So much organic chemistry

H1 NMR spectrum steps

1. How many signals/unique proton enviornments

2. determine chemical shift of eenvrionment

3. determine integration, how many protons in each environment

4. coupling/splitting determines number of neighboring protons

ppm=

observed shift from TMS/operating frequencydow

downfield shift causes

1. Induction (more inductive= greater downfield shift)

   1. deshielded= naked

2. anisotropic effect

   1. super shielding of ring protons, resonance to infinity

downfield v upfield

downfield= 8 ppm, upfield= 1 ppm

Integration

area under each signal tells relative number of protons, divide by smallest integration value

splitting rule

N+1= splitting

N= total number of enighboring protons

couplets

singlet= no neighbors

doublet= 1 neighbor

triplet= 2 neighbors

quadruplet= 3 neighbors

quintet= 4 neighbors

sextet= 5 neighbros

septet= 6 neighbors

coupling represents

number of possible electrical environments that can be produced by a proton type

Rules for splitting

1. equivalent protons do not split each other

extra helpful tip

DOU calculation= how many rings or pi bonds to find different groups that may be present

IR rules

1. Is there a strong broad signal at 3400 inverse centimeters/ —> OH group

2. Are there signals just above 3000 inverse centimeters? Yes —> Csp2

3. Are there signals just below 3000 inverse centimeters? Yes= CSP3

4. Are there signals at 2200 inverse centimeters? Yes= triple benzyl

5. Are there strong signal at 1700 inverse centimeter? Yes —> carbonyl group

Extras for fun

carbonyl with OH 2200-3600

Amine 3340-3600

Csp= 3300

if compound has a chiral center, then 2 protons of CH2 group are chemically:

unequivalent

upfield= electron

RICH

Downfield= electron

poor!

carbonyl

2-2.5

alchohol

1-5

alpha to halogen

3-5

alpha to oxygen

3.3-5

aromatic ring

6.5-8

carbonyl

9.7

carboylic acid

10-13

quartet triplet 2H 3H

ethyl

Septet doublet 1H 6H

isopropryl

change in enthalpy

products-reactants

exothermic (-) vs endothermic (+)

change in free enrgy

endergonic (+) vs exergonic (-)

delta g= delta h - t times delta s

thermodynamically favorable products have

more release of energy

kinetically favorable reactants have

lower activation enegy

transition state vs intermediate

transition: on humps, intermediates; in grooves

Nucleophilic attack

nucleophile attaking an electrophile, draw arrow from nucleophile to back of electrophile

Loss of a leaving group

breaks a bond

Proton transfer

electronegative grabs a proton which will leave its electrosn behind, causing prpotonation of a product and deprotonation of another

Rearrangmenets

carbocations

hydride shifts

migrational H

methyl shift

migration of methy lgroup

two can happen simultaneously

loss of leaving group and replacement with another group

Good leaving groups

Cl, Br, I

transition state representation

dotted line to show both nucleophile attack and leaving group

faster rate of SN2= lowering of transition state

faster rate=

methyl group or primary carbon

Nucelophilicity depends on:

charge and polarizability, which goes down for bigger atoms (ability to distort electron clouds to be more negative)

Strong nucleophiles

I, Br, Cl, HS, RS, NdoubleC, OH, RO

Weak nuclophiles

OH groups

E2 rates

must faster for tertiary: more substitiution, is more stable, lower energy alkenes, lower transition state

Regioselectivity

preference to make double bond in certain region

Zaitsev product

more substituded, lower energy

Hoffman product

less substituted, more accessible alkenes

I-

nucleophile

HS-

Nucleophile

Br-

Nucleophile

RS-

nucleophile

R2S

nucleophile

CL-

nucleophile

CN-

nucleophile

nucleophile PKAs

are generally less than 10 (strong acid= weak base)

KOtbu

base

NaH

base

NaOH

good base/nuc

NaoME

good base/nuc

NaOet

good base/nuc

H20

meh

E1

unimolecular elimination

S1

unimolecular substitution

RDS

formation of carbocation

asses relative stability of intermediates

LG order

I- before Br- before Cl-

ON EXAM REMEMBER

resonance stablilized carbon is not primary because there is no fixed charge on it

MeOH

meh

EtOH

meh

SN1/E1 will not happen on a

primary carbon

OTS is a

good leaving group