Title and Highlight Right side: NOTES! Topic: EQ: Date Date NOTES: Write out the notes from my website. Use different types of note-taking methods to help you recall info (different color pens/highlighters, bullets, etc) Reflect Question: Reflect on the material by asking a question (its not suppose to be answered from notes) When I lecture we will add more info, so leave spaces in your notes DRAW ALL PICTURES, FIGURES, AND WRITE OUT ANY PRACTICE PROBLEMS/QUESTIONS. WE WILL ANSWER THEM TOGETHER. So LEAVE SPACES SO WE CAN ANSWER QUES. Summary (end of notes) : 1-2 Sentences of what you learned
Left side: Skillerbuilder problems Title and Highlight Skillbuilder #4.1 Write out Question only We will practice the skill in class after lecture
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 19 th 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 a competition between two chains of equal length, then choose the chain w of substituents. Substituents are branches connected to the par ent chain: Correct (3 substituents) Incorrect (2 substituents) If the parent chain is cyclic, add the cyclo is prefix used to cyclo indicate the presence of a ring in the structure of an alkan ompounds are called cycloalkanes: Cyclopropane Cyclobutane Cyclopentane
Copy table left side
Practice with Skillbuilder 4.1 (p.135) Write out Question only ( Learn the skill ) We will practice the skill in class together after lecture
Rule #2: Identify and name the substituents Copy table Substituents end in yl instead of ane.
A ring can be either a parent chain or a substituent depending on the number of carbons
Practice with Skillbuilder 4.2 (p.136) Write out Question only ( Learn the skill ) We will practice the skill in class together after lecture
For substituents with complex branches 1 2 3 4 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)
Some branched substituents have common names Two types of propyl groups Three types of butyl groups
Practice with Skillbuilder 4.3 (p.139) Write out Question only ( Learn the skill ) We will practice the skill in class together after lecture
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
Practice with Skillbuilder 4.4 (p.141) Write out Question only ( Learn the skill ) We will practice the skill in class together after lecture
Bicyclic compound contains two fused rings. To name a bicyclic compound, include the prefix bicyclo- in front of the parent name
o bridgeheads, which are the two carbon atoms wh The two carbons where the rings are fused are bridgehead carbons Bridgehead Bridgehead There are three paths (carbon chains) connecting the bridgeheads. nt paths connecting these two bridgeheads. For each uding the bridgeheads themselves. In the compound a r path has two carbon atoms, and the third (shortest pa bers, ordered from largest to smallest, [2.2.1], are th ded by brackets:
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 2 1 1 1 3 2 1 2
Practice with Skillbuilder 4.5 (p.144) Write out Question only ( Learn the skill ) We will practice the skill in class together after lecture
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 C 6 H 14, 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
Practice with Skillbuilder 4.6 (p.147) Write out Question only ( Learn the skill ) We will practice the skill in class together after lecture
READ pg. 147-148 then take notes
Relative stability of isomers can be determined by measuring heat of combustion No branches 2 branches 4 branches For an alkane it is the reaction of an alkane and oxygen to produce CO 2 and H 2 O What do you notice about the ΔH of combustion the branches the alkane has?
ΔH o 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, -ΔH o 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.
Practice with Skillbuilder 4.7 (p.151) Write out Question only ( Learn the skill ) We will practice the skill in class together after lecture
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 CH 3 -----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 CH 3 -----CH 3 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/CH 3. 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/CH 3 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.
Practice with Skillbuilder 4.8 (p.156) Write out Question only ( Learn the skill ) We will practice the skill in class together after lecture
READ pg. 158-160 then take notes
Towards the end of the 19 th 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 sp 3 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 CH 2 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 ringopening 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 highenergy 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.
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 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.
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