organic 7

Question 1

double bonds cant do what

Answer 1

double bonds cant rotate

Question 2

what does P love bonding to

Answer 2

P likes bonding to O

Question 3

the witting reaction forms what

Answer 3

the witting reaction forms double bonds

Question 4

witting reaction steps:

Answer 4

bromo cyclohexane + PPh3
: on P attack the C and kicks off Br.

now u have PPh3 cyclohexane but its (+)

N butyl lithium is a strong base and removes the H on the C thats bonded to the PPh3

the e- go to the carbon and give it a (-) this is called a YLIDE!! positive P, negative C

then the negative goes to the C-P bond to neutralise the P and the C and forms a double bond PHOSPHORANE!!

Question 5

YLIDE AND PHOSPHORANE AREEE

Answer 5

basically resonance structures of themselves

Question 6

PPh3 (+) and C (-) is a

Answer 6

ylide

Question 7

what can a ylide do

Answer 7

the (-) on the C can attack carbonyl C’s.

the (+) PPh3 can accept electrons from the carbonyls double bond

this forms a small ring of C C O P called oxaphosphetane

draw arrow Past P and an arrow past the HR to collapse the ring

this gives an alkene + phsophine oxide

Question 8

describe phosphine oxide

Answer 8

R R R P = O

Question 9

what forms an alkene

Answer 9

R-Br + PPh3 + aldehyde

one R from each (the thing the PPh3 attacked and the R from the aldehyde)

either cis or transsss

Question 10

what is more likely to give a z alkene // same side

Answer 10

an unstable ylide

Question 11

what is more liekly to form an E ylide ,, diff side

Answer 11

a stable ylide

Question 12

stable ylide

Answer 12

E

Question 13

unstable ylide

Answer 13

Z

no time to rotate and give an E (more stable due to steric repulsion) due to being so unstable and being so reactive.

Question 14

describe an unstable ylide

Answer 14

R - C - PPh3

where R is an alkyl
C is (-)
PPh3 is (+)

its unstable bc the R is an alkyl which means it donates e- density towards the already - C

Question 15

describe a stable ylide

Answer 15

R C PPh3

where R is a carbonyl,, so the (-) can be resonated onto the carbonyls O,, forming an O- and it being more stable.

where C is (-)
and where PPh3 is (+)

Question 16

unstable ylide what does it give and why

Answer 16

gives a z alkene
gives a SYN OXAPHOSPHETANE (box made up of C C O PPh3 )

syn bc both H’s are wedged // dashed

this is bc we want the R group as far away from the PPh3 as possible,, to prevent steric hinderance and instability.

Question 17

are aldehydes and ketones more electrophilic than esters

Answer 17

yessss

bc they only have inductive effect making the C more (-)

and inductive effect is weaker than resonance.

Question 18

stable ylides,, what they give and why

Answer 18

stable ylides give E isomers (think that thry have more time to rotate and give the preferred isomer)

they also give an ANTI OXAPHOSPHETANE,, where one H is dashed and the other one is wedged.

this is bc we dont rlly care about the steric hinderance between the R (carbonyl) and the PPh3,, we care more about the 2 electronegative groups being close together (the o and the carbonyl)

so we line the carbonyl up with the R group.

Question 19

describe a sulfonium salt then describe what happens when u react it with a base

Answer 19

S (+) Me Me Me

base: removes a H,, u get S staying (+) and a CH2 being (-)

SULFONIUM YLIDE BC C IS - AND S IS +

Question 20

describe a sulfoxide and what happens when u react it with a base

Answer 20

ketone but with an S instead of a C

u remove an alpha H and it gives u a (-) on that carbon

Question 21

describe a sulfone and how it reacts with a base

Answer 21

R S = O = O - R

with a base u remove a H near the sulfur,, which gives u a (-)

Question 22

julia olesenation

Answer 22

when u have the r s =o =o - r thing and u use a base to remove a H and give u a (-)

this attacks a C on a different carbonyl and forms an O-

the lone pair on the O- then attacks another carbonyl C and forms another O- (u remove the R aswell tho so u get an ester)

B removes a H by the S and a double bond is formed and the ester is removed

treating it with Na removes the sulfur and its o’s. giving u a double bond and a (-) on one end of the doiuble bond

u work it up with H3O+ to give an alkeneee (an E alkene in this case) E alkene is the major product..

Question 23

julia mech simplified

Answer 23

u use the r s=o =o - r and a base to attack a carbonyl.

then use that carbonyl to attackabother one and form an ester

u then use a base to form a double bond and remove that ester.

u then use Na to remove the sulfur bit of it (gives off a smally gas) u get a double bond with a (-)

then u work it up with H3O+ to give an alkene