Page 1 C HAPTER 4 HW S OLUTIONS : A LKANES CATEGORIZATIONS 1. Classify each carbon atom pointed to below as 1˚, 2˚, 3˚ or 4˚. 2˚ 2˚ 3˚ 4˚ 1˚ 3˚ 2. Draw an alkane that contains only: 1˚ and 4˚ carbon atoms 2˚ carbon atoms 1˚ and 2˚ carbon atoms 1˚ and 3˚ carbon atoms or or or or HEAT OF COMBUSTION DATA 3. The combustion data for propane and cyclohexane is shown in the table below. Explain why it is incorrect to say (based on the data) that cyclohexane is 413.8 kcal/mol (944.4 - 530.6) higher in energy than propane. H comb (kcal/mol) + 5 O 2 3 CO 2 + 4 H 2 O -530.6 + 9 O 2 6 CO 2 + 6 H 2 O -944.4 Although the full combustion of cyclohexane gives off 413.8 kcal/mol more heat energy than propane, it’s incorrect to attribute all of this to differences in energy between cyclohexane and propane. The two alkanes have different numbers of C-H bonds being broken, different amounts of O 2 being broken, and they form different amounts of CO 2 and water in the combustion reactions. The H values represent the changes in energy considering all of the bonds that are broken and formed, and it’s incorrect to pick out one part of the process (e.g. the alkane part) and attribute the entire energy difference to just that.
Transcript
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CHAPTER 4 HW SOLUTIONS: ALKANES
CATEGORIZATIONS 1. Classify each carbon atom pointed to below as 1˚, 2˚, 3˚ or 4˚.
2˚ 2˚ 3˚
4˚ 1˚ 3˚
2. Draw an alkane that contains only:
1˚ and 4˚ carbon atoms
2˚ carbon atoms 1˚ and 2˚ carbon atoms
1˚ and 3˚ carbon atoms
or
or
or
or
HEAT OF COMBUSTION DATA 3. The combustion data for propane and cyclohexane is shown in the table below. Explain why it is
incorrect to say (based on the data) that cyclohexane is 413.8 kcal/mol (944.4 - 530.6) higher in energy than propane.
DHcomb (kcal/mol)
+ 5 O2 à 3 CO2 + 4 H2O -530.6
+ 9 O2 à 6 CO2 + 6 H2O -944.4
Although the full combustion of cyclohexane gives off 413.8 kcal/mol more heat energy than propane, it’s incorrect to attribute all of this to differences in energy between cyclohexane and propane. The two alkanes have different numbers of C-H bonds being broken, different amounts of O2 being broken, and they form different amounts of CO2 and water in the combustion reactions. The DH values represent the changes in energy considering all of the bonds that are broken and formed, and it’s incorrect to pick out one part of the process (e.g. the alkane part) and attribute the entire energy difference to just that.
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4. Use the heat of combustion data below to determine which organic compound is lower in energy, and by how much. Support your calculation with a diagram.
DHcomb (kcal/mol)
+ 25/2 O2 à 8 CO2 + 9 H2O -1306.3
+ 25/2 O2 à 8 CO2 + 9 H2O -1303.0
The only difference between these reactions is the organic part, so the DH values can be used to measure the relative energies of the organic compounds.
The reactions “finish” on the same energetic line since the products are the same. Reaction 1 releases 1306.3 kcal/mol, which is a larger amount than 1303.0, so it must start off at a higher energy. This means that the alkane of reaction 1 is 3.3 kcal/mol higher in energy than the alkane in reaction 2 (1306.3-1303.0 = 3.3 kcal/mol).
ALKANE AND ALKYL HALIDE NOMENCLATURE 5. (Multiple Choice) Which is the correct IUPAC name for each compound?
Name 4-iodo-4-propylheptane (1-ethylpropyl)cyclohexane 1-chloro-2-cyclopropyl
cyclooctane
7. Each compound below is named incorrectly. Give the correct name for each.
(Heptane isn’t the longest chain.) (Pick the chain w/ most branches) 3-ethyl-6-methylnonane 3-ethyl-2,2-dimethylhexane
ICl
Br
Br Br
F
Cl
Cl
I
Cl
2-propyl-5-ethylheptane
1 6
9
3-t-butylhexane
1
36
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NEWMAN PROJECTIONS 8. Determine which Newman projection in each set is lower in energy. Explain your answers.
LEFT conformation is lower in energy because it is staggered while the right one is eclipsed. Staggered is lower in energy because the H-H and CH3-H distances are longer (groups on the front are further apart from groups on the back), thus minimizing
electronic repulsions.
RIGHT conformation is lower in energy because it is anti while the left one is gauche. Anti is lower in energy because it puts the largest groups (methyl and isopropyl) furthest apart, lowering electronic repulsions.
RIGHT conformation is lower in energy. The most important repulsion to minimize is between the two large isopropyl groups (methyl group is smaller). Having those groups opposite (anti to each other) is lower in energy than having them gauche.
9. Draw the following Newman projections of:
Hexane, looking down the C1-C2 bond
Hexane, down the C2-C3 bond
Hexane, down the C3-C4 bond
2,2-dimethylheptane, down the C2-C3 bond
10. Draw the line structure of each compound, shown below in a Newman projection.
H
H H
HH
CH2CH2CH2CH3
1
3
CH3
H H
HH
CH2CH2CH3
2
4
CH2CH3
H H
HH
CH2CH3
2
3
5
CH3
CH3 CH3
HH
CH2CH2CH2CH3
2
4
CH2CH3
CH3 CH2CH3CH3
HHH3C
1
122 4
4 5
5
(hidden is #3)H
HBr
H
Br1 1
4
4
6
6
(hidden is #5)
H
H HCH3
HHH
H
H
CH3
HHa.
b.
H
HCH3
HH
H HCH3
HH
c.
CH3
H
HH
CH3 H
HH
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11. Consider the compounds 1-bromopropane and ethylene glycol (HOCH2CH2OH). a. Draw the anti and gauche conformations of both compounds.
Anti
1-bromopropane Gauche
1-bromopropane Anti
ethylene glycol Gauche
ethylene glycol
b. Explain why the anti conformation of 1-bromopropane is lower in energy than the gauche
conformation.
The two large groups (Br and CH3, which are larger than all the H’s) have the lowest energy when they are farther apart (minimized electron repulsions). In the anti conformation they are 180˚ from each other, which is as far apart as they can be. In the gauche conformation, they are only 60˚ apart.
c. Ethylene glycol is unusual in that the gauche conformation is lower in energy than the anti conformation. Offer an explanation.
When the two OH groups are gauche, they are close enough together to form an intramolecular (within a molecule) hydrogen bond. This is a stabilizing interaction, which compensates for any electronic repulsion these groups may experience.
RING STRAIN
12. It is approximated that each CH2 group in a linear alkane has a DHcomb of -157.4 kcal/mol. Cyclooctane has an experimental !DHcomb of -1269.2 kcal/mol. a. Calculate the amount of ring strain in cyclooctane.
The expected ∆Hcomb is 8 × (-157.4) = -1259.2 kcal/mol
Experimental – Actual = 1269.2 – 1259.2 = 10.0 kcal/mol
This means the actual cyclooctane releases 10.0 kcal/mol MORE energy than predicted,
or has 10.0 kcal/mol of ring strain.
b. Compared to other ring systems, would you classify this amount of strain as high, moderate or low?
26 kcal/mol = high; 6 kcal/mol = lowish (5 + 7 ring), so this is probably moderate.
Br
H H
HH
CH3
Br
H H
CH3H
H
OH
H H
HH
OH
OH
H H
OHH
H
O
H H
OHH
H
H
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13. Describe two sources of ring strain in cyclobutane. • Non-ideal bond angles (90˚ instead of 109.5˚) leading to “bent” bonds with poor
orbital overlap.
• Eclipsing interactions of the groups on the ring (in this case H atoms), which
repel and increase the energy. The ring is too small to twist and allow the bonds to be staggered.
14. Explain why 1,1-dimethylcyclopropane has more ring strain than ethylcyclopropane. Hint: draw Newman projections.
Newman projections of cyclopropane structures are completely eclipsed due to the constraints form the small ring.
In 1,1-dimethylcyclopropane (left) there are two bad CH3-H eclipsing interactions, which increase its energy. In ethylcyclopropane, there is only one bad interaction (ethyl-H) to increase its energy. Ethyl groups aren’t much larger than methyl groups (see their A-values), so the one with the two bad interactions is higher in energy than the one with only one bad eclipsing interaction.
15. Which should release more energy with combustion, cyclopentane or methylcyclobutane? Explain. Include a diagram with your answer.
Methylcyclobutane should release more energy.
Both compounds have the same molecular formula but methylcyclobutane has more ring strain. This means cyclobutane would start at a higher energy and release more energy when burned to produce 5 CO2 + 5 H2O.
H
CH2
H
H CH2
H
H
CH2H
CH3
CH3
H
CH2
H
CH2CH3
H
5 CO2 + 5 H2O
7.5 O2 ++ 7.5 O2
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CHAIR CONFORMATIONS 16. Practice drawing chair conformations:
Instructions Drawing of chair
a. Draw in the equatorial C-H bond (using the correct angle) on the carbon atom labeled #1.
parallel to bolded bonds
b. Draw all axial and equatorial C-H bonds for the front 3 carbon atoms of both chairs. Draw the equatorial C-H bonds using the correct angles.
c. Draw all axial and equatorial C-H bonds for the back 3 carbon atoms of both chairs.
d. Draw 1,2-dibromocyclohexane with both bromine atoms axial.
e. Draw 1,3-dimethoxycyclohexane with one group axial and one group equatorial. (“Methoxy” is –OCH3)
f. Draw an isopropyl group equatorial on the dot. Then perform a “ring flip” and draw the second chair conformation beside it.
g. Draw a fluoro group axial on the dot. Draw the second chair conformation.
h. Draw a tert-butyl group axial on the left dot and a methyl group equatorial on the right dot. Draw the second chair conformation.
i. Draw an iodo group equatorial on the dot. Draw the second chair conformation.
H1
H H
H H
H
HH H
H H
H
H
H
H H
HHH
H
HH
H HH
Br
Br
CH3O
OCH3
1
6 5 4
32
1 6 5
432
F
F1
6 5 4
32
1 6 5
432
CH3
CH3
1
6 5 4
32
1 6 5
432
I
I1
6 5 4
32
1 6 5
432
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AXIAL AND EQUATORIAL POSITIONS 17. Consider ethylcyclohexane.
a. Draw the Newman projections of axial-ethylcyclohexane and equatorial-ethylcyclohexane.
Axial Equatorial
b. Why is axial-ethylcyclohexane higher in energy than equatorial-ethylcyclohexane?
Axial-ethylcyclohexane has a gauche interaction between the CH2CH3 group and two of the CH2 groups in the ring (carbons 4+6, you can only see one in the Newman projection that’s drawn). When the CH2CH3 group is equatorial these interactions are anti and further apart. Thus there are more electronic repulsions (increased energy) in the axial conformation.
OR
In the axial position, the ethyl group experiences electronic repulsions with the other axial groups (the H on carbons 4+6), which increases its energy. It is not as close to any other groups when equatorial.
18. The two chair conformations of Compound M are equal in energy, while the two chair conformations of Compound N are different energy. Explain.
In either conformation of M, you have two Br groups axial and two equatorial. That makes these conformations equal in energy.
The two conformations of N are different energy because different groups are axial. In one conformation two Br’s are axial and in the other two CH3 are axial. Br and CH3 have different sizes
(different A values) so repel the other axial groups to different extents.
19. Explain why a tert-butyl group has a larger “A” value than an ethyl group.
Tert-butyl groups are gigantic: no matter how they rotate they have huge repulsive interactions with the H’s across the ring (diaxial interactions). Ethyl groups don’t have a much higher A-value than methyl groups: they can rotate away so that only 2 H’s (and not the rest of the ethyl group) experience diaxial interactions.
H
H
H
H
CH2CH3
H
H
H
H
H
H
H
H
CH2CH3
H
H
HCH2CH3
H
1 12
24 46
6
HH
H
12
46
CH2CH3 1
24
6
Br
BrBrBrBr Br
BrBr
M
Br
BrCH3CH3Br Br
CH3H3C
N
HH
HH
C CH3HHCH3
H3C
H3C
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CHAIR TO CHAIR INTERCONVERSIONS (“RING FLIPS”) 20. For each compound, perform a “ring flip” and draw the second chair conformation. Then determine
which chair is lower in energy. (Use a table from lecture for A values.)
Lower energy Both groups are equatorial
A = 1.74 A = 0.75
Lower energy smaller group (smaller A)
is axial.
A = 0.46 A = 2.15 Lower energy
Lower energy Both groups are equatorial
A = 1.75 + 0.75 = 2.50 A = 0.94 + 1.70 = 2.64
Larger total “A” means bigger, worse axial Lower energy (by a little)
CH3
CH2CH3
CH3CH2 CH3
CH3
OCH3
CH3
OCH31
1
2 23
3
I
I
1
2 1 2 33
ClCl1
12 23 3
OCH3
HO
HO
OCH311
223 3
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21. In each set, determine which isomer is lower in energy, and explain your answer. (Remember that cyclohexane rings are not flat!)
Cyclopentane rings are quasi flat, at least it’s OK to think of them like that. Trans (right) is lower energy because the Br’s are on opposite sides, so don’t repel as much.
Trans (right) is lower in energy. In its lowest energy conformation, both ethyl groups are equatorial (far right conformation). Note: the trans compound favors the di-equatorial
conformation, since that’s lower energy. Therefore, it makes sense to say the trans compound is lower energy than cis, because its “main” conformation is lower in energy than any cis conformation.
Cis (left) is lower in energy. In its lowest energy conformation (bottom left) both groups are equatorial.
Lower energy
Lower energy
CH2CH3
H
CH2CH3
H
CH2CH3
HCH2CH3
H
CH2CH3CH2CH3
CH2CH3
CH2CH3
HH
H
H
CIS TRANS
*Main conformation
c.CH2CH3 CH2CH3
d.
CH3
CH3
CH3 CH3
CH3
CH3
e.CH3CH3 CH3CH3
CH3 CH3
a.Br
Br
Br
Br
b.CH2CH3
CH2CH3
CH2CH3
CH2CH3
CH2CH3
CH2CH3
H
H
CH2CH3
CH2CH3
H
H
CIS TRANS
*Main conformation
H
H
H
H
H3C
H
CH3
H
CH3 H
CH3
CH3
H
CH3
(left structure) (right structure)
H3C
CH3
(left structure) (right structure)
CH3
HH
H
H3C
CH3H
HH
CH3
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STEREOISOMERS 22. Define and give an example of:
a. Constitutional Isomers
Compounds that are isomers (same molecular formula, but are different) because their atoms are connected in a different order.
b. Stereoisomers
Isomeric compounds whose atoms are connected in the same order, but are different in space (3-dimensions).
23. Convert each chair into a line structure (hexagon), using dashed and wedged bonds.
24. Are the following statements concerning the molecules below true or false? Briefly explain each answer.
a. Both molecules are cis. True
Both Cl’s are pointed “up” in each.
b. A cis isomer always has the groups axial-axial or equatorial-equatorial. False
The example shows this isn’t true. The left structure has axial-axial groups, but the right structure has axial-equatorial groups, and yet both are cis.
HO
OH
OCH3
OH
HOOH
OH
OH
H
H
1
1
2
2
3
3
Br
CH3
Br
CH3
H
H1
1
2
2
HO OH
HO
OH
4
H
H
1
1
4
Cl Cl ClCl
H H
H
H
up up upup
Cl
Cl
Cl
Cl
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25. Give the IUPAC name for each compound, including cis/trans designations.
Structure
Name trans-1,3-
dimethylcyclobutane cis-1-chloro-2-
isopropylcycloheptane cis-1-bromo-3-
isopropylcyclohexane
Structure
Name trans-1-ethyl-3-
methylcyclohexane trans-1,2-
dichlorocyclohexane cis-1-t-butyl-4-
ethylcyclohexane
Structure
Name cis-1-iodo-3-
methylcyclohexane trans-1-fluoro-2-
propylcyclohexane
26. Classify the relationship between each pair as constitutional isomers, stereoisomers, identical, or not
isomers.
Pair
Relationship Stereoisomers (cis/trans) Stereoisomers (left = cis, right = trans)
Pair
Relationship Constitutional isomers (both C6H12) Not isomers : C6H12 and C6H14
Pair
Relationship Identical Constitutional isomers (1,2 vs. 1,4)
