gauche newman projection

60 degrees

anti staggared

180 degrees apart

red=axial 

blue=equatorial

dashed=down, solid=up

boat less stable

more stable when larger substiuents are in equatorial position

R and S system

[image]

Rank atoms directely connected to asymmetric center by mass

heaviest gets #1, lightest gets #2

lowest priority needs to be farthest away (dashed side)

draw arrow from 1-2

Clockwise=R

Counterclockwise=S

diastereomers

can not relate by mirror

one is opposite of one another

SN2

Strong anionic nucleophile in DMSO

if carbon is chiral will produce inverted stereochem 

SN1

leaving group leaves

nucleophile comes attaches to carbocation

possible carbocation rearrangement

unhappy camper may be present

will form racemic mixture if the carbocation is chiral

e2

in one step

proton removed

double bond formed

halogen leaving group leaves

keep antiperiplanar in mind

h is removed from beta carbon with least amount of h (zaitsev rule) or most h (anti-zaitsev rule) 

E2

track alpha and beta carbons

positive charge follows the alpha carbon

added step loss of leaving group to carbocation may need to rearrange

1. halide dissociates forming carbocation

2. rearrange if needed

3. base removes proton from beta carbon with least amount of h

anti periplanar 

H and X need to be in the same plane to generate pi bond

CH1 use the newman projection, only forms cis or trans not both

CH2 cis and trans are formed, trans will predominate

CH3 cis and trans don't exist

anitperiplanar cyclo

both halogena and hydrogen need to be in axial positions

dissolved in CH2Cl2=chemically inery 

HCl, HBr, HI

carbocation rearrangement possible

marks rule applied

acid catalyzed addition in H2O

catalyst H2SO4

H3O is the electrophile

carbocation rearrangment possible

Marks rule to derive major product

In ROH same process but ending with an RO and not OH

solvent CH2Cl2=chemically inert

vicinal dihalide: 2 halogens attached to adjacent C

Cl2 and Br2

Carbocation rearrangement not possible

halohydrin

H2O nucleophilic solvent participates in reaction

Halogen ends on Marks carbon (with most H), OH on the other one

CC rearrangement not possible

Same steps for ROH nucleophilic solvent but ends wit RO and not OH

Step one of Oxymurcuration reduction (addition of water)

happens with Hg(OAc)2, H2O, and THF

rxn is two steps

product at end 1 organomercury compound

water attaches to the one with least amount of H

Hg stays with carbon with most amount of H

hydroboration-oxidation (reverse addition of water)

happens in two stepsthe product at the end of step 1 is trialkylborane

the boron is released in the second experimental step under harsh oxidizing conditions

carbocation rearrangement not possible

net result anti-mark alcohol

hydroboration step 2

boron is replaced by OH group

hydrogenation (addition of H2 across double bond) converts alene to alkane

common catalysts include: Pd/C, Pt/C, Ni

carbocation rearrangement not possible

epioxidation-converts an alkene to an epioxide (cyclic ether)

meta chloroperoxybenzic acid (MCPBA) is a commonly used peroxyacid (R-CO3H)

no carbocation rearrangements

in Ch2Cl2 

forms meso achiral compound

syn addition

dihydroxilation 

converts an alkene to a vicinal diol

syn addition

carbcation rearrangement not possible

check for meso (achiral) compound

groups added on the same side

lends to cis product

1. replace ane with yne

2. use longest continuous chain method that contains the carbon carbon triple bond

3. number in the direction that gives the triple bond the lowest possible number

4. substituent recieves lowest possible number if the same number fort the triple bond is obtained in both directions

1. find the longest continuous chain with all multiple bonds

2. diene, triene, tetraene

3. count in direction that minimizes number for all multiple bonds

4. if the two functional groups are a double and a triple bond, the chain in numbered in the direction that produces a name containing the lower number

#-nameen-#-yne

addition of hydrogen halide across triple bond

HCl, HBr, and HI

not in excess stops at alkene stage-> halo-substituted

assume Marks Rule

positively charged carbon where positive charge is located on the double bond

secondary vinylic cation more stable than primary, tertiary does not exist

symmetric across double bond

marks rule applied in second step

one geminal dihalide made

unsymmetric alkyne makes two geminal halides formed

marks follows second addition

1. anti addition: halogens are added to opposite sides      ->dihalo-sub-alkene produced (stops here if X2 isnt in excess)

2. (if in excess) X2 will add to other sides forms tetrahalo alkane

[image]

keto and enol differ only in location of the double bond and a hydrogen

ketone is more stable

enol: unstable double bond moves to O and H moves to where doube bond used to be

tautomerize=enol changes to keto

same molecular formula

ketone=C attached to two carbon containing groups (R)

aldehyde=One R group and one H

hydroboration-oxidation of internal alkynes (addition of water) 

enol is produced that tautomerizes to the more stable ketone

BH3 commonly used

make sure to count carbons

hydroboration-oxidattion of terminal alynes

(sia)2BH used (disiamylborane),

same regioselectivity as seen in the borane additon to alkenes

boron electrophile adds to the sp carbon bonded to the H (marks rule) 

enol will form that tautomerizes to the more stable aldehyde 

complete hydrogenation

alkyne is fully reduced to alkane when excess H2 is in the presence of an active metal catalyst (Pt/C, Pd/C, Ni)

semi-hydrogenation

Lindlar's catalyst acts as a poison able to stop at alkene stage

stops at cis alkene stage

semi-hydrogenation

stops at alkene stage

disolving metal reduction

formation of acetylide ions from deprotonation of terminal alkynes followed by alkylation

NH2- (NaNH2) always used

NH2 pulls off proton on triple bond (deprotonization) 

acetylide ion made=good anionic nucleophile

base used to deprotonate has to be stronger than the conjugate base

good anionic nucleophile with primary alkyl halide substrate

back side attack

SN2 chem

goes from terminal->alkalate->internal alkyne

chain elongation (increase in C count)