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I.
ORGANIC CHEMISTRY: Introduction to Naming,
Drawing, Structure, and Properties
The Organization of Hydrocarbons
1. Read Section 1.2 (p. 11 – 23).
2. Additional Notes
Organization of Hydrocarbons
Hydrocarbons
(carbon and hydrogen atoms only;
backbone of all organic compounds)
Aliphatic
Compounds
Aromatic
Compounds
(non-aromatic)
(contain alternating
double bonds)
Open Chain
Cyclic
Open Chain
(aka acyclic)
(acyclic, ex.tail of
Vitamin.A)
Straight Chain
Cyclic
(multiples
of benzene)
Branched
(unbranched)
Saturated
(single bonds
only in parent
chain)
Unsaturated
(at least one
multiple bond
in parent chain)
Substituted
Organic Family
Linkages
(side groups off
parent chain ex.
-Br, -Cl, -OH,-NH2)
(cpd named after one
prominent side group
ex. alcohols, ketones)
(parent chain is
interrupted by a
non-carbon atom
ex. ethers, esters)
Drawing Organic Compounds
The method for drawing organic compounds used in the textbook is a hybrid method
of two types of chemical formulas: the structural formula and the condensed structural
formula. While this method is commonly used as a short cut by experienced chemists, it is
important that you first learn to draw organic compounds using structural formulas only to
reveal all atoms and bonds. You will learn how to draw condensed structural formulas as a
separate skill.
Go to the Learning Tip on p. 15. The compound here is represented in the following
ways: (a) a hybrid of a structural formula and a condensed structural formula; (b) a spacefill
diagram; (c) a wireframe diagram (the most convenient method for drawing large
biochemicals; and (d) a ball and stick diagram (called a bond and angle diagram in your
text), which most closely represents a structural formula as you are expected to draw them.
The only shortcuts you are allowed are (1) leaving out the hydrogens attached to carbons
(showing sticks only), and (2) showing hydroxyl as OH and not O–H.
Ex.
1-ethylcyclohexane
Structural
Formula
“Hybrid”
Formula
Condensed Struct.
Formula
Molecular
Formula
Functional Groups: Read Section 1.1 (p. 8 - 10)
Additional Notes: Functional Groups vs. Linkages
A functional group is a site of chemical reactivity within an organic molecule. It can
be a side group or a feature of the parent chain, such as a multiple bond. Chemical
reactivity depends upon atoms of differing electronegativities. An alkane contains no
unsaturation or polar groups; therefore, it contains no functional group. If a molecule is
unsaturated or contains some element besides carbon and hydrogen, the molecule contains
at least one functional group.
Ex.
CH3CH2CH2CH3
CH3CH2CH=CH2
CH3CH2CH2CH2OH
no functional group
functional groups
Often initially classified as functional groups, linkages not only offer reactivity to the
molecule, but also introduce a non-carbon atom to the parent chain. Longer molecules are
often made from smaller ones through linkages.
Ex.
Esters
CH3CH2C-O2-CH2CH3
ethylpropanoate
Ethers
CH3CH2-O-CH2CH3
ethoxyethane
(diethyl ether)
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Comparison Chart of the Characteristics of C-C Single, Double, and Triple Bonds
Characteristic
Rel. Bond
Length
C-C Single
Bond
longest
C-C Double
Bond
Shorter
C-C Triple
Bond
shortest
Overall
Bond
Strength
weakest
Stronger
strongest
# sigma
bonds
# pi bonds
stability of
bond
structure
1
1
1
0
high
1
Lower
2
lowest
reactivity
low (highly
unreactive)
Higher
highest
Reasoning
attraction between carbon bonds
increases as number of shared pairs
increases
as the number of bonds between
atoms increases, the energy required
to separate the atoms increases as
well
a sigma bond forms before any pi
bonds can form
a pi bond is more unstable than a
sigma bond because the electron pair
is shared by the sideways overlap of p
orbitals; these shared electrons are not
as tightly held by either nucleus as they
would be in the end-on-end overlap of
sigma bonds
very little energy is required to remove
the electrons from a pi bond because
the electrons are further away from the
nuclei of each atom and are not tightly
held by each nucleus
“Like Dissolves Like”
Why is it true that polar substances dissolve in polar solvents and non-polar
substances dissolve in non-polar solvents? There are two important tendencies to consider.
First, a substance will dissolve at all if the attraction between solute and solvent particles is
stronger than the attraction between the solute particles. This makes sense with polar
solvents dissolving polar compounds. Highly soluble compounds are easily hydrated by
water because water particles are very polar, and fluid, and are able to pull solute particles
towards them and surround them relatively easily. In fact, this process usually releases
energy.
But how does this explain the circumstance when non-polar solvents dissolve nonpolar substances? The forces of attraction among any of these particles are too weak for the
explanation of greater attraction to be adequate. The key here is twofold. First, it is
important to note that polar solvents have relatively no effect on non-polar solutes by way of
attraction. Having said that, it is now important to consider the second tendency: all
particles seek greatest randomness. It is not that non-polar solvents have a stronger
attraction for non-polar solutes, but rather that there is no specific attraction at all, which
allows solute particles to diffuse readily according to the 2 nd law of thermodynamics – all
particles seek greatest randomness. It is the lack of attraction among particles, then, that
actually allows non-polar solute particles to mix readily with non-polar solvent particles. Like
dissolves like!
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The Effect of the Hydroxyl Group, -C-O-H, vs. the Carbonyl Group, -C=O, on the
Boiling Point of an Organic Compound
The hydroxyl group, -C-O-H, the principal functional group of the alcohol organic
family, can form hydrogen bonds with other alcohol molecules. This is the strongest
intermolecular force possible between particles. Carbonyl groups, -C=O, the principal
functional group of aldehydes and ketones, are strongly polar, but can still only form dipoledipole forces of attraction between neighbouring molecules. These are not as strong as
hydrogen bonds. Therefore, alcohols tend to have higher boiling points than aldehydes and
ketones of comparable length.
Similarly, alcohols can be more soluble in water than aldehydes and ketones of
comparable length.
Practice:
- Section 1.1 Q’s: #1, 2 (redraw as structural formulas first) (p. 10)
- #1-3, 5-8 (pp. 15-21) (use the following nomenclature rules)
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Organic Chemistry Nomenclature Rules
SUMMARY
1. Find the longest hydrocarbon chain in the molecule and use the appropriate prefix to designate the
number of carbons in this parent chain (ex. meth-, eth-, prop-, etc.). If an aliphatic ring system is
present, it must be chosen as the parent chain.
2. Choose the appropriate ending to indicate the classification of the molecule (ex. alkane, -ane; alkene, ene; alkyne, -yne; alcohol, -ol; aldehyde, -al; ketone, -one; carboxylic acid, -oic acid; ester, -oate; etc.).
3. Hydrocarbon branches are named according to their size (using the appropriate prefix) and end in –yl
(ex. 2 carbon hydrocarbon branch, ethyl). You must indicate their location on the parent chain by
numbering them according to the carbons to which they are attached. Other functional groups are
added and numbered in the same fashion.
4. For molecules containing multiple bonds, the main chain must contain the double or triple bond; the
numbering is assigned so that the multiple bond has the lower number.
5. Numbering:
a. numbering the positions of side groups along a parent chain must start from the end that
results in a set of position numbers with the smallest total.
Ex. 3,4-dichloromethylpentane is identical to 2,3-dichloro-methylpentane but the former adds
to 7 and the latter adds to 5; therefore, the latter is correct and the former is incorrect.
b. when two or more of the same side group are present in the same molecule, you must add a
prefix to the group name that indicates how many there are (ex. di, tri, tetra, etc.).
Ex.
c.
2,2,3-trichlorohexane and not 2,2,3-chlorohexane
for two different side groups that have the same positions in a molecule, the order is
determined by decreasing chemical reactivity or complexity. If these are similar or not known,
then a numerical order will suffice.
Ex. 2-nitro-4-methylpentane and not 2-methyl-4-nitropentane
d. for aliphatic ring systems, the prefix “cyclo” must precede the side or parent chain name.
6. Written format: Commas between position numbers, hyphens between numbers and words, and no
spaces between words (i.e. last listed side group and parent name)
Ex. 2,3-dichloro-2-ethyl-3-methylheptane
Note: If you were to draw the above compound, you would discover that its name should be changed. What should it be changed to?
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How Aromatics are Different from Ordinary Unsaturates
The situation of alternating double bonds throughout a cyclic or acyclic organic compound is
unique in that a pi bond forms between each successive carbon throughout the molecule (or
section of the molecule, as in the case of Vitamin A). This allows p-orbital electrons to
become delocalized above and below the plane of the molecule to form “pi-clouds” (the
plane of the molecule is formed by sigma bonds).
Multiple Names
Figure 7 (p. 19) shows the condensed structural formula of the molecule methylbenzene,
which can also be named phenylmethane – both names being acceptable by IUPAC. Its
common name is toluene. On a separate page, draw both the structural and condensed
structural formulas for this molecule, and list the names below each one. Copy the captions
from the text below each one as well.
Complete the following:
1. Commit Table 2 (p.12) to memory!
2. Learn the common, non-systematic prefixes used for the skeletal isomers in Figure 5
(P. 13) (n, iso, s, t); write out the corresponding IUPAC names for each isomer.
3. Answer: #1 (b-d) and 2 (p.16)
#3 – 6 (p. 18-19)
4. Draw Fig. 8 (p. 19) into the space below. Write corresponding names below the
systematic names using the classical system described in the Learning Tip on p. 20.
5. Answer: #7 (redraw each molecule using proper structural formulas),
#8 (p. 21)
#1 (not a typo in (6), it should be “iso”),
#2 (draw a structural formula first, then answer),
#3, 4 (p. 22-23)
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