CHAPTER 3 HW SOLUTIONS: INTERMOLECULAR FORCES

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1 APTER 3 W SLUTINS: INTERMLEULAR FRES ENERGY DIAGRAMS 1. Label and answer questions about the following energy diagram. Energy * I * I * small E a3 a. ow many steps are in the overall reaction? 3 b. Label each transition state with an asterisk (*). c. Label each intermediate with an I. Rxn coordinate d. Label the activation barrier for each step with a double-headed arrow. 2. The energy diagrams of two reactions are plotted side by side below. a. Which reaction is more exothermic? B (more downhill) b. Assuming equal concentration of reactants and equal temperatures, which reaction would proceed faster? Δ Rxn A Rxn B A (lower activation barrier) Rxn coordinate BND DISSIATIN ENERGIES 3. Given that the bond dissociation energy (BDE) of -F in 3 F is 109 kcal/mol and the BDE of -I in 3 I is 56 kcal/mol, which bond is stronger, -F or -I? Briefly explain. BDE refers to the amount of energy needed to break a bond in half, into radicals. The higher the BDE, the more energy it takes to break the bond. Therefore, -F is a stronger bond, since its BDE is higher. Page 1

2 4. onsult a table of bond dissociation energies, and use them to determine the D for each reaction. Also state whether each reaction is favorable or not (in terms of D ). a. 4 + I 2 à 3 I + I D = [ I-I ] [ I + -I] = [ 4(104) + 36 ] [ 3(104) ] = [ 452 ] [ 439 ] = +13 kcal/mol Unfavorable (positive D and reactant bonds are stronger) + I I I + I b. 2 = 2 + F à 3 2 F + F F D = [ = + F ] [ F] = [ 4(98) ] [ 5(98) ] = [ 680 ] [ 685 ] = -5 kcal/mol Note: these 4 - are the ones listed as 2 =- Somewhat favorable (negative D and product bonds are stronger) c /2 = à 2 == /2 + D = [ /2 = ] [ 4 = ] = [ 6(98) /2(119) ] [ 4(192) + 6(119) ] = = [ ] [ 1482 ] = kcal/mol Very favorable (very negative D and product bonds are stronger) d. Match the previous reactions to each energy diagram below, considering the calculated D and their relative magnitude. Note: energy diagrams are not perfectly to scale, choose the closest option. b c a Page 2

3 FUNTINAL GRUPS 1. Identify the main functional group in each compound below. Structure N 3 3 Funct Gp Aromatic Alkene Alkane Ketone Amide Ester Structure N Funct Gp Ether Alkyne Alcohol Amine Aldehyde arboxylic acid 2. Identify the functional groups in each compound. alkane aromatic ester 3 3 amine N 3 3 ether amide N carboxylic acid aromatic aromatic N amine Ibuprofen (analgesic) Darvon (analgesic) Melatonin (hormone thought to induce sleep) alkene alkyne N amine alkene alcohol alkyne alkene ester three more alkenes ketone istrionicotoxin (poison secreted by S. American frog) Pyrethrin I (potent insecticide from chrysanthemums) Page 3

4 INTERMLEULAR FRES 3. Multiple choice: The electrons pointed to in this water molecule are part of a: a. ovalent bond. b. ydrogen bond. c. Dipole-dipole force. d. Both b+c. 4. Identify the type of intermolecular force (London dispersion force, weak/strong dipole-dipole force, or hydrogen bond) pointed to in each interaction. l l dipole-dipole force (hard to say weak/strong) LDF strong dipole-dipole force hydrogen bond 5. For the substances below, Structure Br N 3 Draw and label the IMF δ+ δ Br Br dipole-dipole force (hard to say weak/strong) δ+ δ δ+ δ+ δ N δ+ hydrogen bond δ N Place d +, d - labels on the appropriate atoms in any polar bonds. Draw two of the same molecules interacting (using dashed lines), to show how the molecules orient relative to one another in the liquid form. Point to and identify the type of intermolecular force present. 6. Describe how a London dispersion force (LDF) arises, and where the force of attraction comes from. An atom s electron cloud is often evenly distributed, but because electrons move, they can at times be lopsided. When lopsided, there is a buildup of electrons on one side, which is a partial negative area (d - ). The other side is a partial positive area (d +). This is a temporary dipole. The temporary dipole causes nearby electron clouds in other molecules to polarize (create temporary d -, d + areas), and the attraction between the temporary dipoles in different molecules is a LDF. Page 4

5 7. Draw a molecular picture of the substances below, showing how the molecules orient relative to one another in the pure liquid form. Draw at least two molecules interacting and use dashed lines to show the strongest type of intermolecular forces (IMF) in each. Structure l 3 N Draw the IMF δ+ δ l δ+ δ l 3 N N 3 Strongest type of IMF Dipole-dipole forces (weak) LDF ydrogen bond Structure Draw the IMF 3 3 δ δ+ δ+ δ δ+ δ δ+ δ 3 Strongest type of IMF Dipole-dipole forces (strong) ydrogen bonds Dipole-dipole forces (strong) Page 5

6 8. Explain why compound A experiences weak dipole-dipole forces in the pure liquid phase while compound B has stronger dipole-dipole forces in the pure liquid phase A Two reasons why B is strong and A is weak: B Trigonal planar geometry in B allows for close contact of the dipoles, and stronger interaction. A has tetrahedral geometry on the atoms that are involved in the dipole and doesn t allow for as close contact. The net dipole is greater in B. In A, the individual dipoles somewhat cancel each other out, leading to a smaller net dipole. In B there is one dipole with nothing counteracting it. 9. In each set, rank the compounds in order of increasing strength of intermolecular forces. Explain your ranking. a. 3 l middle strongest weakest 3 molecules can form -bonds in the pure liquid phase, which are strong. 3 l molecules have weak dipole-dipole forces in the pure liquid phase, which are stronger than the intermolecular forces of 3 3 (which has only weak London dispersion forces). All molecules are of similar size and have similar LDF s. b. 3 Br 3 l 3 I middle weakest strongest These compounds differ in the halogen, and halogen size/mass goes from smallest to biggest l < Br < I. The bigger the atom, the further the electrons are from the nucleus, causing the atom to be more polarizable and able to distort, thus creating stronger LDF s. With this logic, 3 I has the strongest LDF s (and strongest IMF s) of the series. c. strongest middle weakest The compounds are the same with respect to shape and molar mass, meaning they have equivalent LDF s. owever, the first two compound have additional forces, causing them to have stronger forces than the last compound. The first compound can also form -bonds (the strongest IMF), and the second compound can also form strong dipole-dipole forces (stronger than just LDF). Page 6

7 TRENDS IN BILING PINT 10. In each set, rank the compounds in order of increasing boiling point. Explain your ranking. a. I I 3 I middle highest lowest The only difference between these compounds is chain length and therefore size. The largest one (or one with highest mass) has the strongest LDF s, and so has the highest boiling point. This is because as the carbon chain increases there is increased surface area and more potential electron clouds to polarize and create LDF s. b. highest lowest middle These compounds have the same formula ( 7 16 ) and same mass, but differ in their shape caused by branching. Increased surface-to-surface contact between molecules increases the strength of LDF s and raises boiling points. The branched molecules have less surface area than the linear ones, so have weaker LDF s and a lower b.p. c. lowest highest middle f the two hydrocarbons, the first has the lower boiling point as it is more branched so has weaker LDF s. The alkene (far right structure) and ketone (middle structure) both have stronger LDF s with less branching. Their LDF s are roughly equivalent as they have nearly the same mass and surface area (5 ~ 4+). owever, the ketone has the highest boiling point as it also has strong dipole-dipole forces, which are a stronger IMF than only LDF. d. N 3 N 3 3 middle lowest highest The left two compounds have nearly the same mass (5 ~ 4+N) and the same shape, so have the same strength of LDF. The amine (far left) has a dipole, so can participate in weak dipole-dipole forces in the liquid phase, making it have a higher boiling point than the middle compound. The far right amine has the strongest IMF and highest boiling point due to two factors: it has an N bond so can form hydrogen bonds in the liquid phase, and also has a linear structure so has slightly stronger LDF than the other two compounds. Page 7

8 10 continued e. 3 middle highest lowest strong dip-dip hydrogen bonds weak dip-dip Each compound is the same mass and shape, so all 3 compounds have essentially equivalent LDF s. The trend in boiling point follows the strength of the other IMF s present; the stronger the IMF, the higher the boiling point. From weak to strong: weak dip-dip < strong dip-dip < hydrogen bonds. f. lowest middle highest The alcohol (right) forms strong hydrogen bonds in the liquid phase so has the highest b.p. The two ethers each have weak dipole-dipole forces in the liquid phase, so have a lower b.p. but the difference between them is in their branching (their masses are the same). The flatter, longer ether will have more surface contact, so stronger LDF s and a higher b.p. than the other ether. 11. Explain the following statements. a. Propane ( ) is a gas while pentane ( ) is a liquid at room temperature. Propane exists as a gas at room temperature because its boiling point is below room temp. Pentane has a higher boiling point (above room temperature) so exists as a liquid at room temperature. Pentane has a higher b.p. because it is a much longer compound than propane. This causes it to have stronger London dispersion forces (LDF s) due to its higher mass and greater surface area. b. ompound has a lower boiling point than compound D. D Both +D have similar LDF s in the liquid phase because they have the same mass and shape. ompound D has strong hydrogen bonds in the liquid phase, while compound has strong dipoledipole forces. (Remember that compound does NT hydrogen bond with the drawn because that is bound to a carbon and is a nonpolar region.) Since strong dip-dip are weaker than bonds, has a lower b.p. Page 8

9 11 continued c. ompound E has stronger London dispersion forces than compound F. 3 2 E F LDF s arise from the temporary polarization of electron clouds, and are correlated to mass and shape. ompound E has a higher mass than compound F (6 carbons > 2+) so has a greater electron cloud to potentially polarize, and has stronger LDF s. (Note compound F actually has a higher b.p. due to its strong bonding ability, but this is hard to predict when the LDF are inequivalent). d. ompound G has a higher boiling point than compound. G The only IMF s present in the liquid phase of either compound are LDF, since neither G nor is polar (they are hydrocarbons). They have approximately the same molecular formula (6), but G is linear while is branched. The flatter compound G has a greater surface area, and since LDF s increase with surface-to-surface contact, G has a higher b.p. The fact that G has an alkene makes little difference in this analysis. Page 9

10 WATER SLUBILITY 12. Draw each compound as it would exist in a dilute aqueous solution (assuming each dissolves). Label the intermolecular forces. Structure N 2 l Molecular Picture of Aqueous Solution N 2 δ+ δ l δ δ+ Type of IMF ydrogen bonds ydrogen bonds Wk. Dipole-dipole forces Structure l 3 l Molecular Picture of Aqueous Solution δ δ+ δ+ l δ l δ 3 Type of IMF ydrogen bonds Wk. Dipole-dipole forces ydrogen bonds 13. There are two main criteria necessary for a substance to be water soluble. List them below. ( Water soluble can be defined as when greater than 3g of solute dissolves in 100 ml of water.) The compound must be able to hydrogen bond with water. It must also have a small nonpolar region (have fewer than 5 carbon atoms, if there is one polar area). Page 10

11 14. In each set, rank the compounds in order of increasing water solubility. Explain your ranking. a. l middle least most soluble The ketone (right) can form -bonds to water (shown), making it the most water soluble. The other two cannot -bond with water, so are insoluble. The alkyl chloride is slightly more water soluble than the alkane because it has a weak dipole, allowing for a slightly stronger interaction with water than the LDF s experienced between the alkane and water. b. least most soluble middle All three compounds can -bond with water, but the diol (center) can -bond in two regions, so has the most favorable interactions with water and is the most water soluble (most polar). The alcohol (left) has a larger number of carbons than the ketone (right)- 7 versus 5- so the alcohol is less overall polar and less water soluble. c. most soluble middle least The alkene (right) can t -bond with water so is the least water soluble. Both the ether (left) and the alcohol (center) can form -bonds to water through their oxygen atoms, but the ether has fewer carbons so is more overall polar and more water soluble. 15. Predict whether each molecule should be water soluble or insoluble. l l l 3 N Insoluble Soluble Soluble Soluble 3 N 3 3 Insoluble Insoluble Page 11

12 16. T is the active ingredient in marijuana and ethanol is the alcohol in alcoholic beverages. Propose a brief explanation why drug screenings are able to detect the presence of T but not ethanol weeks after these substances have been introduced in the body Ethanol Ethanol is very water soluble (can -bond with water, and has <5, polar) so it is easily excreted by the body in urine. It is hard to detect it after a little while because it doesn t stay in the body long. 3 T has two spots of polarity, but many carbon atoms so is overall 3 T a nonpolar molecule. This causes it to be harder to excrete, and it stays in our fatty tissues longer. This allows for drug screening to detect T weeks after it has been used. 17. Explain the following statements in detail. a. Pentane is completely soluble in benzene. Pentane Benzene When these two substances are mixed, the LDF s the pentane molecules had with each other are replaced with LDF s between pentane and benzene. Since these LDF s have similar strength, there is negligible enthalpy change in the process. owever, there is an increase in entropy during dissolving, as there are more ways to arrange a solution than the pure solvents. D S dominates the D G equation, and the two substances mix because of the favorable change in entropy associated with mixing. b. Ethylene glycol is insoluble in hexane. Ethylene glycol strongly -bonds with itself, and when it theoretically dissolves in hexane, it is not favorable to Ethylene glycol exane replace its bonds with LDF interactions with the hexane. The glycol therefore maintains its -bonds, forming a very ordered, rigid cage around the hexane molecules, which is unfavorable with respect to entropy (it has restricted connections, or is an ordered system). The unfavorable D S prevents the two from mixing. c. Diethyl ether and 1-butanol have similar solubility properties in water, but 1-butanol has a much higher boiling point. Diethyl ether 1-Butanol Both compounds can form -bonds to water through their atoms (shown), and they are of similar size (4+) making their water solubility similar. Boiling point is determined by the strength of IMF s between molecules in the liquid phase: the alcohol can form -bonds between molecules of itself while the ether has weaker dipole-dipole attractions. There is almost no difference in the IMF with water so they have the same water solubility, but there is a big difference in the IMF with themselves, so their b.p. are different. Page 12