Atomic Structure Since Molecules Are Constructed from Atoms, We Will Quickly Review the Structure of Atoms. What Do Atomic Orbit

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Properties of atoms Beauchamp Atomic Structure Since molecules are constructed from atoms, we will quickly review the structure of atoms. nucleus ecore evalence Volume of electron cloud compared to volume of nucleus electron clouds V r 3 e 4 e 3 15 determine the = 3 r = (1.33)(3.14)(100,000) = 4 x 10 p,n Vn n overall volume of atoms

mass protons mp protons and neutrons = = 1800 mass electrons me 1 determine an atom's mass 1 p = protons = constant # that defines the element 100,000 n = neutrons = varies = defines the isotope e = electrons = varies, depending on bonding patterns associated term if electrons = protons (same # of e's and p's) atom if electrons < protons (deficiency of e's) cation if electrons > protons (excess of e's) anion

(eval) = valence electrons = The outermost layer of electrons, which determines the bonding patterns. The usual goal is to attain a noble gas configuration. This is accomplished by losing e's (forming cations) or gaining e's (forming anions) or sharing e's (covalent bonds)

(ecore) = core electrons = The innermost layer(s) of electrons (usually full shells or subshells). These e's are held too tightly for bonding (sharing) and not usually considered in the bonding picture. These e's cancel a portion of the nuclear charge (called shielding) so that the valence e's only see part of the nuclear charge, called Zeffective.

Zeffective = (# protons) - (core e's) = the effective nuclear charge. This is the net positive charge felt by the valence e's (bonding and lone pairs). Zeffective = same # as the column of the main group elements.

What do atomic orbitals look like? a. s orbitals – are spherical in shape. The wave nature of electrons (phase) may change as the distance increases from the nucleus but at any fixed distance from the nucleus the phase nature of s electrons will be the same (think of layers on an onion). Nodes are regions in space where the probability of finding an electron is zero (the phase is zero).

b. p orbitals –have a dumbbell shape with two lobes at 180o to one another with opposite phase nature. All p lobes intersect at the nucleus with zero probability of finding any electron density (= node).

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all three 2p orbitals (2px, 2py, 2pz) o a single 2p orbital, py are oriented at 90 to one another, has node at nucleus represented as px, py and pz. p x c. d and f orbitals have more complicated shapes and are not important to beginning organic chemistry. However they are important for understanding the inorganic world and many biochemical molecules involve the chemistry of d orbitals (like iron in hemoglobin, cobalt in vitamin b12, and zinc, copper, manganese and others in proteins.). We will not discuss these in our course.

z z z dyz dxz z z dxy dx2-y2 dz2 We rarely have occasion to discuss d orbitals. They are y yy yyhelpful in discussions of sulfur, phosphorous and a few transition metals. Even in x x x x x those discussions you can mainly look at them as bigger lobes in lobes in lobes in lobes in lobes along versions of p-like orbitals. the yz plane the xz plane the xy plane the xy plane the z axis along axes

Bonds represent an overlap of orbitals, which are a region where negative electron density is present. Electrons allow positive nuclei to remain in close proximity (come together). In our course covalent bonds will represent two shared electrons between two bonded atoms. We propose two types of covalent bonds: sigma, where electron density is directly between the two bonded atoms and pi, where electron density is above and below (or in front and in back) of a line connecting the two bonded atoms. Sigma bonds are always the first bond and pi bonds are the second (and third) bonds overlapping a sigma bond. Pi bonds are represented by sideways overlap of two p orbitals.

A B A H pi bonds - the electron density is above and A B  bond below the two bonded atoms  bond  bond (or in front and in back), possible sigma bonds - the electron density none is directly between, is directly between the two bonded atoms, always second or third bonds always the first bond

Problem 1 – What kinds of bonds are between the nonhydrogen atoms below?

a b c d H H H e H H f g H H

HCN HCC H CO CC CN CO H CCH H H H H H H

Atomic Configuration provides a representation of the electrons about a single isolated atom. Atoms represented are NOT in molecules. The usual atomic orbitals (s,p,d,f) are used and electrons fill into these atomic orbitals according to a few simple rules. The electrons closest to the nucleus are held the tightest and called the core electrons (full shells), while those in the incompletely filled outermost shell are called valence electrons. The valence electrons largely determine the bonding patterns and chemistry of the atom in order to gain a Noble gas configuration.

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3d 3d 3d 3d 3d Electrons fill into atomic orbitals according to the following rules. 3s 3p 3p 3p 1. Pauli Exclusion Principle - only two electrons may occupy any 2s 2p 2p 2p orbital and those electrons must have opposite spins 2. Hund's Rule - electrons entering a subshell containing more than 1s one orbital (p,d,f) will spread themselves out over all of the available +z orbitals with their spins in the same direction until the final subshell n=3 shell is over half filled. n=2 shell 3. Aufbau Principle - orbitals are filled with electrons in order of n=1 shell increasing energy (lowest PE to highest PE) nucleus

Periodic Table (another view) – Atomic orbitals (and electrons) are found in the following locations of the periodic table. Z = +1 +2 effective +3 +4 +5 +6 +7 +8 Elements in the upper 1s 1s n = 1 right corner have the 2s 2p n = 2 tightest hold on electrons 3s 3p n = 3 because they have the Elements in the lower largest Zeff (towards the 4s 3d 4p n = 4 left corner have the right) and are closest to weakest hold on electrons 5s 4d 5p n = 5 the effective nuclear charge because they have the 6s 5d 6p n = 6 (towards the top). smallest Zeff (towards the 7s 6d 7p n = 7 left) and are farthest from the effective nuclear charge 4f (towards the bottom). 5f

Table 1 – First ionization potentials of atoms (equals the energy to remove an electron from a neutral atom)

ionization electron potential is lost (kcal/mole) Atom Atom + electron

- greater PE (less stable) Atom e final energy Ionization always has an Potential energy cost to strip an Energy Energy = electron from an atom. Atom lower PE (more stable) starting energy

Energy to ionize an electron from neutral atoms = IP1 (units are kcal/mole). Compare rows and compare columns. Group 1A Group 2A Group 3A Group 4A Group5A Group 6A Group 7A Group 8A H +314 (He) +568 Li +124 Be +215 B +192 C +261 N +335 O +315 F +402 (Ne) +499 Na +118 Mg +177 Al +138 Si +189 P +242 S +239 Cl +300 (Ar) +363 K +99 Ca +141 Ga +138 Ge +182 As +226 Se +225 Br +273 (Kr) +323

Z = +1 Z = +2 Zeff = +3 Z = +4 Zeff = +5 Z = +6 Z = +7 Zeff = +8 eff eff eff eff eff

The general trend, across a row, is that the ionization potential gets larger and the hold on electrons is stronger. Why? The answer is found in the size of Zeffective. As we move from Li (IP1 = +124) to carbon (IP1 = +261) to fluorine (IP1 = +402), the valence shell stays the same (n = 2), but the effective nuclear charge holding those electrons keeps increasing, from +1 to +4 to +7. Because each extra electron goes into the same shell y:\files\classes\0 Organic Topics - latest\315 topics\20 315 lecture notes, 7-9-15\314 Review Problems\01 review, Zeff of atoms 19p, answers, 6p.doc Properties of atoms Beauchamp (n = 2), there is essentially no shielding of the nucleus by any of the additional electrons (the electrons are doing their best to avoid one another). It’s a lot harder to pull an electron away from a +7 charge than a +1 charge. It seems reasonable that fluorine has a stronger attraction for electrons than the other second row elements, even when those electrons are shared in a chemical bond. There are a few deviations that we will ignore. In a column, all of the atoms have the same Zeffective. What changes in a column is how far away the valence electrons are from that same effective nuclear charge (Zeff). Valence electrons in inner quantum shells are held tighter because they are closer to same Zeff.

Problem 2 - Explain the atomic trends in ionization potential in a row (Na vs Si vs Cl) and in a column (F vs Cl vs Br).

Atomic radii and ionic radii also support these periodic table trends in electron attracting power. Smaller radii, below, indicate 1. a stronger contraction in the same row (shell) of the electron clouds due to higher Zeffective attraction for the electrons and 2. in a column, closer distance of electrons to a particular Zeffective nuclear charge, due to electrons being in a lower principle quantum shell (fewer shells). When an atom acquires a negative charge (becomes an anion), it expands its size due to the greater electron-electron repulsion. Remember, most of an atom’s mass is in the nucleus, but most of its size is due to the electron cloud.

Table 2: Neutral atomic radii in picometers (pm) = 10-12 m [100 pm = 1 angstrom] H = 53 He = 31 Li = 167 Be = 112 B = 87 C = 67 N = 56 O = 48 F = 42 Ne = 38 Na = 190 Mg = 145 Al = 118 Si = 111 P = 98 S = 88 Cl = 79 Ar = 71 K = 243 Ca = 194 3d elements Ga = 136 Ge = 125 As = 114 Se = 103 Br = 94 Kr = 88 Rb = 265 Sr = 219 4d elements In = 156 Sn = 145 Sb = 133 Te = 123 I = 115 Xe = 108

Problem 3 - Explain the atomic trends in atomic radii in a row (Li vs C vs F) and in a column (C vs Si vs Ge).

Table 3: Cations (on the left) and anions (on the right) radii in picometers (pm) = 10-12 m [100 pm = 1 angstrom] Li +1 = 90 Be +2 = 59 B+3 = 41 C = N -3 = 132 O -2 = 126 F -1 = 119 Na +1 = 116 Mg +2 = 86 Al +3 = 68 Si = P = S -2 = 170 Cl -1 = 167 K +1 = 152 Ca +2 = 114 3d elements Ga +3 = 76 Ge = As = Se -2 = 184 Br -1 = 182 Rb+1 = 166 Sr +2 = 132 4d elements In +3 = 94 Sn = Sb = Te -2 = 207 I -1 = 206

as cations as anions

Problem 4 - In the tables above:

a. Explain the cation distances compared to the atomic distances (Li, Be, B). (Tables 2 and 3) b. Explain the anion distances compared to the atomic distances (N, O, F). (Tables 2 and 3) c. Explain the cation distances in a row (Li, Be, B). (Table 3) d. Explain the anion distances in a row (N, O, F). (Table 3)

Atoms behave differently in bonding situations when attempting to acquire a Noble gas configuration (ionic, polar covalent and pure covalent). This is often described as the “octet rule,” (8 valence electrons) for main group elements (1A through 8A). Almost always true for second row elements (exceptions possible for higher rows because of the d orbitals).

Electronegativity measures the attraction an atom has for electrons in chemical bonds. There are many proposed scales to do this, we will use the Pauling scale. The larger the number the greater is the attracting power of an atom for the electrons in chemical bonds. Electronegativity will determine nonpolar, polar and ionic characteristics of bonds, and when shapes are included it determines the same attributes in molecules. Compare rows and compare columns in the following table. The largest numbers are found in the upper right corner (strongest attraction for electrons) and the smallest numbers are found in the lower left corner (weakest attraction y:\files\classes\0 Organic Topics - latest\315 topics\20 315 lecture notes, 7-9-15\314 Review Problems\01 review, Zeff of atoms 19p, answers, 6p.doc Properties of atoms Beauchamp for electrons). Group 8A elements don’t have a value listed because the ones toward the top don’t make bonds.

 (chi) is the symbol used for electronegativity, which is the property the indicates an atoms attraction for electrons in chemical bonds with other atoms. Approximate electronegativity values for some main group elements.

Group 1A Group 2A Group 3A Group 4A Group 5A Group 6A Group 7A Group 8A Z = +3 Zeff = +1 Zeff = +2 eff Zeff = +4 Zeff = +5 Zeff = +6 Zeff = +7 Zeff = +8 H = 2.2 Li = 1.0 Be = 1.5 B = 2.0 C = 2.5 N = 3.0 O = 3.5 F = 4.0 He = none Na = 0.9 Mg = 1.2 Al = 1.5 Si = 1.9 P = 2.2 S = 2.6 Cl = 3.2 Ne = none K = 0.8 Ca = 1.0 3d elements Ga = 1.6 Ge = 1.9 As = 2.2 Se = 2.5 Br = 3.0 Ar = none Rb = 0.8 Sr = 0.9 4d elements I = 2.7 Kr = 3.0 Simplistic estimate of bond polarities using differences in electronegativity between two bonded atoms.

A B bond polarity based on  = A B   0.4 considered to be a pure covalent bond (non-polar) 0.4 <  < (1.4 - 2.0) considered to be a polar covalent bond (permanent charge imbalance) considered to be an ionic bond (cations and anions) (1.4 - 2.0) < 

Elemental hydrogen, fluorine, oxygen and nitrogen are examples of atoms sharing electrons evenly because the two atoms competing for the bonded electrons are the same. Each line drawn between two atoms symbolizes a two electron bond. These examples also illustrate that pure covalent bonding can occur with single, double and triple bonds and shows that some elements have lone pairs of electrons.

HH FF O O N N hydrogen molecules with a fluorine molecules with a oxygen molecules with a nitrogen molecules with a single bond, each H has a duet single bond, each F has a octet double bond, each O has an octet triple bond, each N has an octet of electrons (like helium) of electrons (like neon) of electrons (like neon) of electrons (like neon)

If the two bonded atoms are not the same, then there will be different attractions for the electrons. One atom will have a greater pull for the electrons and will claim a greater portion of the shared electron density. This will make that atom polarized partially negative, while the atom on the other side of the bond will be polarized partially positive by a similar amount. Because there are two opposite charges separated along a bond, the term “dipole moment” () is used to indicate the magnitude of charge separation. Bond dipoles depend, not just on the amount of charge separated, but also the distance by which the charges are separated, as indicated by their bond lengths.

The symbols + and -- represent qualitative charge separation forming a bond dipole. Alternatively, an arrow can be drawn pointing towards the negative end of the dipole and a positive charge written at the positive end of the dipole. Two qualitative pictures of a bond dipole are represented below. B is assumed to be more electronegative than A.  + - AB or AB ...... B > A d d amount of distance bond  = charge x between = units of Debye (D = 10-18 esu-cm) separated charges in cm bond  = ( e ) x ( d ) = bond dipole moment

The absolute value of a unit charge on e = electrostatic charge (sometimes written as q) an electron or proton is 4.8x10-10 esu This is often given in angstrums (A = angstrum), but d = distance between the opposite charges -8 converted to cm for use in calculations (1A = 10 cm) y:\files\classes\0 Organic Topics - latest\315 topics\20 315 lecture notes, 7-9-15\314 Review Problems\01 review, Zeff of atoms 19p, answers, 6p.doc Properties of atoms Beauchamp

Because there are two factors that make up bond dipole moments (charge and distance), bonds that appear to be more polar based on electronegativity differences might have similar dipole moments if they are shorter than less polar bonds. The methyl halides provide an example of this aspect. Fluoromethane has a more polar bond, but shorter bond length, while iodomethane has a less polar bond, but longer bond length. The opposing trends of the halomethanes produce similar dipole moments in all of the molecules. Polarity, polarizability and shape have a large influence on physical properties (mp, bp, solubility, etc). We will briefly review these below.

electronegativity

H = 2.2 C = 2.5 N = 3.0 O = 3.5  = 3.2 F = 4.0 Cl Br = 3.0 I = 2.7  = dipole mement (measure of polarity) d = bond distance in picometers (10-12 m)

o  dC-X molecule mp (oC) bp ( C) density CH -F -142 -78 0.53 g/dm3 (gas) 3 CF = 1.5 139 pm 1.86 D 3 CH3-Cl CF = 0.7 178 pm 1.87 D -97 -24 0.92 g/dm (gas) 3 CH3-Br CF = 0.5 193 pm 1.81 D -94 +3 3.97 g/dm (gas) 1.62 D CH3-I CF = 0.2 214 pm -66 +42 2.28 g/mL (liquid)*

* Notice there is a different phase.

Ion-Ion interactions – There are very strong forces of attraction between oppositely charged ions. The melting points of ionic salts are typically very high. The solubility of many salts in water is often high because water has a concentrated partial positive end (the protons) that can interact strongly with anions and a concentrated partial negative end (the oxygen) that can interact strongly with cations, leading to very high overall solvation energies. However, solvation energy has to be larger than the very large lattice energy (energy of the cations interacting with the anions) or the salt will not dissolve.

The first number is the melting point and the second number is the boiling point (in oC). "sol." = solubility in 100 g water. X is the difference in electronegativity.

-2 Each type of ion is typically surrounded F Cl Br I O on all sides by oppositely charge ions in some sort of lattice structure. Before a 846 / 1717 610 / 1383 550 / 1289 467 / 1178 1570 / 2563 solid can melt the whole network of ions sol. = 1670g sol. = reacts has to break down. Li sol. = 1.3g sol. = 550g sol. = 1510g = 3.0 = 2.2 =2.0 =1.7 =2.5

dec 996 / 1787 801 / 1465 747 / 1447 660 / 1304 1132 / 1950 Na sol. = 404g sol. = 359g sol. = 905g sol. = 1840g sol. = reacts = 3.1 = 2.3 = 2.1 =1.8 =2.6

dec 858 / 1517 996 / 1787 734 / 1435 681 / 1345 >300 / - K sol. = 920g sol. = 344g sol. = 678g sol. = 1400g sol. = reacts = 3.2 = 2.4 = 2.2 =1.9 =2.7

A single neutral hydrogen atom would have a single valence electron. This is not a common occurrence in our world because such a hydrogen atom would be too reactive. However, if one were able to generate a source of hydrogen atoms, they would quickly join together in simple diatomic molecules having a single covalent bond with the two hydrogen atoms sharing the two electrons and attaining the helium Noble gas configuration (duet rule). Such a reaction would be very exothermic.

H H H = -104 kcal/mole H H

If two hydrogen atoms should find one another, they would The line symbolizes a two-electron, pure- from a diatomic molecule with covalent bond based on the calculation below. a tremendous release of energy, about 104 kcal/mole.  =  _  = 2.2 - 2.2 = 0  a b

The atoms of organic chemistry, H, C, N, O, S and halogens, tend to attain a Noble gas configuration by sharing electrons in covalent bonds of molecules. Simple formulas often have only one choice for joining the atoms in a molecule (CH4, NH3, H2O, HF). As the number of atoms increase, however, there are many more possibilities, especially for carbon structures (sometimes incredible numbers of possibilities, C40H82 > 62 trillion isomers!). These possibilities may require single, double and/or triple bonds (sigma and pi bonds will be discussed soon), in chains or rings of atoms. Carbon, nitrogen, oxygen and fluorine can be bonded in all combinations, according to their valencies. We study most of the bond types below. Notice that each atom below only makes as many bonds as electrons needed to complete its octet (C = 4, N = 3, O = 2 and F = 1). N, O and F have lone pairs. H

H C H H N H H O H H F

H H pure covalent bond polar covalent bond polar covalent bond polar covalent bond based on the based on the based on the based on the calculation below calculation below calculation below calculation below _ _ _ _   = a b   = a b   = a b   = a b   = 2.5 - 2.2 = 0.3   = 3.0 - 2.2 = 0.8   = 3.5 - 2.2 = 1.3   = 4.0 - 2.2 = 1.8

2a. Dipole-dipole interactions (partial polarity) – Since the partial charges present in molecules having dipole moments are less than full charges and the bonds are very directional (not “omni” like ions), and attractions for neighbor molecules are weaker than is found in ionic salts. However, polar molecules usually have stronger attractions than nonpolar molecules of similar size and shape. Many polar molecules below are compared with a similar size and shape nonpolar molecules to show how polarity can affect boiling points (an indication of the strength of attractions among neighbor molecules, when other factors are similar). A higher boiling point indicates stronger attractions. Dipole-dipole interactions represent moderate forces of attraction between partially polarized molecules. The molecular dipole moments are indicators of charge imbalance due to a difference in electronegativity , bond length and molecular shape. Fluorine is unusual in that it makes very polar bonds but is not very polarizable and holds on very tightly to its lone pairs, so it sometimes surprises us with lower than expected boiling points. Sometimes atoms that hold onto their electrons very tightly are called ‘hard’ and atoms that have polarizable electrons are called ‘soft’. Melting points are better indicators of packing efficiency and boiling points are better indicators of forces of attraction. Miscellaneous compounds are compiled below, representing a variety of factors discussed in this topic.

more polar nonpolar polar more polar H H polar H N C O O C resonance N nonpolar C resonance C C H C C C H H H H H H H H  = 0.0 D o  = 2.98 D  = 0.0 D  = 2.3 D Tbp = 107 C o o o o o bp = -81 C bp = +26 C bp = -104 C Tbp = 84 C bp = -20 C mp = -84oC mp = -13oC mp = -169oC mp = -92oC H2O sol. = insoluble H2O sol. = miscible H2O sol. = 2.9mg/L H2O sol. = 400g/L pKa = 25 pKa = 9.2 pKa = 44 pK = NA a

nonpolar H polar more polar N C N C resonance C C H3C H2C H3C  = 0.78 D  = 3.92 D T = 104oC bp = -23oC bp bp = +81oC mp = -102oC mp = -46oC H2O sol. = insoluble H2O sol. = miscible

Problem 5 – What is the effect on the dipole moment of switching a hydrogen for a methyl in each pair below? Propose a possible explanation.

O O O O O O O O

C C C C C C C C F H F CH H N H H N CH 3 2 2 3 HO H HO CH3 H H H CH3  = 2.02 D  = 2.96 D  = 3.71 D  = 3.76 D  = 1.41 D  = 1.74 D  = 2.3 D  = 2.68 D o o o o o bp = -29 C bp = +21 C bp = +210oC bp = +222 C bp = +118oC bp = +118 C bp = -20 C bp = +20oC o o o o o mp = -142 C mp = -84 C mp = +2oC mp = +80 C mp = +17oC mp = +17 C mp = -92 C mp = -123oC

polar O polar O polar O O O O resonance resonance resonance C C C C C C H3C H H C H H H H H 3 H3C CH3 H C CH  = 2.3 D  = 2.68 D  = 2.91 D 3 3 bp = -20oC bp = +20oC bp = +56oC mp = -92oC mp = -123oC mp = -94oC H2O sol. = 400g/L H2O sol. = very sol. H2O sol. = very sol. What is the effect of "R" groups? Do they make the molecule more polar, less polar or no different? Electron donation or electron withdrawal through sigma bonds is called an inductive effect.

nonpolar nonpolar nonpolar nonpolar H H H 2 H C C C C H C CH C 3 3 H3C CH2 C H H C  = 0.08 D  = 0.37 D 3  = 0.75 D H C  = 0.0 D o bp = -42 C bp = -47oC bp = -23oC bp = -34oC o H mp = -188 C mp = -185oC mp = -103oC mp = -134oC

H CH3 H NH2 OH F Cl Br I

N  = 0.31 D  = 0.0 D  = 1.53 D  = 1.55 D  = 1.66 D  = 1.54 D  = 1.73 D  = 1.91 D  = 2.37 D o o o bp = 81 C bp = 111 C o bp = +182oC bp = +84oC bp = +131oC bp = +156oC o bp = +115 C o bp = 184 C bp = +188 C mp = -95 C o o o o o mp = 5C mp = -6oC mp = +40 C mp = -44 C mp = -45 C mp = -31 C mp = -31oC mp = -42 C

y:\files\classes\0 Organic Topics - latest\315 topics\20 315 lecture notes, 7-9-15\314 Review Problems\01 review, Zeff of atoms 19p, answers, 6p.doc Properties of atoms Beauchamp 2b. Hydrogen bonds – Hydrogen bonds represent a very special dipole-dipole interaction. Molecules that have this feature have even stronger attractions for neighbor molecules than normal polar bonds would suggest. Solvents that have an O-H or an N-H bond are called “protic solvents” and can both donate and accept hydrogen bonds (because they also have lone pairs of electrons). They generally have higher boiling points than similar sized structures without any “polarized hydrogen atoms”. The reasons for a polarized hydrogen atom’s strong attraction for electron density is that there are no other layers of electrons around a hydrogen atom (hydrogen is the only atom to use the n=1 shell in bonding). When a hydrogen atom’s electron cloud is polarized away from the hydrogen atom in a bond with an electronegative atom, usually oxygen or nitrogen in organic chemistry, an especially strong polarization results, resulting in a strong attraction for a lone pair on a neighbor molecule, or even a pi bond. We call such interactions "hydrogen bonds". A molecule that has such a polarized hydrogen is classified as a hydrogen bond donor. A molecule that has a partial negatively charged region that can associate with such a hydrogen is classified as a hydrogen bond acceptor. Quite often the hydrogen bond acceptor is a lone pair of electrons on another oxygen or nitrogen atom, but fluorine, chlorine bromine or sulfur may provide lone pair acceptor sites as well. Weak hydrogen bonds can even form with C=C pi bonds.

H accepts Hydrogen bonding holds the molecules more tightly O hydrogen to one another. This can be seen in higher boiling H bond points among similar structures where hydrogen H donates O H bonding is possible versus not possible. Many examples hydrogen below show this property. bond H O donates H hydrogen H H Cl donates O bond accepts hydrogen accepts hydrogen hydrogen H bond donates H accepts bond bond hydrogen hydrogen O H O bond bond H O H3C H3C C CCH3 C C C H H O H O H3C CH3 three hydrogen bonds in G-C base pair H two hydrogen bonds in A-T base pair H N O H N N N H O N N H N N DNA N H N N N DNA N N H O DNA N guanine adenine H cytosine thymine O DNA

Problem 6 – Which base pair binds more tightly in DNA, GC or AT, or are they about the same?

y:\files\classes\0 Organic Topics - latest\315 topics\20 315 lecture notes, 7-9-15\314 Review Problems\01 review, Zeff of atoms 19p, answers, 6p.doc Properties of atoms Beauchamp Problem 7 – Provide an explanation for the different boiling points in each column. What is there such a large deviation for NH3, H2O and HF?

4B 5B 6B 7B 8B o 100 H2O boiling points ( C) CH4 NH3 H2O HF He o mp = -182oC mp = -78oC mp = 0oC mp = -84oC mp = -272 C o 50 bp = -164oC bp = -33oC bp = +100oC bp = +20oC bp = -269 C Temp HF  = 0.0 D  = 1.42 D  = 1.80 D  = 1.86 D  = 0.0 D o ( C) H Te  = 0.3  = 0.8  = 1.2  = 1.8  = NA 0 2 SbH3 PH NH H2Se SiH4 3 H2S HCl Ne 3 HI o o o o mp = -185oC mp = -132 C mp = -82 C mp = -114 C mp = -249 C -50 o H S AsH3 o bp = -88oC o bp = -85 C bp = -246oC 2 SnH4 bp = -112 C bp = -60 C HCl HBr  = 0.0 D  = 1.0 D  = 1.0 D  = 0.0 D PH  = 0.0 D -100 3 GeH4  = 0.0  = 0.4  = 1.0  = NA  = 0.3 SiH AsH HBr Ar 4 GeH4 3 H2Se o o -150 mp = -165oC mp = -111oC mp = -66oC mp = -87 C mp = -189 C o CH Kr bp = -88oC bp = -62oC bp = -41oC bp = -67oC bp = -185 C 4 Ar  = 0.0 D  = 0.0 D  = ? D  = 0.8 D  = 0.0 D -200  = 0.2  = 0.0  = 0.4  = 0.8  = NA

SnH4 SbH3 H2Te HI Kr -250 Ne o o mp = -146oC mp = -88oC mp = -49oC mp = -51 C mp = -157 C He o o o o o bp = -52 C bp = -17 C bp = -2 C bp = -35 C bp = -153 C -300  = 0.0 D  = ? D  = 0.4 D  = 0.0 D  = 0.0 D row row  = 0.5  = NA row row  = 0.2  = 0.2  = 0.1 2 3 4 5 Answer: Each dotted line represents the boiling points in the hydrides in a column in the periodic table, plus the Noble gases. In the Noble gases there is a continuning trend towards higher boiling point as the main atom gets larger with more and more polarizable electron clouds. Where bonds to hydrogen become polar, hydrogen bonding becomes important, leading to stronger attractions for neighbor molecules and higher boiling points deviate from the trends observed due to dispersion forces. Deviations are clearly seen with NH , H O and HF. Melting points are not plotted. 3 2

Problem 8 – Explain the difference in boiling points of the following molecules. H2 C O H3C H H O O H3C CH3 H3C CH3  = 0.08 D  = 1.30 D  = 1.69 D  = 1.69 D o bp = -42oC bp = -22oC bp = +65oC bp = +78 C o mp = -188oC mp = -141oC mp = -98oC mp = -114 C H O sol. = 71g/L H O sol. = miscible H O sol. = miscible H2O sol. = 40mg/L 2 2 2

2. Dispersion forces / van der Waals interactions / London forces (nonpolar attractions)

Most of the examples in the table below are gases at room temperature, an indication that there are no strong forces of interaction between the molecules. However, clearly there is some attraction between the molecules, because at room temperature bromine is a liquid and iodine is a solid, and ultimately each gaseous substance in the table does condense to a liquid and solidify to a solid. Also there is quite a range of differences in melting and boiling points among the different compounds.

bp MW mp Dipole o o Phase at room Substance (g/mol) ( C) ( C) Moment () temperature HH 2 -259 -253 0.0 gas NN 28 -210 -196 0.0 gas OO 32 -218 -183 0.0 gas FF 38 -219 -188 0.0 gas Cl Cl 71 -101 -35 0.0 gas Br Br 160 -7 + 59 0.0 liquid puzzle? II 254 +114 +184 0.0 solid y:\files\classes\0 Organic Topics - latest\315 topics\20 315 lecture notes, 7-9-15\314 Review Problems\01 review, Zeff of atoms 19p, answers, 6p.doc Properties of atoms Beauchamp

A plot of the trends in melting points and boiling points in the halogen family and Nobel Gas family helps us to see the differences more clearly (below). The difference melting and boiling points show there are differences in the forces of attraction between the halogen molecules even though all of the molecules are nonpolar. Bromine is even a liquid and iodine a solid at room temperature. The higher boiling points (I2 > Br2 > Cl2 > F2) are due to greater polarizability of the larger atoms, where the electrons are held less tightly.

Phase at room temperature 250 The halogen molecules are similar in shape and F2 = gas He = gas nonpolar. There is a smooth, increasing trend in 200 Cl2 = gas Ne = gas both melting and boiling points. This is suggestive Br2 = liquid Ar = gas bp of some factor increasing the forces of attraction 150 I2 = solid Kr = gas between molecles of the halogen family as they get I2 larger. The smooth trend in melting point is not typical, because it can vary so much with differences 100 mp in shapes. The even change in melting points is observed here because the halogen molecules all 50 have a similar, rigid shape. o Temp. Br room temp  25 C (oC) 0 2 -50

Cl The Noble gas molecules are similar in shape and -100 2 nonpolar. There is a smooth, increasing trend in both melting and boiling points. This is suggestive -150 Kr of some factor increasing the forces of attraction Ar bp between molecles of the halogen family as they get mp larger. The smooth trend in melting point is not -200 typical, because it can vary so much with differences F2 Ne in shapes. The even change in melting points is -250 observed here because the halogen molecules all He have a similar, spherical shape. -273 absolute zero = 0 K

Dispersion forces are temporary fluctuations of negative electron clouds from one direction to another, relative to the less mobile and more massive positive nuclear charge. These fluctuations of electron density induce fleeting, weak dipole moments. Polarizability is the property that indicates how well this fluctuation of electron density can occur about an atom. Within a column (same Zeff), larger atoms are more polarizable, because they do not hold as tightly to their valence electrons as smaller atoms, since the electrons are farther away from Zeff. Thus, atoms lower in a column are more polarizable than atoms higher up. In a row it’s harder to predict. Atoms to the right have a larger Zeff which should make them less polarizable, but it might seem that lone pairs are held less tightly by only one atom, instead of two atoms in a bond. However, fluorine is not very polarizable (it holds on very hard to even its lone pairs) and appears to indicate that larger Zeff is more important. Picture a cotton ball (polarizable electron clouds = sticky) and a marble (nonpolarizable electron cloud = not sticky).

In a nonpolar molecule the and are centered, Dispersion Forces on average. This would seem to indicate that in and are not centered nonpolar molecules there is no polarity or attraction creating tempory polarity. between molecules. So why do such substances liquify Fast moving electrons shift and solidify? Why aren't they always gases? position relative to slow moving nuclei, creating a + - + - temporary imbalance of charge, +Z +Z +Z +Z which induces a similar distortion of the electron clouds in neighbor structures +Z = nuclear protons Weak, fluctuating polar forces and a weak attraction for of attraction between molecules. = electron cloud neighbor molecules.

y:\files\classes\0 Organic Topics - latest\315 topics\20 315 lecture notes, 7-9-15\314 Review Problems\01 review, Zeff of atoms 19p, answers, 6p.doc Properties of atoms Beauchamp While the examples in the table above are not organic molecules, they are simple and we can use this simplicity to learn important ideas that apply to organic molecules. Simple diatomic molecules must have a linear geometry, since two points determine a straight line. When both atoms are the same, the bonding electrons must be shared evenly. There can be no permanent distortion of the electron clouds toward either atom, so there are no polar bonds.

Periodic trends in polarizability, .

C NOF

Zeff = +4 Zeff = +5 Zeff = +6 Zeff = +7

Polarizability is larger with smaller Zeff Cl because the electrons are not held as Polarizability is tightly, so they are more easily greater because distortable. Zeff = +7 there is a weaker hold on the electrons because they are Features that increase polarizability: farther away from Br the same effective 1. smaller Zeff, favors C > N > O > F nuclear charge, so they are more easily 2. valence electrons farther from the Zeff = +7 distortable. nucleus when Zeff is similar I > Br > Cl > F. I

Z = +7 eff Dispersion forces are cumulative, so when the contact surface area is larger, the interactions are stronger (because there are more of them). Higher molecular weight alkanes have more carbon atoms to interact than lower molecular weight alkanes (even though only similar weak dispersion forces are present in both).

y:\files\classes\0 Organic Topics - latest\315 topics\20 315 lecture notes, 7-9-15\314 Review Problems\01 review, Zeff of atoms 19p, answers, 6p.doc Properties of atoms Beauchamp Alkane boiling point Alkane boiling point From the examples above, you can see that even the weak methane, CH4 -162 tridecane, C H 235 dispersion forces of attraction become significant when a 13 28 large number of them are present. ethane, C2H6 -89 tetradecane, C14H30 254 propane, C3H8 -42 pentadecane, C15H32 271 butane, C4H10 0 hexadecane, C16H34 287 CH4 CH4 Larger molecules have pentane, C5H12 36 heptadecane, C17H36 302 more contact surface area hexane, C H 69 octadecane, C H 316 with neighbor molecules. 6 14 18 38 CH3CH2CH2CH2CH2CH3 heptane, C H 98 nonadecane, C H 330 Greater dispersion forces 7 16 19 40 mean a higher boiling point. octane, C8H18 126 icosane, C20H42 343 CH3CH2CH2CH2CH2CH3 nonane, C9H20 151 henicosane, C21H44 356 decane, C10H22 174 doicosane, C22H46 369 CH CH CH CH CH CH CH CH CH CH CH CH undecane, C11H24 196 tricosane, C23H48 369 3 2 2 2 2 2 2 2 2 2 2 3 dodecane, C12H26 216 triacotane, C30H62 450 CH CH CH CH CH CH CH CH CH CH CH CH tetracotane, C40H82 563 3 2 2 2 2 2 2 2 2 2 2 3 350 Boiling Point o 300 Temp ( C) 250 200 150 100 bp of water 50 0 mp of water -50 -100 -150

C1 C2 C3 C4 C5 C6 C7 C8 C9 C10 C11 C12 C13 C14 C15 C16 C17 C18 C19 C20 Straight chain alkanes

In alkane isomers (having the same number of atoms, CnH2n+2), more branching reduces contact with neighbor molecules and weakens the intermolecular forces of attraction. Linear alkanes have stronger forces of attraction than their branched isomers because they have a greater contact surface area with their neighbor molecules. Branches tend to push neighbor molecules away. This is very evident in boiling points, where all of the forces of attraction are completely overcome and the linear alkane isomers have higher boiling points (stronger attractions) than the branched isomers. The strength of these interactions falls off as the 6th power of distance. A structure twice as far away will only have 1/64 the attraction for its neighbor.

H2 H2 H2 H2 H2 CH3 H C CH H3C C CH3 3 C C 3 H3C C C CH3 6 o 1 bp = -42 C o o H3C C CH3 = 1 bp = -0.5 C bp = -0.5 C H 1 bp = -12oC 6 More atoms increase the contact 1 = 1 Less branching increases contact 2 surface area with neighbor surface area with neighbor 64 molecules (not isomers). molecules in these isomers.

Problem 9 – Provide an explanation for the different boiling points in each series.

y:\files\classes\0 Organic Topics - latest\315 topics\20 315 lecture notes, 7-9-15\314 Review Problems\01 review, Zeff of atoms 19p, answers, 6p.doc Properties of atoms Beauchamp a.

F Cl Br I

bp = +85 oC bp = +130 oC bp = +155 oC bp = +188 oC b. OH O bp = -0.5 oC bp = +5 oC bp = +78 oC c. O HO O H2N O O C O bp = -78 oC o bp = -47 oC bp = +20 oC bp = +101 oC bp = +210 C (sublimes)

d. N bp = -23 oC bp = +82 oC bp = -42 oC

e. H2 H2 H2 H H2 CH3 H3C CH3 H3C C C C CH3 H3C C C CH3

H3C C CH3 CH o o 3 o bp = -89 C bp = +36 C bp = +30 oC bp = +10 C CH3 e. CH4 CH3Cl CH2Cl2 CHCl3 CCl4 o o bp = -164 oC bp = -24 C bp = +40 C bp = +61 oC bp = +77 oC f. H2 H2 H2 C C H C H H3C CH3 H3C N H3C O o o bp = -42 C bp = +17 oC bp = +78 C H

y:\files\classes\0 Organic Topics - latest\315 topics\20 315 lecture notes, 7-9-15\314 Review Problems\01 review, Zeff of atoms 19p, answers, 6p.doc Properties of atoms Beauchamp Problem 10 – Match the given boiling points with the structures below and give a short reason for your answers. (-7oC, +31 oC, +80 oC, +141oC, 1420oC)

O O

KCl OH 2-butanone 2-methyl-1-butene propanoic acid potassium chloride 2-methylpropene MW = 72 g/mol MW = 56 g/mol MW = 70 g/mol MW = 74 g/mol MW = 74.5 g/mol

Solutes, Solvents and Solutions

Mixing occurs easily when different substances, having similar forces of interaction, are combined. This observation is summarized in the general rule of “Like Dissolves Like”. To discuss the solution process, there are a few common terms that we need be familiar with.

Solutes are substances that are dissolved in a solvent. There may be as few as 1 solute (ethanol in water) or as many as 1000’s of dissolved solutes (blood).

The solvent is the liquid in which the solute(s) is(are) dissolved. The solvent is usually the major component of the mixture. Sometimes when water is present, it is considered to be the solvent, even when present in only 1% (as is often the case in sulfuric acid/water mixtures).

A solution is the combination of the solvent and the solute(s).

To dissolve a solute in a solvent, the forces of interaction among solute molecules must be overcome and the forces of interaction among solvent molecules must be overcome. These are energy expenses. In return, new interactions develop among solute and solvent molecules. These are energy gains. The balance between these energy expenses and energy gains determines whether a solute will dissolve. A solution is formed when the energy gains are greater than the energy expenses. The important interactions are listed below in decreasing order of energy importance.

Possible energy expenses due to (solute / solute) Possible energy gains from and (solvent / solvent) interactions (solute / solvent) interactions

1. ion-ion (lattice energy) increasing 1. ion-dipole (solvation energy) 2. hydrogen bonds energy 2. hydrogen bonds 3. dipole-dipole 3. dipole-dipole 4. dispersion forces 4. dispersion forces

Solvents are classified as polar or nonpolar based on their dielectric constant (see table below). If the dielectric constant is greater than 15, then the solvent is considered polar. If less than 15, then the solvent is considered nonpolar. This is example of an arbitrary division, created for our convenience. Water and alcohols (= “organic water”) are in the special class of solvents that have a polarized hydrogen atom in a covalent bond (O-H > N-H). Solvents in this class tend to be especially good at interacting with both types of charge. Such solvents are called “protic solvents”, because they can both donate and accept hydrogen bonds. In our course, protic solvents have an O-H bond or an N-H bond and tend to be better at solvating (thus, dissolving) polar and ionic solutes than solvents without this feature.

y:\files\classes\0 Organic Topics - latest\315 topics\20 315 lecture notes, 7-9-15\314 Review Problems\01 review, Zeff of atoms 19p, answers, 6p.doc Properties of atoms Beauchamp R O H R Protic solvents usually have an O-H bond. H+ R R O - - O + H O H H O - + R R + H O - O H O H R R Protic solvents are good at solvating both cations and anions. Many partial negative solvent dipoles (lone pairs on oxygen or nitrogen atoms) can take the place of an anion, while many partial positive solvent dipoles (polarized hydrogen atoms) can take the place of the cation.

Solvents that are classified as polar ( > 15, see below), but do not have a polarized hydrogen are called polar aprotic solvents. They tend to be very good at solvating positive charge, but not so good at solvating negative charge. This turns out to be very helpful when anions need to react in chemical reactions, and the use of such solvents is very common in organic chemistry when that is the goal.

Aprotic solvents do NOT have an O-H (or N-H) bond.

H3C CH3 N CH3 C C - - N N Nitrogen and oxygen + C C + acetonitrile lone pairs can interact (ethanenitrile) strongly with cationic N CH3 N N charge. - - Polar aprotic solvents interact C C poorly with anionic charge, so they tend to be available to react. H3C CH3

Polar aprotic solvents are good at solvating cations and not so good at solvating anions.

Solvents with e < 15 and no polarized hydrogen atom are called nonpolar solvents and generally do not mix well with polar and ionic substances. In biochemistry such substances are referred to as hydrophobic (don’t mix well with water) or lipophilic (mix well with fatty substances). This is a good feature because they can keep different aqueous solutions (blood vs. cytosol) apart and prevent mixing (cell membranes). Three different kinds of solvents that chemists tend to talk about are: nonpolar solvents, polar aprotic solvents and polar protic solvents. Examples are shown below.

y:\files\classes\0 Organic Topics - latest\315 topics\20 315 lecture notes, 7-9-15\314 Review Problems\01 review, Zeff of atoms 19p, answers, 6p.doc Properties of atoms Beauchamp dielectric Common Solvents Formula constant () protic/aprotic O O water H2O 78 protic H H H C H methanol (MeOH) CH OH 33 protic 3 3 water methanol ethanol (EtOH) CH3CH2OH 24 protic O 2-propanol (i-PrOH) (CH3)2CHOH 20 protic O methanoic acid (formic acid) HCO2H 50 protic *ethanoic acid (acetic acid, HOAc) CH3CO2H 6 protic C H C H O H3C CH3 2-propanone (acetone) CH3COCH3 21 polar aprotic formic acid acetone dimethyl sulfoxide (DMSO) CH3SOCH3 46 polar aprotic dimethyl formamide (DMF) HCON(CH ) 37 polar aprotic 3 O acetonitrile (AN) CH3CN 36 polar aprotic N nitromethane CH3NO2 36 polar aprotic C hexamethylphosphoramide (HMPA) O=P[N(CH ) ] 30 polar aprotic S 3 2 3 H3C H3C CH3 acetonitrile ethyl ethanoate (EtOAc) CH3CO2CH2CH3 6 nonpolar DMSO ethyl ether (Et2O) CH3CH2OCH2CH3 4 nonpolar O tetrahydrofuran (THF, cyclic ether) (CH2)4O 8 nonpolar dichloromethane (methylene chloride) CH2Cl2 9 nonpolar O trichloromethane (chloroform) CHCl3 5 nonpolar C CH3 tetrachloromethane (carbon tetrachloride) CCl4 2 nonpolar H N N hexane CH (CH ) CH 2 nonpolar 3 2 4 3 H3C O CH *Ethanoic acid violates our arbitray solvent polarity rules but is a strong DMF 3 nitromethan hydrogen bonding solvent, so is included in the protic solvent group. O H O O H P 2 H2 H2 H2C CH2 C C C (H3C)2N N(CH3)2 C C Cl H C O CH H N(CH3)2 3 3 H3C O CH3 H2CCH2 Cl HMPA ethyl acetate ether THF methylene chloride

Problem 11 – Point out the polar hydrogen in methanol. What is it about dimethyl sulfoxide (DMSO) that makes it polar? Draw a simplistic picture showing how methanol interacts with a cation and an anion. Also use DMSO (below) and draw a simplistic picture showing the interaction with cations and anions. Explain the difference from the methanol picture. Two resonance O structures of DMSO a polar, aprotic solvent O S H H S = anion O H3C CH3 C C = cation H3C H H H Obeys the octet rule, H H methanol but has formal charge. Violates the octet rule, but does not have formal charge. Sulfur has 3d orbitals, which provides a possible explanation for drawing a double bond.

y:\files\classes\0 Organic Topics - latest\315 topics\20 315 lecture notes, 7-9-15\314 Review Problems\01 review, Zeff of atoms 19p, answers, 6p.doc Properties of atoms Beauchamp

Simplistic Sketch of Mixing Solutes and Solvents H H H H H H O H O OH O O H O polar protic solvent = water H H H The necessary energy to dissolve NaCl comes from many, many ion-dipole cost = hydrogen bonding interactions (hydration = water, solvation = solvent). All of these smaller cost = lattice energy interactions between ions and solvent balance the very large ion-ion interactions gain = ion-dipole of the lattice structure and hydrogen bonding of water. If the solvation energy ionic is larger than the lattice energy, the salt will dissolve (NaCl), but if the lattice substance energy is larger than the solvation energy, the salt will not dissolve (AgCl).

Na Cl nonpolar A very large energy input solvent = octane is needed to separate There is no way the nonpolar solvent can (like gasoline) ionic charges from one compensate for the lattice energy of the another (lattice energy). salt. The salt stays in its crystalline form CH3(CH2)6CH3 and sinks or floats based on its density relative to the solvent. Whatever is more cost = dispersion forces dense sinks and less dense floats (think of cost = lattice energy a stick and a rock in water). gain = dispersion-ion (too weak)

H H

H O H O The weak dispersion forces do not compensate for the broken hydrogen bonds. polar protic Also, there is a large entropy cost for water solvent = water to structure itself around the nonpolar solute H H molecules. Water stays separate from the cost = hydrogen bonding nonpolar solute and is on the top or bottom cost = dispersion forces based on the densities of the substances gain = dispersion-dipole H O H O (greater density is on the bottom).

nonpolar substance

nonpolar There are no strong forces to overcome in CH3(CH2)16CH3 solvent = octane either the solvent or the solute. There are (like gasoline) no strong interactions gained in the solution octadecane (only dispersion forces). There is a favorable (waxy substance) CH3(CH2)6CH3 increase in entropy (DS) from the disorder created in the mixing of the solution, so the Only weak dispersion cost = dispersion forces substances mix. forces are present in cost = lattice energy pentane. (weak dispersion) gain = dispersion-dispersion (weak)

Problem 12 – Carbohydrates are very water soluble and fats do not mix well with water. Below, glucose is shown below as a typical hydrophilic carbohydrate, and a triglyceride is used as a typical hydrophobic fat. Point out why each is classified in the manner indicated. O HO HO O O HO O O HO OH = O HO OH O HO OH OH O

glucose (carbohydrate) typical saturated triglyceride (fat)

y:\files\classes\0 Organic Topics - latest\315 topics\20 315 lecture notes, 7-9-15\314 Review Problems\01 review, Zeff of atoms 19p, answers, 6p.doc Properties of atoms Beauchamp Problem 13 – Bile salts are released from your gall bladder when hydrophobic fats are eaten to allow your body to solubilize the fats, so that they can be absorbed and transported in the aqueous blood. The major bile salt glycolate, shown below, is synthesized from cholesterol. Explain the features of glycolate that makes it a good compromise structure that can mix with both the fat and aqueous blood. Use the ‘rough’ 3D drawings below to help your reasoning, or better yet, build models to see the structures for yourself (though it’s a lot of work).

OH O

H H H N synthesized in many, H H many steps in the body H H cholesterol O HO HO OH glycolate O H 1. source for steroid syntheses in the body (bile salt) 2. important constituent of cell membranes 2. transported in blood to delivery sites via VLDL LDL HDL All polar groups are on the same face. Which side VLDL = very low density lipoprotein, has high cholesterol concentration faces water and which side faces fat molecules? LDL = low density lipoprotein, has medium cholesterol concentration (See structures below.) HDL = high density lipoprotein, has low cholestero contcentration

CO2 OH OH representation of bile acid (glycolate) as a long bent shape havingtwo different representation of cholesterol OH faces, one polar and one nonpolar HO as a long flat shape

y:\files\classes\0 Organic Topics - latest\315 topics\20 315 lecture notes, 7-9-15\314 Review Problems\01 review, Zeff of atoms 19p, answers, 6p.doc Properties of atoms Beauchamp

Simplistic picture of atoms’ Zeffective and other periodic properties

+1 +2 Hydrogen, lithium and sodium all have The valenece electrons in Li, C, N, O and F hydrogen The combination of both trends is a Z of +1, but the pull for the are all in the same n=2 shell, and all have the that elements in the upper right helium Z = +1 eff total electrons is weaker in K than Na than Li same number of core electrons (2). The corner of the periodic table attract Ztotal = +2 shield = 0 than H because the outer most electron nuclear charge keeps getting larger when electrons strongest. The electron shield = 0 Z = +1 moving across the row and the pull on the attracting power of atoms in bonds effective is farther from the Z in K than Na than Zeffective = +2 IP = +314 kcal/mole eff valence electrons keeps getting stronger. The is referred to as electronegativity, 1 Li than H. The result from this trend is IP1 = +313 at. radius = 53 pm result from this trend is that elements further where a higher number indicates a that elements higher in a column attract at. radius = 31 pm # bond = 1 to the fight in a row attract electron more higher electronegativity. See values electrons more strongly than elements in the table below. # bond = 0 lone pairs = 0 strongly than elements on the left. lower in a column. lone pairs = 0  = 2.2  = ?

+3 -2 +4 -2 +5 -2 +6 -2 +9 +7 -2 +8 -2 -2 +10 -2

lithium berylium boron carbon nitrogen oxygen fluorine neon Ztotal = +3 Z = +6 Ztotal = +4 Ztotal = +5 total Z = +7 Z = +8 Z = +9 Z = +10 shield = -2 total total total total shield = -2 shield = -2 shield = -2 shield = -2 shield = -2 shield = -2 shield = -2 Zeffective = +1 Z = +2 Z = +3 Zeffective = +4 effective effective Zeffective = +5 Zeffective = +6 Zeffective = +7 Zeffective = +8 IP1 = +124 IP = +261 IP1 = +215 IP1 = +192 1 IP = +335 IP = +315 IP = +402 IP = +499 at. radius = 167 1 1 1 1 at. radius = 112 pm at. radius = 87 pm at. radius = 67 pm at. radius = 56 pm at. radius = 48 pm at. radius = 42 at. radius = 38 # bond = 1 or ionic # bond = 2 or ionic # bond = 3 # bond = 4 # bond = 3 # bond = 2 # bond = 1 # bond = 0 lone pairs = 0 lone pairs = 0 lone pairs = 0 lone pairs = 0 lone pairs = 1 lone pairs = 2 lone pairs = 3 lone pairs = 4  = 1.1  = 1.5  = 2.0  = 2.5  = 3.0  = 3.5  = 4.0  = ?

+11 -2 -8 +13 -2 -8 +14 -2 -8 +15 -2 -8 +12 -2 -8 +16 -2 -8 +17 -2 -8 +18 -2 -8

sodium magnesium aluminum silicaon phosphorous sulfur chlorine argon Ztotal = +11 Ztotal = +12 Z = +13 Z = +14 total total Ztotal = +15 Z = +16 Z = +17 Z = +18 shield = -10 shield = -10 shield = -10 total total total shield = -10 shield = -10 shield = -10 shield = -10 shield = -10 Zeffective = +1 Zeffective = +2 Zeffective = +3 Zeffective = +4 Zeffective = +5 Zeffective = +6 Zeffective = +7 Zeffective = +8 IP1 = +118 IP1 = +177 IP = +138 IP = +189 1 1 IP1 = +242 IP = +239 IP = +300 IP = +363 at. radius = 190 pm at. radius = 145 at. radius = 118 pm 1 1 1 at. radius = 111 pm at. radius = 98 pm at. radius = 88 at. radius = 79 at. radius = 71 # bond = 1 or ionic # bond = 2 or ionic # bond = 3 # bond = 4 # bond = 3 # bond = 2 # bond = 1 # bond = 0 lone pairs = 0 lone pairs = 0 lone pairs = 0 lone pairs = 0 lone pairs = 1 lone pairs = 2 lone pairs = 3 lone pairs = 4  = 0.9  = 1.2  = 1.5  = 1.9  = 2.2  = 2.6  = 3.5  = ? 10 "3d" electrons between these two elements.

-2 -8 +19 -8 +20 -2 -8 -8 +31 -2 -8 -18 +32 -2 -8 -18 -2 -18 +33 -8 +34 -2 -8 -18 +35 -2 -8 -18 +36 -2 -8 -18

potassium calcium gallium germanium arsinic selenium bromine krypton Z = +19 Z = +20 total total Ztotal = +31 Ztotal = +32 Ztotal = +33 Ztotal = +34 Ztotal = +35 Ztotal = +36 shield = -18 shield = -18 shield = -28 shield = -28 shield = -28 shield = -28 shield = -28 shield = -28 Z = +1 Z = +2 effective effective Zeffective = +3 Zeffective = +4 Zeffective = +5 Zeffective = +6 Zeffective = +7 Zeffective = +8 IP = +99 IP = +141 1 1 IP1 = +138 IP1 = +182 IP1 = +226 IP1 = +225 IP1 = +273 IP1 = +363 at. radius = 243 pm at. radius = 194 at. radius = 136 pm at. radius = 125 pm at. radius = 114 pm at. radius = 103 at. radius = 94 at. radius = 88 # bond = 1 or ionic # bond = 2 or ionic # bond = 3 # bond = 4 # bond = 3 # bond = 2 # bond = 1 # bond = 0 lone pairs = 0 lone pairs = 0 lone pairs = 0 lone pairs = 0 lone pairs = 1 lone pairs = 2 lone pairs = 3 lone pairs = 4  = 0.8  = 1.0  = 1.6  = 1.9  = 2.2  = 2.5  = 3.0  = 3.0

Core electrons are full shells inside the outermost valence electrons. They are considered to shield the outermost electrons from Ztotal, leaving a net Zeff to attract the valence electrons. The positive number in the middle of each group of circles is Ztotal. Each negative number inside a circle represents a full inner shell (core electrons) and shields Ztotal. The dots in the outer circle represent the valence electrons of each neutral atom feeling the pull of Zeff.

Problem 1 – What kinds of bonds are between the nonhydrogen atoms below?

a b c d H H H e H H f g H H

HCN HCC H CO CC CN CO H CCH

H H H H H H 1 x sigma 1 x sigma 1 x sigma 1 x sigma 1 x sigma 1 x sigma 1 x sigma 2 x pi 2 x pi 1 x pi 1 x pi 1 x pi 0 x pi 2 x pi

Problem 2 - Explain the atomic trends in ionization potential in a row (Na vs Si vs Cl) and in a column (F vs Cl vs Br).

Na (+118), Si (+189), Cl (+300): The ionization potential (IP1) increases across a row because Zeff is larger.

F (+402), Cl (+300), Br (+273): The ionization potential decreased down a column because the electrons are farther away from the “same” Zeff (in this case, Zeff = +7 for the halogens)

Problem 3 - Explain the atomic trends in atomic radii in a row (Li vs C vs F) and in a column (C vs Si vs Ge).

Li (167 pm), C (67 pm), F (42 pm): All of the electron clouds in these examples are using the n=2 shell (similar sizes), but the Zeff gets larger as one moves to the right, which shrinks the electron cloud (gets smaller)

C (67 pm), Si (111 pm), Ge (125 pm): Each atom in a column adds an extra shell of electrons, which gives lower atoms a larger radius.

as cations as anions

Problem 4 - In the tables above: a. Explain the cation distances compared to the atomic distances (Li, Be, B). (Tables 2 and 3)

Li+ (90 pm), Be+2 (59 pm), B+3 (41 pm): All of the cations hold on to their remaining electron tighter, making smaller radii, and the most electron deficient hold the tightest (smallest radius)

Li (167 pm), Be (112 pm), B (87 pm): All of the neutral atoms have a larger radius of their electron clouds than the cations and B with the largest Zeff has the smallest radius of the three in the n=2 row.

N 132 pm), O (126 pm), F (119 pm): All of the anions have greater electron-electron repulsion, which expands the electron clouds over the neutral atoms, N with the greatest negative charge expands the most b. Explain the cation distances in a row (Li, Be, B). (Table 3)

Li+ (90 pm), Be+2 (59 pm), B+3 (41 pm): All of the cations hold on to their remaining electron tighter, making smaller radii, and the most electron deficient hold the tightest (smallest radius) c. Explain the anion distances in a row (N, O, F). (Table 3)

N 132 pm), O (126 pm), F (119 pm): All of the anions have greater electron-electron repulsion, which expands the electron clouds, N with the greatest negative charge expands the electron cloud the most

Problem 5 – What is the effect on the dipole moment of switching a hydrogen for a methyl in each pair below? Propose a possible explanation. In every example, the methyl for the hydrogen increases the dipole moment. The methyl must give up more electron density to the oxygen relative to the hydrogen. This methyl (R group in general) is referred to as inductively electron donating. Hyperconjugation is an alternative explanation using molecular orbitals (HOMO and LUMO). The effect is smallest for the amide because the nitrogen is such a good resonance donator that the inductive is not as significant.

O O O O O O O O

polar O polar O polar O O O O resonance resonance resonance C C C C C C H3C H H C H H H H H 3 H3C CH3 H C CH  = 2.3 D  = 2.68 D  = 2.91 D 3 3 bp = -20oC bp = +20oC bp = +56oC mp = -92oC mp = -123oC mp = -94oC H2O sol. = 400g/L H2O sol. = very sol. H2O sol. = very sol. Each additional "R" group (methyl) makes the carbonyl bond (C=O) more polar, indicated that the "R" group is inductively donating relative to the hydrogen.

The sp2 orbital of an alkene is more electronegative (33% s) than an sp3 orbital (25% s). The methyl substituent gives up some of its electron density because of this (inductively donating). The sp orbital (50% s) of an alkyne is even more electronegative because of the tighter hold by s electrons over p electrons so takes even more electron density from the inductively donating methyl group (has a higher dipole moment).

y:\files\classes\0 Organic Topics - latest\315 topics\20 315 lecture notes, 7-9-15\314 Review Problems\01 review, Zeff of atoms 19p, answers, 6p.doc Properties of atoms Beauchamp

H CH3 NH2 OH H Inductive donation Both aniline and phenol The nitrogen atom by the methyl gives inductively withdraw in pyridine toluene a small electron density from the inductively withdraws dipole moment. aromatic ring, BUT both from the carbon atoms nitrogen and oxygen are making a larger dipole resonance donors (N > O) moment. and add electron density N  = 0.0 D  = 0.31 D  = 1.53 D  = 1.55 D  = 2.37 D o o o to the ring. Usually bp = 81 C bp = 111 C bp = 184oC bp = +182 C resonancedominates over bp = +115oC o o mp = 5C mp = -95 C mp = -6oC mp = +40 C inductive effects, but not mp = -42oC always.

F Cl Br I

The dipole moment depends on both charge separation (larger for F > Cl > Br > I) and bond length (larger for I > Br > Cl > F). The dipole moments are very similar because of opposing effects. The differences in boiling points is due to the greater polarizability of the larger atoms (I > Br > Cl > F).

 = 1.66 D  = 1.54 D  = 1.73 D  = 1.91 D bp = +84oC bp = +131oC bp = +156oC bp = +188oC mp = -44oC mp = -45oC mp = -31oC o mp = -31 C

Problem 6 – Which base pair binds more tightly in DNA, GC or AT, or are they about the same?

three hydrogen bonds in G-C base pairs bind tighter than two hydrogen bonds in A-T pairs H two hydrogen bonds in A-T base pair H N O H N N N H O N N H N N DNA N N N H N DNA N N H O DNA N guanine adenine H cytosine thymine O DNA

Problem 7 – Answer given with table.

Problem 8 – Explain the difference in boiling points of the following molecules. a c d H2 b C O H3C H H O O H3C CH3 H3C CH3  = 0.08 D  = 1.30 D  = 1.69 D  = 1.69 D o bp = -42oC bp = -22oC bp = +65oC bp = +78 C o mp = -188oC mp = -141oC mp = -98oC mp = -114 C H O sol. = miscible H2O sol. = 40mg/L H2O sol. = 71g/L H2O sol. = miscible 2

Structure a, b and c are similar in size. Structure a only has dispersion forces, while structure b has dispersion forces and polar bonds, while structure c has dispersion forces, polar bonds and hydrogen bonds. Intermolecular forces of interaction increase in order of dispersion forces < polar bonds < hydrogen bonds, so c has the highest bp, followed by b and lowest is a. Compound d is like c except it has an extra methyl, thus greater dispersion forces, thus a higher boiling point.

F Cl Br I

bp = +85 oC bp = +130 oC bp = +155 oC bp = +188 oC Similar molecules. The highest boiling points go with the largest dispersion forces (greatest polarizability, I > Br > Cl > F)

b. OH O bp = -0.5 oC bp = +5 oC bp = +78 oC Structure 3 has H bonding and polarity (strongest attractions), structure 2 has polar bonds and structure 1 only has dispersion forces (weakest attractions)

c. O HO O H2N O O C O bp = -78 oC o bp = -47 oC bp = +20 oC bp = +101 oC bp = +210 C (sublimes)

Structure 1 only has weak dispersion forces, structure 2 had polar bonds (stronger), structure 3 has H bonds too (stronger), structure 4 has the strongest polarity and H bonds with both N-H bonds (highest). Finally CO2 has polar bonds, but its symmetry cancels out its polarity. The oxygen atoms hold so tightly to their electrons that they are less polarizable than structure 1 and sublimes at a lower temperature.

d. N bp = -23 oC bp = +82 oC bp = -42 oC Structure 1 (propane) only has weak dispersion forces, and has a bent shape that prevents closer contact with neighbor molecules. Structure 2 (propyne) only has weak dispersion forces, but has a linear shape that allows closer contact with neighbor molecules, so stronger attractions. Structure 3 has the shape of propyne, and relatively strong polarity which makes stronger attractions and a higher bp.

e. H2 H2 H2 H H2 CH3 H3C CH3 H3C C C C CH3 H3C C C CH3

H3C C CH3 CH o o 3 o bp = -89 C bp = +36 C bp = +30 oC bp = +10 C CH3 More carbons have greater contact surface area so stronger dispersion forces (C2 < C5). Straight chains can approach closer than when branches are present (branches tend to push neighbor molecules away = weaker dispersion forces). More branches weaken those attractions even more (D < C < A).

e. CH4 CH3Cl CH2Cl2 CHCl3 CCl4 o o bp = -164 oC bp = -24 C bp = +40 C bp = +61 oC bp = +77 oC Chlorine is a larger atom (than C) and has polarizable lone pairs. More chlorines = more dispersion forces = higher bp. B, C and D have polarity too. A and E are not polar, but E has four chlorine atoms and larger dispersion forces dominate here (highest bp).

f. H2 H2 H2 C C H C H H3C CH3 H3C N H3C O o o bp = -42 C bp = +17 oC H bp = +78 C Bp's follow H bonding possibilities. Structure 3 has the strongest H bonds because O-H is more polar. Structure 2 has some H bonding possibilitiies so a higher bp than structure 1 which only has dispersion forces. All 3 molecues are similar in size, so comparisons are fair.

y:\files\classes\0 Organic Topics - latest\315 topics\20 315 lecture notes, 7-9-15\314 Review Problems\01 review, Zeff of atoms 19p, answers, 6p.doc Properties of atoms Beauchamp Problem 10 – Match the given boiling points with the structures below and give a short reason for your answers. (-7oC, +31 oC, +80 oC, +141oC, 1420oC)

O O

KCl OH 2-butanone 2-methyl-1-butene propanoic acid potassium chloride 2-methylpropene MW = 72 g/mol MW = 70 g/mol MW = 74 g/mol MW = 74.5 g/mol MW = 56 g/mol

1420oC -7oC o +141oC +80 C +31oC Structure 4 has ionic bonds, by far the strongest forces to overcome and has an extremely high bp. Structure 3 has strong H bonding possiblities and the second high bp. Structure 1 has polar C=O bond so higher than structures 2 and 5, which only have weak dispersion forces. Structure 2 is larger and has a larger surface area and more dispersion forces than structure 5, so has a higher bp.

Problem 11 – Point out the polar hydrogen in methanol. What is it about dimethyl sulfoxide (DMSO) that makes it polar? Draw a simplistic picture showing how methanol interacts with a cation and an anion. Also use DMSO (below) and draw a simplistic picture showing the interaction with cations and anions. Explain the difference from the methanol picture. = cations are smaller because they have lost electrons and hold onto the remaining electrons tighter. = anions are larger because they have extra electron density that is repulsive and expands the electron clouds

polar H CH3 H3C H O O C H3 CH3 H O O O H CH3 Methanol is good at solvating positive and negative charge. H H O H3C H H3C H O

H C CH strong bond dipole H3C 3 3 S SO H3C CH3 O S O H3C DMSO is very good at solvating S O positive charge but it is poor at CH3 H3C CH3 solvaing n egative charge. O O S

S CH3 H3C CH3

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Problem 12– Carbohydrates are very water soluble and fats do not mix well with water. Below, glucose is shown below as a typical hydrophilic carbohydrate, and a triglyceride is used as a typical hydrophobic fat. Point out why each is classified in the manner indicated. O HO HO O O HO O O HO OH = O HO OH O HO OH OH O

glucose (carbohydrate) typical saturated triglyceride (fat)

Problem 13 – Bile salts are released from your gall bladder when hydrophobic fats are eaten to allow your body to solubilize the fats, so that they can be absorbed and transported in the aqueous blood. The major bile salt glycolate, shown below, is synthesized from cholesterol. Explain the features of glycolate that makes it a good compromise structure that can mix with both the fat and aqueous blood. Use the ‘rough’ 3D drawings below to help your reasoning, or better yet, build models to see the structures for yourself (though it’s a lot of work).

OH O

H H H N synthesized in many, H H many steps in the body H H cholesterol O HO HO OH glycolate O H (bile salt)

blood Glycolate has a nonpolar, hydrophobic H2O H O face that can cover the inside of a fat ball HO 2 and a hydorphilic face that can point OH blood blood outward toward the aqueous blood, which HO H O allows fats to be transported throughout H2O OH 2 the body to reach fat storage cells and other HO OH essential locations. O2C CO2 blood nonpolar blood fats inside H2O O C H O 2 2 HO CO2 OH HO blood blood OH H O HO 2 H2O OH blood blood H2O

y:\files\classes\0 Organic Topics - latest\315 topics\20 315 lecture notes, 7-9-15\314 Review Problems\01 review, Zeff of atoms 19p, answers, 6p.doc