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organic Y2 S1 > organic 4 > Flashcards

organic 4 Flashcards

1
Q

how do we form an enolate

A

we remove an alpha H and move e- to the carbonyl

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2
Q

what can we do with an enolate

A

we can use it to form bonds to other carbonyls!!!

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3
Q

is an enolate more reactive to a ketone or an aldehyde + why

A

an aldehyde!!

bc less inductive effect moving the e- towards the partially positive carbonyl

bc it only has 1 R group!!

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4
Q

what do end equivalents do

A

they help control the regioselectivity and reactivity of enolates and enols!!

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5
Q

what stops us from generating enols

A

if there’s no alpha protons to remove by the carbonyl

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6
Q

benzaldehyde + LDA reaction product and conditions

A
  • 78*C

enolate but with Li bonded to the O!!!! and no negative charge

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7
Q

what is LDA

A

a strong and powerful base

kinda hindered so it helps us control the reaction

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8
Q

when
ketone is reacted with LDA,,, can we stop at the enolate

A

yessss

the reaction is rlly slow due to the low temp of -78*C!!!

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9
Q

what is LDA

A

lithium
di isopropylamide

N with 2 claw machine things and Li

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10
Q

describe the reaction between LDA and a ketone

A

Li-N bond attacks Alpha H
double bond attacks LI

u get an enolate type thing but with Li on the oxygen,, u still get the double bond aswell

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11
Q

LDA after reacting with a ketone isss

A

N claw claw H

bc u removed the Li but attacked the alpha H.

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12
Q

what does LDA attack

A

the least hindered alpha H

bc it’s a powerful but bulky base!!!

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13
Q

product of LDA a ketone and then the reaction with a carbonyl

A

Li- N attacks alpha H
double bond attacks Li
u get enolate but with Li on the neutral O.

u get H N claw claw as ur product

the Li enolate type thing then uses it’s lone pair and double bond to attack a carbonyl C.

the O will then be (+) bc it’s bonded twice to C and once to Li.

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14
Q

what is the lithium enolate called

A

lithium enolate

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15
Q

how does the powerpoint say lithium enols react

A

they say that the O -Li bond reforms the carbonyl

and that the double bond attacks the Carbonyl C

then the other Carbonyls double bond bonds with the Li.

this forms a little circle between the O’s and the Li.

this is then reacted with acid to get OH and a carbonyl on the original carbonyl.

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16
Q

ketone + TMS + Et3N

A

O bonded to TMS (single bond)
double bond between where the H was removed and the carbonyl

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17
Q

ketone + TMS + Et3N thermodynamic major product

A

most substituted double bond

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18
Q

ketone + TMS + LDA major kinetic product

A

TMS bonded to O (single bond)

double bond between the least hindered H that was removed and the carbonyl

LDA THINK LOW TEMP

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19
Q

kinetic product

A

least hindered H is removed

LDA IS USED SO LOW TEMP TOO

20
Q

thermodynamic product

A

most stable double bond is formed

21
Q

aldol reaction using the TMS enolate

A

lone pair on O reforms the carbonyl double bond

double bond attacks Carbonyl C

22
Q

if ur enol product has a carbonyl on one end and smt else on the other,, what can u do to make the random group into an OH

A

u add H3O + and do a work up

23
Q

how can we make benzaldehyde more reactive

A

add TiCl3

the carbonyl O is weakly basic so will attack TiCl3

TiCl3 is a strong lewis acid (accepts lone pair l)

24
Q

benzaldehyde and TiCl3

A

makes benzaldehyde more reactive

u still have a carbonyl but the O is positive as it’s bonded twice to the C and once to the TiCl3

other things can then react with it. (enolates!!)

25
Q

most important use of LDA

A

making lithium enolates
gets least hindered H
its a bulky base

26
Q

what is a silyl enol ether

A

enolate but with SiMe3 bonded to the O (which is neutral)

27
Q

ether functional group

A

R ‘ - O - R ‘2

where it’s a single bond

28
Q

enol functional group

A

protonated enolate

OH near connected to a double bond

29
Q

how to make a silyl enol ether

A

carbonyl + Me3SiCl (bc they’re hard)

the O attacks the Si
the Cl kinda just goes away.

hard + strong bond so favoured!!

30
Q

enol description

A

double bond OH

31
Q

enolate description

A

double bond O-

32
Q

lithium enolate description

A

double bond O - Li

33
Q

silyl enol ether description

A

double bond O - SiMe3

34
Q

making a silyl enol ether (3 ways)

A

enolate lone pair attacks Si

kicks off Cl

Et3N is then used to remove the enol H to neutralise the O

———————————————————

lone pair on O on a carbonyl attacks the Si of the Me3SiCl

kicks off Cl
Et3N uses its lone pair to take off an alpha H,, which is then resonated to give a neutral oxygen

———————————————————

lithium enolate bond attacks Si and removed Cl
forms the silyl enol ether just like that

35
Q

do silyl enol ethers react with all carbonyls

A

nopeeee

the other carbonyl needs to be ‘activated’ using TiCl4 or another lewis acid

the lewis acid needs to be making the other carbonyls Oxygen positive

36
Q

NEt3 will do whattt

A

it’s a base so it will steal a alpha H

it steals the one that will give the most stable double bond

37
Q

the way the teacher draws the silyl enol ether being made

A

the O- attacks the Si - Me3

it kicks off the Cl

38
Q

if u have like a diester,, so like 2 carbonyls on molecule but further apart,, the symmetric looking ones,, how is a hydrogen bond created

A

EtO- // base can attack the alpha H’s

this can form a double bond and make O-.

this can then attack a H to form OH. hydrogen bond forms between the carbonyl O and the OH

so it forms a little ring type thing with dotted lines where the hydrogen bond is.

39
Q

bonding 2 diff carbonyls together + removing an OH from the added one!

do it with di carbonyl and benzaldehyde: with benzaldehyde being the other carbonyl!! so this would drop down

A

knoevenagel reaction

Eto removes an alpha H
forms a double bond and O-

the O- and double bond attacks another carbonyl and forms O-

O- forms OH

Eto removes remaining alpha H ,, forms a double bond and an O-

O- and double bond is used to form another double bond and OH is removed (LG) kinda a bad LG so we had to go the long way.

NaOH and H3O+ can be added to make the esters caerboxylic acids,, then H bonding between carbonyl O of 1 carb acid and OH of another carb acid. forms a ring type thing

then arrows from double to H thats H bonded. OH bond to C from carbonyl and then the other arrow to remove the carboxylic aciddd.

can tauterise it to give the ketone version using arrows: double bond attacks H,, OH bond makes carbonyl.

40
Q

doetner modification meaning

A

malonic acid : carboxylic acids on both sides

with an aldehyde as another reactant

u add a new R group and remove the old one as its a CO2

41
Q

decarboxylation

A

u get rid of some of the carbonyl groups

42
Q

deotner modification fully

A

di carboxylic acid type thing
A hs are removed and it attacks a carbonyl

OH bonds are formed betwee nthe old carboxylic acid ends!! with the new thing sticking out but its OH and not O-.

then the bond between the original OH is moved to the C of the C=O near it,, then the middle bond is broken and given to the C connected to the R group and a double bond is formed. the OH of the new thing is also removed
💗
this gives u the beginning of the new thing with a c=c and the old carbonyl group at the endddd.

pyridine and piperidine is the base thing and H donor.

remember that this gives an allene and then the OH e- go to the C-O to give C=O,, and the double bond of the allene takes the H

43
Q

making a carboxylic acid from an ester,,

A

remove the alpha Hs,, get the double bond to keep attacking Bromines

this will remove e- densigty from thr C of the carbonyl

an OH- can attack it and it moves to the carboyl O giving O-,, this is ussed to remove the bromines to give a carboxylic acid

44
Q

if u see a long carbonyl,, aka a long chain with 2 carbonyls as a reactant,, and the product is a chain with a different R group and a double bond,, what do we do and what should we think. yk this girl 💗

A

okay so its a carbonyl so we know we probs neeed to form some sort of enolate,, and bc its a different R group we know that that is probably from the thing we reacted to. and bc enolates react with carbonyls we know that itll be ‘C=O - new.R’ and bc its a double bond we know we have to remove a water.

and to get rid of the old part of the chain,, we need to form a hydrogen bond and then kick CO2 off 💗

this would give us an allene which then needs to attack the H and move the OH e-to reform the C=O bond

45
Q

ketone to carboxylic acid

A

attack alpha H with base
form an enolate and use this to react with Br-Br.

keep doing this until all H’s are Br.

attack the C in C=O with OH,, then use the O- to kick off the CHBr3 to give a carboxylic acid.

organic Y2 S1 (9 decks)

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