ORGANIC CHEMISTRY 307
Fall 2010
NOTES
Lectures 3-6
R. Boikess
I. Overall Organization and Systematization
As we have seen, focusing on functional groups, is one of the approaches
we use to systemize the enormous amount of information about the chemical
behavior of organic compounds. But even more basic than that we must have
a way to describe, designate, and communicate about organic compounds .
a. Remember why there are so many compounds (C-C bonds and
chains). So one focus is to describe the carbon skeleton, which
consists of a “main” chain (the longest continuous chain) of C atoms
with various additional C atoms or groups of C atoms (smaller chains)
attached at various points.
b. Think of it as a “connect the dots puzzle” with “branches” allowed.
Let’s draw a big grid of dots and then connect.
Do for 3 dots, 4 dots, 5 dots and 6 dots. Each different pattern of
attachment corresponds to a different carbon skeleton, which is the
starting point for describing and naming all organic compounds. Note
it’s the pattern of attachment that counts, how many dots a given dot
is connected to, not how we draw it, straight, zig-zag, or bent. See
that a right angle bend doesn’t matter for propane and butane etc.
Notice that we could connect dots in closed loops (called rings),
which are different because the pattern of attachment is different.
(More later). For 10 dots there are 75 ways, for 25 dots almost 36.8
million ways (not counting rings). That’s why this is a whole-year
course.
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b. How do we go from a carbon skeleton to a compound? Every C must
have 4 bonds. Most carbons in most compounds get to 4 bonds by
bearing the sufficient number of hydrogens.
d. Introduce 1°, 2°, and 3° carbons, related to the number of C’s to
which a given C is attached. Again, go from Kekule to condensed
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formulas and to condensed formulas with parentheses.
[CH3(CH2)4CH3 and CH3C(C2H5)2CH3]
e. Go from dots to skeletal structures. (omit terminal dots (carbons);
bend lines so that a vertex of an angle represents a dot. (carbon) as
does the end of a line. Show for propane, butane and isobutane Note
branches off the main chain. Constitutional isomers, discuss.
f. Most org cpds have at least one C that gets to 4 in some other way
than maximum H’s. In other words they have a least one functional
group. One obvious way is with a multiple (double or triple) bond,
which uses 2 carbons. Show also that it can’t happen in some
structures (neopentane) and there can be no triple bonds in many
others. Discuss pentavalent carbon, which is impossible under
ordinary conditions.
g. Another way to get to 4 is to use atoms other than H. These are most
commonly (in order) O, N, X (halogen). S, and P. Generally there
aren’t many of these atoms compared to the number of C’s and H’s,
but they are an important determinant of properties.
h. But the carbon skeleton is also very important. Ex: methanol and
ethanol. Compare to each other and also to methane and ethane.
i. The idea of a homologous series is very useful. When a group of
compounds differ only in the number of –CH2- groups in the carbon
chain we have a homologous series. The physical properties of the
members of a homologous series change in a regular way with each
additional CH2.
II. Nomenclature
In order to communicate we must be able to give every organic compound
a name that will unambiguously identify the compound. One name must
equal one compound. Unfortunately the reverse is not always true, except
within in a naming system.
a. Systematic and Common names: Once you learn the system you will
be able to name any structure and draw the structure corresponding to
any name. But in addition, chemists use many nonsystematic
(common) names. Those names must be learned individually in the
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same way you have to learn vocabulary in a foreign language. If you
don’t understand the common names, you won’t understand the
language. Wade often gives common names. You need to know them,
unless we indicate otherwise.
b. What must a systematic name do?
i. Indicate the number of C atoms
ii. Indicate the carbon skeleton
iii. Designate the type and location of the functional groups.
iv. When necessary, indicate any specific 3-D relationships (more
later).
c. Start with compounds that have no functional groups other than
carbon-carbon multiple bonds. These compounds have only C and H
and are called hydrocarbons. If they also have no multiple bonds they
are called alkanes or saturated hydrocarbons. General formula
CnH2n+2. You can see why from the grid. A C in the middle has two
H’s (and two C’s), a C at the end has 3 H’s (and 1 C) and there are
two ends. For C’s in the middle that are 3° or 4° it comes back
somewhere else in an extra 1°. Once we name these we can then use
their names as the basis for naming everything else by naming and
locating functional groups on their carbon skeleton. These names are
systematic but some are based on old common names and don’t seem
very systematic. The system is called the IUPAC [discuss] system. It
has rules that enable us to name any compound we can imagine. It
must have a rule for every situation. It is not possible for most
chemists to know all the rules. But there are reference works and even
experts who can be consulted for new or very complex situations.
1. Start at an end and identify the longest continuous chain, called the
main chain. Referring to the grid of dots, the ends are the dots with
only one line. Remember that the longest chain depends only on
how the atoms are connected and not how we draw them. Go back
to the grid of dots and regard it as a puzzle. Start at an end and
trace the longest path to another end without lifting your pencil
from the paper or hitting the same dot twice. Make sure that there
are no other ends (at the start or finish) that might produce a longer
path.
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2. Indicate the number of C atoms in the chain with a numerical
prefix. Memorize the first 12 (not the first 20 given in your book),
of these prefixes: [Flashcards]
i.
ii.
iii
iv
common but now systematic: meth, eth,
prop, but
Greek: pent, hex, hept, oct, dec
Latin: non
Latin + Greek undec, dodec
3. Add the suffix “ane” to designate no functional groups (alkane).
4. If there are no more C atoms (no branches) other than the ones in
the main chain, we are finished. Homologous series (CH2) of
straight chain or (n-alkanes).
5. If there are branches, we must name and locate them. Same idea
for other substituents as we will see. System is arbitrary, usually no
real chemical reasons for the names.
6. Name of Branch: The suffix indicates a branch. Drop “ane” add
“yl” Meaning of “yl” is one less H (so it can be attached). This
species is called a radical R (in this case an alkyl radical). More
generally, a radical can be any collection of C and H and even
other atoms that has one fewer H atom than the corresponding
compound. Key is point of attachment, which is shown by a line
CH3CH2- etc. Radicals derived from n-alkanes with missing H
(point of attachment at the end of the chain) take name (systematic
= common) from parent: show methyl, ethyl, propyl, butyl (but not
isopropyl or s-butyl because the attachment is not at the end of the
chain). For branched radicals naming is more complex, will come
back to it.
7. Location of branch. Number the main chain, according to certain
rules. Put number in front of branch. Consider 2-methylpentane
and then 3-methylpentane.
2-methylpentane
3-methylpentane
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Notice that the exact shape of the drawing is determined by the
Chem Draw software. (Remember: You can download this
software by following the directions on the course web page.)
Lowest number. [Note: no 1-methyl, 2-ethyl etc Why?]
Punctuation: Separate numbers from words or parentheses by
hyphens and from each other by commas. IUPAC names are
usually one word.
8. More than one branch. Multiplying prefixes for same branches, di,
tri, tetra, penta, Show 2,4-dimethylhexane.
Every branch has its own number. Different branches are listed
alphabetically. For this purpose multiplying prefixes and sec and
tert don’t count. But iso, neo, and cyclo are alphabetized.
9. Numbering is done from the end that gives the lowest number to
the first substituent or first different numbered substituent.
So 3,4, 6 is better than 3,5,6 and 2,7,8 is better than 3,4,9 (decane).
2,7,8-trimethyldecane
Note it’s not the sum. . Simply restated: Start numbering from the end
of the chain closest to a substituent.
If there is a tie then the lower number goes to the lower substituent
alphabetically: 3-ethyl-5-methylheptane. (not 5-ethyl-3-methyl)
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Reminder: The IUPAC name is one word. Numbers in the word
are separated from each other by commas and from letters by
hyphens
10. Branched alkanes; overlap of common and systematic for the
alkanes themselves and as parents of branched radicals.
Consider the branched alkanes with common names. Remember
common works only for simple cases and gives the entire
compound a name, rather than naming the pieces.
Thus isobutane = 2-methylpropane, isopentane = 2-methylbutane,
isohexane = 2-methylpentane. Get the idea, meaning of iso, not
much advantage to the common name in this case. Note the “iso”
alkanes form a homologous series.
Show neopentane= 2,2-dimethylpropane, neohexane = 2,2dimethylbutane etc. Slightly more useful than iso.
.
11.Branched radicals. Here common names when they exist are more
useful and more often used. Again simple cases: Show, isobutyl,
isopentyl, isohexyl etc and what iso means here. Compare to nalkyl and note 1° C as the point of attachment.
Then a few uniquely named radicals: isopropyl (contrast with
isobutyl), sec-butyl (s-butyl), and tert-butyl (t-butyl), can be
extended but usually isn’t and neopentyl, which also isn’t generally
extended. See Figure 3-2 of Wade. All the common names in that
figure are acceptable in this course and can be used, if you
wish, in IUPAC names. You may also use neopentyl. So given
any of these names you must be able to write the structure.
12.Discuss again meaning of sec and tert and lead into 1°, 2°, and 3°
for radicals, functional group substituents, and even H. Note it is
always based on C (except for amines, where it is based on N).
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13.Systematic names of branched radicals. Like naming branched
alkanes with two differences. (a). the suffix is “yl” not “ane”. (b)
the chain is numbered with a 1 at the point of attachment. [That
means that we can have 1-methyl or 1-ethyl as part of the name of
a branched radical]. Also the entire name of the branched radical is
put in parentheses to avoid confusion. Branched radicals are
alphabetized by the first letter of the entire name in parentheses.
Show getting to systematic names: isobutyl = (2-methylpropyl),
isopentyl = (3-methylbutyl) etc. Isopropyl = (1-methylethyl) note
the 1 here but not on alkanes. s-butyl = (1-methylpropyl), t-butyl =
(1,1-dimethylethyl), and neopentyl = (2,2-dimethylpropyl). You
can see why the common names are popular for these radicals, but
also see that once you learn the system that’s it. You must learn the
system, but you still need to know the common names for
communication purposes.
Here is a challenging structure to name (You can use common or
systematic names for the branched radicals)
E. Some Simple Functional Groups
Here is a simpl way to systematize some functional groups. Start with
some organic derivatives of the binary hydrides of important nonmetals.
Systematize: replace one or more H’s with R’s. [Reminder about R] Could
also approach these as substituted alkanes
HX is a hydrogen halide (F, Cl, Br, I) and RX is an alkyl halide.
HOH is water. Replace one H to get ROH, the alcohols, replace both H’s
to get ROR’ the ethers (the R’s may or may not be the same).
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NH3 is ammonia. Replace one H; RNH2 is a primary amine. Replace two
H’s; RNHR’ is a secondary amine. Replace all three H’s; RN(R’)R” is a
tertiary amine. Note R’s can be the same or different. Contrast this use of 1°,
2°, 3° with that for C which applies to halides and alcohols. Compare
isopropyl alcohol (2°) with isopropyl amine (1°) This is an important
exception to the rule that 1°, 2°, and 3° refer to R (or C).
The structures of these compounds reflect the structure of the parent
inorganic compound. There are of course chemical and physical differences
that arise from changing H to R.
a.
Nomenclature
Two main ways to name compounds with these functional groups:
systematic (IUPAC) and common. In simple cases they may be the same.
Sometimes there is more than one common name. Sometimes the common
name is much simpler and is widely used.
IUPAC names as a substituted alkane, designating the type and location of
the functional group(s). Common names are generally derived from the
parent inorganic hydride.
1. Halides RX.
a. If R has a simple or common name then the common name is often
used. It is the name of the radical followed by halide (F, Cl, Br, or
I) Methyl fluoride, ethyl bromide, isopropyl iodide, neopentyl
chloride etc. These names are derivatives of HX (hydrogen
chloride-methyl chloride) and are two words
b. If not, use the IUPAC system, which treats the X as if it is an alkyl
radical, the names are fluoro, chloro, bromo, iodo. (Could just as
well be methyl) and they are alphabetized and have the same
numerical preference as R. Names are one word. Consider these
two names: 3-ethyl-4-fluorohexane and 3-chloro-4-ethylhexane.
Draw the structures and understand the numbering and order of the
substituents.
F.
Physical Properties of Alkanes
Review HP 11.1-11.3, 11.5-11.7, 12.1, 12.3-12.4
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Major unifying theme: Physical properties are determined primarily by
polarity, secondarily by size and shape. A major factor can be H-bonding.
1. Alkanes are nonpolar.
a. Like dissolves like thus alkanes are not soluble in polar solvents.
b. Intermolecular forces in alkanes are all due to polarizability, but
they can be very important if the alkane gets big. In a homologous
series of alkanes the BP increases by about 29 K/CH2 and the ΔH
of vaporization increases by a bit more than 4 kJ/C. So an alkane
with more than about 80 C’s can’t be boiled. Why? [C-C bond
energy is 347 kJ/mol]
c. Shape can also influence boiling point: pentane = 36 °C,
isopentane 28 °C, and neopentane 9 °C. snakes and balls. Or
greater contact area = higher boiling point
d. melting point depends even more on shape because of packing, as
well as contact area. Good packing = higher melting point if there
are comparable intermolecular forces. Notice that odd numbered nalkanes melt slightly lower than expected. (Figure 3.4) This
observation can be explained by packing differences.
The closer molecules can approach each other in a crystal, the
better the packing, and the higher the melting point. The odd
numbered n-alkanes do not pack as well as the even numbered
ones. [This explanation is not in the book] Consider an n-alkane as
a zig-zag chain (a good approximation of its shape) and draw what
happens when two zig-zags are next to each other.
n-hexane
or
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n-pentane
or
In the one with the odd number of carbons the methyl at each end
is facing the same way and they bump when the molecules try to
get close.
2. Most functional groups are polar so properties are determined by
competition between nonpolar alkyl part and polar O or N or X. The
larger the alkyl part the less the effect of the polar group on overall
properties. Solubility: 3 or 4 C atoms still can be soluble in water.
3. H-bonding (HP 11.6) especially important in determining physical
properties of organic substances in those instances where it is
possible. (OH and NH bonds)
How do we make the study of the physical and chemical properties of
millions of compounds manageable and comprehensible? Physical and
chemical properties depend on molecular structure, which has two
components, electronic (distribution of electron density in bonds and over
the molecule) and steric (the shape, 3-D arrangement of the atoms in space).
Both are very important.
G.
Representations of 3-D Structure (very important)
1. Can do it in 3-D with models. There are different types of models: ball
and stick, space filling, framework. Your kit is basically a ball and
stick type.
2. Projections: a 3-D structure in 2-D, not all atoms need be shown by
symbols, just as they are not in condensed formulas. Different types
are useful for showing different things. Unlike the “bond-line
condensed formulas we have been using, projections often show the
H’s when they play a role in the 3-D structure.
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a. Dotted line-wedge (already presented), easy and good for
designating where things are relative to each other. But not so good
for helping us to see the actual shape and spatial interactions.
b. Newman Projection; Note location of the eye and representations
of the C atoms. Note how flattening appears to change the angle.
Don’t forget to interpret the 2-D back to 3-D. Note how the
vertical on the front C is up. [Figure 3-5]
c. Sawhorse, like a perspective drawing. Eye is above and to the right
of a 3-D framework model. We will use these projections
extensively when we look at rings. They are good for seeing spatial
interactions. [Figure 3.6]
H.
3-D Structure of some simple compounds. As we will see it’s
usually not as simple as it appears.
1. Methane, this one is simple. Ideal tetrahedral, bond angles, draw
dotted line wedge and sawhorse.
2. Ethane, not so simple. Do all this with models and draw sawhorse and
Newman projections [Section 3.7B in your text] Look at the pictures
in this section very carefully for a long time.
a. First look at each C. It will be very close to ideal tetrahedral.
b. But what about the CH3 groups with respect to each other? Does it
matter? Explain why. These structures are all related to each other
by rotation around a single bond. They are defined as
conformations of the compound. A particular conformation is
called a conformer. The least stable conformer is called the
eclipsed. The relative instability of the eclipsed is the result of
repulsions between the electrons in C-H bonds on the adjacent
carbons. Relative instabilities related to the shape of a molecule are
due to what is called strain. This particular type of strain is called
torsional strain. The destabilization (or torsional strain) is the
result of bringing bonding electron pairs in C-H bonds too close to
each other.
c. Energy differences between conformers [Figure 3.7]. Understand
why. Note and name two extremes eclipsed and staggered.
d. Free rotation around C-C single bond. Not totally free so it is T
dependent. Note barrier. If we lower the temperature enough we
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can slow down the rotation greatly. The reason we can treat
different conformations as a single compound is because the
temperature on the surface of the earth is what it is, high enough
that most rotations around single bonds are very rapid. In outer
space (outside the satellite) staggered and eclipsed ethane would be
isomers.
3. n-Butane is more complicated. Focus on rotation around the 2-3 bond.
Methyl is much bigger than H. Thus there are two staggered
conformations of different energy and two eclipsed conformations of
different energy. [Figures 3.10-3.11]
The two staggered conformations are called gauche and anti. Energy
difference between them of about 4 kJ/mol. This number is important
because we will use it to help explain relative energies of other
compounds. It is called a gauche butane interaction. Both are much more
stable than the eclipsed conformations, which can be ignored. The
relative instability of the gauche is due to what is called steric strain, the
result of atoms or groups of atoms approaching each other too closely.
III Cycloalkanes
Consider an imaginary process in which a cleavage of a C-H bond at
each end of a chain occurs and then the two ends come together to form a
new C-C bond. Such a compound has two fewer H’s (because there are no
ends), CnH2n. Its key feature is of course a ring. When there are no functional
groups the compound is called a cycloalkane and the chemistry is much like
that of alkanes with a very few exceptions that we will explain.
A. Representing Cycloalkanes.
1. Skeletal structures are used almost universally to show rings,
very important, we understand that there are enough H atoms to make
every C tetravalent. Regular polygons are used for the skeletal
structures of the cycloalkanes. Show examples and identify the
missing H’s. Except for the three carbon ring, none of the
cycloalkanes have the actual structure of the regular polygon. We will
see why soon.
B. Nomenclature
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1. The presence of a ring is signaled by the prefix cyclo. The
number of atoms in the ring is designated in the usual way starting (no ring
with 2 atoms) with 3 = prop, 4 = but etc. The absence of any functional
groups is signaled by the suffix ane.
2. A substituted cycloalkane can be named in one of two ways.
(a) By identifying and locating the substituents on the
ring. Same basic method as for acyclic (define) alkanes, but if there is only
one substituent no number is needed because all positions are equivalent.
Example, methylcycloheptane. If there is more than one substituent, then the
first one gets position 1and we follow the same rules as for chains. Consider
1, 2-diethylcyclopentane and number in the right direction. The rule of
numbering in the direction that gives the lower number at the first point of
difference. So 1,2,4- rather than 1,3,4- is chosen overriding the alphabetizing
of substituents. Alphabetizing only comes into play when there is a tie. So 1ethyl-2-methyl is preferred over 2-ethyl-1-methyl.
(b) By naming the ring as a cycloalkyl radical and
treating it like any other alkyl radical. Thus the name 3-cyclopropylhexane.
If the ring has additional substituents the point of attachment is always 1.
For example: (3-methylcyclobutyl) or (2.2-dimethylcyclopentyl).
(c) Which way do we do it? (a) or (b)? Use common
sense in situations where one name is much simpler than the other. Two
rules to follow in order
i. maximum number of substitutions on a single
unit of structure [1-ethyl-2-methylcyclohexane, not (2-methylcyclohexyl)
ethane. And dicyclobutylmethane, not (cyclobutylmethyl)cyclobutane
ii. smaller unit as a substituent on a larger [3cyclopropylpentane (3 on 5), not (1-ethylpropyl) cyclopropane (5 on 3)]
3. Free rotation around C-C bonds in cyclic structures is not
possible, because the ends are tied together. This restriction has a number of
consequences. One is that a ring has a top and bottom. Of course this makes
no difference (because you can just turn the ring over in space) except when
there are two or more substituents on two or more carbons of the ring.
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Consider 1,2-dimethylcyclopentane.[Figure 3-13] We could
put both methyls on the same side of the ring (the top or the bottom) or we
could put the methyls on opposite sides of the ring. Because there is no free
rotation around the C-C bonds of the ring, the only way these two
arrangements can be interconverted is by breaking a C-C bond, which does
not occur anywhere near room temperature. Therefore we are dealing with
two different compounds, two isomers, which do not differ in connectivity
(what’s attached to what) but only in the 3-D arrangement of the atoms.
Notice the difference becomes apparent only when you build a model or use
a projection drawing so that you represent the third dimension.. Such
isomers are called stereoisomers.
We need a way to name different stereoisomers. In the case of
disubstituted cycloalkanes, we use the prefix cis to designate the isomer with
both groups on the same side of the ring and trans to designate the isomer
with the groups on the opposite side of the ring.
So the two isomers would be named cis-1,2-dimethylcyclopentane
and trans-1,2-dimethylcyclopentane.
C. Strain
1. The 3-D constraints that result from closing a ring of carbon atoms
have more ramifications than simply the creation of another kind of
isomerism. These ramifications come under the broad label of strain. Strain
is a destabilization of a species that results from unfavorable interactions due
to its 3-dimensional structure or shape.
2. Torsional Strain. We have already encountered one type of strain
called torsional strain that results from eclipsing of C-H bonds. As we shall
see, the restricted rotation in rings can also lead to C-H eclipsing and
therefore torsional strain.
3. Angle Strain. The geometry of rings can cause a destabilization
called angle strain. An sp3 hybridized carbon atom prefers bond angles of
109.5°. But if you think of a ring as a regular polygon, its angles are not this
ideal value, but something different. The greater the difference between the
actual angle in the ring and 109.5°, the greater is the angle strain.
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4. Steric Strain. Sometimes the geometry of the ring forces relatively
large groups of atoms (in this context even CH2 is relatively large) too close.
We have already seen this type of strain in gauche butane. The resulting
electronic repulsions cause a destabilization called steric strain.
5. Strain Energy. We can find the total destabilization that results
from these three kinds of strain in a given cycloalkane using thermochemical
measurements. As we have seen, the heat of combustion provides a
convenient way to measure relative stability of compounds. We can measure
the heat of combustion of a cycloalkane and compare it to a value we can
calculate. The measured heat of combustion will be more exothermic than
the calculated value.(Be sure you understand why; this is a critical point)
The difference between the measured and calculated values is called the
strain energy.
One simple way that we can calculate the heat of combustion is by
using the measured value for the combustion of a CH2 group in alkanes,
where there is no strain. This value is 659 kJ/mol CH2. Thus the calculated
value for cyclopropane is 3 CH2 x 659 kJ/mol CH2 = 1977 kJ/mol. The
measured value is 2091 kJ/mol. So the strain energy of cyclopropane is 2091
– 1977 = 114 kJ/mol. We can repeat these calculations and measurements
for all the cycloalkanes. The results are listed in Table 3-5 of your text. You
should focus on the last two columns, which show the way in which strain
changes with ring size.
5. Let’s look at the structures of some of the cycloalkanes and try to
understand the source of their strain energy.
a. Cyclopropane. The three carbons of this molecule must lie
in a plane and the bond angles (as measured by the lines
connecting the nuclei) must be 60°. So these bond angles are
very difficult for sp3 orbitals to achieve and instead the
overlap of the orbitals is not along the internuclear line and
therefore not as effective. This poor overlap forms a
relatively weak bond and is treated as angle strain. [Figure
3-15] In addition, Cyclopropane has substantial torsional
strain because the flat structure forces all the H’s to be
eclipsed. [Figure 3.16] A model will reveal this clearly. In
fact any flat structure will have complete H-H eclipsing.
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b. Cyclobutane; two problems: Angle strain, geometry predicts
90° so not as bad as cyclopropane; 4 pairs of C-H bonds that
are eclipsed leads to even more torsional strain than in
cyclopropane. To reduce torsional strain (at the expense of
more angle strain) cyclobutane puckers, reducing the CCC
bond angle to 88° (bad) but reducing the torsional strain
(good) gaining more than it loses. [Figure 3-17] Notice that
we use sawhorse projections for rings. [Examine a Model]
(Remember the molecule always does what’s best) From
Table 3-5 not as much strain energy as cyclopropane.
c. Cyclopentane; less angle strain but more eclipsing than
cyclobutane. Geometry predicts bond angles of 180- 360/5 =
108°. Thus very little angle strain but 5 pairs of eclipsing
hydrogens. Ring puckers to 4 C’s in the plane and the 5th
bent up like an envelope flap. The CCC angles are about
105° and ring is dynamic. Each C bends up giving 5
conformations that are rapidly interconverting. Thus some
angle strain and still some torsional strain because we can’t
get all the H’s staggered in the envelope conformation.
[Figure 3-18]
d. Cyclohexane; has no strain and it is by far the most
important cycloalkane. Many of its derivatives are also very
important. Seems like it should have strain. The angle of a
regular hexagon is 180 – 360/6 = 120° and there are 6 pairs
of H-H eclipsings in a flat ring. So the ring is not flat. It
assumes nonplanar conformations in some of which the
angles are very close to 109.5°. These conformations have
virtually no angle strain. Remember conformations are
related to each other by partial rotation around single bonds.
(the rotations are not always so obvious). While there are a
number of conformations without angle strain, we are
especially interested in those conformations that also do not
have H-H eclipsing either.
i. Chair. The chair is the most stable and most
important conformation of cyclohexane. There are two
chairs.They have no angle strain and no H-H eclipsings.
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Build, play with, and study models of chairs. [Figure
3.19] Understand why the two chairs are different
(although the same in energy for unsubstituted
cyclohexanes). Emphasize axial and equatorial.
[Figures 3.22 and 3.23]. Study the instructions on how
to draw chairs on page 112 of Wade.
ii. Understand ring flip and boat. Understand
flagpole and H-H eclipsing in boat. [Figure 3.20 and
3.21]
iii. Monosubstituted cyclohexanes. Back to axial
and equatorial Explain why eq is better than ax, two
equiv ways: gauche butane and 1,3-diaxial.[Figure 3.243.26]. Either way these interactions between atoms or
groups of atoms that are too close are steric strain. Each
gauche butane interaction generates about 3.8 kJ mol-1 of
strain.
e, Conformational analysis
Disubstituted cyclohexanes (dimethyl) discuss cis and
trans in cyclohexanes, which have a top and a bottom even
though they are not flat. Define configuration and contrast with
conformation. Then relate to e and a. Then do conformational
analysis on all (not all in book) three dimethylcyclohexanes
including ring flips. If two conformations differ only in the
number of gauche butane interactions we can calculate the
approximate energy difference between them based on the
observation that each gauche butane interaction generates about
3.8 kJ mol-1 of strain.
i. Discuss and show the two 1,4-diMe’s.
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is e,a and has two gauche butane interactions = 7.6 kJ mol-1 and no
change on ring flip
has no gauche butane interactions in the di-equatorial(e,e) and 4 in the
di-axial(a,a). Energy difference between the two chair conformations
of this configuration is 15.2 kJ/mol.
ii. cis-1,2 is e,a and has 3 gauche butanes, two of the
axial methyl with the ring and one methyl-methyl. No change
on ring flip. (draw and discuss) Build a model.
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trans-1,2 is e,e and has only one gauche butane Me-Me.
Ring flip converts to a,a which has 4 gauche butanes with the
ring, but the methyls are anti. Build a model.
iii. cis-1,3 is e,e and has no gauche butanes (the Me’s are
too far to influence each other). Ring flip converts to a,a very
bad, four gauche butanes plus a 1,3-diaxial. Build a model.
trans-1,3 is e,a and ring flip doesn’t change it. Two
gauche butanes from the axial but the Me’s are too far apart to
interact. (draw and discuss) Build a model. (See Wade pg 116)
Note the more eq and fewer ax the more stable.
iv. The difference in energy between an axial and an
equatorial substituent depends on its effective size. The bigger
it appears to the cyclohexane ring, the greater the energy
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difference. Some data are shown in Table 3-6 of your text.
Look at the table and note that CH3 and indeed any alkyl group
is relatively large. Note that, as expected, the energy difference
for Br is greater than for Cl. But measurements indicate that
the energy difference for I is smaller than it is for Br. How
could you explain that observation?
Suppose you had a disubstituted cyclohexane with two
different substituents and suppose it has an ea configuration.
You can predict correctly that the larger group would prefer to
be equatorial.
v. A very large substituent, such as t-butyl, almost
freezes the ring flip. (K>5000). Note that the data in Table 3-6
in the text are obtained by studying the equilibrium between the
two configurations (isomers) of a given 4-substituted
t-butylcyclohexane. Remember the t-butyl must be equatorial.
5. Larger Rings. Ring sizes from 7 to 12 are called medium
rings and they all have strain. Some are shown in Table 3-5. The strain is
primarily steric strain because all of these rings pucker to relieve angle strain
and torsional strain from eclipsed hydrogens. As a result atoms on opposite
sides of the ring often approach each other too closely.
Cyclodecane, for example has 59 kJ/mol of strain. You can see the
steric strain in this picture of an important conformation of the molecule:
6. Polycyclic Ring Systems
Many of the most interesting organic compounds are those with more
than one ring. While you might guess that such compounds simply have two
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or more rings attached to a carbon backbone, actually most polycyclic
compounds have two or more rings that have some atoms in common. Such
rings are said to be fused or bridged depending on the way in which they
share atoms.
In fused rings, two adjacent carbon atoms are shared between two
rings. In bridged rings the carbon atoms that are shared are nonadjacent.
Many well known compounds have bridged rings. Examples are
cocaine, camphor, codeine, morphine, and strychnine. The steroids have a
basic structure that includes four fused rings.
Let’s look at a few relatively simple, but important polycyclic
ring systems
a.
Decalin (common name), A fused ring system, two six
membered rings sharing two adjacent carbons. [Figure
3-27]. Note the 10 carbons and the common name.
Many important compounds (e.g. steroids) have a
decalin as part of their ring systems or structure. Note
that for this system the rings are large enough so that
the ring fusion can be cis or trans. Will not be the case
in smaller rings. If the ends aren’t close, the cis is more
likely because the ends won’t reach each other in the
trans. Note that the trans isomer of decalin is more
stable than the cis because all the C-C bonds between
rings are equatorial in the trans.
b.
Norbornane (common name) is a typical bridged rather
than fused system. Related to terpenes. Note the atoms
that are shared by the rings. They are not adjacent.
Note the 1 carbon bridge and how it must be joined cis.
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c.
Naming ring systems in which two rings share two
atoms. (sketchy in text)
i.
ii.
iii.
iv.
v.
vi.
Two rings are signaled by prefix “bicyclo”
Number of carbon atoms in the ring system
signaled in the usual way. Decalin is a
bicyclodecane and norbornane is a
bicycloheptane.
Identify the shared atoms and the three paths
to go between them. Count the number of C
atoms in each path. It is 2, 2, and 1 for
norbornane and 4, 4, and 0 for decalin.
Put these numbers in [] between bicyclo and
alkane, separate by periods.
Names are bicyclo[2.2.1]heptane and
bicyclo[4.4.0]decane
Two more examples: bicyclo[1.1.0]butane
and bicyclo[3.2.1]octane
bicyclo[3.2.1]octane
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