Molecular Structure: Introduction and Review Structures and Representation of Molecules: Lewis Structures: Note: this is not my PDF, but I have included it for those who need to review this. I do suggest it; this section of the document outlines some essentials we might have forgotten since taking Chemistry 14A &B although we should be practicing these skills in our everyday Chemistry 14C homework But just incase:

LEWIS STRUCTURES AND VSEPR Electronegativity - the power of an atom in a compound to draw electrons to itself. In the periodic table Electronegativity increases up a group and across a row from left to right. Noble gases have very little Electronegativity. I.e. F is the most electronegative atom and Cs has a very low Electronegativity.

Basic Rules Step 1: Determine the total number of valence electrons in the molecule or ion. Note the group number gives the number of valence electrons contributed by each atom.

1A 2A 3A 4A 5A 6A 7A 8A Li Be B C N O F Ne Na Mg Al Si P S Cl Ar Etc. (If the charge is negative add electrons, and if the charge is positive subtract electrons.)

Step 2: Write the skeleton structure of the molecule. The least electronegative atom will be central. Hydrogen is probably not central.

Step 3: Use two valence electrons to form each bond in the skeleton structure between the central and the outer atoms.

Step 4: Try to satisfy the octets (duet for H) of the atoms by distributing the remaining valence electrons as nonbonding electrons. It is usually best to start with the outer atoms.

Additional Rules

1 Thou shalt not violate the octet rule for C, N, O, and F. (C, N, O, F)

2. B and Be may have less than an octet because they are not very electronegative.

3. Third row elements or lower in the periodic table (ie P, S, Cl) may exceed the octet rule.

4. Always try to satisfy the octet rule first, then if electrons are left over, place the excess electrons on 3rd

period atoms. Formal Charges

1) Formal Charge = (Valence electrons) – (assigned electrons) (on the free

atom) Assigned electrons = lone pairs + ½(the bond pairs)

In other words for formal charge purposes, an atom owns all the electronbs in the lone pairs on that atom, and the atom shares the electrons in bonds with the atom that it is bound to .

2) Sum of the formal charges must equal the overall charge. For neutral molecules the overall

charge is zero. 3) If several structures can be drawn (these are resonance contributors and do not represent the molecule in actually) the best structures are those with the fewest formal charges and with any negative charges on the most electronegative atoms.

Wedge Dash Structures: You will see dashes and wedges in acyclical and cyclical molecules it is important to understand what these different structures mean and what they would look like in 3D These aren’t arbitrarily drawn and therefore it will matter how you interpret and where you place the attachments on the molecule and it can effect anything from the structure, the isomer you might be dealing with, or the various conformations a molecule could take on. Here's a trick: Wedges: They come UP out of the paper they are in front as said below. Dashes: Go DOWN into the paper they are behind the plane of the page.

Up, Down, Equatorial and Axial. Earlier in this section the Up Down nature of wedges and dashes was emphasized. This is because when looking at cyclic molecules this is quite important especially on frequently seen cyclohexane. Notice in the figure below, a wedge, which is always coming out of the plane, UP, is not always equatorial or axial, and a dash, which is going away from you into the paper and is always DOWN, isn’t always definitively equatorial or axial. This is because the axial and equatorial positions alternate up and down

In this figure above I have tried to show you how up and down on a cyclohexane ring alternate. This is important to note, as up and down are different then axial and equatorial. Cis and Trans and why this relates to the faces of the ring: We will address this sooner than later as it is something that is quite crucial when determining isomers later. Cis and Trans depend on whether substituents lie on the same face of the ring or opposing faces of the ring. As we know: When looking down on the overhead view in the first figure those Wedges coming up and out of the page lie above the topside of the ring. The dashes that we see lie on the bottom side of the ring. Therefore: Cis: 2 Substituents (that differ from the others this isn’t applicable if all of them are H’s) that are both on the same side i.e.: BOTH are on dashes, OR, BOTH are on wedges are Cis to each other. Trans: If two substituents lie on opposing sides of the ring, one on a dash and down and the other on a wedge and up, then this is the trans conformation.

Bond Types: Properties and Visuals Single Bond:

A covalent bond of bond order 1. This bond consists of a sigma bond and does not have a pi orbital. There is free rotation around this bond and this is an important quality of single bonds as they allow for free rotation around a bond. We will see why this is essential later when discussing conformational isomers.

Double bond: One sigma and one pi bond, The hybridization, which can accommodate one double bond between two atoms, is sp2; this consists of 3 sp2 orbital and one pi orbital on each atom involved. For a double to exist these p orbitals MUST be adjacent. Double bonds do not have free rotation,

Why? Double bonds do not have free rotation around them, as there must be adjacent pi orbital overlap to ensure the two electrons shared within a double bond are distributed between pi orbitals. This inhibits free rotation around the double bond. Practice problem: This picture is representative of Problem 59 page 17 in the think book. This shows what ethylene would look like if the p-orbitals were perpendicular to each other and if they were coplanar.

Triple Bond: A triple bond might or might not have free rotation. It is highlighted in the course reader under problem 59 as well, there are two arguments one that thinks there is free rotation because as soon as rotation occurs pi overlap that was destroy is immediately reconstituted due to the other perpendicular orbital that starts to overlap. The other argument says that it takes too much energy to allow for free rotation.

Types of Strain:

Where this will be applicable -Isomers: stability and the analysis of different isomers -The geometry of a molecule: What will the molecule look like? Planar, not planar and why? - Energy within the molecular structure: Energy whether it be relievable or a fixed energy and why it occurs

Angle Strain: strain that results when the bond angle differs from the preferred tetrahedral bond angle of 109.5 degrees. When does this occur?

When certain larger groups take up more “space” as they have more electrons around them and therefore there is more repulsion and they push other attached groups AWAY from them

When an electron pair distorts the bond angle due to electron repulsion. When there is a double, triple, or more than 4 bonds attached to a central atom Why does it have to be 109.5? So random right? Well not necessarily this allows the electrons from each involved lone pair or atom which are within a molecule (remember the electrons do NOT belong to one atom this is a bad way to view a molecule) to be spaced as fair away from each other as possible and therefore cause less electron-electron repulsion and therefore less energy and more stability Steric Strain: Strain, which cannot be relieved by rotation, this strain, is caused by Van der Wals interactions between molecules, larger neighboring molecules will have more strain.

Torisional Strain: This strain unlike steric strain can be relieved but occurs when groups are eclipsed or must pass by each other. It is overcome by rotation around a bond.

Careful with this type of strain, it can be present often, and is applicable when determining

the planarity of the molecule in question. We will see this come into play

Ring Strain: ring that results in the presence of the ring. Usually results from angle or Torisional strain (this is when the ring can change conformations but there is still an angle strain, (always a little angle strain since a ring can’t form the 109.5 angles that are preferred) OR it could be because of angle strain and Steric strain, this occurs when the ring cannot relieve the strain by rotating or changing shape therefore attached groups interact. Now that we got all those wonderful strain types done lets talk hybridization: Well all know that nasty VSPER theory. Now lets modify our elementary understanding of hybridization, geometry and the concept of stability to something a bit more sophisticated!

Now this chart above is just a general guideline for you. The angles WILL vary if there are larger groups or electron pairs around the central atom this is very important to note, as this VSPER model is not definitive. Later on in this study guide we will see it “violated”, What is important to take away is the general idea and the geometry. Molecules tend towards maximum stability; minimization of energy and it is this that determines the hybridization and the arrangement in space or geometry. For most, but not all molecules the VSPER theory provides an idea of what an atom with a certain amount of bonded groups (electron pairs or atoms) would spatial arrange itself and hybridize. This is because when NO conjugation and NO aromaticity can be achieved, and therefore no higher level of stability, the VSPER theory geometry arranges the electrons as far away from each other as possible to reduce electron repulsion and increase stability.

Resonance: The significance of a dragon, unicorn, and a rhinoceros… Resonance Contributors: First and foremost, they help us visualize the places where electrons can delocalize, where electron density is spread throughout, as it isn’t restricted to one individual molecule or area. The resonance contributor are not real, they do not definitively represent the molecule at hand rather they represent a structure which shows one depiction of where electrons can be distributed. Significance of contributors and why does it even matter? Electrons tend towards minimization of energy and maximum stability as said early. Although these resonance contributors aren’t “real” they do represent possible positions electrons can be in within the actual molecule, better represented by a resonance hybrid. Each individual resonance contributor can be ranked for significance and this gives us an idea of what electron positions are the most stable and therefore each more significant resonance contributor better represents what the molecule actually looks like. Drawing resonance contributors and when resonance is present: Preface: Please recognize that resonance isn’t a set of rules that we follow to draw them or rank them, missing out on the conceptual aspect of this section will come back to haunt you when you get to then next sections pertaining to conjugation and aromaticity. Where resonance is possible….

To have resonance there needs to be areas where electrons can move from and too without violating the octet rule.

(Unless the atom in question has an exception to the octet rule (if this seems like a foreign concept then review B, Be and Third row elements at your leisure S – sulfur comes up most often out of these in Ochem) in this class we will be dealing with mainly 1st row elements such as CNOF which should NEVER violate the octet rule)

Where are these areas that electrons can move around? Places where there are assumed but also possible pi orbitals: - You may ask what is a “possible pi orbital”? Since hopefully you have done the section on conjugation we understand hybridization of atoms with

four attachments, in which at least one is an electron pair, can adopt sp2 hybridization instead of sp3 hybridization if it benefits stability.

While this concept isn’t directly applicable it is still important to recognize now, if we see an electron pair as a position where a p orbital has a possibility of occurring then we can identify it as an area where resonance can occur. We need three of these areas adjacent to each other for resonance to be present.

Why is this concept important? This is important because when you can’t remember every electron shift rule it is important

to recognize the general pattern that electron density will be shift between these areas among difference resonance contributors.

Resonance Structures

When to use them and what they mean Resonance structures are needed when the line structure for a compound doesn't accurately portray the bonds present in the molecule. This most commonly occurs when the p orbital on a charged atom or radical overlaps with one or more pi bonds next to it, spreading the charge around and stabilizing the molecule. Resonance structures are always shown in brackets, and have a double-headed arrow between them.

Resonance structures Actual compound

+1/2 +1/2 The pi bond is actually spread over all three atoms The + charge is shared by both of the atoms on the end

Resonance structures are an approximation of the actual molecular orbitals; the true characteristics of the compound are a blending of the individual resonance structures.

The true picture involves molecular orbitals in which pi bonds are spread over three or more atoms. As a result, pi bonds and lone electron pairs (and the resulting positive or negative charges) may also be spread over more than one atom.

Three original p orbitals

+ +

three new pi orbitals

the top (antibonding) pi orbital is empty the nonbonding pi orbital where the third electron should be is empty - the positive charge is on the two end atoms

both electrons go to the lowest energy pi orbital which includes all three atoms

Since there is a positive charge, one electron has been lost.

Resonance structures may only occur when the p orbitals are next to each other. If there is an sp3 atom in between, no overlap can happen.

no resonance p orbitals can't overlap Compounds with resonance structures are always more stable than similar compounds that don't have resonance structures.

no resonance structures less stable stabilized by resonance

Resonance structures most often occur cations, radicals, and anions which are allylic or benzylic, in aromatic reaction intermediates, or in enolates.

H OH O

Br Cl

NO2

allylic Ch 15

benzylic Ch 16

electrophilic aromatic

substitution Ch 16

nucleophilic aromatic

substitution Ch 16

enolates Ch 18

Equal, greater, or lesser resonance contributors If all of the resonance structures are equal in energy, then they all contribute equally to the actual compound.

Resonance structures Actual compound

-1/2 O O C C

H O H O

O

C H O

-1/2

bond shared equally, giving 1 1/2 bonds between both C-O's

electron pair spread across both O atoms, both have -1/2

both have a C=O both have a negatively charged O equal resonance contributors

charge

If one or more of the structures are lower in energy (more stable), they will contribute more to the actual compound; that is to say, the actual compound will be more like them. These are called "greater resonance contributors". The structures which are higher in energy (less stable) will contribute less to the actual compound; it will be less like them. They are called "lesser resonance contributors".

Resonance structures Actual compound

electron pair more here; O O

C H C H

H C H C

H H

more than -1/2 charge O less than 1 1/2 bond

more that 1 1/2 bond C H

H C

negatively charged O more stable

negatively charged C less stable

electron pair less here; H less than -1/2 charge

C=O about the same as C=C pi bond with lone electron pair is larger on O

lesser resonance contributor

greater resonance contributor

pi bond across all three is larger on the two C's

Difference between resonance and equilibrium The compound represented by two or more resonance structures is not going back and forth between these structures; rather it is a blending of the structures. On the other hand, an equilibrium involves an actual change in the molecule; it is represented by two arrows in opposite directions, rather than a double-headed arrow, and has no brackets.

Resonance structures

H H

Equilibrium O H

O

H H H H

H H H H

H C H C C

H H H H

H C H H C C

H H

In an equilibrium, atoms change positions; in a set of resonance structures, only electrons appear to move (they are actually spread out). Atoms cannot occupy two places at once, while electrons are waves which can be spread over several nuclei.

These Rules are avalible in the Thinkbook but if you want to have them here I added them to the document : Credit to Dr. Hardinger Resonance Contributor Preference Rule: These rules allow us to estimate the stability of the resonance contributors. Contributors that violate the majority of these rules are less stable and will not appear in the resonance hybrid as strongly as the contributors that violate only a few of these rules. 1. Contributors that have atoms with full octets are more stable than the ones with open octets. * The above rule has priority over the rest of the rules. * A contributor that violates two of these rules is still more stable than the ones that violate the first rule. 2. Contributors with the maximum number of covalent bonds are more stable. 1 . p.13, Steven Hardinger, PhD. The Think Book. 2 P.17, Steven Hardinger, PhD. The Think Book. * When counting the number of covalent bonds do not forget to count each Pi bond as two sigma bonds. 3. Contributors with the least number of Formal Charges are more stable. * For calculating the Formal Charge of each atom one has to subtract the number of nonbonding electrons and _ of shared electrons from the atom’s group number. 4. Contributors that have negative formal charges on the more electronegative atoms and positive charges on the least electronegative atoms are more stable. * This is only if formal charges on the resonance contributors cannot be avoided. 5. Contributors that have bonds between atoms in the same row (especially C, N, O, and F) of the periodic table tend to be more stable. * The violation of this rule is usually more important than the electronegativity consideration. Take away concepts: Electrons are delocalized and stabilized. Resonance is the first level of greater stability and these concepts are important to understand BEFORE you try to understand conjugation. Understand, when the electron density is spread out over multiple bonds and atoms it allows the electrons to delocalize, space out, and minimize interactions. This increases stability and minimizes energy.

This section is something I posted on the discussion board it could possibly help you visualize and comprehend why resonance and conjugation occurs and on which atoms it occurs. Connecting Conjugation: A comprehensive connection of hybridization to conjugation Resonance: In resonance contributors these electron shift patterns are trivial, in space electrons aren’t moving one at a time via arrows from lone pair to valence shell to pi bond act. These electron shifts simply show us the places that can differ in electron density from contributor to contributor. Then we also know that these contributors are …well not real (not more real than a dragon – unicorn as Professor Hardinger said in lecture), they are simply a way to show the possible places for electron density and find out which places are more likely through ranking importance of contributors. From all this we deduce a resonance hybrid to represent the delocalized electrons and formal charges within a molecule. Through this resonance hybrid per say we achieve that electron delocalization and the molecule achieves prized stability! Well now there’s this new level that can further increase stability, CONJUGATION.

Electron delocalization occurs when electron density can be delocalized or spread beyond the confines of one particular atom or bond. This occurs in molecules with resonance, as we know. Stability increase as energy and unfavorable interactions are minimized. By allowing electron density to have a large space to distribute over there is less electron-electron repulsion

Hybridizat ion:Hybridizat ion:

So in previous chemistry classes hybridization has been about counting the lone pairs and the bonded atoms and adding that getting some Steric number which corresponds to some chart.

While that model is useful for the elementary concepts up to this point, now, one has to think about this in relation to the energy and stability within actual 3D molecules instead of sticks and dots on paper.

First thing to understand: P- orbitals don't exclusively appear on those molecules that have three atoms bonded to it or one atom two electron pairs or ones that have one electron pair and 2 atoms bonded. The point I am attempting to make is that sp2 hybridization isn’t limited to the atoms, which have it according to the VSPER chart. P-orbitals that can participate in resonance, conjugation, and aromaticity can come from: -Open valence shells (sp2 hybridized) -Sp2 hybridized atoms from double bonds or (they only have three bonded atoms this happens frequently with boron and nitrogen) (we can just assume those are sp2 hybridized) -And lone pair electrons (these adopt a sp2 hybridized structure {assuming there isn't a pi bond between an adjacent atom (double bond) that is occupying the P orbital in the same plane of the their pi bonds)

How this Connects to Conjugation

As we know the electrons like to be delocalized, and conjugation allows this through adjacent p orbital and partial pi bonds, which allow for new places for the electron density to be. In conjugation, the electron density can be spread between these adjacent planar p orbital and when 3 or more adjacent p orbital are /could be present in the molecule presented (or if asked to find the isomer one of the molecules contributors) this is preferable as the molecule achieves the lower energy and more stability. So the priority is lower energy and greater stability. Keeping this idea in mind then we don’t need to make the assumption that the N is sp2 hybridized, it is! In VSPER theory we assumed that an atom was sp3 because it had 4 things bonded onto it whether those things are electron pairs or atoms (the angles might have differed but it was still sp3 hybridized). In this case the sp3 hybridization and tetrahedral structure positioned the electrons as far away from each other as possible and reduced the electron repulsion and provided for the greatest stability. In most cases that assumption holds, but now with conjugation as a possibility IF those electrons aren’t committed to a covalent bond (AKA the lone pairs) then it is beneficial for the atom in question to adopt a sp2 hybridization and a trigonal planar structure which allows for a adjacent p orbital to lie in the plane of the other adjacent p orbitals and for these all to participate in conjugation. The electrons, if they can will adopt what promotes stability, its for the molecules benefit so it isn’t going to obey some VSPER chart just to make it easy (okay so the electron doesn’t think about this but its cool to think about how this happens this way) So yes the lone pairs, open valence (sp2 hybridized) and double bonds are able to adopt this sp2 hybridization and be conjugated. So when looking at a molecule remember the geometry that the molecule and the hybridized atoms within the molecule are adopting is dependent on what is going to decrease energy and increase stability in most cases.

This also occurs in Aromaticity and I have provided two pictures in order to demonstrate this concept.

Conjugation:

Chem 360 Jasperse Ch 15 Notes. Conjugation 1 Ch. 15 Conjugated Systems

The General Stabilization Effect of Conjugation (Section 15.1, 2, 3, 8, 9)

1 Cations

Conjugated (more stable)

Isolated (less stable)

Notes:

2 Radicals

3 Anions

4 Dienes

5 Ethers

sp2, not sp3!! O An N or O next to a double bond becomes sp2. 3

O sp3 An isolated N or O is sp

O

N N sp3 H

7 Esters O

O sp2

O O sp3

8 Amides O sp2

N H

O H N sp3

Very special, chapter 23, all of biochemistry, proteins, enzymes, etc.

9 Oxyanions

(Carboxylates) O

O sp2

O sp3

Very special, chapter 21

10 Carbanions O O Very special, chapter 22

(Enolates) sp2

sp3

11 Aromatics Very special, chapters 16 + 17

Conjugation: Anything that is or can be sp2 hybridized is stabilized when next to π bonds.

• oxygens, nitrogens, cations, radicals, and anions Notes: 1. Any atom that can be sp2 will be sp2 when next to a double bond 2. “Conjugation” is when sp2 centers are joined in an uninterrupted series of 3 or more, such that an

uninterrupted series of p-orbitals is possible 3. Any sp2 center has one p orbital

Impact of Conjugation 4. Stability: Conjugation is stabilizing because of p-orbital overlap (Sections 15.2, 4, 7)

• Note: In the allyl family, resonance = conjugation

One p Two p’s Three p’s Four p’s Six p’s in circuit Unstabilized π-bond Allyl type Butadiene type Aromatic

Isolated C=C

C=O O

C=N

O

O O

O NH2

O OH

O OR 5. Reactivity: Conjugation-induced stability impacts reactivity (Sections 15.4-7)

• If the product of a rate-determining step is stabilized, the reaction rate will go faster (product stability-reactivity principle) o Common when allylic cations, radicals, or carbanions are involved

• If the reactant in the rate-determining step is stabilized, the reaction rate will go slower (reactant stability-reactivity principle) o Why aromatics are so much less reactive o Why ester, amide, and acid carbonyls are less electrophilic than aldehydes or ketones

6. Molecular shape (Sections 15.3, 8, 9)

• The p-orbitals must be aligned in parallel for max overlap and max stability • The sp2 centers must be coplanar

All four sp2 carbons must be flat for the p's to align

7. Bond Length: Bonds that look like singles but are actually between conjugated sp2 centers are

shorter than ordinary single bonds

1.54 A normal single

1.33 A normal double

1.48 A = Shortened and Strengthened conjugated single

Shortened and Strengthened

Shortened and Strengthened

Shortened and Strengthened

O NH2

O OR

O OH

• In amides, esters, and acids, the bond between the carbonyl and the heteroatom is shortened •

8. Bond Strength: Bonds that look like singles but are actually between conjugated sp2 centers are stronger than ordinary single bonds

9. Bond Rotation Barrier: Bonds that look like singles but are actually between conjugated have

much larger rotation barriers than ordinary single bonds • Because in the process of rotating, the p-overlap and its associated stability would be

temporarily lost

1.54 A normal single

1.33 A normal double

1.48 A = Shortened and Strengthened conjugated single

Shortened and Strengthened

Shortened and Strengthened

Shortened and Strengthened

O NH2

O OR

O OH

10. Hybridization: Conjugated sp2 atoms have both sp2 and p orbitals. You should always be able to classify the hybridization of lone pairs on nitrogen and oxygen.

• Isolated oxygens or nitrogens: sp3 atom hybridization, sp3 lone-pair hybridization, and tetrahedral, 109º bond angles

• Conjugated nitrogens: sp2 atom hybridization, p lone-pair hybridization (needed for conjugation), and 120º bond angles

• Conjugated oxygens: sp2 atom hybridization, one p lone-pair hybridization (needed for conjugation), one sp2 lone-pair, and 120º bond angles

Aromaticity:

In this section it will not cover the basics as on the Chem 14C website there is already another great tutorial on Isomers and Stereoisomerism. This is a concept, which is difficult to grasp as some of the vocabulary such as conformational, constitutional, and configuration, sound ever so similar. In this section I will try to clarify each one, provide a guide to recognizing these isomer types, and include pertinent vocabulary to each section. I suggest you know each type of isomerism well and how and when they are present in a molecule. (By the way use this resource it will make a huge difference on the exam, some definitions online or even in the book are vague or difficult to understand, if you aren’t using the glossary you are missing out and I wouldn’t be surprised if you get confused by the types of isomers because even other people who write websites have varying definitions about what is what. USE THE ILLUSTRATED GLOSSARY OF ORGANIC CHEMSITRY): Note: Before I proceed I am splitting each type if isomerism into a section with definitions and examples, please realize multiple types of isomerism can occur within one molecule and because of this I will have a final section and links to some resources which can help you visualize these and recognize why they occur Conformational Isomers: Isomers that have the same connectivity sequence and can be interconvert by rotation around one or more single (σ) bonds.

Recognizing this type of isomerism is fairly easy. The difficulty comes when determining stability of the conformations and converting between wedge dash and Newman projections that depict the conformers.

Eclipsed:

This is the eclipsed conformation: The dihedral angle is 0 o this conformation has large Torisional strain which is relived by rotation around the single bond that connects the two carbons of ethane within this particular example Note: please recognize this is one example there are various other ones that have other attachments than hydrogen, which are eclipsing. This conformation will be the least stable and is not the preferred conformation for the molecule for this reason.

Staggered:

Two atoms and/or groups whose dihedral angle is 60o, or close to 60o. In other words, their bonds to the axis of rotation are not aligned. Also a conformation, which contains this arrangement.

Some Important Distinctions: Anti and Gauche Conformations and interactions

Gauche: The relationship between two atoms or groups whose dihedral angle is more than 0o (i.e., eclipsed) but less than 120o (i.e., staggered). A conformation which has one or more gauche interactions is can be called a gauche conformation. Understand there are gauche interactions and the gauche conformation; understand the gauche interactions within a molecule are between those larger groups in this case these groups are methyl groups (Me). While there might be several gauche conformations the important thing to recognize when ranking stability of the staggered conformations, especially those in which some gauche interactions are not avoidable, is that the conformation in while there are more gauche interactions will be less stable. Anti and Anti Periplanar: Anti is a conformation in which the methyl groups are anti-Periplanar from each other we designate Anti as the name for this conformation of butane. Anti-Periplanar: When one is trying to explain that two groups are opposite from each other on the Newman conformation these are Anti-Periplanar. Drawing Conformations from wedge dash. Owls 7: This problem tests ones ability to draw a wedge dash structure ”

So I drew these pictures to help you. So it’s really easy to get tripped up with these wedge dash line drawings and transferring them into Newman conformations. The reason those groups cant neighbor is because as well as there being conformational isomerism there is also chirality within this molecule, there are two stereocenters and therefore the 4 attachments on each carbon are arranged a particular way in space. They are attached in the same way, same formula, but different arrangement in space so we can't just assign the attachments a place all wily nilly. You can see an example of this in the link I attached where they show examples of how a molecule can have conformational and constitutional isomerism Now to discuss the way to understand how to transfer from one representation, in this case Wedge-dash to, to another, the Newman conformation. So first let’s just try to get an arrangement by drawing one of the staggered conformations... (Remember we can rotate around the single bond to get the others I am just trying to show you how they are attached. First those lines represent the parts of the molecule that lie in the plane of the paper. In this molecule I have colored those groups a lime green. Second the Wedges are coming out of the paper towards us. It might be hard to see but if we were looking at the molecule on a model from the side those would be on the same side, not opposite from each other in the Newman conformation. Then we have those dashes, which show those attachments, which go into the paper, and those would be on the other side of the molecule next to each other. So let’s look at the first carbon, we will say this is the one with the Ph, H, and NH2 attached. -The NH2 is down in the Newman conformation, this is how it is displayed in the OWLS answers and on the wedge dash it lies in the plane of the paper. -Then we have the Phenol group, which is coming towards us, which is the wedge. And in this case -Lastly the H is going away from us on the opposite side (not anti just the neighboring opposite side) of the front or first carbon

Now let’s view the carbon behind this one the second carbon. This carbon has the COOH, H and an OH group attached. Again let’s go through the same process. The CO2H group is on the top it is anti from the NH2 in this case and that is because they lie in the same plane and are opposite from each other. Then we have the H, which in this case is coming towards us and is coming out of the paper. Then finally we have the OH which is going away from us into the paper and is neighboring opposite from the H that is coming towards us (again not anti as it’s on the same carbon) Now let’s try to see this three dimensionally. -The H on the first carbon and the other H on the second carbon must be anti Periplanar from one another, as this is staggered. The H on the first carbon represented by the dash is going away from us and then the H on the Second carbon is depicted on a wedge that is coming towards us. We couldn’t have a wedge that comes towards us and a dash that comes towards us too right? Well they are anti-Periplanar from each other then. This is the same for the OH and the Phenol. Phenol on the wedge and the OH on the dash are anti Periplanar. The attachments that lie in the plane when transferring from the wedge-dash to the Newman conformation will be anti periplanar. Now once you have this Newman conformation, try rotating the back carbon with its three attachments and you will see there is no conformation, which will achieve COOH and the benzene ring anti-periplanar to one another and have OH and NH2 anti-periplanar to one another.

Confirgurational Isomerism: Stereoisomer: “ One molecule in a set of isomers that differ by the arrangement of atoms in space but are not constitutional isomer” – illustrated glossary of organic chemistry So some vocabulary to go over: Stereocenter:Stereocenter: An atom that is tetrahedral bonds to 4 different attachments. I caution in your interpretation of this word “different”, it is important to recognize that different doesn’t the atomic composition must be different. These attachments could each be singly bonded to the stereocenter and branch off while some having the same molecular formula but a different structure; this is enough to constitute four different attachments. Here is an Example:

This is an example from the Stereocenter Tutorial. This is a good example as it can be easily missed when there isn’t or is a stereocenter due to structural differenced in the attachments. In both figures, two attachments have the same molecular formula but in figure a we have a constitutional difference in one of the 3 carbon attachments. In Figure B both three-carbon attachments, although it might look deceiving, have a continuous 3-carbon chain and therefore have no constitutional differences. Stereocenters: Not Always Nitrogen’s As we know a lone pair can but does not always count as a contributing attachment, if it does count it is of the lowest priority as it has the lowest atomic mass since it is simply 2 electrons. Nevertheless a lone pair can but will not ALWAYS count. By now you should be recognizing a trend of breaking our set “rules”. We see this case of lone pairs not counting for the Nitrogen Atom. This is due to the concept of nitrogen inversion. This is directly from the Illustrated Glossary of Organic Chemistry but I realize that some of you might not take the time to go between this page and the illustrated glossary but please check it out it will help you. Nitrogen inversion: A process in which the lone pair of an sp3 nitrogen atom migrates from one face of the atom, travels through the nucleus (i.e., tunnels), and reappears on the other side. This causes the nitrogen atom's substituents to move in a manner like an umbrella inverting in a strong wind. The two stereo chemical configurations of the inverting nitrogen atom are in equilibrium. Nitrogen inversion does not occur if the inversion processes causes too much strain, or if the nitrogen atom lacks a lone pair (because of resonance or because it is an alkylammonium cation).

A

B

Nitrogen inversion causes the inter-conversion of the S and R enantiomers of ethylmethylamine. Keq = 1 and the rate of inversion is so fast that the enantiomers cannot be separated. Therefore so

this molecule exists as a racemic mixture.

Quinuclidine does not undergo nitrogen inversion because the nitrogen inversion transition state structure has too much ring strain. Test this with a model.

Tetramethylammonium cation does not undergo nitrogen inversion because it has no nitrogen lone pair.

Test your self onTest your self on Page 96 of the Thinkbook number 10 is all about this concept. Also the stereochemistry tutorial is very useful and good to take a look at before trying to understand this section. This Section I wanted to introduce determination of Labeling: The Chemistry 14C website has a tutorial that I would put in this spot within the study guide: This is the link: http://www.chem.ucla.edu/harding/tutorials/stereochem/rsez.pdf This is a PDF I have found to help you visualize Stereochemistry. This isn’t mine but I hope it helps you grasp some of these harder concepts: here is where you can find this.

http://staff. thhs.qc.edu/science/Organic_Chem/Stereochem_Primer.pdfhttp://staff. thhs.qc.edu/science/Organic_Chem/Stereochem_Primer.pdf 1. Visualization Skills 2. Chirality and Enantiomers 3. Diastereomers 4. Finding Meso

1. Visualization Skills

Proper use of stereochemical concepts requires an ability to perceive 3-dimensional objects rendered on a two dimensional sheet of paper or a board.

You need to be able to “see” how rotating a molecule in space changes the positions of groups around a stereocenter. You also need to “see” how reflecting a molecule in a mirror plane

changes the relative positions of groups around a stereocenter. A stereocenter (also called a chiral center or stereogenic center) is an atom connected to four different groups.

Rotate a molecule about an axis - changes the molecule’s orientation. Rotation about an axis does not change a molecule’s shape, conformation or any stereocenter’s configuration.

A B A B B A

C D

C D D C

C D A B D C

B A C D A B

object rotated

rotate about a vertical axis in the page

CONVINCE YOURSELF THAT

object rotated rotate about a horizontal

axis in the page

object rotated

rotate about an axis perpendicular to the page

- ALL 3 ROTATED STRUCTURES ARE THE SAME AS THE OBJECT Reflection a molecule in a plane – not a physically achievable motion (try to convert your left hand into your right hand ). Reflection changes every stereocenter’s configuration.

A B B A A B C D

A B A B C D A B C D C D

C D D C

object reflection

reflection in vertical plane perpendicular to the page

object reflection reflection in plane defined by the page

object reflection reflection in horizontal plane perpendicular to the page

CONVINCE YOURSELF THAT - ALL 3 REFLECTED STRUCTURES ARE THE SAME - ALL 3 REFLECTED STRUCTURES ARE NOT SUPERIMPOSABLE ON THE OBJECT

2. Chirality and Enantiomers A chiral entity is not superimposable on its mirror image. The above object and its reflection are chiral. Neither is superimposable on the other (its mirror image).

The above molecule (object) and its reflection are one example of stereoisomers. Two molecules are stereoisomers if they have the same composition, constitution and connectivity but different 3-dimensional shapes that cannot be superimposed in any conformation without breaking bonds.

There are two main classes of stereoisomers: enantiomers and diastereomers.

Enantiomers: two molecules are enantiomers if they are stereoisomers and mirror images. (Alternate definition: two molecules are enantiomers if they have the same connectivity, are mirror images but are non-superimposable).

object reflection rotated reflection

180 rotation

Starting with the object [ (R)-3-methylheptane ], reflection in the plane generates a new structure (reflection). A 180o rotation about a vertical axis generates the rotated reflection. Although the heptane chain of the object and the rotated reflection can be perfectly overlayed, the 3-CH3 group of the object points toward the viewer whereas the 3-CH3 group of the rotated reflection [ (S)-3-methylheptane ], points away from the viewer. The third carbon is a stereocenter.

The object and the reflection (and rotated reflection) are mirror images and non-superimposable. Thus, they are enantiomers (and chiral by the previous definition).

Consider the two alcohol molecules below. Both have two stereocenters. The structure of the alcohol on the right was generated by reflecting the alcohol on the left in the dotted mirror plane. If the two molecules are not superimposable, they are enantiomers and each is chiral. To check whether they are superimposable, rotate the right structure by 180o about a vertical axis.

Note, this rotation moves

1 2 Me Me 3

OH HO

1 carbons 2 and 3 from behind the page to in front of the

Me page. If you compare the left OH and right most structures, the

ring carbons and OH groups 3 2 can be perfectly overlayed but

Me

the methyl group (Me = CH3) is in different places on the ring. Thus, the two molecules are not superimposable. The original molecule and its mirror image are enantiomers. (left most molecule is (1R,3R) 3-methylcyclohexanol. CONVINCE YOURSELF OF THIS.

3. Diastereomers The Methyl and Hydroxyl (OH) group in the above two structures are trans to each other. If you can’t see this, visualize the hydrogen atoms connected to the ring carbons bearing the Me and OH group. The Methyl group is above the hydrogen attached to the same carbon atom of the ring. The OH group is below the hydrogen attached to the same carbon atom of the ring. Two substituents are trans if one is above its partner H and the other is below its partner H. Two substituents are cis if both are above or both are below their partner H.

Me H

H OH

Let’s look at 3-methylcyclohexanol in which the Me and OH groups are cis. Using a similar analysis as for the trans isomers, the reflected structure does not superimpose on the left most structure. Thus, these two structures are non-superimposable mirror images. They are each other’s enantiomer.

1 1 2

Me OH OH HO Me

Me

3 2 Convince yourself that the left most structure is (1R,3S) 3-methylcyclohexanol.

What is the relationship between the cis and trans molecules. If we put a trans isomer next to a cis isomer, can they be superimposed? Although, the ring Me carbons and the hydroxyl group overlay perfectly, the methyl OH groups are attached to different carbons. Furthermore, the Me OH group is axial in one isomer and equatorial in the other. Clearly, cis and trans isomers are not superimposable. (1R, 3R) (1S, 3R)

Are cis and trans isomers mirror images? I’ve rotated the right hand structure so that the ring carbons and hydroxyl group appear as object and mirror image (vertical plane perpendicular to the page). The Methyl groups on the two structures clearly do not exhibit a mirror image relationship.

These two molecules are diastereomers. Two molecules are diastereomers if they are stereoisomers but are not mirror images. (Alternate definition: two molecules are diastereomers if they have the same connectivity, are not superimposable and are not mirror images. We just tested the latter two characteristics.) In a structure with multiple stereocenters, the enantiomers have opposite configurations at all stereocenters. The diastereomers have opposite configurations at some, but not all, stereocenters. Check this below.

(1R, 3R) Diastereomers

(1R, 3S)

Enantiomers Diastereomers

Enantiomers

(1S, 3S) Diastereomers (1S, 3R) Two molecules can be diastereomers even if they have no stereocenters. These molecules constitute a special case of diastereomers called geometric stereoisomers. Two alkenes may be diastereomers if one is (Z) and one is (E). cis and trans 1,4-disubstituted cycloalkanes lacking stereocenters are diastereomers.

F

(E) (Z)

F

cis trans

Pairs of diastereomers that lack stereocenters (geometric stereoisomers). If a structure has N stereocenters, you will be able to generate a maximum of 2N unique stereoisomers. If a structure is not symmetrical, there will be exactly 2N stereoisomers and each stereoisomer will have one enantiomer.

4. Finding Meso

What happens if a structure is symmetrical? Well first, we have to define what we mean by the term. A simplistic definition says that an object is symmetrical if it can be cut into two mirror image parts by a plane passing through the molecule. Consider one stereoisomer of butan-2,3- diol (a). This molecule has two stereocenters. We can rotate the molecule into a high energy eclipsed conformation (b) in which the two methyl groups are eclipsed and the two hydroxyl groups are eclipsed. Can you see that the dotted plane in (b) cuts the molecule into two mirror image parts? This molecule is symmetrical.

HO OH HO OH

HO OH

OH

(a)

OH (b)

(c)

mirror plane used to generate reflection

If you compare the eclipsed conformer with its mirror image (c), you should be able to see that the two structures are superimposable. An object that is superimposable on its mirror image is not chiral. Instead of two enantiomers, there is only one symmetrical isomer. A structure that has

stereocenters but is not chiral (because it is symmetrical) is called a meso compound. Meso compounds reduce the number of unique stereoisomers below 2N.

(A) How do you determine whether one of the stereoisomers of a particular structure will be a meso compound? In other words, how to you determine if a structure is symmetrical?

(B) How do you determine whether a specific stereoisomer is a meso compound?

Consider the following approaches and examples for the above two tasks: Will any of the stereoisomers of 1,2-difluoropentane be a meso compound? F F Draw the molecule as a flat, regular pentagon with two fluorines (no stereochemistry)

(A-1) Does this structure have more than one stereocenter? Although no stereochemistry is indicated in the flat structure, you should see that two carbons are each connected to four different groups. All meso compounds have more than one stereocenter.

(A-2) Is there a plane that divides this structure into two mirror image parts? Yes. You can draw a vertical plane, perpendicular to the page that splits the upper C-C bond and passes through the lowest ring carbon. This plane divides the molecule into two mirror image parts. This tells you that a stereoisomer is a meso compound.

(A’-1) An alternative approach to determining whether a molecule is symmetrical: Can you identify two DIFFERENT carbons that, if selected as carbon # 1, give the same name for the compound?

For 1,2-difluoropentane: Select the upper left carbon as number 1: the name is 1,2-difluoropentane Select the upper right carbon as number 1: the name is 1,2-difluoropentane

These two names are the same. This structure is symmetrical. There will be a 1,2- difluoropentane that is a meso compound.

Try this for the following compounds:

2,5-difluorohexan-3-one F If you start numbering from the right end, the name is 2,5-difluorohexan-4-one These two names are different. This molecule is not symmetrical. None of its O F stereoisomers will be meso compounds.

4,7-dimethyldecan-3,8-diol OH If you start numbering from the left or the right, the methyl groups are numbered 4,7 and the hydroxyl groups are numbered 3,8. This molecule is symmetrical. At least one of its stereoisomers will OH be a meso compound.

Approaches (A-1,2) and (A’-1) should give you the same result.

(B) How do you determine whether or not a specific stereoisomer is a meso compound?

For starters – if the molecule has one or fewer stereocenters, it can not be a meso compound. For starters – the molecule must be symmetrical (A’-1) or (A-1,2).

(B-1). Draw the specific stereoisomer of interest and its mirror image. If possible, draw the molecules such that the carbon CHAIN or RING exhibits a mirror image symmetry. Carefully add in the substituents and functional groups. Pay special attention that you add the stereocenters correctly. Determine the original and mirror image superimpose. If they do, the two structures are the same and are a meso compound.

(B’-1). Determine the absolute configurations of the stereocenters. Name the compound, including absolute configurations, starting from the two chain ends or starting at two parts of the ring. If the absolute configurations associated with each number switches when you number the two different ways, the molecule is a meso compound. If the configuration associated with each number stays the same, the molecule is not a meso compound.

OH

Example 1 :

OH Approach B-1 Draw the molecule so that the carbon framework is symmetrical. Add the stereocenters carefully. Is the left half of the molecule a mirror image of the right half? NO !! Both functional groups on the left point back. The corresponding groups on the right point forward. This molecule is not a meso compound.

OH OH

Approach B’-1. Name starting at the left: (3S, 4S, 7S, 8S) 4,7-dimethyldecan-3,8-diol

Name starting at the right: (3S, 4S, 7S, 8S) 4,7-dimethyldecan-3,8-diol The two names are the same – the molecule is not a meso compound. --------

OH OH

How about the molecule to the left? Approach B-1 The left and right halves are clearly mirror images. Approach B’-1

Name starting with the left end carbon as #1: (3R, 4R, 7S, 8S) 4,7-dimethyldecan-3,8-diol Name starting with the right end carbon as #1: (3S, 4S, 7R, 8R) 4,7-dimethyldecan-3,8-diol

The absolute configuration associated with every stereocenter (3, 4, etc.) switches when you start numbering at the left or at the right. The molecule appears to be its own enantiomer. That makes no sense. The molecule can not be chiral. The molecule is a meso compound. ------------------------------------

Stereocenters are just one aspect of Confirgurational isomerism Here are some other vocabulary words your should know Chiral: Any object that is NOT super imposable on its mirror image!

For example your hand! Achiral: Any object that is super imposable on its mirror image R/S determination: we use this to notate the absolute configuration of enantiomers and diastereomers. D/L (+/-): NOT the same as D/L for carbohydrates~ this is a bad misconception students will easily make. This is applicable to the way a molecule rotates plane polarized light Remember not all molecules rotate plane-polarized light Classifying R and S This has no correlation to d (+) or l (-) directly. BUT if we are given one enantiomer with R (+) the other enantiomer is S (-), If we were given one enantiomer with R (-) we would have the other one which said S (+).

Stereogenic Centers within a molecule.

1) Find Stereogenic Center 2) Rank the Attachments using Caghn Prehold System. 3) Either visualize the least important attachment back and then determine if it is R or S. OR trick: If they least important group is on a wedge coming towards you determine if its R or S, then flip the R to S or the S to R. This works as if the lowest priority group is in front coming towards us we would have to flip the molecule around to determine R or S. By determining it with the lowest priority in front and flipping it we get the same result.

+/- or d-dextro and l-levo Not the same as D or L for carbohydrates and amino acids.

This occurs in enantiomers and sometimes diastereomers.

This will be given to you since we do not have equipment to test rotation of plane polarized light on the test. What you should recognize is that given one isomer R (+) then you should be able to determine the other one is S (-) or vis versa. If a molecule has an internal plane of symmetry in any direction and it is determined to be a meso compound and will not rotate plane polarized light

Diastereomers: This is a configuration isomer or a stereoisomer which does not have a mirror image and isn’t an enantiomers.

Be careful, this can apply to the diastereomers which are classified in their relationship to enantiomers but these also can be CIS and TRANS

When classifying these see the definition in the organic glossary. Diastereomers don’t only apply to those isomers of enantiomers but also those molecules that are Cis and trans as they also are stereoisomers in which attachments differ by their arrangement in space but they are not enantiomers.

Cis and Trans EZ system for importance:

Making those Enantiomers: When asked to draw an enantiomer either build a model or visualize it.

This process on paper is more difficult.

1) Find the stereocenters, all of them 2) Making sure you understand the

orientation of all atoms attachment to the stereocenters whether it is in a ring or acyclical. Understanding what the wedges and dashes are indicating, which attachments lie in the plane of the page and which are coming out and or going into the page.

3) To create an enantiomer you must invert or swap each of the stereocenters

Click here for how to draw Enantiomers Drawing Diastereomers (There must be multiple stereocenters)

Same as above, much easier to build a model for many people

1) Find stereocenters. 2) Don’t invert all of the

stereocenters but some. 3) This is Diastereomers as it will

NOT have a mirror image Equations for how many stereocenters and Diastereomers

2^n= number of stereoisomer - n = number of stereocenters

For each stereoisomer with a stereocenter (these are non CIS/TRANS) there is one pair of enantiomers (2). The rest are diastereomers. To determine if there is a meso compound or compounds look for any internal plane of symmetry.