addition Flashcards

1
Q

addition rxn

A

addign two groups to a double bond which destroys the double bond

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
2
Q

regiochemistry

A

where the 2 groups are being installed on the double bond
-only matters when adding two different groups to an unnsymmetric alkene

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
3
Q

symmetric alkene or adding the same groups

A

regiochemistry is irrelevant

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
4
Q

regiochemistry types

A

mark and anti mark

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
5
Q

mark

A

h gets less sub

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
6
Q

anti mark

A

h gets more sub

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
7
Q

stereochem

A

-syn or anti
-mechanism will explain stereochemistry
-does not matter is there are no stereocenters being made (not 4 unique groups) (here only one product)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
8
Q

syn

A

groups being added on the same side of the plane

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
9
Q

anti

A

groups each being added on opposite sides of the plan

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
10
Q

regio vs stereo

A

-regio described the order in which they are added / which group is added to which atom
-stereo describe how they are geometrically added to the mocule/ determines that the product is going to look like

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
11
Q

both carbons sp2

A

tigronal planar
-must rotate the molecule to do either syn or anti

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
12
Q

products of syn and anti

A

each represent a pair of isomers
-4 products total

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
13
Q

non stereospecific

A

expect all 4 products
-if stereo specific it would be 2 enan from syn or 2 enan from anti

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
14
Q

stereochemistry irrelevant

A

-when there are no stereocenters made
-when there is one stereocenter made but no others int he molecule

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
15
Q

one stereocenter

A

2 possible products
-will get racemic mixture (enan pair)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
16
Q

if it creates a plane of symmetry

A

meso
-only need to draw one because there is only one product they will be the same compound

17
Q

meso

A

are not enantiomers they are just two drawings of same molecule
-choose one
-can have the same group along the plane of symmetry with a dash or a wedge dont have to match
DO NOT WRITE PLUS ENANTIOMER
-can also say D2

18
Q

anti adition

A

one dashed one wedges

19
Q

hydrogenation

A

adding h and h across alkene
-regiochemistry irrelevant
-uses a metal catalyst (pt, pd or ni)
-hydrogen interacts with the surface of the metal catalyst which breaks the h-h bond and leaves sinlge h bonds to the metal that are ready to interact with the alkene bond
-will be a syn addition

20
Q

adding hydrogen halide

A

-2 steps (proton transfer and nucleophilic attack)
-proton transfer gets rid of the double bong leaving a carbo cat and then the halide/nu will attack
-regiochem irrelevant if symmetric
-not symmetrical: make the most stable carbocation, the proton from protonation goes to the less stable carbocat then the halid will attack the carbocation, ending up on the more substituted carbon thus mark
-stereochem (syn or anti) irrelevant bc there is not two stereocenters
-shift may occur after protonation to make the carbocat more stable
-hbr withour roof is still mark

21
Q

hydrogen halide rxn that can be anti mark

A

ONLY HBR and peroxides (ROOR)
-BR ends up on LESS sub (ANTI)
-Does NOT involve br- (ionic) involves Br as a radical intermediary
-radicals formed by peroxide bonds breaking

22
Q

adding water across a double bond

A

hydration

23
Q

hydration

A

can be mark or anti mark
-depends on the reagents

24
Q

mark hydration

A

-reagent can be water and an acid source or in brackets (h2so4), brakets mean the h is not consumed in the rxn it is a catalyst

25
Q

mark hydration name

A

acid catalyzed reaction
-because the h is the catalyst and oh is the product

26
Q

acid catalyzed hydration

A

-first is protonation that will yield carbocation
-nu (h20) attacks
-the product will be CHARGED
-extra step to remove charge and do it by using water as a base (deprotonate w water)
-use reversible arrows bc the product/ equilibrium can be controlled by contorlling the amount of water used
-use dilute acid and more water to favor alcohol, push equilibrium right

27
Q

anti mark addition of oh and h
-hydroboration oxidation
-2 step synthesis

A

-oh ends up on less substituted (regiochem)
-stereochem= syn rxn
-uses bh3 and THF
-THF will stabalize the empty b orbital of bh3 by donating electron density
-bh3 is an electrophile and looking to fill empty orbitals/ wants electrong so the pi bond attacks it and that will trigger a hydride shift
-adding h and bh2 across the double bond (BH2 goes anti mark because it is bbb) (step 1) then comes oxidation
-step 1= hydroboration
-there is an alkyl shift which places oxy between b and r but stereocenter is preserved
-final product will be an alcohol

28
Q

not creating two stereocenters

A

-stereochem is irrelevant

29
Q

br-br addition

A

-regiochemistry irrelevant (same molecule)
-creates two new stereocenters sometimes and results in enantiomers
-anti addition
-the alkene functions as the nu
-br is the electrophile bc it has a partial positive becuase there is a covalent bond btwn br and br that creates a dipole with the pi bond
-alkene attacks the “electron poor” nu
-if water is the solves an oh group will be on the product bc water will act as a nu too
-last step is deprotonation (when water is the solvent)

30
Q

water as solvent for br br

A

there will be regiochem
-oh goes on the MORE sub c
-halohydrin rxn

31
Q

two oh groups

A

can be added anti or syn

32
Q

oh oh anti

A

two steps
-first step involves peroxyacid (RCO3H) to form epoxide
-step two add water like acid catalyst rxn (protonates then form a charged intermediate (not carbocat) and then water attacks)
-water attacks from the back therefore anti

33
Q

oh oh syn

A

syn dihydroxylation
-KMnO4 cold is often used
-oso4 also
-pair of enantiomers