Problem Set 1
Solution Set
Chapters 1 and 2, Structure, Bonding, Alkanes
Due: Monday, September 19, 2011
John Dalton (1766-1844) John Dalton's formulation of an Atomic
Theory in the first decade of the
19th century provided a theoretical basis for understanding
chemical behavior. In addition to defining the Law of
Multiple Proportions, he also formulated the Rule of
Greatest Simplicity, which held that water was a binary
compound, OH. (Note: Dalton did not use our modern symbols,
which came to us from Berzelius,
but rather
circles that were distinguishable
from one another.) Dalton established the combining masses
of H to O in water as ~1:6. This ratio was later refined to
1:8. Dalton postulated that in a molecules comprised of two
different atoms, the simplest one in the series would be
binary. While this rule applied to CO and CO2, it
did not apply to the pair, water and hydrogen peroxide.
Thus, water, according to Dalton, was OH. The Rule
of Greatest Simplicity, which was
at odds with Gay-Lussac's
Law of Combining Volumes of Gases that demonstrated the
volume of hydrogen produced upon electrolysis of water was
twice that of oxygen, was dismissed by Dalton as a faulty
result. Moreover, although there was agreement regarding the
combining masses of atoms in the first half of the
nineteenth century, there was
disagreement as to the unit mass
of the common atoms encountered in organic chemistry:
hydrogen (1), carbon
(2x6 or 1x12), oxygen (2x8 or
1x16). Since hydrogen was the lightest of the elements, it
was assigned a mass of one (Prout's
Hypothesis), a notion that is
unrelated to today's mass of hydrogen owing to the presence
of a single proton in the hydrogen nucleus. Berzelius's
proposal of a mass scale based upon O = 100 would have
worked as well. For a Brief History of Organic Chemistry
(PowerPoint), click
here.
1. Identify the functional groups in the red circles. The front inside cover of your text will be of use. Complete this problem on a copy of this page and attach it to your homework.
2. Draw resonance structures (if they exist) for the following species. Include all formal charges.
3. Identify the hybridization (sp, sp2, sp3) of each of the non-hydrogen atoms in each of the following structures.
a) Allene (1,2-propadiene) has linear arrangement of the three carbon atoms. The planes of the atoms H-C-H are orthogonal to each other. The picture presented does not provide an accurate 3D perspective.
b) The oxygen of acetone is sp2 hybridized.
c) The carbons of the alkene portion of acrylonitrile are sp2 hybridized, just like ethylene. The remaining carbon and nitrogen, connected by a triple bond, are sp hybridized.
d) Methyl groups are sp3 hybridized. The nitrogen of amines are close to sp3 hybridized. The amine has three sigma bonds and an electron pair.
4. Name the alkane shown here (dynamic view). [Read the Jmol instructions on how to manipulate the structure.] For a static view, click here.
The longest chain is C13 (tridecane). The ethyl group at C4 is closer to the end of the longest chain. Start numbering at the ethyl end. Alphabetize groups: ethyl, isopropyl, trimethyl. The prefix tri- does not count toward alphabetizing but iso- does.
5. Determine the heat of combustion of n-decane by estimating its heat of formation. For assistance, read about Hess's Law and utilize the Heats of Formation Tables in the Thermochemistry Module of the Study Aids. [Hint: Notice the pattern in the heat of formation of n-alkanes as they increase in mass by one methylene (-CH2-) group. Show work!
The Heats of Formation Table does not have a value for n-decane. You will notice that the series of n-alkanes differ by -5 kcal/mol as the chain length increases. Since n-octane has ΔHfo = -49.8 kcal/mol, then n-decane is estimated as ΔHfo = -59.8 kcal/mol. The total heat liberated by burning 10 moles of graphite and 11 moles of hydrogen is -1691.3 kcal/mol. But the combustion of n-decane does not start at the standard state, but at -59.8 kcal/mol. Thus the heat of combustion of n-decane is (-1691.3) - (-59.8) = -1631.5 kcal/mole.