CHEM 343 EXAM 1

Hyperconjugation in alkenes -hyperconjugation basic meaning? -mechanism ex? H role? -effect on stabilization? -why does stabilization occur? -why is it stabilized by more C's?

"connected" -> electrons from filled orbitals are shared into empty orbitals -interaction of bonding electrons on neighboring groups with antibonding orbital on alkene. -electrons in C-H bonding orbitals that are attached to one of the center C's begin to spread out and fill into the anti-bonding orbitals of the center C. Must be from CH's that are directly connected to the C. +does not occur with just H because they are too far to share electrons! H cannot hyperconugate! -very small stabilizing effect on alkenes -stabilization occurs because electrons have more space to occupy and therefore charge is spread out and wave function flattens. -more C's attached to center C -> more opportunities to donate electrons to anti-bonding orbitals (hyperconjugation) -> increasing # of stabilizing effects

Experimental design for determining mechanistic hypothesis -formation of alkane from normal alkene, good or? what could be better? -experimental design includes what three ideas?

-basic alkene may not be ideal to evaluate the possibility of a CC intermediate -> does not revel anything stereochemistry because alkane can have different conformations of the same molecule that rotate to be the same molecule (consistent with other 2 mechanisms based on electrons).... better choice -> ring structure (take advantage of fact that Br selectively goes on "top" or "bottom" of p-orbitals) and P cannot be rotated between due to rigid backbone 1. selection of R with intended products (R will only give certain P)... products will not be conformations (alkanes) -chosen to assess stereochemical outcomes 2. Products (P) that can specifically support or falsify a mechanistic hypothesis... explain why the P support/falsify mechanistic hypothesis -P would be different for each mechanistic hypothesis 3. Description of possible outcome and how they will be interpreted -I can directly use product outcomes to falsify specific mechanistic hypothesis

Hyperconjugation -what bonds can contribute to hyperconjugation? -how many bonds can contribute at one time? -in normal alkenes?

-bonds coming off of directly connected adjacent carbons can contribute to hyperconjugation. Can be CH bonds or any other bonds like C-CH3 or C-OH or.... -only one bond on adjacent C can contribute to hyperconjugation at a time yet all bonds can contribute at separate times from one another if the bond between the C+ and adjacent C can rotate! -same logic goes for basic alkenes (not with other C in double bond tho?)

Exam 1 review -how many enantiomers does a chiral molecule have and why? -how many diastereomers does a chiral molecule have? -reaction sequences; consider? -predicting starting alkenes from products and mechanism to do so?

-every chiral molecule has one enantiomer (making a pair) because all chiral enters must be flipped -every chiral molecule has 2^n diastereomers where n is the # of chiral centers in the molecule -reaction sequences --> steps must be in right order (more substituted alkenes react faster so put those steps first) +where things are bonded (C with more or less alkyl groups) +rearrangement of lack there of +stereochemistry -find mechanism that gives you products with different stereochemistries and then use the product stereocchemistry to determine what the starting alkene stereochemistry was. ex--> anti would show Z and syn would show Z but use model kit to check... what mechanisms could work tho??

Ozonolysis full mechanism? -textbook answer (on separate slide too!)

1. electrons from the O- attack alkene and break the pi bond. pi bond electrons attack O. O electrons from pi bond in O3 move to O+ 2. lone pair electrons attack C-O to form C=O. Leads to broken sigma bond between C's and form another C=O. electrons in other O-O bond (weak) move to central O 3. O electrons attack C in smaller O structure. Smaller O structure pi electrons attack C in larger O structure and pi bond electrons in large O structure attacks central O. 4. Me2S or Zn, H2O is introduced (Me2 for example) 5. lone pair S electrons attack the O and O-O sigma bonds attack other O 6. C-O electrons attack O and releasing to form DMSO. O- electrons attack C-O to form C=O and group releases and forms. C-O electrons attack other C-O and break group off to form all 3 final products. -Textbook answer --> split alkene down middle and attack O's to both parts of broken double bond. +indicate 2 steps (1. O3 and 2. Me2S by arrow). -look at picture!

Rules for which resonance structures contribute most?

1. octet matters the most --> must have octets filled first 2. minimize formal charge 3. if you do have formal charge, - charge should be placed on the most electronegative atom possible. -positive charge on least electronegative atom

Bond line formula

1. only show atomic symbols (letters) for heteroatoms, not C! 2. Only show H on heteroatoms, not C! 3. every change in direction or bond order is new C atom in structure -( ) for short hand -changes direction to signify carbon in single and double bonds -DOES NOT change direction to signify carbon with triple bond.

Hydroboration-oxidation; basics -basic 2 step mechanism (full mechanism on another slide); which binds to more highly substituted group? -syn or anti? -what is BH3 replaced with?

1. pi bond electrons attack the B which binds to the C with less alkyl groups. B-H bond electrons than go to C with more alkyl groups and leads to H bonding there 2. Then treated with H2O2 and NaOH and OH replaces the BH2 (mechanism on another slide!) -syn addition -> OH and H end up syn -BH3 is replaced with OH in presence of H2O2 and NaOH

Hydration (acid-catalyzed) of alkenes mechanism -mechanism? -where do OH and H bond ? mechanism type?

1. pi bond electrons attack the H in H3O+ and the electrons in the H-O bond go to the O 2. OH2 electrons then attack the CC and bonds to it 3. H2O electrons attack the OH2 on the C and OH2 on C transfers electrons from O-H to O. This leads to H breaking off and bonding to H2O to reform the H3O+ (catalyst for reaction) and leave the OH on the C! -OH bonds to the C with more alkyl groups -H bonds to the C with less alkyl groups -carbocation forms!!

ionic hydrogenation -mechanism and steps?

1. pi bond electrons in alkene attack the H on the acid; electrons in the O-H bond on the acid go to the O 2. tri-ethyl silicon electron in Si-H bond then attack the C+ and H bonds to it -in notes on sheet!

# kCals for change in almost any reaction constant -H-bonds vs covalent bonds strength? -selective drug release example?

1.4 kCals leads to 10x change in almost any reaction constant -ex: 1.4 kCals added leads to 10x faster reaction (log scale like pH) -pH of 4 has 100x higher [H+] than pH of 6 -Covalent bonds are much stronger (90 kCals) than H-bonds (3 kCals) -drugs may be designed so that medicine is attached to alkyl chain with some atom (often O) and linkage between drug and alkyl chain is set to break at certain pH due to higher acid and H+ activity. +ex: cancer tissue has higher pH and therefore cancer meds may selectively release and kill that tissue and not other noncancerous tissue that has a higher pH

Alkenes General Structure -bond angle? -C hybridization and geometry? -double bond length, composition, strength, and strength of parts? -rigid or flexible? -saturation? -bonding and anti-bonding orbitals?

120 - C is sp2 and trigonal planar -C=C is shorter and stronger than C-C; C=C has sigma and pi bond. sigma bond is stronger (90kcals) than pi bond (64 kcals) and thus pi bond is more reactive -rigid and cannot rotate about the double bond (stereochemistry!) -unsaturated due to double bond presence and H's not completely occupying the C's -bonding orbitals formed by additive reactions of 2 p-orbitals in which they are in phase and join to hold 2 electrons in a bond. orbitals mix! +antibonding orbitals formed by negating reactions of 2 pi orbitals in which they are out of phase and do not join (nodes between them) and hold no electrons in them. Orbitals do not mix!

Resonance Structures (More from lecture) -how to view them?

2 different hypothetical molecules that give idea of what real resonance hybrid is like -ex: parents give children that is mix of them, not a form at equilibrium that is constantly changing! - more than 1 valid lewis structure -sigma bonds (connectivity) must be same --> same molecule -pi bonds and lone pairs move around -same net charge; formal charges can change

halogenation -mechanism? -EN and attachment to which C rule?

2 steps! X-X with alkene 1. pi bond electrons in alkene attack X and electrons in X-X bond go to unattacked X. Attacked X then also attacks the other C to form a pi-complex 2. X- attacks the C with more alkyl groups (more + charge on it) and electrons in bond between attacked C and X (in pi-complex) go to X (in pi-complex) and forms the final product. -more EN atom tends to attach to the C with more alkyl groups -less EN atom tends to attach to the C with less alkyl groups.

oxymercuration reduction -mechanism? -no what formed and see what with pi complex? -where do OH and H end up being bonded?

Avoids rearrangement so that more specific products can be formed -no carbocation -pi complex -> still see regioselectivity with added groups -OH ends up bonded to the C with more alkyl groups -H ends up bonded to the C with less alkyl groups

Drawing chair conformation for cyclohexane

C6H12 represented 3D structure using 2D drawings -start with two staggered lines. From line that reaches further on each side, draw line going out and toward the other line then connect the lines. Lines should be in parallel sets -axial substituents --> think of arrows pointing up or down to place them. put line at tip of where arrow points -equatorial substituents --> draw according to diagram and have lines parallel to 2 of the other lines in the base!

Organic acid strength (discussion)

CARDIO - all relate to stability of conjugate base! 1. Charge effect -> acidity increases as charge of molecule increases because the conjugate base that forms is often neutral which is much more stable than a - base. 2. Across periodic table acidity increases due to increased electronegativity of atoms which stabilize - charge of conjugate base better. -down periodic table -> acidity increases because atoms become larger and larger atoms have more space to delocalize - charge and become stable 3. Resonance - look for lone pair of electrons to be able to form resonance with pi bond nearby. -resonance --> increases stabilization of conjugate base because the - charge is delocalized and therefore strength of the acid increases 4. Electronegativity and inductive effects -electron withdrawing atoms (high EN) can increase acidity of nearby atom which increases more as acidity of EN atom increases and decreases in strength when EN atom moves farther from the acidic atom. -EN atoms pull electrons from partial - atom which helps to stabilize the conjugate base by spreading out - charge. 5. orbitals - increasing the S-character of a bond to H increases the acidic because it stabilizes electrons in orbitals better -s-orbital electrons are held more tightly than p-orbital electrons and thus are stabilize more effectively by + nucleus. -sp > sp2 > sp3 in order of increasing acidity

Addition in carbocations -CC involve what? -electrons are added to where on CC? -picking R to do what?

CC involve an empty p-orbital with a + charge -electrons are added into an empty p-orbital from either side of the "dumbbell." Hyperconjugation is stabilizing the CC -if we pickour R's (starting material) carefully we could get products that give us information into the reaction mechanism and help us support or refute the evidence of a carbocation intermediate. --> CC can create 2 different structures depending on where the group enters the p-orbital dumbbell

Properties of enantiomers vs diastereomers -enantiomers? chemical properties? reactivities? racemic idea? +difficulty? -diastereomers? chemical properties? reactivities? +advantage? -mirroring general idea?

Enantiomers --> have the same chemical properties (melting, boiling, IR and NMR signals, etc...) -in ACHIRAL environments they will react the same! Will react different in chiral environment. (react same in racemic mixture tho because there is enough achiral elements to react with) +hard to separate and isolate one from the other! Diastereomders --> have different chemical properties (melting, boiling, IR and NMR signals, etc...) -have different reactivities +easy to separate! -mirroring --> flipping 180 onto its face

hydrohalogenation -most commonly used halogens ? why? -flashcard idea? -steps?

H-X reaction in which H is H and X is a halogen (Cl, Br, I, or F) -Cl and Br are most commonly used halogens because F is too reactive and dangerous and I is not reactive enough (due to the weak C-I bonds it forms) -flashcards-> show P and R and predict other R +shows R and R and predict P +show R and P and predict other R -2 steps 1. pi bond electrons attack the H and H-X electrons go to the X -> forms CC. Check for rearrangement!! 2. X attacks the CC and bonds to it to form final product

conformational analysis and elimination reaction example? -idea behind displacing Cl with the CH bond after OH removes the H from the CH bond?

OH comes in and displaces the H from the molecules. -electrons from the CH bond then need a place to go. -antibonding orbital of the CCl bond lies directly coplanar to the CH bond. Therefore, the electrons from the CH bonds are able to occupy this CCl antibonding orbital and form a CC double bond (add pi bond). -Cl is then displaced because C has an exceeded octet and Cl will form the most stable anion. -when electrons come into the pi bond, the hydribization of C changes to sp2 and the orientation of the molecules slightly changes. -electrons in CH broken bond could attempt to displace the other groups via their antibonding orbitals; however, the antibonding orbitals do not lie anticoplanar to them and do not form stable leaving groups -H is only removed and Cl is only displaced because H in CH bond is anticoplanar to the Cl which allows the electrons to occupy the antibonding orbital and become a pi bond.

Halohydrins (alkenes)

Results in OH on C with more alkyl groups 3 main steps 1. pi bond electrons attack one of the X (halogens) and the electrons in the X-X go to the unattacked X. X that is attacked also counterattacks with electrons to form a pi complex 2. electrons on H2O come and attack the C with more alkyl groups; electrons in pi complex going to X then move from bond between C (being attacked by H2O) to X and X (in pi complex) fully bonds to the C with less alkyl groups 3. separate H2O attacks H in H2O bonded to C and electrons in H2O bond (on structure) move from O-H to O and H is released to attacking H2O to form H3O+ and C structure is left with OH on C with more alkyl groups and X on C with less alkyl groups. -consider stereochemistry for more detail!

Organic chemistry three ideas for practicality? How do they relate and what order do chemists often pursue them in?

Structure - Property (Function) - Synthesis -chemists often start by thinking of a desired function/property and then make predictions and test different structures. Once the structure that yields the desired structure is found, chemists must determine how to synthesize the compound and make it readily available. -ex: taxol - cancer drug ex: dimethyl fumerate - MS drug -ex: LSD - recreational drug

rate of formation of carbocation (CC) -transition state (TS) looks more like CC or R? -for scenario with CC, rate of formation is linked to what? -flow of rate idea with CC? -TS1? -Product determining step (PDS)? ex?

TS looks more like CC than R because it is closer in energy to CC than R (Hammond's Postulates) -for scenario with CC intermediate --> rate of formation is linked to the relative stability of the CC intermediate +more stable CC --> lower TS --> faster reaction and more products formed from that pathway -more stable CC -> more stable TS1 -> higher rate of formation --> more product resulting from that reaction pathway -TS1 is the TS of the RDS (in this reaction its the first step where H is attached) -PDS -> step determining which product is formed --> different from the RDS!! Based on carbocation intermediate formed -> + charge determines where the Br- will bond and structure that will form.

Bonding and electrons goal?

To know where the electrons are and where they can go? -different orbitals and what not.

formal charge

[# of valence electrons on atom] - [non-bonded electrons + number of bonds] -assigns charge to certain atoms based on the # of electrons that atom "owes" vs its valence electrons. 1. determine valence electrons for atoms 2. determine electron count: how many electrons the atom "holds 3. FC = [# of valence electrons on atom] - [non-bonded electrons + number of bonds] -only signify formal charge once: either directly next to the atom or outside the brackets

Lewis structure

accounts for or places all valences electrons in a molecule or ion. Valence electrons are ones that do chemistry -shows bonding and non bonding electrons -structural formula in which electrons are represented by dots; dot pairs or dashes between two atomic symbols represent pairs in covalent bonds. -shows connections in 2D and therefore structures that may appear as different are actually the same due to limited nature of lewis structure representation. -atoms on same plane -no information about WHERE bonds go, just the plane they are on!!

Drawing Newman projection cyclohexane

align C's on the left and right side -look at left and right side separately and then connect the middle -substituents are not shown in for the center C's... rotate if desired to put center C's on sides and show substituents. -use model kit!

lewis acids and bases

all about the electrons -acid --> molecule that gets attacked by (accepts) electrons -base --> molecule that attacks with (donates) electrons -electrons move from source to sink

Bronsted-Lowry acids and bases

all about the protons (H+) -acids --> donate H+ -bases --> receive H+ -AH + B- --> BH + A- -electrons in B attack the H in AH causing the AH bond to break. B then forms a bond with H forming BH and A retains the electrons from the AH bond and becomes A-. -AH is acid -B- is base

conformational analysis and penicillin example -amides and stability? -unstable rings ? -lone pair on nitrogen and potential displacement of O bond with resonance?

amides are very stable compounds; yet the amide in penecillin tends to be a bit less stable. -half life depicts amount of time to wear half of the bond away (more or less) -penecillin has a small square ring that is unstable and wants to break apart when possible which decreases the stability of its overall amide complex. -lone pair on N does not directly align with N lone pair and therefore the antibonding orbital is not a good option for the electrons of N to occupy. Also a double bond added to the N O bond in the ring would make the ring very unstable and unlikely to form!

carbocation and stability as number of C attachments increases? -why? -primary, secondary, and tertiary order of stability? -relate to rate of formation?

as the number of C's attached to the carbocation increases, the stability of the carbocation increases because there are more opportunities for hyperconjugation and therefore more electrons can spread into the empty + C p-orbital and help stabilize it by partially filling it. -connect relative stability of carbocations to the rate of formation of carbocations and stereochemistry of products... more stable carbocations are formed faster because they have a lower transition state energy which leads to them forming more than the alternative molecule. -primary -> carbocation with one C attached to it -secondary -> carbocation with 2 C's attached to it -tertiary -> carbocation with 3 C's attached to it. -most to least stable: tertiary > secondary > primary

chiral; occurs when? -achiral; occurs when? -testing?

asymmetry such that 2 objects are non-superimposable mirror images of one another; occurs when C has 4 different groups attached to it and is sp3 hybridized -achiral -if you make mirror image you actually have just recreated the exact same thing as original, superimposable on mirror image; atom does not have 4 bonds or have 2 of same groups attached to it. -test by trying to align mirror image with original molecule --> chiral if cannot align and achiral if can align

octet rule

atoms are most stable when they have 8 valence electrons in their valence shell -common exceptions: H and B and elements past P on the PT

Stereoisomerism in alkenes -how to change direction? -is not a what and why? alkanes vs alkenes ? -why cant alkenes rotate freely? -cis/trans terms ? -E/Z terms ? -relation of ideas to products?

change in directionality in 3D space. -usually fixed and must break bonds to change where things are facing in space -distinct isomers and are NOT conformations! Must break bonds to change directionality between them due to rigid pi bonds. -alkanes can rotate but alkenes cannot due to pi bond -cis (relative term) CH3 is cis to CH3 . while CH3 is trans to H -trans (relative term) CH3 is trans to CH3. while CH3 is cis to H absolute terms --> E/Z; E has high priority groups on oppositve side and Z has high priority groups on same side. Priority = size and weight generally -how will this stereoisomerism change to the products; use stereochemistry of product (in relation to reactant) to predict the mechanism it took to get there!

stereoisomers

changes direction of connectivity but not actual connectivity -maintain same order of atoms but change order that bonds point

relative positions on alkenes -cis and trans idea? problem? -ex? -solution?

cis -> same side trans -> opposite side -problem -> ambiguous and does not specify where groups are on group -ex (in picture) -> on left the X are trans to each other and the X is cis to the H + on right the X are cis to each other and the X is trans to the H -solution -> use the E/Z classification!

bonding orbitals

created by head-to-head or tail-to-tail overlap of atomic orbitals of the same sign and are energetically favorable -lower energy than original atom orbitals

antibonding orbitals -replacement of atom through utilization of antibonding orbital?

created by head-to-head or tail-to-tail overlap of atomic orbitals that have opposite signs and are energetically unfavorable -are opposite of bonding orbitals -always form for bonds and are higher in E than normal atomic orbitals -atoms attached to central atoms can be displaced when another atom comes and bonds to the central atom via the antibonding orbital created from the original bond. -ex: C and Cl have a bond that created a antibonding orbital opposite from the bonding orbital. Iodine (I) can come and distribute electrons to the antibonding orbital and possibly displace the Cl via the antibonding orbital if it is energetically favorable. -occurs if the formation of a new bond releases more energy than the energy absorbed from breaking the original bond... but only to limited extent and other factors must be considrered

Keq and delta G

delta G = -RTln(Keq) = -2.3RTlog(Keq) Keq = 10^(pKbh - pKah) where pKbh is the pKa of the conjugate acid and pKah is the pKa of the original acid) -delta G is Gibbs free energy. If - the reaction is spontaneous. If + the reaction is nonspontaneous -Keq is the equilibrium constant. If 1 the reaction is at equilibrium. If > 1 the reaction is product favored. If < 1 the reaction is reactant favored +Keq = [Products] / [Reactants]

Resonance hybrid: hybridization and distribution of electrons ?

different resonance structures depict different hybridizations.... raises question as to which hybridization occurs in the resonance hybrid? -depends on relative electronegativities and structure of hybrid -ex: CCNOH -> lone pair from N can interact with pi bond on the C to O double bond in hopes of spreading out the electrons and stabilizing the molecules. Interplays with antibonding orbitals and attempts to spread out electrons to decrease energy and increase stability. -penalty of close electron interaction can be offset by resonance stabilization. -important in biochemistry as polypeptides can have increased stability resulting from sp3 N orbitals having a trigonal type shape and properties.

Hyperconjugation in carbocation and stability: mechanism? -why does it increase stability? -H role? -how much is stability affected?

electrons in bonding orbitals like C-H that are connected to the C with the + empty p-orbitals can spread out their electrons into the empty carbon p-orbital... must be from CH's that are directly connected to the + C -this stabilizes the molecules because electrons are able to spread out and flatten their wave functions. -H cannot spread its electrons into the + p-orbital of the carbonation because it is to o far away... H cannot hyperconjugate!

racemate (racemic mixture) -alkene and Br attacking Carbocation example? why?

even 50/50 mixture of 2 enantiomers -Br attacking a carbocation (CC) forms an enantiomer pair of molecules; one has Br on top and other has Br on bottom; C with 4 groups must show stereochemistry +occurs because can attack the empty + p-orbital in the CC from either the top of the bottom ; no preference for top of bottom attack usually and thus both enantiomers usually form

Addition reactions that involve pi-complexes -examples? -usually what kind of addition and why? -what C does second group usually attack and why? -rearrangement? why? -how many steps? where is RDS? -reaction rates increase with? why? -why do pi-complexes usually form (and not CC's)?

ex: halogenation of alkenes (Br2 or Cl2) -halohydrine formation (adding Cl2 or Br2 in presence of ROH/excess solvent) -oxymercuration reduction (using Hg2+ to form pi-complex, then to attack water, and then as a reducing agent) -usually an anti-addition (added groups are opposite to each other) because first group (electrophile) "blocks" one face of the alkene and the second group must attack from the other side (anions are bulky)... shows stereoselectivity -second group (attacking group/nucleophile) usually attacks C with more alkyl groups... shows regioselectivity +rationale -> 2 C's have different amounts of + charge from the pi complex -> C with more alkyl groups have higher partial + than other C which attacks the - second group more via electrostatic attraction -avoid/no CC rearrangement since full CC is not formed -usually 2 step 1. pi bond electrons attack the X and the electrons in X-X bond transfer to other X. Attacked X also attacks the other C and forms pi-complex 2. X- attacks C with more alkyl groups and electrons in partial bond between attacked C and X (in pi-complex) go to X which fully bonds to the other C +first step is usually the RDS -reaction rate increases with greater alkene substitution because + charge in pi-complex can be better stabilized (via increased hyperconjugation) and therefore reduce the TS and increase reaction rate -pi-complexes usually form because the anion being attacked by the alkene is bulky and has excess lone pairs and thus "blobs" over side of alkene (unlike H which has no lone pairs and forms CC)

Absolute stereochemical assignment: E/Z system -helps solve what? -applicable to all alkene? -basic idea? E/Z ? -3 steps? consider what first?

-helps solve the cis/trans unclear problem -universally applicable to any alkene that can have stereoisomerism (each C of alkene has 2 different groups) -split molecule in half down the C=C and assign priorities to groups coming off of the C's in the C=C; "E is for extended apart from one another and Z is for zame zide" +E -> has high priority groups on opposite sides (usually more sable) +Z -> has high priority groups on same side (usually less stable) Use Cahn-Ingold-Prelog for assigning priorities/rankings to groups -> first consider the groups that are directly attached to the C's in the C=C 1. go 1 atom at a time 2. use atomic number (higher number = higher priority/rank) 3. break ties as soon as possible -if tie -> write both atoms and list the atoms attached to the atoms in order of highest priority. Go down the line crossing off atoms of same priority until you find an atom with higher priority... first atom with higher priority determines that group is the higher priority (ex in notes and phone)

Factors affecting Acid strength (4) -if same atom attached to H (2) ? -if different atoms attached to H (2) ?

If same atom attached to H 1. resonance stabilization --> if resonance stabilization is present than the molecule will be able to better stabilize the - charge that is present when the H+ leaves thus resulting in a stronger acid 2. Attachment of EN atoms elsewhere --> EN atoms attached elsewhere on molecule leads to negative charge being distributed more equally and inductive effects (+charge shields the H+ from returning) and thus increases acidity If different atoms attached to H 3. If atoms are in same period / row --> atoms farther to the right are more EN and stabilize negative charge better when H+ is gone and thus are more acidic. 4. If atoms are in same group --> atoms father down are larger and therefore have more room to stabilize negative charge and thus are more acidic -do NOT intertwine between groups and periods when comparing! -memorize pKa's on white handout sheet to be able to predict Keq using formula of Keq = 10^(pKbh - pKah) +pKbh is the pKa of the conjugate acid and pKah is the pKa of the original acid.

diimide mechanism ? -N2H2 reacts with alkene

N2H2 reacts with alkenes 1. alkene pi bond electrons attack the H; N-H bond electrons than go to the N=N to form an NN triple bond and other N-H bond electrons go to other C in alkene and H bonds to the C to form syn alkane and N2 gas with triple bond

Potential addition reaction mechanisms for alkenes with HBr example -things to consider for deciding which is best mechanism?

1. electrons from pi bond attack the H in HBr and bond with it forming a carbocation intermediate and Br-. Br electrons than attack the + carbocation and form a bond with it. -2 steps with carbocation intermediate 2. electrons from pi bond attack H in H-Br and bind to it. pi complex is formed as H hovers in center of alkene between carbons and "blobs" over them. Br- electrons than attack + area and form bond with C, H+ bonds to other C. -2 step with pi complex 3. electrons from pi bond attack H in HBr and form bond with it and at same time the Br- forms a bond with the other C -1 step with no intermediates -which is more plausible: consider which carbon is more highly substituted (has more C's attached) and therefore is faster. Also consider, regioselectivity

Addition reactions that involve concerted Syn-addition -ex? -are groups syn or anti and why? -steps in reaction? -what influences regiochemistry when two groups are different? -reaction rate?

ex: hydrogenation, hydroboration-oxidation, epoxidation, and ozonolysis (ends up looking weird) -syn addition because 2 connected groups are added to the alkene at the same time, and they cannot help but be added to the same face of the alkene -concerted -> "one step reaction" -> often associated with syn arrangements -when 2 groups are different -> electronic and steric factors influence the regiochemistry +electronics is often more important; during transition state C that is not binding originally to the less EN atom ends up gaining a partial + as electron denstiy changes... this partial + charge is on more highly substituted C and therefore this is where second group will end up binding because it is attracted by the partial +. -reaction rate tends to increase with greater alkene substitution -> more alkyl groups allows for more hyperconjugation and therefore stabilizing of the partial + charge that occurs during reaction due to unequal electron density in transition state

stereoisomers -conformational? -constitutional?

molecules that have the same pattern of atomic connectivity but differ in permanent direction in which bond points (3D arrangement in space) -if you dont show any stereochemistry then 2 drawings look the same; dash-wedge allows for the molecules to look different (vs the old flat projection) -stereoisomers --> break bonds and attach to same atom to form a new molecule ex: enatiomers -conformational -> same molecule, just rotated! -constitutional -> break bonds between molecules and change atom connectivity; atoms are connected differently!

Relative stability of alkenes based upon number of (C) substituents? -why? -does length of alkyl groups matter?

more C substituents/alkyl groups attached to the center C's --> more stable because there are more opportunities for hyperconjugation and therefore electrons can be spread out more and - charge can be delocalized leading to a flattened wave function and more stability. -length of alkyl groups plays a very minor role and slightly increases stability as groups get longer but not a lot! # of alkyl groups is way more important!

hydroboration oxidation mechanism -mechanism showing how BH3 converts to OH? -what converts the H2O2 to the anion to work the react? -how is partial positive initially stabilized?

on pic! -partial positive is initally stabilized by being on the C with more alkyl groups and the partial + occurs because of unequal electron density during the transition state +look in notes!

Reaction coordinate diagram

shows hypothesis of the energy changes that take place during the course of a reaction -reactants at start and products at end -transition states --> high points in middle -reaction intermediates --> low valleys in middle (between 2 transitional states) -activation energy (Ea) forward -> distance between reactants and transition state (could be multiple).... distance between products and transition state if (Ea-1 backwards reaction) -delta E for reaction is difference between R and P... relates to Keq and pKa because K = [P] / [R] -# of elementary steps = # of peaks / transitional states on diagram -rate determining step (RDS) is the step with the largest Ea (highest peak on graph) and determines how fast the reaction proceeds..... first reaction could include all reactants or some.

Resonance identification and focus -benzene with group coming off?

sp3 atoms are resonance stoppers! -be careful with lone pairs because resonance changes hybridization look for X with lone pair with single bond and then atom with a double bond. X (lone pair) - C = C look for double bond to atom to single bond to atom to double bond to atom. C=C-C=C-C-C -only can move electrons to next atom over -benzene resonance with bonded group coming off of it --> double bond moves to group going off and rearrange 2 double bonds within ring!

enantiomer -converting between pair? -ex? and what is not? -similarities and differences between pair? -determining with R/S?

specifically for a chiral molecule, each mirror image; if chiral --> set of molecules that are mirror images of one another but are different -each chiral molecule has one enantiomer to form an enantiomer pair -must break bonds and rearrange to convert between pair -ex: chiral molecules ! +NOT E/Z because mirror images are the same molecule -posses same energies and physical properties; difference occurs when interacting with other another chiral environment +ex: cannot put left foot in right shoe +ex: certain enzymes only accept certain chiral forms (R or S) -all R/S assignments will be different on each chiral C in the molecules

diastereomer -determining with R/S?

stereoisomers that are not enantiomers -pattern of connectivity is same but we dont have special pair of (non-superimposable) mirror images -atleast one R/S assignment will be the same on a chiral C between the 2 molecules

ozonlysis; basic mechanism

syn -concerted one step

epoxidation alkene

syn -concerted one step (full mechanism on another slide)

Ozonolysis mechanism -completely shown

syn! -concerted one step

Carbocation rearrangement -what to look for? when to consider? supports what? -what happens? -why does it happen? -how fast is the rearrangement step? -preference of shifting groups? -RCD? -ex: with HBr --> number of products formed and predicted mechanisms to get to those products? which product is most prevalent?

-look for formation of a primary or secondary carbocation by a C with more alkyl substituents +consider with all reactions that have carbocation intermediate! +evidence of rearrangement shows that CC intermediate was likely present -1,2 shift occurs and alkyl or H group shifts (basically hyperconjugation to extreme and group takes over) -happens because resulting carbocation after the shift is more stable due to its higher degree (often tertiary) because more hyper conjugation can occur and thus stabilize the positive charge better -rearrangement step is VERY FAST (low TS), usually faster than the addition of the halogen to the CC -Hydrides (H) shift faster than C-based groups (also want to leave C behind for more hyperconjugation) -RCD -> has 3 TS; 1 TS is the CC taking on the H with its pi bond electrons; 2 TS is the rearrangement of the carbocation to a more stable form; 3 TS is the addition of the final group/halogen to the CC -ex: HBr (hydrohalogenation) with alkene shows that multiple products form however the one with the most stable CC is the most prevalent --> less stable product could be most abundant because it had a more stable CC +main formations -> 2 normal ones from alkene additions plus 1 with rearrangements (other insignificant ones could form but not considered)

Dash-Wedge Notation

-solid normal lines represent bonds and atoms on the plane of the paper -dashed wedges represent bonds and atoms going away from you into the paper -solid wedges represent bonds and atoms coming toward you off of the paper -spatial arrangement in 3D

Meso compounds -determining with R/S?

molecules that are achiral overall because of their symmetry but also have chiral centers within the molecules; 2 compounds formed by flipping are actually the exact same molecule -must also consider the symmetry of the molecule and know that it is overall achiral with 2 or more chiral portions. -R/S assignments are the same on each chiral C when the C is mirrored +CONSIDER that molecule has flipped when comparing the C's and their R/S +for instance C1 is on left and C2 is on right, but when the molecule is mirrored C1 is on the right and C2 is on the left --> compare C1 with C1 and C2 with C2 for each molecule

3 products that can form between 3 different reaction mechanisms? -does same mechanism apply for all alkenes with specific reagent? -what can we do before we know mechanism and what gives evidence to specific mechanism? choose what kind of reactants? -what gives evidence about the mechanism?

1. Carbocation -> forms syn and anti addition -also has potential to rearrange -second group can enter from either side of the empty p-orbital 2. pi-complex --> forms only anti addition -side of C structure is blocked by large group blobbed on it and therefore the second group must enter from the opposite side 3. concerted --> forms only syn addition -reaction occurs in one step and therefore only a syn arrangement is possible -same reaction mechanism will generally apply for all alkenes when they react with a specific reagent. ex: if one alkene mechanism is C+ with certain reagent then other alkenes will react with that reagent via a C+ intermediate -we can only hypothesize before we know a specific mechanism. Experimental evidence and products formed gives insight into the reaction mechanism! +choose reactants that will form products that allow you to determine the mechanism at hand (ex: ring structures for syn, anti, or both and alkenes that can rearrange for C+) -both products and LACK of products gives information about the mechanism that occurred! +ex: lack of syn and only anti can show pi-complex

Expected products from 3 different potential reaction mechanisms using ring structure to determine hypothesis?

1. Carbocation intermediate -> forms 2 different P depending on where the group enters the empty p-orbital. Forms both anti-addition and syn-addition products -more likely both products formed from CC intermediate than from partial pi complex and concerted rxn mechanism 2. pi complex -> tends to give only opposite side addition (anti-addition) 3. concerted alkene addition (1 -step) -> tends to give only same side addition (syn-addition)

Drawing Lewis Structures Steps

1. Count total valence electrons. - # of valences electrons = group number 2. Layout atoms for symmetry - condensed structures convey info! 3. Commit one bond to connect atoms -recount valence electrons 4. fill octet outside in with lone pair electrons until you are out of valence electrons 5. satisfy octet rule with multiple bonds if needed 6. assign formal charge

Rules for resonance structures

1. Do not break sigma bonds! 2. Net charge must stay the same or you made a mistake 3. Maintain same number of unpaired electrons (usually). 4. Do NOT exceed the octet rule for 2nd row elements! -use little arrows to show how resonance structures convert between one another

Factors affecting carbocation stability

1. Hybridization of carbocation -what orbital is empty -carbocations are generally sp2 and are more stable in this state than in sp3. sp3 would from a very unstable carbocation (be careful with benzene, has 4 electrons in sp2 with + charge and 2 electrons in p orbital, and other molecules potentially forming it) -high to low energy of orbitals: p > sp3 > sp2 > sp > s -sp3 would have 6 electrons in sp3 orbitals and + charge there. -electrons are placed into sp2 orbitals and positive charge sits in empty p orbital when stable as sp2. 2.Adjacent atom's inductive effects -> how electron withdrawing atoms effect carbocation stability. -more electronegative adjacent atoms --> less stable because EN atoms pull electrons away from the + C and make the carbocation even more unstable 3. Adjacent orbital interactions --> hyperconjugation and resonance stabilization -tertiary > secondary > primary (C+ bound to one alkyl) rated most to least stable. CH bond electrons (from adjacent alkyl groups) can delocalize into carbocations empty p-orbital and help stablilize the molecule (hyperconjugation) -resonance stabilization leads to most stable molecule via delocalization of charge throughout molecule.

R/S naming system -4 steps? -R/S are? converting between them? -complex molecules? if H comes out?

1. Identify a chiral atom that has an sp3 center and 4 different groups attached to it 2. Rank 4 attached groups based on their priorities; 1 is highest and 4 is lowest -highest priority -> has highest atomic number -go more than one atom out to settle ties (look at atoms attached to tie atoms!) 3. Arrange molecule with C in front of the lowest priority group -often in a Newman projection or dash wedge projection 4. S -> priority of groups decreases counterclockwise R -> priority of groups decreases clockwise -R/S are the two different chiral molecules that make up an enantiomer pair; convert between them by changing connectivity of 2 of the groups bonded to the central atom -CIG (like E/Z) -works for complex molecules like penicillin as well! +just consider local chiral center one at a time and same process +if H comes out instead of going away from you -> can either flip the molecule and look other way or assign priorities without flipping and just change the priority to opposite because H comes out (Ex: if it clockwise with H going away it would be R but if it is clockwise with H coming out then it would be S)

Features impacting acid strength (4)

1. If atoms are same (XH and X loses H+) then look elsewhere for resonances stabilization. Resonance stabilization can result in delocalization of negative charge (electrons) in product or + charge in reactants -how does resonance influence species stability? -resonance stabilization makes conjugate bases more stable due to delocalization of negative charge and therefore increases the strength of the acid 2. If atoms are the same (XH and X loses H+) look for resonance stabilization. -if no resonance stabilization --> look for electron withdrawing (high EN) groups/atoms. Groups pull electrons toward them with inductive effects. -groups don't move electrons as much as resonance stabilization and therefore is not as strong as resonance stabilization 3. If atoms are not the same (XH vs YH) and 2 atoms are in the same group on PT, the larger atom forms a more stable anion. Moving down group, acid strength increases because conjugate base is more stable because it can delocalize negative charge better. -ONLY when in the same group 4. If 2 atoms are not the same (XH vs YH) and 2 atoms are in the same row/period then further right on PT leads to higher EN which leads better stabilizing of negative charge and stronger acid. -moving right on PT --> stronger acids -do not mix between groups and periods/rows! -pKa's on white sheet to memorize!!

alkene addition reaction -know? -assume? -want to know?

1. Know: C=C pi bond is broken and consumed. -H-Br bond is broken. -C-H bond forms and C-Br bond forms 2. Assume: completely establishes structure of the starting material and the product (connectivity, stereochemistry, and composition of mixtures). -can measure [ ] of reactants and products throughout and therefore can determine rate dependence and rate law -cannot directly observe or measure [intermediate] or transition state. 3. Want to know: what is the mechanism? -what factors determine product structure? -what factors influence reaction rate?

ozonlysis - what occurs? -2 steps you must indicate when showing basic answer? -2 different "reactors" you could use to form products? -where does 3rd O from ozone go?

forms a ring with 3 oxygen and 2 carbons; the carbon-carbon bond is then broken to form two separate molecules -in alkenes, the double bond is broken (down the middle for simplicity) and then O's are attached to either side of the double bond -must include 2 steps by arrow between reactants and products; 1. O3 and 2. Me2S (or Zn, H2O) -2 different "reactors" with O3 could be Zn, H2O or dimethylsulfate (Me2S) -3rd O from O3 goes with Me2S or with the Zn, H2O

Absolute stereochemistry? ex? -relative stereochemistry? ex?

gives definite relations between groups in molecule -ex: E/Z and R/S based on priority and order of groups -gives relative relations between 2 groups based on position compared to one another. -ex: cis/trans and syn/anti

formation and structure of carbocations? ex with HBr -how do they form simply? and with electron transfer between p-orbitals? -where could Br attack on + p-orbital? will this always result in the same molecule formed?

in 2 step alkene addition reaction, the pi bond electrons attack the H on the HBr and then the H bonds to the C with less alkyl groups with the Br- on the side in the first step. -C that does not bond to the H is left with a + charge because the double bond was removed so the other C could bond to the H. This molecule is called cation -electrons shared in pi bond transfer to one C p-orbital. These 2 electrons than attack the H and form a bond with it. The other C p-orbital is left with a + charge +Br could attack top or bottom of + p-orbital because both parts are empty! >based on there the Br attacks the p-orbital, different alkanes could be formed (conformations and arrangement of atoms)

Identifying chirality Centers -enantiomers?

look for sp3 carbon with 4 different groups attached to it -enantiomers -> non-superimposable on their mirror images! Chiral C; 2 groups will not line up when superimposed on mirror image and 2 other groups will likely line up -be careful with rings! -look more than one atom out to determine if entire groups are different from one another

Epoxidation alkene mechanism -syn or anti? -# of steps? -reactants? products? -common peroxy acid?

mechanism shown in picture -syn -concerted one step -alkene and peroxy acid form an epoxide plus a carboxylic acid -mCPBA

hydrogenation; syn or anti? -mechanism basics? -use of catalyst? -number of steps? parts of steps? -need H2 to do it?

metal is used as catalyst; commonly Pd/C (palladium on Carbon); concerted one step -H's add syn? -2 parts of one step 1. H2 bond is hard to break so H2 interacts with the Pd/C and H-H bond breaks and H's bond to the Pd/C surface. 2. Alkene then comes by and binds to 2 H's sticking out of the Pd/C surface -can be done without H2 and other groups as well!

pi bonds as bases in alkenes -basics: what attacks? stability depends on ? -3 things that could occur when pi bonds attack a molecule with + and - regions -what happens with electrons and pi bond when + region bonds to alkene ?

pi bond electrons attack a positive charge and stability depends on the amount of positive charge left (on the carbocation generally). 1. pi bond electrons attack + region and bond it to C with less alkyl groups --> - region remains unbonded to anything. Other C on product with + charge is attacked by Br- and forms bond. -2 step with carbonation intermediate. 2. pi complex - pi bond electrons attack positive region and positive region is very large so it sits in the center of the 2 pi orbitals and "blobs" over them with - region on side unbonded to anything. Resonance idea (not resonance) but unstable strained triangle. - region than binds to + carbon area and + molecule bonds to the carbon -2 step with intermediate pi complex 3. pi bond electrons attack the + region which binds to the C with less alkyl groups causing other C to gain + charge. - region immediately counterattacks with excess electrons (taken from + region when it left) and forms bond to + C. -1 step with no intermediate -2 pi bond electrons transfer to one p-orbital in the pi bond (made of 2 p-orbitals). this - region attracts the + region that binds to it.

Alkene addition reaction: basic mechanism? -major product features ? -minor products features? -reaction rate? -experiment not falsifying hypothesis? 2 reasons? -regioselectivity?

pi bond electrons attack the + H region and electrons in H-X bond go to the X. H then bonds to the C in the alkene with less alkyl groups (more H's). + carbocation then bonds to the -X group. X groups bonds to C with more alkyl groups. -major product --> has Br on the more highly substituted (has more alkyl groups) C atom of the alkene and H on C with less alkyl groups and more H's -minor product --> has Br on C with less alkyl groups and H on C with more alkyl groups... less stable cation! -reaction rate increase with alkene substitution (more alkyl groups attached to C).... more groups = faster because of hyperconjugation that occurs stabilizing the carbocation which by Hammond's postulates we know the more stable the carbocation the lower the transition state and thus the faster the reaction! -if experiment does not falsify the hypothesis either the experiment was poorly run or the hypothesis was not thorough and detailed enough! -bias of products to be in a certain form

stereocenter; test? -common types?

region of molecule that can give rise to steroisomerism; if you change permanent direction of bonds than you create a new molecule -test by breaking bonds and rearranging and seeing if it is a new molecule (can't be aligned with old molecule); bonds must be broken to convert between two molecules but atoms are still bonded to same center common types --> chiral C's and E/Z alkenes (must have 2 different groups attached to each C in the alkene) +E/Z alkenes --> break pi bond and rotate molecule to convert between E and Z +chiral C's --> break sigma bonds are rearrange atoms on molecule

constitutional/structural isomers

same molecular formula but different order in which atoms are arranged and put together -different connectivity... bonds are broken to convert between the 2. -different connectivities and different properties -look at each central atom and consider what was bonded to it and what has changed... naming

Conformational isomer

same molecule; some bonds have just been rotated -generated by rotating around bonds --> can only rotate around sigma bonds... cannot rotate pi bonds (or cyclic bonds?) -different energies because some conformations generate strain (less stable). Groups in molecule may come in different proximities. Repel is less stable and attract is more stable -different reactivities due to different placement of electrons -some conformational isomers are so rare that we don't consider them

Resonance structures

same structure with different arrangement of electrons -drawn with brackets and with double arrow between structures -delocalization of electrons --> molecule has more than one valid lewis structure that differ only in the distribution of electrons -no one structure is sufficient... must make hybrid of all structures -Concept idea -> contributing resonance structures help us describe true resonance structure of hybrid -does not jump between structures, not equilibrium, but rather a mix of both

carbocation rearrangement; what occurs? -when must you consider? -why is it favorable to change for secondary to tertiary? Reaction coordinate diagram and effect on rate? Which transition state affects the rate? -ex: with hydride shift and electrons? -hydride vs alkyl, which ones leads to major product if both lead to carbocation of same stability? More stable final product? -carbocation ring expansion? Most important energetic driver of 2 main ones? Occur if carbocation was originally tertiary? -what to consider when predicting most stable alkene product? Is H lost when reforming alkene?

the rearrangement of a carbocation to a more stable carbocation; group on carbocation shifts to form a more stable carbocation (often to change carbocation from secondary to tertiary or relieve strain in carbocation ring) -secondary to tertiary allows for more hyperconjugation and therefore more stabilty... slight transition state resulting from rearrangement but does is not the rate-determining step and therefore does not affect the rate. +only highest transition state affects the rate! -must consider if you have a carbocation intermediate in a reaction! -electrons in CH bond move to + C in carbocation and H movest too (H shifts spots) -hydride (H shifts) and alkyl (C group shifts) -> if both lead to carbocation of same stability the hydride shift will form the major product because it is an easier/faster reaction and uses starting material faster, forming more. +alkyl shift will lead to a more stable final product (tetra-substituted alkene) than hydride shift (tri-substituted alkene) but will be present in lesser amounts still because earlier hydride shift in mechanism occurred much faster than alkyl shift! -ring expansion -> changes carbocation from secondary to tertiary to stabilize it with increase hyperconjugation; however, the main energetic driving force in a ring expansion if the release of ring strain (ex 4->5 membered ring)!! +reaction would still occur if original carbocation was tertiary because the main driving force is the release of strain from the ring! -when predicting major products with alkenes -> consider the amount of substitution in the alkene. More highly substituted alkenes are slightly more stable due to higher amounts of hyperconjugation in p anti-bonding orbitals +H+ is lost when reforming alkene (lost from carbon that would lead to more highly substituted and stable alkene forming)

Drawing Newman projections for acyclic molecules

think about looking down axis of projected bond -look at it straight on! -2 atoms line up and one is behind the other -indicate front atom with a dot -indicate back atom with a circle -most stable when groups are separated as much as possible to reduce repulsion -energies are different due to different lengths between atoms -eclipsed --> less stable than staggered and depicted with both groups showing, however one group truly is right behind the other

SP hybridization and resonance steps (steps)

to determine which hybridization is most likely given different resonance structures.... 1. draw out all resonance structures and label what hybridization each structure "looks like" based on regions of electron density. 2. pick resonance structure with lowest sp count --> most stable via resonance stabilization (ex: pick sp2 over sp3).