Joseph Mordarski9/13/2007

Organic Chemistry Dr. Erb

I. Orbital = house of electrons where the e’s reside; also called electronic cloud and subshellsa. There are mainly four types of orbitals

i. S, P, D, Fii. --------- energy

II. Where do these orbitals start?a. S starts from first orbitb. P starts from 2nd orbit (no 1p)c. D starts from 3rd orbit (no 1d, 2d)d. F starts from 4th orbit (no 1f, 2f, 3f)

III. How many electrons max can these orbitals have?a. S = 2, P = 6, D = 10, F = 14

i. Each one goes up by +4b. Pauli’s exclusion principle: an orbital can have only two electrons with

opposite spinc. Shapes

i. S = golf ball/tennis ball sphereii. P = dumb bell with node in the center where the electron density is

zero

IV. Order of filling electronsa. 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 5p, 5d, 5f … etc b. 26 Fe = 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d6

c. Total of 26 electrons: outermost electronic cloud is 4s2, therefore number of valence electrons is 2

d. 17 Cl = 1s2, 2s2, 2p6, 3s2, 3p5e. VE = 3s2 + 2p5 = 7f. note

i. for all group A elements, the group # is the # of valence electronsii. For He, only 2 e’s are in the outermost orbit

V. Octet Rule – to have eight electrons in the outermost orbit is the most stable configuration. All near-by elements will try to get that configurationa. Element: Cl Ar Kb. Atomic# 17 18 19 LOTTOc. Config: 2,8,7 CLOSE 2,8,8 PERFECT 2,8,8,1 CLOSE

STEAL FROM THE NEIGHBOR

VI. Ionic Bonding – in this type of bonding, one or more electrons are completely transferred from one atom to another atoma. Example of Ionic Bonding – 11Na + 17Cl Na (+) Cl (-)b. 2,8,1 2,8,7 2,8 2,8c. Na: 11-10 = +1 Cl 17-18 = -1 (+1) (-1)d. These bonds are called ionic because they form charged particlese. NOTE: because of the transfer, both elements get octet configuration

i. Product is ionic compoundii. Usually, these types of compounds are formed between elements

of 1A or 2A (ready to give e’s) to 6A 7A (ready to take electrons)

VII. Covalent Bonding - are formed by SHARING of electronsa. When two electrons are shared, a single bond is formed b. When four electrons are shared, a double bond is formedc. When six electrons are shared, a triple bond is formedd. BE CAREFUL WITH WORDING

i. 4 electrons = 2 pairs… duhii. 3 pairs = 6 electrons, DUH again

VIII. Rules for writing covalent bonds (Lewis)a. Count the total number of valence electrons

i. Example: CO = 4 + 6 = 10b. Write a skeleton structure, using non-hydrogen, non-oxygen (‘cept with

water) as central atomi. C-O

c. Count the total used and unused electronsi. CO: Used: 2, unused, 8

d. Distribute the unused electrons two at a time on each atom starting at each end

i. Note: 1. Do not put any electrons on Hydrogens2. Do note put more than eight electrons around any one atom

Ex. C—OTotal: 10 XX XX

Used: 2 C—O XX

Unused: 8 XX

e. Check whether all atoms have octet configurations. If any do not have an octet, convert a long pair of electrons into a fucking double or triple bond, bitch.

Ex. C—O

i. Convert two lone pairs on oxygen to two pi-bonds, forming a triple bond and satisfying both octet rules

XI. How to write the lewis structure for an ion?a. NOTE:

i. (-) ion means MORE electrons (add electrons)ii. (+) ion means LESS electrons (subtract electrons)

1. DUH.iii. SO4 ^-2 (negative two ion)

1. Count electrons: 32 electrons2. Count used: 8 electrons used3. Count unused electrons: 24 unused4. Draw skeletal structure5. Distribute in pairs6. You get this, bitch:

IX. Electronegativity (EN) – The ability of an atom to attract a shared pair of electrons to itselfa. A : B – Even electronegativity; no chargeb. A: B – A = more EN, more electron density delta – charge on A and

delta + charge on Bc. A :B – B = more EN = same thing opposite wayd. Due to the difference in EN, there will be some charge separation

i. The more EN atom becomes slightly negativeii. The less EN atom becomes slightly positive

X. Polar Covalent Bondsa. The covalent bond may be 100% covalent of partially polar due to EN

differenceb. Question: State whether the following are 100% covalent or partially

polar?i. Cl 2

ii. HCliii. O 2 iv. N 2 v. H2SO4

XI. Formal Charges (FC) – A formal charge on a particular atom can be calculated as followsa. Fc = (Group #) – (# of non-bonding electrons) – (1/2 shared electrons)b. What is the formal charge on this molecule’s oxygen atom?

XX

c. 6 – 2 – 3 = +1d. Hydrogen formal charge = 1 – 1 = 0; neutral formal charge

XII. Resonancea. Note: Resonance contributors are not real, only the resonance hybrid is the

real structureb. All resonance structures do not contribute equally to the resonance hybridc. Resonance hybrid is an in-between structure

XIII. Empirical formula (EF) – it is the shortest formulaa. Example: C6H12O6 = molecular formulab. C6H12O6 ‘s empirical formula is CH2Oc. How to get EF

i. Divide the given # by the atomic weight of elementii. Divide results by smallest #

iii. Make everything whole #’s

XIV. Line angle formulasa. Carbon atoms are not usually shown; implicit at anglesb. Hydrogen atoms bonded to carbon are not shown; hydrogen attached to

any other element is shownc. Around carbon, if anything is missing, it is hydrogend. All other atoms are shown

There are six carbons, six hydrogens in benzene’s aromatic ring

XV. Acids and Bases

Type Arrhenius Bronsted Lowry Lewis

Acid: Gives H+ Gives H+ Accepts e-

Base Gives OH- Accepts H+ Donates e-

a. The higher the pka, the more basicb. The lower the pka, the more acidicc. Duhd. Note: pka can be used for bonds also; pka goes up to 38e. Predicting acid/base reactions from pka

a. pKa of HA (leading component of first reactant) should be lowestb. ex 1

i. K-C=C-H +OH- - H-C=C- + H2Oii. PKA= 25 PKA = 18,74

iii. Will this reaction proceed?iv. No, because the pKa of HA (leading reactant) is not the highest in

the Rxn

c. ex IIi. CH3COOH + OH- - H2O + CH3COO-

ii. PKA = 8.2 PKA = 15.74iii. Will this reaction proceed?iv. Yes, because the pKa of HA (leading reactant) is the highest found

in the reaction

XVI. Conjugate Acids/Basesa. An acid and its conjugate base differ only by a H+

i. Acid Con Baseii. HCl Cl-

iii. NHO3 NO3-iv. NH3 NH4+ or NH2-v. H2SO4 HSO4-

b. Rules to remember for conjugate acids/basesi. If the acid is strong, its conjugate base will be weak

ii. If the acid is weak, its conjugate base will be strongiii. Goes for bases tooiv. Strong acids:

1. HCl HBr HI2. Acids in ascending strength3. Cl- Br- I-4. Conjugate bases in descending order of strength

CHAPTER II

I. Remember orbital shapes:

II. Molecular Orbital: If the orbitals of two different atoms interact, a molecular orbital is formeda. Used for bond formation between atoms

III. Atomic orbital- If orbitals of the same atom interact, an atomic orbital is formed a. this determines the shape of the molecule

IV. Sigma bonding (σ) – direct overlap of s or p orbitals. All compounds have σ bonds. They are much stronger than pi bondsa. Methane has ONLY single sigma bonds

V. Pi bonding – When p orbitals laterally overlap, a pi bond is formeda. Pi bonds are only in double and triple bonds; never in singleb. Pi bonds are weaker than sigma bondsc. Methyne has two pi bonds and three sigma bonds

VI. Hybridization – orbitals with different energies mix to give a set of orbitals with identical propertiesa. Like a milkshake because you put a bunch of stuff in it that’s different at

first and when you take it out of the blender its all one consistencyb. Note: if total of “n” orbitals with different energies take part in the

hybridization, then “n” # of hybrid orbitals with identical properties are formed

c. For Hybridization purposes TREAT DOUBLE AND TRIPLE BONDS AS SINGLE BONDS

d. Examples:i. One S + two P = three Sp2

ii. One S + One P = Two Spiii. One S + three p + two d = six sp3d2iv. One S + three p + one d = five sp3d

e. Bonds around central atom Hybridi. 2 Sp

ii. 3 Sp2iii. 4 Sp3iv. 5 Sp3d (rare)v. 6 Sp3d2 (rare)

f. Lone pairs count as single bonds for hybridization purposesg. Remember that H’s aren’t shown in line angle formulas

VII. Shapes

Number of Atoms Around “Central

Atom”Shape  

2Linear

180 degree angles

 2 Bent (2 lone pairs)

3Trigonal pyramidal

(1 lone pair)120 degree angles

 4Tetrahedral

No lone pairs109.5 degree angles

I. Isomers – are different compounds with the same molecular formula are called isomersd. Example: glucose and fructose the same molecular formula C6H12O6

but are different compoundse. Types

i. Structural ( ) isomers – they differ in connectivity of atoms or groups

ii. Stereo isomers- These isomers are seen around carbon = carbon double bonds or in a ring where free rotation is not possible

iii. Cis = the same atom or group are on the same side (at least one set)iv. Trans = The same atom/group are on opposite sidesv. Not Isomers:

vi. No double bond usually means constitutional isomer

II. Partial Polar- How to predict whether a compound containing more than two atoms is partially polar or 100% covalent1. Know structure2. The following will be 100% covalent

a. AX4

b. If it is a trans isomer where same groups are on opposite side, then it is 100% covalent

c. Ex: Forces cancel each other out

III. How to measure partial polarity1. Dipole Moment (Mew) = little delta (charges) times d (distance between charges)

a. Note: if dipole moment is zero (mew = 0) then it is 100% covalentb. If mew is not equal to zero, it is partially polar; higher the value of mew,

the more polar it is, the lower the value of mew, the less polar it is

IV. Hydrogen bonding – it is an intermolecular attraction between two moleculesA. Intra = within one moleculeB. Inter = between other molecules C. Effect of Hydrogen bond

a. Because of hydrogen bonding, water is a liquidb. Q. Why is H2O a liquid while H2S, H2Se and H2Te are gases even though O,

S, Se and Te belong to the same group (6A)?i. H2O forms hydrogen bonds due to its electro-

Ngativity and relative sizeii. Attempted hydrogen bonding between the larger molecules fails due to length of bond

D. Those solvents that form hydrogen bonds are called protic solventsE. Those solvents that don’t form hydrogen bonds are called Aprotic solventsF. Protic solvents have higher boiling points

a. Look for –OH and –NHb. The more –OH’s and –NH’s, the higher the boiling points will bec. Question: Which has the higher boiling point?

i. CH 2 CH 2 OH ii. CH3OCH3

iii. CH4

1. Hydrocarbons – are compounds that consist of only hydrogen and carbona. Alipathic hydrocarbons – do not contain any aromatic rings

i. Note: it may contain aliphatic ringsb. Aromatic hydrocarbon – Contains aromatic ring (like benzene ring)

Alipathic Cyclohexane Aromatic Benzene

2. Alipathics are divided into Three groupsa. Alkane – fully saturated, no C = C double or triple bonds

i. Chain formula = CnH2n + 2

ii. Cyclo formula = CnH2n

iii. No functional Group

b. Alkene = At least one C=C double bondi. Chain formula = CnH2n

ii. Functional Group = double bondiii. From 4 carbons on, the position of the double bond must be

mentionediv. Always give smallest # possible for double bond

1-Butene 2-Butene (Trans) 2-Butene (Cis)

v. Cyclo formula = CnH2n – 2

vi. if There is only one double bond in a ring, it is always on the #1 Carbon Atom

#1 Carbon

c. Alkyne = C – C TRIPLE Bondi. Functional group = triple bond

ii. Follows same numerical rules as Alkenes

I. Functional Groupsa. Alcohol: R – OHb. Ethers: C – O – C “Oxygen trapped between two carbons with single bondsc. Carbonyl Containing Compounds: Carbonyl is R-C=O-Z; Depending on what Z is

dictates which carbonyl containing functional group you’re looking atd. Aldehyde: When you have a carbonyl attached to a Hydrogen (Z = H)

i. R – C double bond O attached to Hydrogen

f. Ketone – When you have a carbonyl attached to another R groupa. R – C double bond O attached to another R group:

g. Carboxcylic acid: When you have a carbonyl attached to anotherR group Z = OH

a. R – C double bond O single bond to the carbon OH

g. Ester: When you have a carbonyl attached to an O – R group a. R – C double bond O another O attached to C single bond R group

h. Amide – When you have a carbonyl attached to a Z = NH2 a. Yea just look at the pic

alkene

alkyne

arene

halide - chloride, fluoride, bromide, iodide

alcohol

ether

amine

nitrile

nitro

Sulfide

{sulfoxide, sulfone}

thiol (aka mercaptan)

aldehyde

ketone

carboxylic acid

ester

amide

acid chloride

acid anhydride

I. Alkanes – a. General formula straight chain alkane: CnH2n + 2

b. General formula for cycloalkane: CnH2n

II. Nomenclature Rulesa. 1. Find longest continuous carbon chain and choose it as the parent (or

backbone)b. If there are two possibilities with the same length of carbon chains, use the

chain that contains the greater number of subtenantsc. From which end the number of carbon atoms starts:?

i. Whichever way gives the substituants the slowest possible numbers

d. Write name of substituants followed by the parenti. Note: alphabet before numerical position, ALWAYS

e. If the same substituants numbers are obtained in both directions, the lower alphabet gets the lower #

III. Properties of alkanesa. Note: as the number of carbon atoms increases, the alkanes transfer from

gases liquids solidsb. Melting points: Of two similar-weighing alkanes, the alkane with the even

number of carbon atoms has a higher melting point than the alkane with an odd number of carbon atoms

c. A branched alkane has a higher melting point than a non-branched alkane with the same molecular weight

IV. Conformations – different arrangements formed by rotation about carbon – carbon single (sigma) bondsa. Called conformersb. Cannot be isolated in a lab (because they’re constantly rotating)c. Example: ethane

i. Eclipsed = least stable conformer = 0 degree dihedryal anglesii. Staggered = most stable conformer = 60 degree dihedryal angle

I. Properties of alkanesa. Note, as the number of carbon atoms increases, the alkanes transfer from

gases liquids solidsb. Melting points – Alkanes with even number of carbon atoms and have the

same molecular weight have higher melting points than odd number carbon atoms

c. C15H32 has a lower melting point than C14H30 d. Branched alkanes have a higher melting point than non-branched alkanes

with same molecular weight

I. Nomenclature for Cyclo Alkanes a. If cyclo alkane and longest side chain are equal in the number of carbons

they have, the parent is the cyclo alkaneb. If longest chain has less carbons than the ring, the cyclo alkane is the

parentc. If longest chain surpasses the number of carbons that the cyclo alkane has,

the parent is the longest chaind. In mono substituant cyclo alkanes, the substituant gets the #1 positione. If the cyclo alkane is substituded twice by the same substituant, just go for

lowest possible numberingf. If substituants are different, lower alphabet gets lower position; like ethyl

beats out methyl

*When you write, finally use alphabetical order!*

II. Stability of cyclo alkanes

a. Bond Angle Strain = 109.5 – the angle in structure (abs value)b. Cyclo-propane: 109.5 – 60 = 49.5c. Cyclo-butane: 109.5 – 90 = 19.5d. Cyclo-pentane: 109.5 – 108 = 1.5e. Cyclo-hexane: (ideally) 109.5 – 120 = 10.5

III. Cyclo-Hexane sucks because it’s not planer; it is 3-dimensional

a. Red bonds = axialb. Blue bonds = equatorialc. Note:

i. Axial bonds go up and down alternating in adjacent carbon atomsii. Equatorial bonds go up and down alternating in adjacent carbon

atoms too, but they’re more at an angle (blue)iii. All bonds are STAGGERED; 109.5 degree bond angles across all

so angular strain = 109.5 – 109.5 = ZERO (0)iv. Chair forms are the most stable form of cyclohexane due to the

zero bond angle strain

I. Other conformers of cyclohexanea. Boat Form (less stable than the chair)

b. Why is boat form less stable than chair form i. Flag pole hydrogens on C1 and C4 are too close together (and

since they “hate each other”)1. Note: in between the chair and boat form, there are some

other conformations possiblea. Skewedb. Half chairc. (we don’t care really about those)

c. Conformers of mono-substituted cyclohexane:i. Rules

1. A substituant in equatorial position is more stable2. If there are more than one substituant, the bigger

substituant in equatorial is more stable.

equatorial methyl-cyclohexane is more stable than axial methyl-cyclohexane

II. Di-substituted cyclohexanea. Two Categories

i. When both substituants are the same, the most stable is when both are in equatorial position, then one equatorial one axial then the least is both axial

ii. When one substituant is bigger than the other, the most stable is when both are equatorial, next is when the bigger one is equatorial and the smaller one is axial, next is when the smaller one is equatorial and the bigger one is axial, and the least stable is when they’re both axial

b. Figuring out Cis and trans trick:

A(xial) U D U D U DE(qautorial) D U D U D U

Use it for:

A(xial) U D U D U DE(qautorial) D U D U D U Therefore it’s CIS

Chapter 4

I. Activation Energy – Ea, the extra energy needed to start the reaction:

a. Think of activation energy as a speed bump, the smaller the activation energy, the faster the reaction will proceed; the bigger, the slower

a. Note that a transition state is different from an intermediate because a transition only exists for maybe a picosecond while bonds are being broken/formed

b. Primary, secondary, tertiary carbona. Primary carbons are attached to only one other carbonb. Secondary carbons are attached to two other carbonsc. Tertiary carbons are attached to three other carbonsd. Qutinary carbons are attached to four other carbons and are really raree. Likewise a primary alcohol is one that is attached to a primary carbon

c. Amines – children of ammoniaa. Primary amine = RN:H2 Highest Boiling point due to NH bondsb. Secondary = R2N:Hc. Tertiary = R3N:

d. Carbo-cations: C+ ionsa. Primary Carbo-Cation same idea as carbonsb. Stability = Tertiary > Secondary > Primary > Methylc. Why? Alkyl groups are electron-donating groups

e. Carboanionsa. Note: carbon anions are used as nucleophilesb. Stability is opposite from C+ ionsc. Methyl > Primary > Secondary > Tertiary

f. Radical – in the outermost orbit, there will be only seven electrons (neutral/no charge)

a. Denoted by a single dotb. Stability follows carbon: Tertiary > Secondary > primary > methyl

I. Vinyl carbons – are carbons that are involved in C=C double bonds and are Sp2 hybridized.

II. Allylic carbons – are neighboring carbons that are involved in a C=C double bond and are Sp3 hybridizeda. Allylic cations are stable because they are stabilized by resonance!b. Allylic Radicals are also stabilized by resonance

III. Use of NBSa. Rule: if you see NBS, replace hydrogen with bromine in the Allylic

positionb. Allylic bromination is a radical reaction:

c. NBS = N-bromo saccinimide

IV. Types of organic reactionsa. There are mainly four types of reactions

i. Addition reactions – In this reaction, a group or atom adds to the substrate (reactant) to give a single product. This usually happens in pi bonds (alkenes/alkynes)

ii. Substitution reactions – is a reaction where one atom/group is replaced by another atom or group; usually happens to Sp3 hybridized carbons

iii. elimination reactions – Elimination reactions – a group is eliminated from a compound. Usually, in elimination reactions, a hydrogen atom and a halogen atom are eliminated from neighboring carbons

iv. rearrangement reactions – this reaction takes place when atoms or groups move from one place to another

1. Note: the rearrangements are usually seen whenever a carbo-cation is formed (C+)

v. Bond breaking – is taking place in reactants; it is an endothermic reaction delta H = positive (+)

vi. Bond formation – is taking place in products; it is an exothermic reaction; delta H = negative (-)

vii. Bond Breaking Has two types:1. hemolytic cleavage (even divorce), symmetrical cleavage

No charges develop USE HALF HOURS

2. Heterolytic cleavage – uneven divorce, unsymmetrical cleavage

Polar reactionsWith ChargesUSE FULL ARROWS

viii. There are two types of bond forming1. Homognic – (opposite of hemolytic cleavage) each radical

gives one electron for bond formation2. Heterogenic (opposite of heterolytic cleavage) one atom

(region) gives two electrons for bond formationa. Note this is the basis for all chemical reactions

ix. Nucleophile – (nucleus-loving) – is one with higher electron density; nucleophiles start reactions by attacking electrophiles

1. Note: nucleophiles attacks where e- density is the lowest2. Nucleophile can be the negative ion or neutral

a. CH-, NH2-, Cl-, F-b. Neutral nucleophiles must have lone pair (like

H2O:)

x. Electrophile (electron-loving; deficient in electrons) – These are electron deficient; they want to accept electrons from nucleophiles

1. An electrophiles can be a positive ion or neutrala. Positive: C+, Cl+, I+

xi. Nucleophilicity – the power of nucleophilie to attack an electrophiles is called nucleophilicity

1. Note: all nucleophiles are not of the same strength, The power of attack depends on many factors like solvent, basicity, etc

2. Solvent = “home field advantage”3. Structure and acidity, basicity, etc EN: and UP4. ----------------------------------5. C N O F EN increases

xii. CH4 NH3 H2O HFxiii. ---------------- Acidity is increasing! (following electronegativity)xiv. Con bases work the oppositexv. CH3- NH2 OH- F-

xvi. -------------Basicity is increasing in the opposite manor, not following electronegativity

xvii. Nucleophile and basicity – The stronger the base, the stronger the nucleophile = faster attack

1. Q: which is stronger nucleophile CH3- OH-2. Halides: HI > HBr > HCl > HF3. Strength -----------------------4. note: this is contrary to electronegativity!! HALIDE

ACIDITY DOESN’T FOLLOW ELECTRONEGATIVITYxviii. con bases: I- < B- < Cl- < F-

1. ----------------- Strongest2. Cl- + electrophiles R – I leaving group (Cl – is more

powerful than I)3. Note: stronger bases are stronger nucleophiles4. Weaker Bases are better leaving groups

I. What is spectroscopy?a. Tools to figure out structure of compoundsb. Wavelength, denoted by lambda which is an inverted y, is defined as the

distance between two adjacent peaksi. Units cm, nm, m (distance)

c. Frequency denoted by Nu, is the number of times a full wave crosses a middle point in one second.

i. Units are cycles/second or hertz

II. What happens when a molecule is attacked by…a. UV Ray – It destroys the moleculeb. IR ray – it vibrates the moleculec. Microwave – it rotates the molecule

i. Polar molecules rotate faster hence why you can’t stick metal in the microwave

d. Vibrations in IRi. There are different kinds of vibrations possible

1. Stretching2. Bending3. Rocking

ii. If a non-linear molecule has n # of atoms, then generally there will be [3n -6] vibrations

iii. 3n – 6 for Methane (CH4) 5 atoms [5*3 – 6] = 9e. IR active or IR inactive – If a molecule is symmetrical and has a dipole

moment of zero, then that compound will be IR inactivei. Example: H – C = C – H where the = is a triple bond = inactive

ii. But H – C = C – CH3 is IR active at about 2200 inverse cm

Higher frequency, Lower Frequency Higher energy Lower Energy

III. Note (molecular vibrations)a. In a group of similar bonds, those bonded with heavier atoms absorb at

lower frequencies, the right sideb. The bonds with lighter atoms absorb at a higher frequency, on the left of

the IR spectrai. C-H will absorb to the left of C-D because C-H’s C is bonded to

the lighter H than Dc. Stronger bonds absorb at a higher frequency (LEFT SIDE)d. Weaker bonds absorb at a lower frequency (Right Side)

C triple bond C C = C C – Cstrongest, furthest right-----middle right-----------weakest, furthest left

IV. Common absorption patterns for IRa. C – C @ 1200b. C = C @ 1600c. C triple bond C @ (Sp2) 2200d. C = C – H @ 3000e. C triple bond C – H @ (Sp) 3300f. R – O – H strong @ 3300g. Acid strong and broad @ 3300h. Amine = sharp V @ 3300 (N – H)i. Alkane = multiple stretches at like 3300j. Carbonyl compounds R-C=O strong absorption at 1700 if they’re involved

in resonance k. Conugation – alternate carbon, carbon or carbon oxygen single and double

bonding gives extra stability

l. Carbon triple bond N comes around 2200 cmm. Plus points of IR

i. Very easy to find functional groups like alcohol, acid, carbonyls, amines, double bonds and triple bonds

ii. Note: by the absence of these peaks it can be assumed that they are absent

n. Minus points of IRi. Only the functional group can be known, but not the exact

compoundii. If a carbonyl is present, it may be many different things

I. NMR – Nuclear magnetic resonance1. Certain nuclei have a spin of + ½ or – ½

a. These can show NMRi. Proton

ii. C-13iii. F-19iv. P-31

2. Why not H2 or C-12? Unfit because to be an NMR candidate, the # of protons should not equal the number of neutrons

3. Theorya. In the absence of an external magnetic field, the nucleolus spins are

randomly oriented b. In the presence of an external magnetic field, the plus + ½ align along

the external magnetic field clockwise c. Negative ½ spin align against the external magnetic field, counter

clockwised. When a sample is subjected to an external magnetic field, some nuclei

absorb energy and go to beta spin state. This is in radio frequency region (Rf)

e. These absorb energy and when they come back, they emit signals different groups emit different signals (different Rf)

f. All these signals are recorded by the machinei. Fowier transfer (FT) math-based computer program that converts the

data from intensity vs time to intensity vs frequency

4. Shielding – middle men = shielding a. A nucleus is surrounded by an electric cloud. It shields the nucleus from

the external magnetic field. If shielding is more, the nucleus experiences less external magnetic field and vise versa

b. Beta effective = betaapplied – beta local

a. Note: if the shielding is high, less magnetic field reaches the nucleus

b. If the shielding is low, more magnetic field reaches the nucleus

He – High Frequency Lou – LOW frequency Laid – Less shielding / LEFT Messed – MORE shielded Down – DOWN FIELD Up – UP FIELD RIGHT

Up and down field are opposite of up and down town

I. The number of signals in H-NMRa. Chemically equivalent protons = protons that are in the same chemical

environment are chemically equivalentb. NOTE

i. Each set of chemically equivalent protons give ONE signal eachii. Look for symmetry!

c. How many signals are there:i. CH3CH2CH2Br NO symmetry! 3 signals

ii. CH3CH2CH3 Symmetry! Two signalsd. How many sets are there? NOTE:

i. The sets are named A, B, C etc starting with MORE SHIELDED protons

ii. CH3 is more shielded and comes up field, so CH3’s will always be A if they’re present, do them first

iii. CH2 is slightly downfield of CH3, do them secondsiv. CH is least shielded and comes more downfield to the right, do

them last

v. Protons close to electron-donating group like alkyl group move up-field and are more shielded

II. Chemical shift (little delta) measured in ppm is the distance downfield from TMS divided by the operating frequency of the machine you’re usinga. Reference: TMS tetra-methyl-silane; this is taken as the standard and its

chemical shift is (0) zerob. CHCl3 is Chloroform c. CDCl3 is deuterated chloroform w/a deuterium instead of a normal

hydrogend. Note:

i. Units are hertz/megahertz or ppmii. Chemical shifts are independent of the machine

e. Question: If a methyl group of your compound comes at 1 ppm in a 60 mHz machine, where will it come in a 400 mHz machine?

i. 1 ppm because chemical shifts are INDEPENDENT of the machine you use and NEVER change, ever.

I. Mass Spectroscopy (MS)a. Theory: molecule is bombarded with electronsb. Molecule (e-) M- (Radical cation) + 2e- + mad fragmentsc. All these fragments are analyzed and the recorder records it

d. Neutral species (radicals) are pumped out; radicals don’t reach recorder; only plus cations reach recorder

e. remember that big molecules, like mac trucks, have to make wide turns

II. Mass / charge (charge on electron) = only mass because charge always = 1III. Base peak = tallest peak (or signal); it is considered 100%

a. Note: base peak need not be the positive cation of what you’re looking atb. Exception: this may be M+1 or M + 2

V. Isotopes in mass spectraa. (M+1) peak – it occurs due to C-13

i. Uses of M+1 peak?ii. Ratio of C-12 to C-13 = 98.9 to 1.1

b. By measuring the intensity (height) of M+1 peak, we can find out how many carbons are in the molecule

c. # carbons = (intensity of (M+1) Peak) / (.01 times intensity of M Peak)d. Always round this result e. Note: this calculation does not work if there is nitrogen in the moleculef. Uses of (M+2) peak

i. If M+2 peaks are there, then the molecule will have a Cl, a Br or a S

ii. By looking at the M to M +2 peak we can have a good understanding of whether it is Cl or Br

iii. 3:1 M to M + 2 ratio is indicative of Cliv. 1 : 1 ratio of m to m + 2 is indicative of Br

g. presence of Iodinei. peak @ 127 is indicative of the presence of iodine (I)

h. Nitrogen rule – if the M+ is an odd #, then the molecule may only have an odd # of nitrogens

i. Eg. CH3CH2N = ODD numberii. Eg. H – N – CH2 – N – H = EVEN number

VI. Fragmentation – cleavages / breaksa. Electron is first knocked out from the hetero atom @ the begning to form

a M+ radical ionb. Fragmentation resulting from a radical cation is of two types

i. C – Hetero atom bond is cleaved heterolitically; this is shown by a full arrow when the two electrons BOTH go to the hetero atom

ii. C – C bonds break homolyticallyiii. The C1 – C2 carbon does not cleve (end two carbon bonds) only

inner C – C bonds cleave, because a primary radical and methyl radical is less stable

iv. C – C – C – C – C – C – C only middle carbons cleavev. During alpha cleavage, the largest alpha substituant is the one that

is readily cleavedc. Only the species with a (+) charge gives m / z peak (again, m / z is mass

over charge, which is really only the charged. Radicals do not give peaks, so ignore them

Don’t CLEAVE

I. Initial ion gives mass = 92II. Alpha CH3CH2 radical cation gives mass of 57

Sample Exam Question: How do you account for 57, 63, 65 and 92 for above compound

i. Main cation is formed by electron bombardmentii. Hetero-cleavage forms CH3 CH Cl cation

1. 35 + 24 + 4 = 632. 37 + 24 + 4 = 65

iii. take Cl away from the above compound and you get something that weighs in at 57

1. 92 – 35 = 57II. Alcohols show m – 18 peak ( H2O )

1. Note: primary and secondary OH shows M+ peak2. Tertiary OH does not show M+ peak3. the main cleavage of alcohol is alpha cleavage 4. Molecule with OH ends up being the cation

III. How to get m – 181. The H is the gama carbon and OH loses a molecule of H2O2. Ketones: give intense M+ peak; this undergoes alpha cleavage3. again part with O is the cation

I. Stereo chemistrya. Stereo isomer – they have the same bonding sequence but they differ in

orientation in spaceb. Stereo isomers include:

i. Cis/transii. Enantiomers

iii. Diastereomersiv. Meso compounds

II. Cis / Transa. To be isomer it must be that each sp2 carbon be attached to two different

things; you can’t have one sp2 carbon with two of the same thingsb. If one block fails, its not an isomerc. Cis isomer – if the same atom or group is on the same side it’s CISd. Trans – if same atom or group is on opposite side, it’s Transe. Hydrogen counts a group!

III. Chiral – these compounds have left handedness and right handednessa. Example = hands

IV. achiral – these compounds do not have left handedness or right handednessa. example = golf ball

V. Enantiomers – stereo isomers where one is the mirror image of the othera. These are all chiralb. Not super imposable

i. Example: hands againVI. Diasteriomers – stereo isomers which are not enantiomers

a. Note: they are not SAME compoundVII. Meso compound- these contain a plane of symmetry WITHIN the compound

a. Not enantiomers; enantiomers are two different compounds, meso’s are one compound with a plane of symmetry and internal mirror images

VIII. Chiral Carbon – if a carbon is attached to FOUR DIFFERENT atoms or groups, it is called chiral; also asymmetric carbons and it’s denoted by a star

IX. Number of possible isomers is always equal to 2 ^ N where N is the number of chiral carbons

a. If you have two chiral carbons, your compound has 4 possible isomers

X. How to find {R} and {S} configurationsa. Each chiral carbon is assigned an R or S configurationb. RULES

i. Assign a priority to each atom or group based on atomic number (if a group, based on the first member of that group)

1. look for the first atom that is connected to chiral carbon2. If there are isotopes of the same element, the heavier isotope wins out in priority3. If there is a tie, you go to the rest of what is attached4. O wins over CH2CH3 but CH2CH3 wins over CH35. Treat double bonds as though they were separate

a. C=C means that the C is attached to “2” C’s and 3 for triple bondXI. Overview

a. Chiral = carbon with four different things attachedb. Assign priority based on atomic numberc. In the case of a tie, go til you get a tie breakerd. Pi bonds treated as multiples of what they’re attached to

XII. There are two types of models / projections used in stereo chemistrya. Newman projection = two things off to the side one coming at ya and one

away from you

b. Fisher projection – the vertical things are away from you

XIII. Newman projection – put #4 priority on the dotted line and go from 1 2 3a. If it is clockwise, it is (R) b. If it is counterclockwise it is (S)

d. If the lowest priority is in the dotted position, you need only do above; assign priorities and go with what you get

c. in Newman projection, if the lowest priority does not occupy the dotted position, you have to switch the lowest priority with whatever is in the dotted position and make an imaginary compound. From there, identify whether the imaginary compound is R or S. Once you’ve done that, REVERSE it.

I. How to find R or S from Fisher Projectiona. There again are two casesb. Case 1 – if # 4 is on the VERTICAL (SAME AS DOTTED) line,

i. Go from 1 2 3

ii. Clockwise = Riii. Counter = Siv. Note: it’s ok to go over the #4 position

R configuration

c. Case 2: if # 4 is on horizontal line (WHICH IS MOST OF THE TIME)i. 1. find configuration and reverse it

ii. See example:

S configurationII. Common question type is to give a pair of these projections and ask if they’re

the same compounda. Note: to be the same compound, all four groups must be the same

i. If all same and both are R, it’s the same compoundii. If all the same and both are S they’re the same compound

iii. If all are the same but one is R and one is S, they’re enantiomers

III. Optical isomers a. Plane of Polarized Light:

i. Dextro (+) ii. Laevo (-)

iii. Note: if the R isomer of a compound is dextro (+), the S will be laevo (S); if the S isomer of a compound is laevo (S), the R isomer of a compound has to be dextro (+)

iv. There is no connection, however, the R can be either dextro or laevo, but if R is one, S must be the other

v. If the light passes straight through without bending either way, the sample you’re looking at is not an isomer (no chirality or R/S

vi. Also, if you have even amounts of R and S isomers of the same compound, the light will go straight through as well

IV. Optical purity is defined as…a. Optical purity = [observed rotation] / [rotation of purest compound] * 100b. Example: observed rotation was 32 and expected was 40; 32/40 * 100 =

80%V. Enantiomeric excess – is another method for finding the purity of the

compounda. Enantiomeric excess = { [d – l] / [d + l] } * 100 b. Example: in an experiment, six grams of R isomer is accidentally mixed

with eight grams of S isomerc. (8 – 6) / (8 + 6) * 100 = 1/7 * 100 = .143* 100 = 14.3%

VI. Isomers w/more than 1 chiral centera. Many organic compounds have more than one chiral centerb. Remember that the # of isomers = 2 raised to the N, where N equals the

number of chiral carbons in that atom.

VII. Meso isomersi. if one half of the compound is the mirror image of the other half of

the compound, it is a meso isomer

ii. NOTE:1. there is a plane of symmetry INSIDE a molecule2. enantiomers and meso isomers are DIFFERENT3. If top is R, bottom = S4. if top is S, bottom = R5. If you pass a plane of polarized light through a meso

compound it will go straight through

iii. How do you convert Newman to fisher?1. find whether it’s R or S2. construct same configuration3. put # 4 for priority on the TOP for converting Newman to

Fisher projection4. Put # 4 priority on the dotted line for converting from

Fisher to Newman projection

Rule for meso isomers: 2^n (n = # of chiral carbons) – 1!

Erymiro = hydrogens on the leftThereo = hydrogens on the right

Chapter VI

I. Substitution and elimination reactions

a. Nucleophiles are rich in electrons; they attack where electron density is least (bill gates)

i. Types of nucleophiles:1. May be negatively charged (STRONG)2. May be neutral (weak)

ii. Note: a Nucleophile can not have a positive charge, everb. electrophiles – poor in electrons; they are attacked by nucleophiles

(college student)i. positive charge = strong electrophile

ii. neutral (w/carbonyl) = weak electrophileiii. Note: all chemical reactions start by a nucleophile attacking an

electrophilec. Nucleophilicity – the power of nucleophile / ability to attack depends on

i. Nucleophile strength (negative or neutral)ii. Leaving group

iii. Solventd. HF < HCl < HBr < HI NOTE: Goes against electronegativity e. Weak Acid strength strong

Con Base F- > Cl- > Br- > I- Note: goes with electronegativity Strong Base Strength Weak

II. Solventsa. RULE 1: Stronger bases are better nucleophilesb. RULE 2: weaker bases are better leaving groups quitters !c. NOTE: these comparisons are done assuming that all reactions are done in

an Aprotic solventi. If no solvent is mentioned, it is assumed to be aprotic

d. protic solvent – is a solvent that forms hydrogen bonding with water mixes well with water

i. Has OH or NH groups!ii. EX: H2O, CH3CH2OH

e. Aprotic solvent – is a solvent that does not form hydrogen bonds with H2O; doesn’t mix well with water (makes suspension)

i. EX: CH2Cl2, THF, CH3CH2OCH2CH3, DMF, DMSOf. NBS – replace hydrogen in Allylic position with Br

i. Vinyl = involved in carbon = carbon double bondii. Allylic = carbon neighboring vinyl carbons

iii. This is a radical reaction

III. Nucleophylic substitution reactiona. R – LG b. Two types:

i. SN1 – substitution nuclephilic unimolecularii. SN2 – substitution nuclephilic bimolecular

IV. SN2 Reactionsa. Second order reaction

b. Alkyl halide + base c. R – LG + nucleophiled. Rate of reaction depends on both the alkyl halide and the nucleophile

i. Rate = k[Alkyl halide][nucleophile]ii. Mechanism: Nucleophile attacks Carbon from behind the leaving

group, kicking it offiii. SN2 means 2nd order, but it only has ONE step!iv. This is back side attack; the nucleophile attacks the carbon from

the back side of the leaving groupe. General notes about SN2 reactions

i. 2nd orderii. One step process

iii. Back side attackiv. Resulting product has INVERTED configuration due to the back-

side attackv. If it starts R, it’ll be S at the end of the reaction

vi. If you start S and do two SN2 reactions, you’ll end up S againf. Sovent effect in SN2

i. All reactions assumed to take place in aprotic solvents unless otherwise stated

Aprotic ProticF- STRONG WEAK

Cl- strong weak

Br- weak strong

I- WEAK STRONG

Just write F through I as is on the periodic table, then remember AP UP DOWN

The nucleophile has to be stronger than the leaving group, so the leaving group will quit.

Examples:

(A) Cl - + CH3CH2CH2I DMF DMF = Aprotic, therefore Cl – is stronger than I-, so the reaction will proceed

(B) Br- + CH3CH2F H2O H2O = protic, therefore Br- is stronger than F- so the reaction will proceed

(C) Br- + CH3CH2CH2I H2O = protic, therefore I- is stronger than Br-, so the reaction will NOT proceed because the nucleophile is weaker than the leaving group

V. Steric effecta. Since the nucleophile has to attack in a backside manor, in SN2 reactions,

the reaction speed will differ

Nu- + CH3CH2I Nu- + CH3-CHICH3 Nu- + tertiary halogen Fastest Slowest

Methyl halogens go fastest, tertiary don’t go at all in SN2

If you’re using a SN2 mechanism on a tertiary halogen, there will be no reactionTertiary halogens go by SN1 always

I. Elimination reactions are of two typesa. E1 – Elimination unimolecularb. E2 – Elimination bimolecular

II. E1 is like Sn 1 – first order, two stepsa. Start with an alkane, end with an alkeneb. Note: the formation of a carbocation in the second step makes

rearrangement possiblec. After carbo cation is formed, there is a chance for the solvent to attack it,

making it SN1 instead of E1d. There will be competition between SN1 and E1 (highly messy)

III. E2 – elimination bi-molecular (like SN2)a. Dehydrohalogenation (elminated from neighboring carbons)b. Functional group = alpha carbonc. Elmination reactions are also called beta eliminationd. Note: the H present in the beta carbon is eliminated, along with the leaving

groupe. The hydrogen present in the same carbon is NOT eliminatedf. Concerted = one step process

IV. What happens if there is more than one beta hydrogen present?a. Zatseff rule = IRS ruleb. In elimination reactions, the beta hydrogen is lost from that carbon which

has less number of hydrogen atomsc. Usually the alkene that has more substitutions along the carbon carbon

double bond will be the major product

V. EXCEPTIONS TO ZATSEFF RULE – but there are three exceptions!a. Anti ` seff product will dominate if the base is bulky

i. IE Tert-butoxide

b. if Fluorine is the leaving group it goes anti-zetseffc. if antizetsef product produces conjugation IE -C-C=C-C=C-C=C-, go with

it as the major product

VI. elements of Unsaturationa. double bond = 1 unsaturationb. ring = 1 unsaturationc. triple bond = 2 unsaturationsd. FORMULA

i. Unsaturations = ½(2X + 2 –H)e. What if there is a hetero atom?

i. Treat halogens as hydrogenf. what if there is oxygen

i. fuckin’ ignore itg. What if there is nitrogen?

i. Treat nitrogen as ½ carbonii. Example C4H9N

iii. ½ (2(4 carbons + ½ nitrogen) +2 – 9)iv. ½ (2 (4.5) + 2 -9)v. ½ (2) = 1 unsaturation!

VII. If all atoms across a Carbon, Carbon double bond are different, use E/Z configurationsa. Assign priorities #1 and #2 to each carbon substituantb. If the heavy things are on the same (ZAME) side, it’s Zc. If they are on Either side, it’s E

VIII. Energy differences in alkenesa. Trans isomer is more stable than cis isomer alwaysb. More-branched alkene is more stablec. During combustion, less stable will give higher heat; more stable will

produce less heat

IX. Alkene reactionsa. Electrophilic addiction reactions in alkenes (start with high electron

density, leaves with furniture b. Markognikov’s rule (rich get richer) – only works if Sp2 carbons are

unsymmetrical and the E and N are also unsymmetricali. CH3CH=CH2 + E – N

ii. Hydrogen = $$$$iii. Goes to that SP2 carbon that has more hydrogen atoms (richer)iv. During markognikov’s addition, most substituted alkene will be

formedv. Note: since the first step is the formation of a carbo cation,

rearrangements are possible

X. Reactionsa. Hydration (addition of water

i. Conditions in acid catylist (H2SO4)ii. What is added H, OH

iii. M’s rule = yesiv. Anti-additionv. CH3CH=CH2 (H2SO4/H2O)

b. Addition of alcohol in acid catylisti. Whats added? H, OR

ii. M’s rule? Yesiii. Addition? Anti

HOLY SHIT FINALS

Conditoin Added? M’s Rule? Addition

H+/H20 H, OH YES ANTIH+/ROH H, OR YES ANTIHg(OAc2)/H20 H, OH YES ANTIHg(OAc2)/ROH H, OR YES ANTIHBR / THF H, BR YES ANTIBH3/THF or H2O2 H, OH NO SYNB2H6HBR ROOR H, Br NO N/A only with HBRBr2 Br, Br N/A ANTIBr2 / H2O Br, OH Yes ANTIH2 H, H NA ANTIKMnO4 OH, OH NA SYNOsO4 OH,OH NA SYNPeroxy AcidRCO3H FORMS EPOXIDE!

I. Cleavage of double bonds: Two conditionsa. KMnO4 Hotb. Ozonolysis

i. Oxidative ii. Reductive

Used for locating the position of carbon, carbon double bond in molecule

Note: aldehydes are further oxidizedKetones are not

Aldehydes are made into acids under KMnO4 Hot conditionKetones stay the same

Ozonolysis is the same as KMnO4 HOT except for reductive: w/ Zn/HCl or (CH3)2S aldehydes stay the same

Review of rules:

II. Zetseff = IRS rule for Elimination. The H is popped off the beta carbon with less Hydrogen (the poorer beta carbon)

III. Disobeyed whena. Bulky base is involved (tert butoxide )b. If F is the leaving groupc. If the anti-zetseff product yields a conjugated organic compound

Markcognicoff’s rule = Rich get richer rule for addition. The H is added to the vynal carbon that has more hydrogen.

Spring Semester

I. Alkynesi. Functional group Carbon – Carbon triple bond

ii. Two types of Alkynes exist:1. Internal Alkyne: R – C –trip- C – R2. Terminal Alkyne: R – C –trip- C – H

iii. Common Alkynes1. Acetylene: H – C –trip- C –H2. Propyne: CH3C –trip- CH3. Phenyl acetylene: ph – C –trip- C – H

iv. Physical properties1. lower #’s are gases2. Acetylene – Gas used in welding; not soluble in water3. Test for – C –trip- C group

a. Alkyne will decolorize in Br2 / CCl4 (Brown colorless)b. IR absorption at 2100

v. Hybridization rules1. sp3: 1:3 (25/75) (Carbon single bonds usually)2. sp2: 1:2 (33/66) (Carbon double bonds usually)3. Sp 1:1 (50/50) (Carbon triple bonds usually)4. Note: As percentage of p increases, bond becomes longer

II. Acidity in Alkynesi. H present in terminal alkyne is acidic (low pka / pH)

ii. The H can be removed by a strong base like…1. NaNH2, Na, NaOH2. R – C –trip- C – H (NaNH2)/solvent R – C –trip- C(-) Na (+) + NH33. This is unique because you get a carbo-anion 4. This can be used as a carbon nucleophile. This can attack any

electrophile and a new compound can be formed5. Note: this nucleophile can be used to attack aldehydes and ketones6. By changing the ‘R’ and the electrophile we can build as many carbons

as desired in a molecule

III. Synthesis of alkynesi. Elimination reactions

ii. From a dihalide 1. vicinal dihalide = X’s on adjacent carbon atoms2. Geminal = X’s on the same carbon atom3. dihalide (strong base) R – C –trip- R (Geminal works the same way)4. Example:

a. CaC2 H2O Ca(OH)2 + H – C –trip- C – H (WWI smoke screen used by navy)

iii. Reactions of alkynes1. ABsorbtion – fully consumed

2. ADsorbtion – taken on surface 3. Higher surface area = better4. catylitic hydrogenation 5. What’s added? H, H; syn addition6. Mechanism:

a. R – C –trip- C – R + H2, Pd [R – HC = CH – R] b. Go all the way R – CH2 – CH2 – R

7. How do you get this bad boy to stop at the alkene stage?a. Use Lindler’s catylist to get cis alkeneb. Use a metal in ammonia to get a trans alkene

iv. Addition of Halogens1. Condition – Br2/CCl4 (Anti addition)2. What’s added? Br, Br3. R – C –trip- R + Br2/CCl4 R CBr2 – CBr2 – R (goes all the way to a

tetra halide)

v. Addition of HBr, HCl, HI1. Condition: HX / solvent2. What’s added? H and X3. M’s rule is followed 4. note: Addition: terminal alkynes are ok, internal may give many

products

vi. Special case for HBr in presence of peroxide1. Radical reaction2. condition: ROOR or RO2R3. M’s rule IS NOT FOLLOWED 4. R – C –trip- H + HBr/ROOR R – CH = CBr –HR – CH2 – CHBR25. R – C –trip- H + HI/ROOR R – CI2 – CH3 (rich got richer here)

vii. Addiiton of water – hydration (Note: vynal means carbon that is involved in C = C, Allylic means carbon that is neighbor to C = C bonding

1. theory – keto-eno tautomerism2. Rule: The OH present on a vinyl carbon is not stable; it rearranges to a

keto(ne) form3. Vinyl alcohol ketone

viii. Addition of water again…1. Condition; Hg++ or HgSO4 or H2SO42. What’s added? H, OH3. M’s rule: yes4. R – C –trip- CH5.

Fuck I failed a test

I. Ethers – Must have C – O – C linka. Two types:

i. C – C – C – C – O – C – C is a straight chain ether

ii. = is a cyclic ether

b. properties of ethersi. they have low boiling points

ii. they cannot form hydrogen bondingiii. APROTIC solvents

c. What has the highest boiling pointi. CH3CH3OH = forms H bonding, HIGHEST boiling point

ii. CH3OCH3 – mid boiling pointiii. CH3CH2CH3 (NO HETERO atom so this has the lowest boiling

point)d. Characteristics of ethers

i. The ether linkage is stable; not easily attacked by ordinary reagents (you need really harsh conditions)

ii. Ethers are inert, so they are usually used as solventsiii. A + B ---(ETHER)- C

e. note: ethers are dangerous!!! Known for making peroxides

II. Nomenclature of ethersa. CH3OCH2H5 = Ethyl Methyl Ether b. = Methoxy benzene

c. = Ethoxy benzene

d. = epoxidee. = THF

III. Synthesis of ethers – Williamson synthesisa. Condition: strong base (Na – Metal, or NaH)b. R – OH –Na R – O(-)Na(+)c. CH3ONa = Sodium methoxided. R – O Na –LG R – O – R1e. Sn2 mechanism; inversion of configuration if there is a ring

IV. To make cyclic ethersa. Suppose the same compound has an OH and a leaving group; if this is the

case and the OH and LG are separated by four or five carbon atoms, the compound can be made into a cyclic ether:

b. Method #2: Making an ether from an alkenei. Alkoxy mercuration/demercuration

ii. Condition: Hg(Ac)2, R – OH, NaBH4iii. Added: H and ORiv. M’s Rule – Yesv. Anti Addition

c. Question: How will you prepare 2-methyl THF from 4-bromo, 3-methyl butanol?

Fill in the Blanks:

V. Cleavage of Ethersa. Condition: Harsh conditions are needed to cleave ethers

i. HBr/strong heat or HI / Strong heatii. NOTE: look for whether one mole of HBr or HI is given, or is it

given in EXCESS?iii. Look for whether reaction goes SN1 or SN2

1. Look at the two neighbor carbons to the oxygen2. if at least one of these neighbors is a tertiary (3’), then it goes SN13. If none of the neighbors is tertiary, then it goes SN2

iv. If it goes SN1:1. Rule: the halide goes to the 3’ carbon and the hydrogen goes to oxygen

2. If there is HBr in excess this happens

Rule #2

3. If no neighbor is a tertiary carbon, the halide goes to the LEAST substituted carbon!!

If HI is in excess, the alcohol goes all the way to an alkyl halide

VI. Auto-oxidation of Ethersa. Ethers are notorious to take oxygen from atmosphere to form peroxide

i. It can cause explosions1. NOTE: usually white stuff seen around screw cap is peroxide2. Peroxide, remember is R – O – O – R

2/19/2007

I. Conjugated systemsa. C=C double bond

i. Isolated: -C=C-C-C=C (no order, highly reactive, localized)ii. Cumulative: C-C=C=C- (very rare)

iii. Conjugated: C-C=C-C=C-C= (alternating, very stable)1. pie orbitals line up and allow for movement of electrons for

added stability b. Note: for alternate single bond and double bond, the pi electron cloud can

move (called delocalization)c. This delocalization of pi electrons is important for resonance

II. Characteristics of Resonancea. The pi orbitals must be in same plane so they can overlap)b. The pi electrons must be delocalized (mobile)c. If (-) or (+) charge or lone pair of electrons is on the Allylic carbon,

then resonance is possibled. If there is resonance, the compound is stablee. More res. Structures = more stable

III. how do you know how to push arrows in resonance?a. Rules:

i. Start where electron density is highii. Start from (-) charge double bond

iii. Start from a double bond a (+) chargeiv. Start from lone pair double bondv. If there is an SP3 carbon, put the (-) charge on the carbon itself, not

between the carbons, as the SP3 carbon is saturatedvi. Allylic carbo-cations

1. Stable due to resonance

IV. Conjugated Dienesa. Diene – must have two “C=C”’sb. Rule: most stable releases less heatc. Less stable releases more heat

V. Addition reactions to 1,3 buta diene

1,3 buta diene for 1,2 addition product

1,4 addition

VI. Discussion of 1,2 and 1,4 productsa. Kinetic vs. thermodynamic products

i. Kinetic product – product that forms faster – (1,2 prod)ii. Thermodynamic product – product that is more stable (1,4 prod)

iii. Rule – more branches = more stable, around (C=C)iv. Less branches = less stable

b. Rule:i. At low temperature, kinetic products are favored

ii. At high temperature, thermodynamic products are favored

VII. Diels – Alder reaction (cyclo addition reaction, 4+2 addition reaction)a. Note: best method to form a 6 – membered ringb. Two components:

i. Conjugated Dieneii. Dienophile

iii. Transition state must be planer so all pi electrions can overlap

2/21/07

I. Characteristics of D-A Reactiona. The diene should be in the S-Cis position; if it is in S-trans position, it

changes to S-Cis position (if possible) and takes part in the D-A Reaction

Which Reaction goes faster?

3. It is stereo-specific; it retains its stereo chemistry

Example:Cis

Trans

Confusion!

4. In an unsymmetrical D-A reaction, the diene and dienophile usually give 1,2 or 1,4 product (substituent in 1,2 or 1,4 position) NO 1,3 product!

5. Endo Rule a. If a bi-cyclic compound is formed, it forms an endo product, not exo product

1. Comparison of speed in D-A reactionsa. Diene: electron rich, the richer in electrons it is, the faster the reaction will gob. Dienophile – electron poor, as poor as possible for fastest results

d. so you want your diene (first) to be attached to an electron donator and you want your dienophile (second) to be attached to an electron withdrawing group

e. Electron donators (D!): i. R (alkyl)

ii. Anything attached to something with a lone pairiii. Anything with a (-) charge on it

f. Electron withdrawing (W!)i. Halogens

ii. Stuff attached to carbonyls iii. Anything with a (+) charge on it

3. If S-cis is formed easily (or already there) = fastest4. If S-Trans = slowest

a. Note: if a ring stops one form from converting into another…b. Which goes faster?

c. The ring has a 0% chance of converting to trans configuration, so it goes faster.

d. Non-ring has slight chance of being trans5. Bridge concept:

I. Aromatics a. Definition

i. Ring (if not ring, called aliphatic or non aromatic)ii. There must be a free flow of pi electrons inside the ring

(delocalized)iii. All connecting members are Sp2 hybridized, and the pi electrons

must be in the same planeiv. Must obey Huckels rule

b. Huckel’s Rule (4n + 2 Rule) – to be aromatic, the compound must have 4n + 2 pi electrons, where N = 0, 1, 2, 3 or 4

c. Pairs: 1, 3, 5, 7, 9 etc (odd #’s)

N = 3, aromatic N = 2, doesn’t work N = 5, aromatic

Sp3 hybridized, Sp2, works nowNOT aromatic

Question: The H is acidic/easily removed, why? The anion that is formed is helping make the compound become aromatic

More Basic Less Basic

Because in order to be considered a Lewis base, you need to have electrons available for donation. The compound on the right has a lone pair, but it’s involved in the aromaticity, while the lone pair in the compound on the left is free and easily popped off

Less Basic More basic because of no aromatic ringDue to AromaticRing

Less Basic More Basic

II. Reactions of Benzene a. Tropylium cations are aromatic

Question: During electrophylic reactions with benzene, electrophylic substitution is the main product and no electrophylic addition product, why?

Answer: Aromatic rings are more stable, and the substitution reaction yields such a product, whereas the addition loses the aromatic-ness

Chart for Test II of Semester II

Condition What’s Added? Addition? M’s Rule?STRONG base converts alcohol into(Na-Metal, NaH) uneven ether N/A N/AWilliamson synth

Hg(Ac)2, ROH,NaBH4 Alkoxy H and OR Anti YesMercuration/demerc.

HBr / HIStrong Heat Ethers are cleaved halide goes to tertiary if there is one Cleavage of Ethers and H goes to primary; if not halide

goes to least substituted carbon and H goes to the more sub’d… if there is acid in excess, your ROH goes to RX

Conjugated Hydrogen and Bromine Forms carbo cation following m’s

Diene, HBR rule 1st, 2nd: Br is added

91!!! Yay!

Test III

3/12/07

Reactions of Benzene

Note: Always go with the substitution product because it is aromatic still!

There are 5 benzene reactions of the RING that need to be remembered:1. Halogenation: addition of Cl, Br and I

a. Cl and Br are added with either FeCl3/Cl2 and FeBr3/Br2 respectively

b. I is added with I2 / oxidizing agent like KmNO4

2. Nitration: HNO3 / H2SO4

3. Sulfuration – H2SO4 / sometimes with SO3 REVERSABLE by adding H2O, cat H2SO4

Also

4. Friedel Crafts Acylation: AlCl3 / RCOCl or (RCO)2O Adds Ester

Friedel Craft Acylation makes ester substituant

5. Friedel Crafts Alkeration: AlCl3 / RCl

III. Limitations of friedal/crafts reactionsa. If benzene ring is strongly deactivated (at least attached to strong electron

withdrawing group, like NO2 or SO3H) F/C doesn’t go at all

b. F/C ALKYIATION can form carbo-cation rearrangementsc. F/C Alkylation can give multiple products

IV. Nomenclature of substituted benzenea. RESERVED NAMES

Aniline Styrene

Benzoic Acid Benzaldehyde

b. For everything else, last name is benzene

Example: Bromo Benzene c. groups to remember:

Rules: if reserved name is present, it gets priority #1

a. 1,2 is orthob. 1,3 is metac. 1,4 is parad. Go with lowest numbering e. Then go ABC with that numbering in mind

3/14/2007

I. Alternative methods of creating a carbo-cation electrophile a. Alkene: condition: alkene + HF

b. From Alcohol, usually a secondary degree R-OH

II. Reactions of the Side Chain

a. Use of a bulky base after treating the benzyl side chain with NBS, favors an E2 mechanism

b. Note: Whether the alkyl group be, it is oxidized to a –COOHi. Length does not count

c. Side chain Reduction

d. The effect of substituants on reactivity

Middle e- rich ring e- poor ring (ACTIVATED) (Deactivated)

D = electron donating and W = electron withdrawing

Rings with donating groups react faster because the electrophile likes high electron concentrations

D is faster than H which is faster than W

e. There are two effects that deal with activation and deactivation of benzene rings

i. Inductive effect – This operates by difference of electronegativity of the two atoms

ii. Resonance effect – only ortho and para positions are effected by resonenance

iii. Meta positions are not affected by resiv. RULE: if both inductive effect and res effect are operating,

resonance always wins

v. activating groups: R, x: , x-

vi. deactivating groups: x+ NO2, x – C=O

f. Second substitution in benzene ringi. Rule: the group that is already present in the ring where the

incoming E+ should go relatively (ortho, para or meta)

ii. if the extant group is electron donating, ortho (1,2) and para (1,4) positions are favored

iii. if the extant group is electron withdrawing, meta (1,3) position is favored

Examples:

a. You always make 2x more ortho than para because there are two possible ortho positions and only one possible para

b. If the group that is extant or is coming in is big, you get 50/50

3/19/2007

Effect f substituants and PKa

Middle Activation Most activated Least ActivatedMiddle s Highest PKa Lowest PKaMiddle Basic Least Acidic Most Acidic

I. PKa Rules

a. Higher the PKa = more basicb. More electron density means it is less acidic (higher PKa)c. If the benzene ring is activated, then the phenol is less acidic (more

basic) higher PKad. If the benzene ring is deactivated, then it is more acidic (less basic)

and therefore has a lower PKa

II. Which is more activated ?

This one cuz ofLone Pairs

Rule: Resonance effect is mightier than inductive effect

Arrange in order of bromination:

2 1 3

I. Phenols – anything aromatic – OHa. Are o/p directing and the O – H bond is acidicb. You don’t need a really strong nucleophile to undo the bond between the

O and the H as in normal Alcohols because the Phenoxide ion is very stable due to resonance

Easy to pop that acidic Hydrogen off:

A Strong base is not needed as it is here:

II. Rxn’s of Phenols

1. Formation of asprin (acetyl salicylic acid)a. thee ways:

2. Examples of oxidation of phenols

4. Bromination

Note: phenol will decolorize bromine in Br/CCl4 no lewis acid neededBenzene needs a lewis acid to do that

IV. Aldehydes and Ketones

1. Reactivitya. Acid chloride > anhydride > aldehyde > ketone > ester >b. Aldehydes

i. Names: all end in al

2. Reactions of aldehydes / ketones with different nucleophiles a. There are two kinds of nucleophiles

i. Strong: OH-, OR-, RMgX, R-Li+, CH3C(trip bond)C acetylideii. Weak nucleophiles: Cl-, Br-, I-, CN-, RNH2, NH3, H2O

c. remember that if an acid is strong its conjugate base is weak and vise versad. strong nucleophiles don’t need protination of the carbonyl oxygen to

attacke. weak nucleophiles don’t attack carbonyl carbon unless the carbon/oxygen

is protenated

Tetrahedral Complex

3/21/07

I. Aromtaicity reviewa. In order to be considered an antiaromatic compound…

i. Ringii. Pi electrons must be arranged in a way that they can move

iii. Even # of pi electron pairs

b. In order to be considered an antiaromatic compounds…i. Ring

ii. Pi electrons must be able to move aroundiii. (4n+2) pi electrons or odd number of pairs

3/21/07

Reactions of Aldehydes / Ketones with nitrogen nucleophiles

General form:

The O and H2 combine to form water and the C double bonds itself to the N

Aldehyde / Ketone + RNH2Aldehyde / Ketone + Primary Amine Schiff’s Base

Aldehyde / Ketone + Hydroxyl Amine Oxime

Aldehyde / Ketone + Hydrazine Hydazone

Aldehyde / Ketone + Phenyl Hydrazine Phenyl hydrzaone

Aldehyde / Ketone + Semizarbizide Semicarbazone

I. Reduction reactions of aldehydes/ketonesa. Hydride reduction (H- = Hydride)

i. NaBH4 (Sodium Borohydride)

ii. Lithium Aluminum Hydride

iii. if you start off with a ketone, you end up with a secondary alcohol’

iv. if you start off with an Aldehyde, you get a primary alcohol

II. Wolf – Kishner reduction (note, do not confuse with hydrazine reaction)

a. Condition: NH2NH2 + (OH- and extreme heat)

b. NOTE: the OH- and extreme heat are what distinguishes it from the normal hydrazine reaction

III. Baeyer – Villeger oxidationa. Condition – peroxi acid (RCO3H)b. MCPBA

(know it for mcat)

IV. Addition of alcoholsa. There are 4 kinds of products to be known

i. Acetal, Hemi Acetal, Ketal, Hemiketal

ii. Easy Rules:1. Acetal = Hydrogen2. Hemi mono ether . or OH Group

V. Uses of Acetals and ketalsa. Preparation of Ketal

VI. Characteristics

a. These acetals or ketals do not react with oxidizing agentsb. But they are easily destroyed by H3O+c. These acetals/ketals are used as protecting groups

VII. Reactivity of groupsa. Acid chloride > acid anhydride > Aldehyde > ketone > ester > acid >

amide

88…

4/11/07

I. Aminesa. NH3 NH2R and NHRR1 are PROTIC solventsb. NR1R2R3 is Aprotic

II. Naminga. CH3NH2 = methyl amineb. CH3CH2NH2 = ethyl amine

Di-isopropyl amine! Benzyl Amine!

What is more basic?

Tertiary because its attached to three electron donators; MORE BASIC = MORE ELECTRON DONATORS

III. Preparation of Aminesa. From Nitrile R – CN

b. Amine from Amide: R – CO- NH2i. SAME # OF CARBONS:

ii. amine from amide with one carbon less:1. Called Hoffman Degredation2. Hoffamn Degradation is different from Hoffman

elimination!

c. From a Nitro Compound

d. Gabriel Synthesis – final = Primary Amine

I. Reactions of Aminesa. All amines are basic; we go from Red to Blue on lithmus

b. Acids:

Quaternary Amonium Salt

c. primary amine + Aldehyde / ketonei. Yields SCHIFFS BASE

d. acid chloride

i. you get a substituted amine

e. Anhydride

f. Amine RXN

i. Bromination; no need for FeBr3

II. Hoffman eliminationa. Only for quatinary ammonium hydroxide!b. Follows ANTI Zetseff rule (Zetseff rule is IRS Rule, so anti zetseff is “the

robin hood” rulec. Hydrogen from beta carbon is eliminated; the beta carbon that has more

hydrogen loses one Hd. The LG is a 3 aminee. Robin hood!

III. Reactoins with nitrus acid HNO2a. It has to be prepared during the rxn

b. Sand Mayor RXNi. Converts aniline to bromo, chloro or iodo benzene.

4/16/2006I. Carboxylic acids and their derivitves

What are LG’s and what aren’t

a. Order of Reactivitiesa. CL > OR > Anhydride > OH > NH2

b. Naming

(Formic) Methonoic Acid (Acedic) Ethanoic Acid Benzoic Acid

Trans, 3 methyl – cylcopentane carboxylic acid

II. PKA and carboxylic acid; high PKA = high basic, low PKA = acid

Highest PKA Middle PKA Lowest PKAMost Basic -- Most Acidic

CH3COOH // CH2ClCOOH // CHClCOOH // CCL3COOH 4.75 1.48 2.85 .64

Preparation of Acids

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Reactions of Acids

Reactions of Acid Chlorides

c. Acid isn’t very useful in organic synthesis, but an acid chloride is very useful, so the first step is to make an acid chloride.

Reactions of anhydride

I. Esters

a. naming

Ethyl AcetateMethylBenzoate

b. Reactions of Estersi. Hydrolysis (addition of H2O)

ii. This hydrolosis can be done in two conditionsc. Acid conditions

d. Basic Conditions

e. LAH with ester

f. LAH with RCN and nitrile

g. Amides, CONH2, very less reactive

h. Hoffman Elimination

I. Aldehydes

a. OH- and OR- do not attack carbonyl, instead it attacks the alpha hydrogen

b. B-keto esters

c. Haloform RXN – tests for methyl ketones

ONLY FOR METHYL KETONES

HVZ RXN

Hell – Volhard – Zelinski (carboxylic Acid Only)

Uses of L DA (Lithium – Di-isopropyal amide)

I. Aldol Reactionsa. It is done in aldehydes and ketonesb. An alpha Hydrogen is required for this shit to work

These do not have alpha hydrogens, and therefore cannot undergo aldol reactions

c. At least 2 molecules of the aldehyde/ketone are neededi. Make anion on alpha position of last molecule

ii. This anion attacks second molecule (its carbonyl) to give aldol addition product

iii. This addition product loses an H2O molecule to give a condensation product

1. Reason: Condensation product conjugationWrite the products:

II. Addition product = only one productIII. Condensation Product = main product / byproduct

a. More stable due to conjugation IV. Mixed aldol reaction

4/23/07

I. Aldo = intra aldol reaction

V. Claisen Condensation – similar to aldo reaction but starting material is not an aldehyde or ketone but is an ester

Note: if any ester is involved, the base cannot be OH- … why? Because ester will be hydrolyzed to acid (soponificatoin)

II. Beta keto acids / beta – keto estersa. B- keto acid – a keto group (karbonyl) and a COOH gp are in beta

positions (separated by one carbon)Examples:

Beta keto acid upon warming…

III. Which will decarbxylate?a. Look for COOH and carbonyl @ 1,3 position

b. Beta-keto estersc. The last sequence of this reaction is…

Reactions of melonic ester