) = 18 e-. There are two equivalent Lewis structures:

The Lewis structures are not consistent with the observation that the end atoms are equivalent. In the Lewis structures, the end atoms have different numbers of bonds to the central atom and different numbers of lone pairs. Ozone is expected to have delocalized pi e- (you expect delocalization when you can write non-equivalent Lewis structures that differ only in the arrangement of e-). The molecule will be more stable when the e- can spread out over three atom cores. Neglecting delocalization, the bond order between the central O and the end O atoms is either 1 or 2 (depending on whether there is a single or double bond. Considering delocalization, the bond order is the average or 1.5.

(b) O-O is 148 pm, O=O is 120 pm, and the observed bond length for ozone is in between. This makes sense if the bond order is 1.5 (in between the bond orders for O-O and O=O, 1 and 2, respectively).

5.41-5.42 What happens when substances are cut?

Before/after photos are shown below. The lump of Na is soft and can easily be cut with a spatula, while the NaCl crystal is hard and cannot be cut with a spatula. When struck with a hammer, the NaCl shatters into pieces with sharp edges.

Crystals made of + and ≠ ions are difficult to cut because doing so requires that many very strong Coulombic interactions between + and ≠ ions be disrupted (see Figure 2.13).

The situation is different in a metal crystal where all the atoms are the same and are packed together as shown in Figure 5.26. The valence electrons in metals like Na are delocalized throughout the metal crystal. In effect, the positive metal atom cores are immersed in a sea of electrons (see p. 316, Figure 5.27). It is easy to cut Na since the interactions between any individual atom core and the delocalized electrons are relatively weak. When enough force is applied, the atom cores can be pushed apart and the e- orbitals readjust to the new atom core arrangement.

5.43 Three-dimensional packing of spheres

The spheres are packed together with one row of atoms fitting into the åindentationsπ of the row just below it. This is shown in Figure 5.26. A familiar example of this type of close packing arrangement is seen is displays of apples or oranges at grocery stores.

5.44 More properties of metals

Recall from 5.41-5.42 that individual metal atom cores interact with the åsea of electronsπ very weakly. This means that it is relatively easy to move the metal atom cores around, so the metal can be reshaped by pounding and pulling. In contrast, crystals made of cations and anions break or shatter when they are struck with force because they have very strong individual cation-anion Coulombic interactions.

5.45 Interpreting 3-D structures

(a) The structure in (a) of Figure 5.29 most clearly shows the trigonal pyramid structure. The others show the trigonal arrangement of atoms, but they do not show the pyramid as clearly.

(c) Here are sketches of the structures. Use your model kit to build the molecule and rotate it to correlate it with each of the three sketches. The way you sketch the molecule depends on your orientation with respect to it.

5.46 3-D structures

We will draw Lewis structures and 3-D sketches for all of these molecules in lab.

5.47 How do you draw 3-D structures for larger molecule?

(a) The molecule has 3(4) + 6(1) = 18 valence e-. The Lewis structure is shown at left with a ball and stick model at right:

The C atoms can all lie in the same plane. Build a model of the molecule to see that this is true. Manipulate the model to see that there is more than one way to orient it so that all the C atoms lie in one plane.

(b) The molecule has 4(4) + 8(1) = 24 valence e-. The Lewis structure is shown at left with a ball at stick model at right:

Build a model of the molecule and manipulate it to see that all the C atoms can be made to lie in the same plane. As for propene, there is more than one way to manipulate the model to force the C atoms into one plane.

5.49 3-D and condensed structures for the C₅H₁₂ isomers

Make models and compare them to these structures.

5.50 How many hydrogen atoms are bonded to a carbon atom?

None of the structures have non-bonding (lone pairs) on C. Carbon atoms always make four bonds, so if you know how many other atoms a C atom is bonded to, the difference will be made up with bonds to H atoms.

5.52 Skeletal structures for the C₅H₁₂ isomers.

(a) There is a typographical error in the question. The text figure referred to is Figure 5.32. The first left segment of the line represents a methyl (CH₃-) group.

(b)

pentane

2-methylbutane

2,2-dimethylpropane

5.53 Can different skeletal structures represent the same molecule?

Note: The same molecule/different rotations are examples of conformational isomerism.

(a) Make the model of 1-butanol. It can be rotated to the shape shown in 5.53 without breaking any bonds.

(b) Both pentane and 2-methylbutane can be rotated to give different skeletal structures, but 2,2-dimethylpropane cannot. It is very difficult to see this without building models!

5.54 Interpreting skeletal representations

(a)

There are only single bonds in 2-methylpropane, but there is a double bond in 2-methylpropene (so, 1 less H atom).

(b) The condensed structure for retinol is shown below. It is similar to the figure for retinal shown on page 280 except that instead of the alcohol group that retinol has, there is an aldehyde (H-C=O) group. Note that the structure for retinal on p.280 is flipped 180 degrees compared to the structure for retinol shown below.

5.54-5.55 How many C₄H₈ molecular structures are possible?

There are 6 possible structures. Some of these involve rings of atoms while others involve chains of atoms. The skeletal and line (condensed) formulas are:

(CH₂)₄

CH₃CH(CH₂)₂

CH₂CHCH₂CH₃

(CH₃)₂CHCH₂

CH₃CHCHCH₃

The last two structures have the same line formula (are isomers). They are isomers even though they have the same atoms connected to one another because they have different spatial arrangements. The molecule at left has the two methyl groups on opposite sides (so it is the trans isomer). The molecule at right has the two methyl groups on the same side (so it is the cis isomer).

5.57 Cis-trans isomers

(a) We will do this problem in lab.

(b) This one is a little confusing since there are both cis and trans configurations in this molecule. Focus on the center section of the long chain of C atoms and remember that there is either a methyl group or a H atom bound to each of these C atoms One of the trans configurations is marked with red arrows ≠ you can see that the H atoms here will be on opposite sides of the chain.

(c) Below is a diagram from www.vs-c.de/vsengine/printvlu/vsc/de/ch/12/oc/vlu_organik/alkene/alkene_kiel.vlu.html showing the cis to trans conversion of retinal (attached to a protein molecule called opsin). This conversion, important in the vision process, occurs when a photon of light is absorbed by the cis isomer. You can see that the trans isomer is a much straighter chain.

5.58 Identifying isomers

All of the isomers have the formula C₅H₁₀O. Structural isomers differ in connectivity. Stereoisomers have the same connectivity but different spatial arrangements (but are not interconvertible). Structures ii and v are identical (since they can be rotated to become superimposable). Structural isomers are i/iii with iv and ii/v with iv (the difference is the location of the double bond). Structures i and iii are stereoisomers (trans, cis, respectively).

5.59-5.61 Do solutions affect polarized light?

This will be discussed in Week 14 lecture.

5.63 Can you identify optial isomers?

(a) We will discuss lactic acid in Week 14, but to do this exercise you will have to build the models. In one isomer, the ≠OH group is in the 8:00 position. In the other, the ≠OH group is in the 4:00 position.

(b) The square planar arrangement is not possible. One square planar structure can be flipped and superimposed onto its mirror image structure. So, optical isomers of lactic acid in square planar arrangements are not possible.

5.64 Properties of optical isomers

(a), (b) skip

(c) We will do this activity in Week 14. The hexagons are not superimposable, so the stereoisomers of glucose are optically active. This is what we expect since in Investigate This 5.59, we saw that corn syrup (which contains a lot of glucose) was able to rotate plane polarized light.

5.65-5.66 How many C₃H₈O₂ isomers are there?

This will be discussed in Week 14.

5.67 Retinol and retinal

Retinol and retinal are very similar, but retinol has a HO- (hydroxyl) group while retinal has an O=CH (aldehyde) group at the end. Both contain many alkene (C=C) groups.

5.68 What are the interactions between an alcohol and an aldehyde?

(a) Yes, these molecules can hydrogen bond. The H from the (very polar) alcohol group is attracted to the lone pairs on the O of the aldehyde. This interaction (the hydrogen bond) is shown as a dashed line in the figure below. (Note, these figures are NOT drawn in 3D. Make models to see the 3D relationships.)

(b) Figure 5.38 in the text shows an extensive blue region (low e- density or relatively positive) at the H-C-H end of the methanal molecule. A lone pair on the ethanol O (negative center) can interact weakly with the positive region on methanal:

5.69 Identifying functional groups

(a) Retinol has one HO- group and five alkene groups. Retinal has one aldehyde group and five alkene groups.

(b) Structure i has an alcohol and an amine group; its molecular formula is C₂H₇NO. Structure ii has two ketone groups; its molecular formula is C₇H₁₂O₂. Sturectures iii and iv both have one alcohol group, one carboxylic acid group; their molecular formulas are the same, C₇H₁₂O₃ (there are two H atoms on four of the ring C atoms and 1 H atom on two of the ring C atoms).

5.71 Can you construct a molecule-to-order?

(a) Here are the straight or branched chain structures (only the cis isomers are shown, but a trans isomer is possible in each case).

Ring structures are also possible:

Do not worry if you could not generate all these structures.

(b) Here are some possible structures. It is not important that you be able to generate every one, but you should be able to generate most of them.