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Organic Chemistry Chapter 1 (Review of General Chemistry) Homework—
Instructor
Name: ______________________________________________________
1. (24 points) Draw structures for all six constitutional isomers with the molecular formula
C2H5Cl.
2. (10 points) Draw the 3-dimensional structure (using VSEPR) of each molecule below;
indicate the presence of bond dipoles or molecular dipole moments. Label each as
either polar or nonpolar.
CH2Br2This compound has two C–Br bonds, each of which
exhibits a dipole moment. To determine if these dipole
moments cancel each other, we must identify the molecular
geometry. The central carbon atom has four σ bonds so it is
expected to have tetrahedral geometry. As such, the C–Br
bonds do not completely cancel each other out. There is a net molecular
dipole moment, so it is polar.
CO2This compound is linear, with dipoles of
equal magnitude and charge pointing in opposite
directions. The dipole moments cancel, so the molecule is
nonpolar.
3. (10 points) Consider the compound below and tell where each of the σ and π bonds are
found (and how many there are of each within the compound).
Answer: The double bond represents one σ
bond and one π bond, while the triple bond
represents one σ bond and two π bonds. All
single bonds are σ bonds. Therefore, this
compound has sixteen σ bonds and three π
bonds.
4. (15 points) For each pair of compounds below, circle the compound that will have the
highest boiling point and explain your choice.
(a) CH3CH2CH2OCH3 or CH3CH2CH2CH2OH
Answer: The latter compound is expected to have a higher boiling point, because it
has an O–H bond, which will lead to hydrogen bonding interactions.
(b) CH3CH2CH2CH3 or CH3CH2CH2CH2CH3
Answer: The latter compound is expected to have a higher boiling point, because it
has more carbon atoms, and thus more opportunity for London interactions.
(c)
