Chem 220 - Organic Chemistry
Problem Set 3 -Solution Set
Chapter 4, Chemical Reactions
Due: Monday, September 28, 2009
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The prevailing theory of organic structure in the early 19th century was Dualism or the Electrochemical Theory, principally championed by Berzelius. Since inorganic sodium chloride could be considered as Na+Cl-, then an alkyl halide such as RCl could be thought of as R+Cl-. The R group or "radical" of its day, was thought to be immutable, the carbons and hydrogens behaving as though they were an element. Liebig (German) and Dumas (French), influential chemists of the day, published a joint paper (1837), On the Present State of Organic Chemistry, extolling the concept and claiming all that was left to do in organic chemistry was to identify these immutable radicals (benzoyl, ethyl, acetyl, etc.) As the story goes, a Parisian reception at the Tuileries was to change all of this. [The Tuilerie Gardens was one of the images in Mussorgsky's "Pictures at an an Exhibition", composed in 1874 for piano and later orchestrated by Ravel. The Tuileries selection is here. For the history of this composition and background music, go here. (At Yale, try this.] The guests were discomforted by fumes from the burning candles. Dumas was called in as a consultant. He found that the waxes (fatty esters) had exchanged chlorine for hydrogen, the culprit being the by-product hydrogen chloride. [This story is likely apocryphal. It was told by August Hofmann at a eulogy for Dumas (1884). If the event did occur, it is more likely that the bleaching of candle wax involved addition of chlorine to double bonds. Nonetheless, Dumas did investigate substitution reactions.] |
Jean-Baptiste-Andrè Dumas (1800-1884) |
The concept of exchanging electropositive hydrogen for electronegative chlorine was anathema to dualism. Liebig was not enamored with substitution. Why shouldn't he be? After all Liebig and Wöhler had done precisely this in 1832 during their work on the benzoyl radical (C7H5O). They had converted benzaldehyde [(C7H5O)H] into benzoyl chloride [(C7H5O)Cl] by the action of chlorine. So disenchanted was Liebig with the controversies regarding theory in organic chemistry, by 1840 he turned his attention to the practical applications of agricultural chemistry. Thus arose Liebig's beef extract. Dumas's student, Laurent, not one to shirk from controversy, was bold enough to call the process substitution rather than exchange. Thus, Substitution Theory. Moreover, Dumas (1838) was able to substitute three of the four hydrogens of acetic acid for chlorine to form trichloroacetic acid, which had properties to acetic acid. The recognition of these similar properties led to early Type Theory. In 1842, Melsen, a student of Dumas, reversed Dumas's experiment by reducing trichloroacetic acid to acetic acid by the action of zinc metal. The promulgation of Substitution Theory gave the wry wit of Wöhler, a.k.a., S. C. H. Windler, an opportunity to shine. At the beginning of the 20th century free radicals were detected and named free radicals to distinguish them from the older radicals of Radical Theory of the early 19th century. During the 19th century chemists tried to isolate the older radicals to no avail. When Kolbe and Frankland thought they had isolated methyl, they actually had made the dimer of methyl, ethane. The very process of substituting chlorine for hydrogen is a free radical reaction. |
Study the Alkane Module in Organic Reactions Go Online (ORGO).
1. Determine the percent of each of the
constitutional monochloro isomers expected from the chlorination of
2-methylpentane. Show work. 2-Methylpentane
(A) has five constititional isomers.
Positions a-e are labeled in A. The monochloro compounds are
B-F. Relative reactivity for
chlorination: 1o = 1; 2o = 4.5;
3o = 5.5. Multiply the number of hydrogens at
each, primary, secondary or tertiary site by its relative
reactivity. Add these values to obtain 32.5. Divide each
product to obtain a fractional part. Convert to a
percentage. Although a and e are methyl groups, they are
considered seprately because they lead to different
products: a --->
B b --->
C c --->
D d --->
E e --->
F Position #H's Rel. React. #H's x Rel. React. (#H's x Rel. React.)/32.5 % a 6 1 6 0.185 18.5 b 1 5.5 5.5 0.169 16.9 c 2 4.5 9 0.277 27.7 d 2 4.5 9 0.277 27.7 e 3 1 3 0.092 9.2 Sum 32.5 1.000 100
2. Show the initiation and propagation steps for the free radical chlorination of cycloheptane.
a) Using Bond Dissociation Energies (BDEs), calculate the ΔHo for each relevant step and for the overall reaction. Show work.
b) Calculate the heat of formation (ΔHfo)
of chlorocyclopentane. Show work. The initiation step
does not contribute to the heat of the reaction. Low
concentrations of radicals are produced. BDEs are from the
table.
Representative examples are used. Breaking bonds: think
uphill, positive; breaking bonds: think downhill, negative.
Both propagation steps are exothermic, the second more so
than the first. Adding the two propagation steps gives -30
kcal/mol for the reaction. DHo(reaction)
= DHo(products)
- DHo(reactants) x =
DHfo
(chlorocycloheptane) = -36 kcal/mol
The ΔHfo of chlorocycloheptane
can be determined using heats of formation found in the Δ
Hfo
table.
-30 = [x + (-22.1)] - [(-28.1) + 0]
x = (-30) - (-22.1) + (-28.1)
3. Alexander
Borodin (1833-1887), the
accomplished Russian composer, did not quit his day job. He
was a Professor of Organic Chemistry at St. Petersburg
although he is more famous as a composer.
He unknowingly discovered a method for the decarboxylation
of carboxylic acids (via their silver salts) in the presence
of bromine to afford an alkyl bromide with the liberation of
CO2. The reaction proceeds through a free radical
chain mechanism. Borodin did not isolate the alkyl bromide.
That accomplishment rested with Herr und Frau Hunsdiecker
(1942), after whom the reaction is named. The silver
carboxylate and bromine form AgBr and A before the
free radical chain begins. What is A? Write the
initiation and propagation steps for this
reaction. CH3CH2CO2Ag
+ Br2 -----> CH3CH2Br +
CO2 + AgBr
4. When cis-cyclooctene undergoes radical chain bromination, only one monobromination product is isolated.
a) Draw the structures of the possible monobromination products. Using BDEs, explain which one is formed.
b) Write a radical chain mechanism for this reaction.
For this reaction to be successful,
bromine must be present in low concentration.
N-Bromosuccinimide (1, NBS) is
often used for this purpose (Wohl-Ziegler
reaction). Traces of HBr cause the formation of succinimide
2 and bromine as shown below.
c) Carbon tetrachloride is the classic solvent for this reaction.
Benzene may also be used as a solvent but not toluene
(methylbenzene). Explain why carbon tetrachloride and benzene are
acceptable solvents but not toluene.
d) From a preparative viewpoint, why is
this reaction not suitable for the free radical bromination
of 1-methyl-1-cyclohexene 1.
5. After reading "Hess's Law of Constant Heat Summation" and/or "Heats of Combustion, Heats of Formation, and Bond Dissociation Energies (Powerpoint)" in the Thermochemistry Module, answer the following :
a) Draw a Standard State diagram that illustrates how to calculate the ”Hfo of an ethyl radical. Make use of the BDEs and Heats of Formation Tables.
b) How does your answer in a) compare with the value in the BDEs table?
c) Using the value obtained in a), illustrate and calculate the C2-C3 bond energy in n-butaneane.
d) How does your answer in a compare with the
value in the BDEs
table?