Problem Set 2
Chapter 3, Alkanes
Due: Monday, February 4, 2013
Solution Set
The Baeyer Laboratory, Munich, 1893 (This photograph is in the hallway across from 110 SCL) Adolf von Baeyer (1835-1917); Nobel Prize 1905. (center, seated with derby), who was a student of Kekulé, succeeded Liebig at Munich. In the photograph (second row; third from right) is Henry Lord Wheeler (1867-1914); Yale Faculty 1896-1911. As was the custom in the 19th century, many Americans, such as Wheeler, did advanced study in chemistry in Europe. Karl Gimmig is the laboratory assistant. (The only person wearing an apron and no tie; upper left.) In 1885, as an addendum to a paper on
acetylenic compounds, Baeyer proposed that cyclopentane was
the least
strained of the cycloalkanes.
While he accepted the idea that the carbon atoms in
cycloalkanes were tetrahedral, he treated the cycloalkanes
as though they were flat. He argued that there is only one
cyclohexane carboxylic acid, not two (axial and equatorial)
as was predicted by a chair cyclohexane.
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Sir Derek H. R. Barton (1918-1998) 1969 Nobel Prize with Odd Hassel for their work on conformational analysis For a video of Barton talking about conformational analysis, click here. Cyclohexane in the chair conformation |
a. Work through How to Draw Cyclohexanes (PowerPoint)
b. The Conformation Module in the Study Aids will give you a good overview of the subject of conformation.
c. View The Evolution of Formulas and Structure in Organic Chemistry During the 19th Century (PowerPoint).
1. Redraw (line angle formula) and name (IUPAC) the hydrocarbon located on the right. [What if there are two different longest chains? Check here.] The longest chain is 9 carbons, hence a nonane. Numbering begins on the right because the methyl group at C-3 is closer to the terminus of the longest chain than the isopropyl group at C-6. The groups are numbered accordingly. Notice that the alphabetization is e-i-m (see arrows). The di-prefix does not count.
If there is more than one longest chains, pick the one with the greatest number of substituents. This rule minimizes the complexity of the names of the substituents. |
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2. Compound A (MW=170.24), a 1,4-disubstituted cyclohexane soluble in aqueous sodium hydroxide,
has the following composition: C, 70.55%; H, 10.66%. The
difference in conformational energy for the two chair conformations
of A is 0.7 kcal/mol. Using the A-value
data (Energy Differences Between ..... Cyclohexanes), determine the
structure of A. Illustrate and explain. What is the
conformational energy difference for the stereoisomer of A,
---namely A'. Explain and illustrate. Show the chair
comformations of A and A' with the appropriate
equilibrium arrows to illustrate the major and minor conformations.
Label each conformation with its energy.
The first order of business is to determine the molecular formula of the compound. Does compound A contain only carbon and hydrogen? No, because the analysis does not add up to 100%. The difference is oxygen (18.80%).
The formula cannot be C5H9O. First, a cyclohexane must have at least six carbons. Secondly, a compound having only C, H and O cannot have an odd number of hydrogens (Check
here). Double the empirical formula giving C10H18O2. M.W.
calculated: 170.24, which agrees with the given value. Compound A is a 1,4-disubstituted cyclohexane.
Cyclohexane is C6H12. Subtracting two hydrogens
for the positions of the two substituents leaves a cyclohexane
nucleus of C6H10. Subtracting: C10H18O2 -
C6H10 = C4H8O2 for the sum of the
composition of the two substituents. There are two possibilities for the pair of substituents: a) -CO2CH3 (1.2 kcal/mol) and -CH2CH3
(1.9 kcal/mol), b) or -CO2H (1.7 kcal/mol) and -CH(CH3)2 (2.1 kcal/mol). Both pairs have an energy difference of 0.7 kcal/mol but only b) applies because, of the four groups, only the carboxylic acid group is soluble in aqueous sodium hydroxide [RCO2H pKa = ~5; H2O pKa = 15.7; pKa Table]. Since the energy difference is less than either substituent, the groups must be cis. Thus, compound A is cis-4-isopropylcyclohexane carboxylic acid.
The stereoisomer of compound A is A', trans-4-isopropylcyclohexane carboxylic acid.
3. Predict the heat of formation of 2-methyloctane using the data
presented here.
Explain. Look at lower homologs of 2-methyloctane. See the table below. Note that the addition of a -CH2- group decreases (becomes more negative) the heat of formation by ~5 kcal/mol. 2-Methylpropane is not a member of the homologous series; it has no -CH2- group. 2-Methylpropane is an outlier. Thus, the increment per -CH2- group is ~5.0 kcal/mol. The predicted heat of formation of formation of 2-methyloctane is -56.5 kcal/mol.
2-methylalkanes |
ΔHfo (kcal/mol) | Δ( ΔHfo) (kcal/mol) |
|---|---|---|
| 2-methylheptane | -51.5 | - |
| 2-methylhexane | -46.6 | -4.9 |
| 2-methylpentane | -41.7 | -4.9 |
| 2-methylbutane | -36.7 | -5.0 |
| 2-methylpropane | -32.1 | -4.6 |
4. Examine the heats of formation of the four acyclic octanes listed in the
heats
of formation tables.
a)What trend do you notice? Increased branching in the series results in a more negative heat of formation.
b) Draw a diagram that shows the heat of formation and heat of combustion of the two extreme cases: n-octane and 2,2,3,3-tetramethylbutane. Show calculations. The octanes have the molecular formula C8H18 . At STP, calculate the heat liberated by combusting 8 moles of graphite and 9 moles of hydrogen. The total is -1367.1 kcal/mol. Review here. The difference between this value and the heat of formation is the heat of combustion of the structural isomer. The more branched the isomer the less heat liberated upon combustion. Also, the difference in the heat of combustion of the two isomers is equal to the difference in the heat of formation.
c) Consider the cyclic octanes trans- and cis-1,4-dimethylcyclohexane in the heats of formation tables. Explain and illustrate their difference in heats of formation. Both chair conformations of the cis-isomer are identical with an energy of 1.8 kcal/mol. The trans-isomer exists nearly exclusively in the diequatorial conformation with an energy of 0 kcal/mol and is more stable than the cis-isomer by 1.8 kcal/mol. The trans isomer liberates 1.8 kcal/mol less heat on combustion than the cis-isomer. This difference is reflected in the difference in their heats of formation: trans = -44.0 kcal/mol; cis = -42.2 kcal/mol.
5. a) Calculate the heat of combustion of cyclobutane using the
data (ΔHfo of cyclobutane,
CO2 and H2O) in the heats
of formation tables.
Compare your value with
the value in Table 3-5 in your text. Cyclobutane has the formula
C4H8 (ΔHfo = +6.6
kcal/mol). The heat of combustion of four moles of graphite and four
moles of hydrogen is 4 x [(-94.05) + (-68.3)] = -649.4
kcal/mol. The heat of combustion of cyclobutane = -649.4 - (+6.6) =
-656 kcal/mol compares well with the table. Note that cyclobutane is
less stable than the atoms from which it is formed.
b) Calculate the strain energy in cyclobutane given the heat of
combustion of cyclohexane (Table 3-5 in your text) and the knowledge
that cyclohexane is strain-free. Cyclobutane has 2/3 as many
methylene groups as cyclohexane. Therefore, its heat of combustion
should be 2/3 the heat of combustion of cyclohexane or 2/3 x (-944.4)
= -629.6 kcal/mol for strain-free cyclobutane. But cyclobutane has a
heat of combustion of -655.8 kcal/mol. The difference between these
numbers is the strain energy: 26.2 kcal/mol.
6. Draw Newman projections for the eclipsed and staggered
conformations of 2,3-dimethylbutane viewed along the
C2-C3 axis. Calculate the energy of each
conformation, both staggered and eclipsed. It is irrelevant which
eclipsed conformation is taken as 0o. Eclipsed (kcal/mol): H/H, 1.0; CH3/H, 1.3; CH3/CH3, 3.0 kcal/mol. Staggered (kcal/mol): H/H, 0; CH3/H, 0; CH3/CH3, 0.9.
