SUPPLEMENTAL OMEWORK SOLUTIONS WEEK 10 Assignment for Tuesday, March 21 st 9.105 a) ombustion reactions require oxygen (in addition to the hydrocarbon). b) The products are O 2 and 2 O (carbon dioxide and water). 9.106 The boiling point of a hydrocarbon is determined by the strength of the dispersion forces that attract the molecules to each other. exane is a larger molecule than butane, so the dispersion forces between hexane molecules are stronger than the dispersion forces between butane molecules. Therefore, it takes more energy to pull hexane molecules away from another, giving hexane a higher boiling point. 9.118 This is a review of material from hapter 6. a) The formula weight of 4 is 16.042 amu, and the balanced equation calls for one 4 molecule. Therefore, we know that burning 16.042 grams of 4 will produce 213 kcal of heat. Use this relationship to calculate the heat you get from burning 1.00 g of 4 : 1.00 g 4 213 kcal 16.042 g 4 = 13.3 kcal of heat Do a similar calculation for 3. The formula weight of 3 is 44.094 amu, so we know that burning 44.094 grams of 3 will produce 531 kcal of heat. Use this relationship to calculate the heat you get from burning 1.00 g of 3 : 1.00 g 3 531 kcal 44.094 g 3 = 12.0 kcal of heat We get more heat when we burn 1.00 g of methane. b) The equation for the combustion of methane produces one molecule of O 2 (formula weight 44.01 amu) when we burn one molecule of methane (formula weight 16.042 amu). Therefore, burning 16.042 grams of 4 will produce 44.01 grams of O 2. Use this relationship to calculate the mass of O 2 you get from burning 1.00 g of 4 : 1.00 g 4 44.01 g O 2 16.042 g 4 = 2.74 g of O 2 The equation for the combustion of propane produces three molecules of O 2 (total formula weight = 3 x 44.01 amu = 132.03 amu) when we burn one molecule of propane (formula weight 44.094 amu). Therefore, burning 44.094 grams of 3 will produce 132.03 grams of O 2. Use this relationship to calculate the mass of O 2 you get from burning 1.00 g of 3 :
1.00 g 3 132.03 g O 2 44.094 g 3 = 2.99 g of O 2 We get more O 2 when we burn 1.00 g of propane. 10.2 a) Two products are formed in this hydration reaction: 2 2 2 d) Only one product is formed in this hydration reaction, because the two possible ways of adding and to the double bond give you the same compound. and 2 2 2 The answer! is the other option, but it is not a different compound. It is the same molecule, just rotated. 10.10 b) 2-heptanol c) cyclopentanol 10.22 a) 1-pentanol has a higher solubility than propane, because 1-pentanol is the only molecule that contains a hydrophilic group. b) 2-propanol has a higher solubility than 2-pentanol. Both molecules contain a hydrophilic group, but 2-propanol has a smaller hydrophobic region. c) yclopentanol has a higher solubility than cyclopentene, because cyclopentanol is the only molecule that contains a hydrophilic group. 10.47 Ethylene glycol dissolves in water, because it can form hydrogen bonds with water molecules. Ethylene difluoride does not contain nitrogen or oxygen, so it cannot form hydrogen bonds. 10.50 Dimethyl ether molecules cannot form hydrogen bonds with each other, so the molecules have little attraction for each other. Propane is similar to dimethyl ether. Ethanol, on the other hand, can form hydrogen bonds (thanks to its group), so ethanol molecules are attracted to each other more strongly. It therefore requires a lot more energy to pull ethanol molecules away from one another giving the higher boiling point. 10.57 For many of these, there is more than one acceptable answer. I ll give just one answer for each part. a) c) 2 2 2
d) e) 2 2 10.58 a) b) c) 2 2 2 d) and 2 2 2 2 e) 3 and 2 f) 2 g) No reaction (this compound can t be dehydrated) h) i) 2 2 2 j) and 2 2 2 k) and 2 2 2 2 and 2 2
10.60 a) methanol b) 2-butanol c) phenol d) 3-octanol e) cyclohexanol 10.61 a) 2 b) 2 2 c) d) e) 2 2 2 2 2 10.63 a) 2 2 is more soluble in water, because it can form hydrogen bonds, whereas 2 cannot. b) 2 2 is more soluble in water, because it can form hydrogen bonds, whereas 2 2 S cannot. c) 1-pentanol is more soluble in water, because it can form hydrogen bonds, whereas 1- pentene cannot. d) 2-pentanol is more soluble in water. Both 2-pentanol and 2-heptanol contain a hydrophilic group (the alcohol group) that can form hydrogen bonds with water. owever, 2- pentanol has a smaller hydrophobic group than 2-heptanol. e) 2 2 2 2 2 2 2 2 is more soluble in water, because it can form hydrogen bonds, whereas 2 cannot. (The length of the chain only matters when both molecules contain the same hydrophilic group.) f) The second molecule is more soluble in water, because it contains two hydrophilic groups, whereas the first molecule only contains one. 10.66 a) These molecules are constitutional isomers, because they have the same molecular formula ( 4 10 O) but different structures (the is attached to a different carbon atom). b) These molecules are not constitutional isomers. They are just two different ways to draw the same molecule, so they do not have different structures. c) These molecules are not constitutional isomers, because they have different molecular formulas. d) These molecules are constitutional isomers, because they have the same molecular formula ( 4 10 O) but different structures (one has an unbranched carbon chain and the other has a branched carbon chain). e) These molecules are constitutional isomers, because they have the same molecular formula ( 5 12 O) but different structures (one is an alcohol, the other is not). f) These molecules are not constitutional isomers, because they have different molecular formulas. The first one is 5 12 O while the second is 5 10 O. g) These molecules are not constitutional isomers. They are just two different ways to draw the same molecule, so they do not have different structures. h) These molecules are constitutional isomers, because they have the same molecular formula ( 6 12 O) but different structures (the is attached to a different carbon atom). i) These molecules are not constitutional isomers, because they have different molecular formulas. One contains an oxygen atom while the other contains a sulfur atom.
j) These molecules are constitutional isomers, because they have the same molecular formula ( 5 10 O) but different structures. k) These molecules are constitutional isomers, because they have the same molecular formula ( 6 12 O) but different structures. 10.71 There are four possible products: 2 2 2 2 2 2 2 2 2 2 2 2 10.74 Rayelle is correct. If the double bond ends up at the second position in the carbon chain, it can be either cis or trans. There are three possible products for this dehydration reaction: 2 2 2 2 1-pentene cis-2-pentene trans-2-pentene 10.76 b) ompounds A, B, and F cannot be dehydrated. ompounds with attached directly to an aromatic ring (lika A and F) cannot be dehydrated. ompound B has no hydrogen atom on the carbon next to the functional group carbon. 2 2 There is no attached to this carbon atom. c) ompounds and D can form only one dehydration product. In each molecule, only one carbon atom adjacent to the functional group carbon is bonded to hydrogen. 2 2 The only hydrogen-bearing carbon that is adjacent to the functional group. d) ompound E can form two different dehydration products, because there are two hydrogen-bearing carbon atoms adjacent to the functional group carbon.
10.77 phenol alcohol alkyne O 10.85 The reaction is: 2 = 2 + 2 O 2 To calculate mass relationships, it s easiest to change all of the structures into molecular formulas: 2 4 + 2 O 2 6 O The formula weight of ethene ( 2 4 ) is 28.052 amu, and the formula weight of ethanol ( 2 6 O) is 46.068 amu. Therefore, we know that if 28.052 grams of ethene reacts with water, we will form 46.068 grams of ethanol. Using this as a conversion factor 46.068 g ethanol 8.5 g ethene = 14 g ethanol (rounded from 13.959 g) 28.052 g ethene