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Date
Skillbuilder #4.1
Title and
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• We will practice the skill in class after
lecture
Left side: Skillerbuilder problems
READ pg. 133 then
take notes
Hydrocarbons –composed of
hydrogen and carbon
Hydrocarbons are either
saturated (lack pi bonds) or unsaturated
READ pg. 133-146
then take notes
In the early 19th century, organic
compounds were often named on a
whim
Many of these compounds were given
“common names”
In 1892 a group of 34 Europeans
chemist met in Switzerland and
developed a system to naming organic
compounds
The group became known as the
International Union of Pure and Applied
Chemistry (IUPAC).
IUPAC nomenclature - system of
naming organic compounds
IUPAC names include:
1. Parent name (longest carbon chain)
2. Names of substituents
3. Location of substituents
Rule#1: Identify the parent chain -
the longest consecutive chain of
carbons
If there is more than one possible parent
chain, choose the one with the most
substituents (Branches) attached
If the parent chain is cyclic, add the
prefix “cyclo”
134 CHAPTER 4 Alkanes and Cycloalkanes
Names produced by IUPAC rules are called systematic names. T ere are many rules, and we
cannot possibly study all of them. T e upcoming sections are meant to serve as an introduction to
IUPAC nomenclature.
Selecting the Parent Chain
T e f rst step in naming an alkane is to identify the longest chain, called the par ent chain:
Choose longest chain
In this example, the parent chain has nine carbon atoms. When naming the parent chain of a com-
pound, the names in Table 4.1 are used. T ese names will be used very often in this course. Parent
chains of more than 10 carbon atoms will be less common, so it is essential to commit to memory at
least the f rst 10 parents on the list in Table 4.1.
meth
eth
3 prop
4 but
5 pent
hex
7 hept
oct
9 non
dec
undec
dodec
tridec
tetradec
pentadec
eicos
triacont
tetracont
pentacont
hect
TABLE 4.1 PARENT NAMES FOR ALKANES
NUMBER OF
CARBON ATOMS
PARENT NAME
OF ALKANE
NUMBER OF
CARBON ATOMS
PARENT NAME
OF ALKANE
If there is a competition between two chains of equal length, then choose the chain with the
greater number of substituents. Substituents are branches connected to the par ent chain:
Correct
(3 substituents)
Incorrect
(2 substituents)
T e term “cyclo” is used to indicate the presence of a ring in the structure of an alkane. For
example, these compounds are called cycloalkanes:
Cyclopropane Cyclobutane Cyclopentane
Klein3e_ch04_132-180_LR_v3.1.indd 134 11/07/16 12:05 PM
Copy table – left side
Write out Question only (“Learn the
skill”) We will practice the skill in class together
after lecture
Practice with Skillbuilder 4.1 (p.135)
Rule #2: Identify and name the
substituents
Substituents end in yl
instead of ane.
Copy table
A ring can be either a parent chain or a
substituent depending on the number of
carbons
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after lecture
Practice with Skillbuilder 4.2 (p.136)
For substituents with complex branches
1. Number the longest carbon chain WITHIN the substituent. Start with the carbon attached to the parent chain
2. Name the substituent (in this case butyl)
3. Name and Number the substituent’s side group (in this case 2-methyl)
The name of the substituent is (2-methylbutyl)
1 2 3
4
Some branched substituents have
common names
Two types of propyl groups
Three types of butyl groups
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Practice with Skillbuilder 4.3 (p.139)
Rule #3: Carbons in the parent chain
have to be numbered
2-methylpentane means there is a
methyl group on carbon #2 of the
pentane chain
Guidelines to follow when numbering the parent
chain
1. If ONE substituent is present,
number the parent chain so that the
substituent has the lowest number
possible
2.When multiple substituents are
present, number the parent chain to
give the first substituent the lowest
number possible number
3. If there is a tie, then number the
parent chain so that the second
substituent gets the lowest number
possible
4. If there is no other tie-breaker, then
assign the lowest number alphabetically
The same rules apply for cycloalkanes
To assemble the complete name:
Put the # and name of each substituent
before the parent chain name, in
alphabetical order
A prefix is used (di, tri, tetra, penta, etc.)
if multiple substituents are identical.
note: “di” or “tri” is ignored when
alphabetizing the substituents
IUPAC Rules - Summary
1. Identify the parent chain
2. Identify and Name the substituents
3. Number the parent chain; assign a locant to each substituent
4. List the numbered substituents before the parent name in alphabetical order
Following the rules, we can
name the following compound:
Parent name:
cyclohexane
Substituents:
1-tert-butyl
2-ethyl
4-methyl
4-methyl 4,4-dimethyl
The name is……. 1-tert-butyl-2-ethyl-4,4-dimethylcyclohexane
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after lecture
Practice with Skillbuilder 4.4 (p.141)
Bicyclic compound contains two
fused rings.
To name a bicyclic compound,
include the prefix bicyclo- in front of
the parent name
The two carbons where the
rings are fused are bridgehead
carbons
There are three “paths (carbon
chains)” connecting the
bridgeheads.
4.2 Nomenclature of Alkanes 143
T e problem is that this parent is not specif c enough. To illustrate this, consider the following two
compounds, both of which are called bicycloheptane:
Both compounds consist of two fused rings and seven carbon atoms. Yet, the compounds are
clearly dif erent, which means that the name of the parent needs to contain more information.
Specif cally, it must indicate the way in which the rings are constructed. In order to do this, we
must identify the two bridgeheads, which are the two carbon atoms where the rings are fused
together:
Bridgehead
Bridgehead
T ere are three dif erent paths connecting these two bridgeheads. For each path, count the number
of carbon atoms, excluding the bridgeheads themselves. In the compound above, one path has two
carbon atoms, another path has two carbon atoms, and the third (shortest path) has only one carbon
atom. T ese three numbers, ordered from largest to smallest, [2.2.1], are then placed in the middle
of the parent, surrounded by brackets:
Bicyclo[2.2.1]heptane
T ese numbers provide the necessary specif city to dif erentiate the compounds shown earlier:
Bicyclo[2.2.1]heptan eBicyclo[3.1.1]heptan e
If a substituent is present, the parent must be numbered properly in order to assign the correct
locant to the substituent. To number the parent, start at one of the bridgeheads and begin numbering
along the longest path, then go to the second longest path, and f nally go along the shortest path. For
example, consider the following bicyclic system:
CH3
12
3
45
6
7
8
In this example, the methyl substituent did not get a low number. In fact, it got the highest number
possible because of its location. Specif cally, it is on the shortest path connecting the bridgeheads.
Regardless of the position of substituents, the parent must be numbered beginning with the longest
path f rst. T e only choice is which bridgehead will be counted as C1; for example:
Correct
12
34
56
7
8
CH3
Incorrect
12
3
45
6
7
8
CH3
Either way, the numbers begin along the longest path. However, we must start numbering at the
bridgehead that gives the substituent the lowest possible number. In the example above, the correct
path places the substituent at C6 rather than at C7, so this compound is 6- methylbicyclo[3.2.1]
octane.
Klein3e_ch04_132-180_LR_v3.1.indd 143 11/07/16 12:05 PM
Count the carbons from the longest path
(carbon chain) to the smallest path
(carbon chain).
Note that the bridgehead carbons
should be the first carbons numbered
and the peak carbons (carbons
protruding from bridgehead carbons)
should be the last carbons numbered
1
1
1
2 1 1 1
3 2 2
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after lecture
Practice with Skillbuilder 4.5 (p.144)
READ pg. 146-147
then take notes
CONSTITUTIONAL ISOMERS
Same number of atoms but
different connectivity of atoms
As the number of
carbon atoms
increases, the
number of
constitutional
isomers increases
When drawing the constitutional isomers of an alkane, make sure to avoid drawing the same isomer twice. As an example, consider C6H14, for which there are five constitutional isomers.me compound.
You can test if structures are the same in two ways:
1. Flip one of the molecules and rotate around its single bonds until it can be placed over the other molecule
2. Name them. If they have the same IUPAC name, they are the same compound
180˚ rotation along the C3 – C4 bond
would make it more obvious these two
compounds are the same
Following IUPAC rules for naming yields
the same name as well
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Practice with Skillbuilder 4.6 (p.147)
READ pg. 147-148
then take notes
Relative stability of isomers can be
determined by measuring heat of combustion
For an alkane it is the reaction of an alkane
and oxygen to produce CO2 and H2O
What do you notice about the ΔH of
combustion the branches the alkane has?
No branches 2 branches 4 branches
ΔHo is the change in enthalpy, associated with the complete combustion of 1 mole of the alkane in the presence of oxygen
For a combustion process, -ΔHo is called the heat of combustion
By comparing the heat of combustion of all constitutional isomers we can determine the most stable isomer = lowest amount of energy released (exothermic – heat given off during reaction) in combustion
Combustion can be conducted under experimental conditions using a device called a calorimeter
Branched alkanes are lower in energy (more stable) than straight-chain alkanes
We will study more about enthalpy in Ch 6
READ pg. 150-152
then take notes
Single bonds rotate, resulting in multiple 3-D shapes, called conformations
There are various ways to represent the 3-D shape of a compound
Some conformations are higher in energy, while others are lower in energy.
In order to draw and compare conformations, we will need to use a new kind of drawing—one specially designed for showing the conformation of a molecule.
Newman projections are ideal for comparing the relative stability of possible conformations resulting from single bond rotation (Some
This how Newman projections are drawn…
Begin rotating it about the vertical axis drawn in
gray so that all of the red H’s come out in front of
the page and all of the blue H’s go back behind
the page.
The second drawing (the sawhorse) represents a
snapshot after 45° of rotation, while the Newman
projection represents a snapshot after 90° of
rotation.
One carbon is directly in front of the other, and
each carbon atom has three H’s attached to it.
A Newman projection is the perspective of looking straight down a particular C-C bond
The point at the center of the drawing represents the front carbon atom, while the circle represents the back carbon.
All the hydrogens in red are coming out of the page and all the blue hydrogens are going to the back of the page.
We will use Newman projections extensively throughout the rest of this chapter, so it is important to master both drawing and reading them.
Write out Question only (“Learn the
skill”) We will practice the skill in class together
after lecture
Practice with Skillbuilder 4.7 (p.151)
READ pg. 152-154
then take notes
These two hydrogen atoms
appear to be separated by an
angle of 60°. (360/6)
The angle between atoms on
adjacent carbons is called a
dihedral angle or torsional
angle. It is 60° in the molecule
below
The dihedral angle changes as
the C-C bond rotates. Which
makes it so that there is an
infinite amount of conformations
However, we only care about the staggered conformation (lowest in energy and most stable) and the eclipsed conformation (highest in energy and least stable)
The difference in energy between these conformations is due to torsional strain. Here, the difference in energy is 12 kJ/mol
All staggered conformations are
degenerate (have the same amount of
energy) and all eclipsed conformations
are degenerate
It’s possible the eclipsed conformation is 12 kJ/mol less stable because of electron pair repulsion between the eclipsing bonds (4 kJ/mol for each eclipsing interaction)
With a difference of 12 kJ/mol in stability, at room temperature, 99% of the molecules will be in the staggered conformation
The difference in energy can also be
rationalized by the presence of stabilizing
interactions in the staggered conformation
A filled, bonding MO
has side-on-side
overlap with an
empty anti-bonding
MO.
The analysis of torsional strain for
propane (below) is similar to ethane
The barrier to rotation for propane
is 14 kJ/mol, which is 2 kJ/mol more
than for ethane
If each H-----H eclipsing interaction
costs 4 kJ/mol of stability, that total
can be subtracted from the total 14
kJ/mol to calculate the contribution of
a CH3-----H eclipsing interaction
READ pg. 154 -156
then take notes
The analysis of torsional strain for
butane shows more variation
Note that there
are multiple
staggered
conformations
and multiple
eclipsed
conformations
The three highest energy conformations are the eclipsed conformations, while the three lowest energy conformations are the staggered conformations.
In this way, the energy diagram above is similar to the energy diagrams of ethane and propane.
But in the case of butane, notice that one eclipsed conformation (where dihedral angle = 0) is higher in energy than the other two eclipsed conformations.
In other words, the three eclipsed conformations are not degenerate.
Similarly, one staggered conformation (where dihedral angle = 180°) is lower in energy than the other two staggered conformations.
Clearly, we need to compare the staggered conformations to each other, and we need to compare the eclipsed conformations to each
Let’s begin with the three staggered
conformations.
The conformation with a dihedral angle of
180° is called the anti conformation, and
it represents the lowest energy
conformation of butane.
The other two staggered conformations
are 3.8 kJ/mol higher in energy than the
anti conformation. Why? We can more
easily see the answer to this question by
drawing Newman projections of all three
staggered Conformations
In the anti conformation, the methyl groups achieve
maximum separation from each other.
In the other two conformations, the methyl groups are
closer to each other. Their electron clouds are repelling
each other (trying to occupy the same region of space),
causing an increase in energy of 3.8 kJ/mol.
This unfavorable interaction, called a gauche
interaction, is a type of steric interaction, and it is
different from the concept of torsional strain.
The two conformations above that exhibit this
interaction are called gauche conformations, and they
are degenerate
Now let’s turn our attention to the three eclipsed
conformations.
One eclipsed conformation is higher in energy than the
other two. Why? In the highest energy conformation,
the methyl groups are eclipsing each other.
Two of the eclipsed conformations of butane are
degenerate.
Each CH3-----CH3 eclipsing interaction accounts for 11
kJ/mol of energy (torsional and steric strain).
In each case, there is one pair of eclipsing H’s and two
pairs of eclipsing H/CH3. We have all the information
necessary to calculate the energy of these
conformations. We know that eclipsing H’s are 4
kJ/mol, and each set of eclipsing H/CH3 is 6 kJ/mol.
Therefore, we calculate a total energy cost of 16 kJ/mol
To summarize, we have seen just a few numbers that
can be helpful in analyzing energy costs.
With these numbers, it is possible to analyze an
eclipsed conformation or a staggered conformation and
determine the energy cost associated with each
conformation.
Write out Question only (“Learn the
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after lecture
Practice with Skillbuilder 4.8 (p.156)
READ pg. 158 -160
then take notes
Towards the end of the 19th century Adolph
von Baeyer proposed a theory describing
cycloalkanes in terms of angle strain - the
increased in energy associated with a bond
angle that has deviated from the preferred
angle if 109.5
Ideal bond angles for sp3 hybridized carbon is
109.5˚
If cycloalkanes were flat, each carbon in the ring
would experience angle strain.
Also, if a ring was flat, then all the C-C bonds would be in eclipsing conformation … causing considerable torsional strain.
However, since Baeyer theory was based on the assumption that cycloalkanes are flat (planar) it did not hold because the carbon ring can position themselves in 3D space to achieve a staggered conformation
The combustion data for cycloalkanes shows
that a 6-member ring is the most stable ring
size (it is lowest in energy per CH2 group)
Cyclopropane
Two main factors contributing to its high
energy:
angle strain (from small bond angles)
torsional strain (from eclipsing H’s)
As a result of the large
amount of strain makes
the 3-membered rings
highly reactive and
susceptible to ring-
opening reactions
Cyclobutane Cyclobutane has less angle strain than
cyclopropane.
1. Angle strain bond angles of 88-90°
2. Has more torsional strain, because there are
four sets of eclipsing H’s rather than just three.
To alleviate some of this
additional torsional strain,
cyclobutane can adopt a
slightly puckered
conformation (has less
torsional strain than a flat
conformation)
Cyclopentane Cyclopentane has much less angle strain than
cyclobutane or cyclopropane. It can also reduce
much of its torsional strain by adopting the following
conformation.
Cyclopentane
1. Very little angle strain - bond angles are nearly
109.5˚
2. Slight torsional strain – adopts an envelope
conformation to avoid most of it
READ pg. 161-162
then take notes
Cyclohexane can adopt many conformations
We will explore two conformations:
Chair conformation
Boat conformation
Both conformations possess very little angle strain
The significant difference between them can be seen when comparing torsional strain.
Chair conformation:
No angle strain – bond angles are 109.5°
No torsional strain - all adjacent C-H bonds are
staggered (none are eclipsed as it can be seen with a
Newman projection)
The other possible conformations of cyclohexane
have some amount of angle and/or torsional
strain (i.e. ring strain)
Boat Conformation:
Has two sources of torsional strain.
1. Many of the H’s are eclipsed
2. H’s on either side of the ring experience steric
interactions called flagpole interactions.
The boat can alleviate some of this torsional strain by
twisting (very much the way cyclobutane puckers to
alleviate some of its torsional strain), giving a
conformation called a twist boat.
Cyclohexane can adopt many different
conformations, but the most important is
the chair conformation - most stable
(lowest energy).
There are actually two different chair
conformations that rapidly interchange via
a pathway that can pass through many
different conformations, including a high-
energy half-chair conformation, as well as
twist boat and boat conformations.
Energy diagram summarizing the relative energy levels of the various conformations of cyclohexane.
The lowest energy conformations are the two chair conformations, and therefore, cyclohexane will spend the majority of its time in a chair conformation.
READ pg. 162 -164
then take notes
But first step we must master drawing them….
The following procedure outlines a step-by-step
method for drawing the skeleton of a chair
conformation precisely for cyclohexane:
When you are finished drawing the chair, it
should contain 3 sets of parallel lines.
If you chair does not contain 3 sets of parallel
lines, then it has been drawn incorrectly.
Each carbon in the ring has two substituents:
1. Axial position - parallel to a vertical axis
passing through the center of the ring.
2. Equatorial position - positioned approximately
along the equator of the ring.
In order to draw a substituted cyclohexane, we
must first practice drawing all axial and equatorial
positions properly.
Let’s practice…Skillbuilder 4.10
Draw all axial and all equatorial positions on a chair
conformation of cyclohexane.
Let’s begin with the axial positions, as they are easier to
draw.
Begin at the right side of the V and draw a vertical line
pointing up. Then, go around the ring, drawing vertical
lines, alternating in direction (up, down, up, etc.).
These are the six axial positions. All six lines are vertical.
Now let’s draw the six equatorial positions.
The equatorial positions are more difficult to draw
properly, but mistakes can be avoided in the following
way.
We saw earlier that a properly drawn chair skeleton is
composed of three pairs of parallel lines.
Now we will use these pairs of parallel lines to draw the
equatorial positions (blue lines). In between each pair of
red lines, we draw two equatorial groups that are parallel
to (but not directly touching) the red lines:
READ pg. 164 -166
then take notes
Drawing Both Chair Conformations
Consider a ring containing only one
substituent.
Two possible chair conformations can be
drawn:
The substituent can be in an axial position or in
an equatorial position.
These two possibilities represent two different
conformations that are in equilibrium with each
other:
Ring flip
The term “ring flip” is used to describe the conversion of one chair conformation into the other.
This process is not accomplished by simply flipping the molecule like a pancake.
Rather, a ring flip is a conformational change that is accomplished only through a rotation of all C−C single bonds.
This can be seen with a Newman projection
Axial substituents become equatorial and vice versa.
Practice with Skillbuilder 4.11 (p.164) – Write it out
Draw both chair conformations of bromocyclohexane:
STEP 1: Draw a chair conformation.
STEP 2: Place the substituent.
Continue…
When two chair conformations are in
equilibrium, the lower energy conformation
will be favored.
Consider methylcyclohexane.
At room temperature, 95% of the
molecules will be in the chair conformation
that has the methyl group (Me) in an
equatorial position. This must therefore be
the lower energy conformation, but why?
When the substituent is in an axial position, there are
steric interactions with the other axial H’s on the
same side of the ring.
The substituent’s electron cloud is trying to occupy
the same region of space as the H’s that are
highlighted, causing steric interactions.
These interactions are called 1,3 – diaxial
interactions, where the numbers “1,3” describe the
distance between the substituent and each of the H’s.
When the chair conformation is drawn in a Newman
projection, it becomes clear that most 1,3-diaxial
interactions are nothing more than gauche
interactions.
The presence of 1,3-diaxial interactions
The steric strain from a substituent being in the axial
position is the result of 1,3-diaxial interactions, which
are are actually gauche interactions
Causes the chair conformation to be higher in
energy when the substituent is in an axial position.
In contrast, when the substituent is in an equatorial
position, these 1,3-diaxial (gauche) interactions are
not present .
For this reason, the equilibrium between
the two chair conformations will generally
favor the conformation with the equatorial
substituent.
However, it all depends on the size of the
substituent.
Larger groups will experience greater
steric interactions, and the equilibrium will
more strongly favor the equatorial
substituent.
READ pg. 166 -170
then take notes
With multiple substituents, solid or
dashed wedges are used to show
positioning of the groups on the ring
Chlorine atom is on a wedge, which
means that it is coming out of the page: it
is UP. The methyl group is on a dash,
which means that it is below the ring, or
DOWN.
Cl atom is above the ring (UP) in both chair conformations, and the methyl group is below the ring (DOWN) in both chair conformations.
The configuration (i.e., UP or DOWN) does not change during a ring flip.
It is true that the chlorine atom occupies an axial position in one conformation and an equatorial position in the other conformation, but a ring flip does not change configuration.
The chlorine atom must be UP in both chair conformations and the methyl group must be DOWN in both chair conformations.
Practice with Skillbuilder 4.12 (p.167) – Write it out
Draw both chair conformations of the following
compound:
STEP 1
Determine the location and configuration of each
substituent.
FYI - It does not matter where the numbers are placed;
these numbers are just tools used to compare
positions in the original drawing and in the chair
conformation to ensure that all substituents are placed
correctly.
Practice with Skillbuilder 4.12 (p.167) – Write it out
STEP 2
Place the substituents on the first chair using the
information from step 1.
STEP 3
Place the substituents on the second chair using the
information from step 1.
Practice with Skillbuilder 4.12 (p.167) – Write it out
Therefore, the two chair conformations of this
compound are:
Comparing the Stability of Chair Conformations:
Lets consider the following
In the first conformation, both groups are equatorial.
In the second conformation, both groups are axial.
In the previous section, we saw that chair conformations will be lower in energy when substituents occupy equatorial positions (avoiding 1,3-diaxial interactions).
Therefore, the first chair will certainly be more stable.
In some cases, two groups might be in
competition with each other. For example,
consider the following compound:
In this example, neither conformation has two
equatorial substituents.
In the first conformation, the chlorine is
equatorial, but the ethyl group is axial.
In the second conformation, the ethyl group is
equatorial, but the chlorine is axial.
In a situation like this, we must decide which group exhibits a greater preference for being equatorial: the chlorine atom or the ethyl group. To do this, we use the numbers from Table 4.8:
Both conformations will exhibit 1,3-diaxial interactions, but these interactions are less pronounced in the second conformation.
The energy cost of having a chlorine atom in an axial position is lower than the energy cost of having an ethyl group in an axial position.
Therefore, the second conformation is lower in energy.
Practice with Skillbuilder 4.13 (p.169) – Write it out
Draw the more stable chair conformation of the
following compound:
STEP 1
Determine the location and configuration of
each substituent.
Practice with Skillbuilder 4.13 (p.169) – Write it out
STEP 2
Draw both chair conformations.
Practice with Skillbuilder 4.13 (p.169) – Write it out
STEP 3
Assess the energy cost of each axial group.
• In the first conformation, there is one ethyl group
in an axial position. According to Table 4.8, the
energy cost associated with an axial ethyl group
is 8.0 kJ/mol.
• In the second conformation, two groups are in
axial positions: a methyl group and a chlorine
atom. According to Table 4.8, the total energy
cost is 7.6 kJ/mol + 2.0 kJ/mol = 9.6 kJ/mol.
• Energy cost is lower for the first conformation
(with an axial ethyl group). The first conformation
is therefore lower in energy (more stable).
READ pg. 170 -171
then take notes
When naming a disubstituted
cycloalkane, use the prefix cis when
there are two groups on the same side of
the ring, and trans when two
substituents are on opposite sides of a
ring
The drawings above are Haworth projections
(as seen in Section 2.6) and are used to clearly
identify which groups are above the ring and
which groups are below the ring.
Each compound above is better represented as an equilibrium between two chair conformations.
cis-1,2-Dimethylcyclohexane and trans-1,2-dimethylcyclohexane are stereoisomers (Starts Ch 5).
They are different compounds with different physical properties, and they cannot be interconverted via a conformational change.
trans-1,2-Dimethylcyclohexane is more stable, because it can adopt a chair conformation in which both methyl groups occupy equatorial positions.
Each compound exists as two
equilibrating chairs, spending more time in
the more stable chair conformation
This is the lowest
energy conformation
for the cis isomer
This is the lowest
energy conformation
for the trans isomer