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Chapter 11 Problems Key
11 Prob-1
5/23/01
Chapter 11 End-of-Chapter Problems Key
REVIEW QUESTIONS
1. Identify each of the following as representing an alcohol, a carboxylic acid, an aldehyde,
a ketone, an amine, an amide, or an ester.
a.
alcohol
ester
b.
c.
aldehyde
d. CH3CH2NH2
amine
e.
carboxylic acid
ketone
f.
g.
Copyright 2001 Mark Bishop
amide
Chapter 11 Problems Key
11 Prob-2
5/23/01
Section 11.1
1. Identify each of the following structures as representing a carbohydrate, amino acid,
peptide, triglyceride, or steroid. (Obj #2)
a.
amino acid
carbohydrate
b.
triglyceride
c.
d.
e.
Copyright 2001 Mark Bishop
steroid
peptide
Chapter 11 Problems Key
11 Prob-3
5/23/01
2. Identify each of the following structures as representing a carbohydrate, amino acid,
peptide, triglyceride, or steroid. (Obj #2)
a.
triglyceride
b.
peptide
peptide
c.
d.
amino acid
e.
carbohydrate
Copyright 2001 Mark Bishop
Chapter 11 Problems Key
5/23/01
11 Prob-4
3. Identify each of the following structures as representing a monosaccharide, disaccharide,
or polysaccharide. (Obj #5)
a.
polysaccharide
b.
monosaccharide
disaccharide
c.
d.
Copyright 2001 Mark Bishop
monosaccharide
Chapter 11 Problems Key
11 Prob-5
5/23/01
4. Identify each of the following structures as representing a monosaccharide, disaccharide,
or polysaccharide. (Obj #5)
disaccharide
a.
b.
c.
monosaccharide
monosaccharide
d.
polysaccharide
Copyright 2001 Mark Bishop
Chapter 11 Problems Key
5/23/01
11 Prob-6
5. Describe the general difference between glucose and galactose. (Obj #4)
Glucose and galactose differ in the relative positions of an –H and an –OH on one
of their carbon atoms. In the standard notation for the open-chain form, glucose and
galactose differ only in the relative position of the -H and -OH groups on the fourth
carbon from the top. In the standard notation for the ring structures, the –OH group
is down on the number 4 carbon of glucose and up on the number 4 carbon of
galactose.
6. Describe the general differences between glucose and fructose. (Obj #4)
Glucose has an aldehyde functional group in the open-chain form, as opposed to
fructose, which has a ketone functional group. Fructose forms a 5-membered ring,
and glucose forms a 6-membered ring.
Copyright 2001 Mark Bishop
Chapter 11 Problems Key
5/23/01
11 Prob-7
7. Explain why glucose is found in two ring forms and one open-chain form. Describe the
general differences between these three structures. (Obj #5)
The open-chain form rearranges to form the ring-form. The rearrangement is
reversible, so glucose in solution is constantly going back and forth between the
open-chain form and a ring form (Figure 11.2). When glucose is in the open-chain
form, the carbon in the aldehyde functional group rotates with respect to the carbon
to which it is attached (Figure 11.3). If the ring forms when the open-chain glucose
is in the second form shown in Figure 11.3, a ring that is slightly different than the
one in Figure 11.2 is created (Figure 11.4). All three forms are shown in Figure
11.5.
8. Is the following structure α-glucose or β-glucose? (Obj #6)
α-glucose
The –OH group on the #1 carbon is on the
opposite side of the #6 carbon.
9. What saccharide units form maltose, lactose, and sucrose? (Obj #7)
Maltose – 2 glucose units
Lactose – glucose and galactose
Sucrose – glucose and fructose
10. Describe the similarities and differences between amylose, amylopectin, and glycogen.
(Obj #8)
Both are long-chain polymers composed of glucose monomers. They are both found
in plants. Amylose is a long-chain molecule composed of glucose molecules.
Amylopectin and glycogen have branches. The branches in glycogen are usually
shorter and more frequent than for amylopectin, but otherwise, they have very
similar structures.
amylose
Copyright 2001 Mark Bishop
Chapter 11 Problems Key
5/23/01
11 Prob-8
amylopectin and glycogen
11. Describe the similarities and differences between starches (like amylose, amylopectin,
and glycogen) and cellulose. (Obj #8)
Starch and cellulose molecules are composed of many glucose molecules linked
together, but cellulose different linkages between the molecules than starch. See the
starch structures above.
cellulose
12. Explain why the starch molecules found in a potato can be digested in our digestive tract
and why the cellulose in the same potato cannot. (Obj #9)
Starch and cellulose molecules are composed of many glucose molecules linked
together, but cellulose has β(1→4) linkages instead of the α(1→4) and α(1→6)
linkages of starch. Animal enzymes are able to break the α(1→4) and α(1→6)
linkages and convert starch to energy-producing glucose, but they are unable to
liberate glucose molecules from cellulose because they cannot break the β(1→4)
linkages (Figure 11.10). Cellulose passes through our digestive tract unchanged.
Copyright 2001 Mark Bishop
Chapter 11 Problems Key
5/23/01
11 Prob-9
13. Explain why glycine amino acid molecules in our bodies are usually found in the second
form below rather than the first. (Obj #11)
One end of the amino acid has a carboxylic acid group that tends to lose an H+ ion,
and the other end has a basic amine group that attracts H+ ions. Therefore, in the
conditions found in our bodies, amino acids are likely to be in the second form.
14. Show how the amino acids leucine, phenylalanine, and threonine can be linked together
to form the tripeptide leu-phe-thr. (Obj #12)
15. Show how the amino acids tryptophan, aspartic acid, and asparagine can be linked
together to form the tripeptide try-asp-asn. (Obj #12)
16. Identify descriptions of the primary, secondary, and tertiary structure of proteins. (Obj
13)
The sequence of amino acids in a protein molecule is called its primary structure.
The arrangement of atoms that are close to each other in the polypeptide chain is
called the secondary structure of the protein. Images of two of the possible forms of
the secondary structure, α-helix and β-sheet, are shown in Figures 11.14 and 11.15.
The overall shape of a protein molecule is called its tertiary structure. The protein
chains are held in their tertiary structure by interactions between the side chains of
amino acids.
Copyright 2001 Mark Bishop
Chapter 11 Problems Key
5/23/01
11 Prob-10
17. Describe how disulfide bonds, hydrogen bonds, and salt bridges help hold protein
molecules together in specific tertiary structures. (Obj #14)
Each of these interactions draw specific amino acids in a protein chain close
together, leading to a specific shape of the protein molecule. Disulfide bonds are
covalent bonds between sulfur atoms from two cysteine amino acids (Figure 11.15).
Hydrogen bonding forms between –OH groups in two amino acids, like serine or
threonine, in a protein chain (Figure 11.16). Salt bridges are attractions between
negatively charged side chains and positively charged side chains. For example, the
carboxylic acid group of an aspartic acid side chain can lose its H+, leaving the side
chain with a negative charge. The basic side chain of a lysine amino acid can gain
an H+ and a positive charge. When these two charges form, the negatively charged
aspartic acid is attracted to the positively charged lysine by a salt bridge (Figure
11.17).
18. Explain why it is more efficient to store energy in the body as fat rather than
carbohydrate or protein. (Obj 15)
There are 37 kJ/g from fat and only 17 kJ/g from carbohydrate and protein.
19. Identify each of the following triglycerols as saturated or unsaturated. Which is more
likely to be a solid at room temperature and which one is more likely to be a liquid?
(Obj #17)
saturated – solid
Copyright 2001 Mark Bishop
unsaturated - liquid
Chapter 11 Problems Key
5/23/01
11 Prob-11
20. Draw the structure of the triglycerol that would form from the complete hydrogenation
of the triglycerol below. (Obj #18)
21. Draw the structure of the triglycerol that would form from the complete hydrogenation
of the triglycerol below. (Obj #19)
Copyright 2001 Mark Bishop
Chapter 11 Problems Key
5/23/01
11 Prob-12
Section 17.2
22. When you wash some fried potatoes down with a glass of milk, you deliver a lot of
different nutritive substances to your digestive tract, including lactose (a disaccharide),
protein, and fat from the milk and starch from the potatoes. What are the digestion
products of disaccharides, polysaccharides, protein, and fat? (Obj #20)
Disaccharides – monosaccharides (glucose and galactose from lactose)
Polysaccharides – glucose
Protein – amino acids
Fat – glycerol and fatty acids
23. Describe how the digestion of protein molecules is facilitated by conditions in the
stomach. (Obj #21)
The acidic conditions in the stomach weaken the links between amino acids that hold
the protein molecules in their tertiary structure. When the tertiary structure of
proteins is relaxed, they are more easily digested. One way this is done is by
disrupting salt bridges due to the removal of the negative charge on aspartic acid
side chains when H+ ions are added to them.
24. Explain why strong acids catalyze amide hydrolysis. (Obj #22)
(Skip this one.)
25. Explain why each enzyme only acts on a specific molecule or a specific type of
molecule. (Obj #23)
(Skip this one.)
26. Describe in very general terms how the enzyme chymotrypsin is able to catalyze the
hydrolysis of peptide bonds in neutral conditions. (Obj #24)
(Skip this one.)
Copyright 2001 Mark Bishop
Chapter 11 Problems Key
5/23/01
11 Prob-13
Section 17.4
27. Describe how Nylon 66 is made. (Obj #25)
Adipic acid (a di-carboxylic acid) reacts with hexamethylene diamine (with two
amine functional groups) to form long-chain polyamide molecules named Nylon 66.
The reactants are linked together by condensation reactions in which an –OH group
lost from a carboxylic functional group combines with a –H from an amine group to
form water and an amide linkage between the reacting molecules. Because the
reacting molecules each have two functional groups, after the first reaction is
complete, new molecules can be added to each end in other condensation reactions.
In this way, many molecules can be linked together in long chains.
Copyright 2001 Mark Bishop
Chapter 11 Problems Key
5/23/01
11 Prob-14
28. Explain why Nylon 66 is stronger than Nylon 610. (Obj #26)
One of the reasons for the exceptional strength of nylon is the hydrogen bonding
between amide functional groups. A higher percentage of amide functional groups in
nylon molecules’ structures leads to stronger hydrogen bonds between them. Thus,
changing the number of carbon atoms in the diamine and in the di-carboxylic acid
changes the properties of nylon. Nylon 610, which has four more carbon atoms in
the di-carboxylic acid molecules that form it than for Nylon 66, is somewhat weaker
than Nylon 66 and has a lower melting point.
29. Describe how Nylon 6 is made. (Obj #27)
Nylon 6 is made from a synthetic amino acid, 6-aminohexanoic acid. See Figure
11.24.
30. Describe how polyesters are made. (Obj #28)
Polyesters are made from the reaction of a diol (a compound with two alcohol
functional groups) with a di-carboxylic acid. The figure below shows the steps for
the formation of poly(ethylene terephthalate) from ethylene glycol and terephthalic
acid.
31. Describe the three steps in the formation of addition (chain-growth) polymers from free
radical initiation. (Obj #29)
In the first stage (the initiation stage), a substance, like a peroxide, is split into two
identical parts (free radicals), each with an unpaired electron. The free radical
initiates the reaction sequence by forming a bond to one of the carbon atoms in the
double bond of the reactant molecule. One electron for this new bond comes from
the free radical, and one electron for the bond comes from one of the two bonds
between the carbon atoms. The other electron from this bond shifts to the carbon
atom on the side of the molecule away from the new bond that is formed. In Figure
Copyright 2001 Mark Bishop
Chapter 11 Problems Key
5/23/01
11 Prob-15
11.26, each half-headed arrow indicates the shift of one electron.
In the propagation stage, the chain grows when the new free radical formed in the
initiation stage reacts in a similar way to the original free radical and in this way
adds two more carbon atoms (Figure 17.26). This process repeats itself over and
over again to form very long chains of carbon atoms. In the termination phase, the
process is terminated when any two free radicals come together to pair their
unpaired electrons and form a covalent bond that links the chains together (Figure
11.26).
32. Describe the similarities and differences between the molecular structures of
low-density polyethylene (LDPE) and high-density polyethylene (HDPE). (Obj #30)
Polyethylene molecules can be made using different techniques. One process leads
to branches that keep the molecules from fitting closely together. Other techniques
have been developed to make polyethylene molecules with very few branches. These
straight-chain molecules fit together more efficiently, yielding a high-density
polyethylene, HDPE, that is more opaque, harder, and stronger than the low-density
polyethylene, LDPE.
Copyright 2001 Mark Bishop
Chapter 11 Problems Key
5/23/01
11 Prob-16
33. Identify each of the following as representing nylon, polyester, polyethylene, poly(vinyl
chloride), polypropylene, or polystyrene. (In each case, the “n” represents some large
integer.) (Obj #31)
Know the following only.
a.
polyethylene
polyester
b.
c.
polypropylene
d.
polystryene
e.
nylon
f.
poly(vinyl chloride)
34. Describe the structural difference between thermoplastic and thermosetting polymers
and use it to explain why thermoplastic polymers can be recycled and thermosetting
polymers cannot. (Obj #32)
It is possible to recycle polymers that can be heated and formed and then reheated
and reformed again (called thermoplastic polymers). Thermoplastic polymers are
usually composed of linear or only slightly branched molecules. Polymers that
consist of molecules with extensive three-dimensional cross-linking (called
thermosetting polymers) decompose when heated, so they cannot be reheated and
reformed. In general, thermosetting polymers cannot be recycled.
Copyright 2001 Mark Bishop
Chapter 11 Problems Key
5/23/01
11 Prob-17
35. Describe the three steps in the recycling of polymers. (Obj #33)
The first step is collection of materials in curbside recycling bins or central
collection areas. The next step is sorting. The recycling symbols found on objects
made of polymers tell the recycling companies what type of polymer was used in the
object’s construction. The last step is the removal of contamination. The objects to
be recycled as chopped up, washed, and separated from the contaminants.
36. Find three plastic objects in your home that are labeled with a recycling code of 1. From
what substance are these objects made? Are objects of this type recycled in your town?
(Obj #34)
The recycling code of 1 identifies substances made with polyester terephthalate,
PET. Objects made of PET include bottles that contain soft drink, furniture cleaner,
leather conditioner, window cleaner, and baby shampoo.
37. Find three objects in your home that are labeled with a recycling code of 2. From what
substance are these objects made? Are objects of this type recycled in your town?
(Obj #34)
The recycling code of 2 identifies substances made with high-density polyethylene,
HDPE. Objects made of HDPE include bottles containing children’s vitamins, 3%
hydrogen peroxide, shampoo, orange juice, milk, spray laundry cleaner, floor
polish, contact lens solution, carpet cleaner, hair spray, bubble solution, brass
polish, and motor oil.
38. Find one object representing each of the recycling codes 3, 4, 5, 6. From what substance
is each object made? Can these objects be recycled in your town? (Obj #34)
The recycling code of 3 identifies substances made with poly(vinyl chloride), PVC or
V. Objects made of HDPE include bottles containing citrus cleaner and degreaser,
furniture scratch cover, lemon-scented wood treatment, and liquid dye.
The recycling code of 4 identifies substances made with low-density polyethylene,
LDPE. Objects made of HDPE include milk bottle caps and storage container lids.
The recycling code of 5 identifies substances made with polypropylene, PP. Objects
made of HDPE include toy train tracks, kitchen storage containers, and containers
for prescription drugs, fish food, silver cream, and wing & rib sauce.
The recycling code of 6 identifies substances made with polystyrene, PS. Objects
made of HDPE include picnic bowls and cups and containers for vitamins and
sunblock.
Copyright 2001 Mark Bishop