******135B- carbonyl barrie Flashcards
In carbonyl chemistry, what are the orbitals that most of the chemistry come from in a C=O functional group
mostly the LUMO dictates the chemistry,
but interested in homo and lumo
what is the chemistry of the C,O and alpha H
-can turn carbonyls from electrophilic to nucleophilic this is because weakly acidic alpha hydrogen can be removed. This forms a conjugate base, which makes the carbon at point α nucleophilic
- Can react with Lewis acid
- Fire nucleophiles at it to react with δ+ carbon
- Or remove H with base and turn it into a nucleophile at α carbon
must be able to identify each of these
2 types of ester: cyclic and non cyclic form
2 types of amide: cyclic and non cyclic form
chemistry across C=O chemistry at the δ+ carbon
nuclophillic addition- no good leaving group
[usually happens with aldehydes or ketones]
What would the addition mechanism look like with this nucleophile and ketone
-include a work up step [adding an acid]
this is addition because there is no good leaving group on the positive carbon
C has gone from sp2 planar, to sp3 tetrahedral structure
what would the substitution mechanism look like with this nucleophile and ketone
this is a substitution because there is a good leaving group on the carbon
the Nu and Cl are unstable next to each other
when a nucleophile attacks a carbonyl group carbon
-what orbital does it attack
-and how does the orbital orientate itself to bond with the nuclphile
-what is the upside and downside of this orientation
-π* orbital is always interacted with
- upside: π* orbitals are originally 90° to double bond, but change itself to 109° so that the nucleophile has better access and to minimise static repulsion
downside: this now means the other groups attached to carbon (X and Y) will effect the rate of reaction if they are too bulky
why is the acyl chloride the most reactive
[inductive effect] Cl has high electronegativity, so draws electrons towards it
-leaves carbon very δ+
why is the anhydride second most reactive
O is very electron withdrawing
-carbonyl group attached to oxygen [in red]. Carbonyl group is also electron withdrawing
electron density pulled from =O to O. then pulled from O to other carbonyl group
-this makes R very δ+
why is the left hand side compound have a more δ+ carbon than the right compound
- The =O is electron withdrawing making the carbon somewhat positive, however the OR has a lone pair which is donates, making the carbon almost neutral
- The =O is electron withdrawing, however N holds its lone pair of electron less tightly than O meaning it adds more electron density into the carbonyl system, making the carbon less positive
why is the acid [conj base form here] the least reactive
puts a lot of electron density into the carbonyl system as it is the only one where the oxygen is negativity charged, so can donate more e-
the better the leaving group the better the carbonyl δ+ carbon can undergo substitution
aldehydes are more reactive than their corresponding ketones
-why is this
-to do with R groups and attack angle of nucleophile [onto δ+ carbon]
- R groups are electron donating, and so push electron density onto the carbon in the C=O [the more δ+ this carbon, the more reactive]
- the aldehyde has only got one small H attached as a group. Ketone has 2 R groups. Sterics slow down substitution
aldehydes and ketones only undergo nucleophilic addition
how do you go from a primary alcohol to a ketone to an acid
-secondary alcohols will also react, but cannot go past a ketone
- Jones oxidation [sodium dichromate, sulfuric acid]
- Potassium permanganate
-usually have jones or permanganate in excess to go all the way to acid
-Cr is electrophile and OH is nucleophile
these are Jones conditions
note- think about at which oxygen does the reaction take place
-tertiary alc cannot be oxidised so reaction must happen at secondary OH
-dichromate is in acidic solution so is protonated
dont know where the water comes from
and dont know why the last step happens [chromate ester becomes as leaving group]
- know that the product must be a ketone because it is a secondary OH being oxidised
-PCC is an alternative way to oxidise an alcohol that isn’t using manganate or the Jones method
what does PCC look like
when oxidising an alcohol to a ketone or acid you can use
-Jones/ permanganate
or
-PCC
-NMO
-PCC is better as it doesn’t require water so you can stop the alcohol from going to an aldehyde. Jones requires water
what are the reactants to go from an alcohol to an aldehyde in this reaction
what is the mechanism for this
[alot of similarities to jones reaction]
Ru is electrophile
OH is nucleophillic
what would this form
will take your primary alc to aldehyde only
and secondary alc to ketone
what would this form
what is the mechanism for this reaction
-youve made a bond to an iodine
-therefore you have to break a bond with iodine
-as acetate group is a good leaving group, that bond is broken
what does this make
what is the mechanism for this
what does this form
what does this form
-radicals are generated
-you need to use an organic solvent that water is miscible in [THF]
know the order of reactivity of these carbonyl groups
why does the ester group react with the nucleophile instead of the amide carbonyl group?
the ester group is more reactive than the amide functional group
at which carbonyl group will the nucleophile substitute at
acyl chloride end is more reactive than the ester end
is a metal hydride an example of an oxidising or reducing agent
reducing agent
metal hydrides are good reducing agents. Below are the most common hydrides used
-order LiAlH4, LiBH4 and NaBH4 from most reactive to least reactive
the more reactive a metal hydride the more/less selective it is
the less selective
shows which carbonyl group the hydrides will reduce
-LiAlH4 is not selective and will reduce most
-BH3 struggles with aldehydes, ketones and esters but is good for amides
which reactions are addition and which ones are substitution
the ester is reduced to the aldehyde, which is then reduced down to the alcohol; doesn’t just go straight to product
what would these make
reduction reactions to make alkane
what does this make
what 2 reactants to go from a nitrile to an acid
sulfuric acid and water
How can you make a C-C bond at a carbonyl using organometallic elements [grignard]/ increase a carbon chain in a molecule
-what conditions for this
usually in the formula R-Mg-X or R-X
-conditions= dry ether because they will react with (deprotonate) H2O
and with acid
what are the products of both of these reactions
how are gingnard reagents formed
halogen group with Mg
what is the wittig reaction
-what is the reagent used
-and what product is formed that is the driving force for this reaction because it has a very strong bond
PPH3
-treat halide with PPH3
-red H is deprotonated by any strong base
forming the O=PPH3 is the driving force of the reaction because it is very thermodynamically favourable
what is the difference between a stabilised ylid and an unstablaised ylid
stabilised- has an EWG adjacent to the phosphorus
unstablaised- doesn’t have EWG/ will have neutral or EDG
-e.g the stabilised has a C=O next to P which is EWG
you can have a stabilised or unstablaised ylid
-which forms the E alkene and which forms the Z
why do you get the different E/Z alkenes from the unstablaised vs stabilise ylid?
-electronics and streics
-what is the mechanism for these reactions
-ylid approaches C=O at 90° angle in both mechanisms
Z selective: streics
-this keeps the largest substituents away from each other
-dotted lines are being formed
E selective: electrostatics and streics
MeO2 is EDG so want to keep that and the =O as far away from each other as they both have high electron density
when making these reagents you cant have any acidic or carbonyl functional groups present [acidic protons]
-why is this
-why are the reactions crossed out not possible
Mg-X group is basic
-the OH is acidic, so even if it formed the grignard, the MgBr groups is basic so it would just react with the acidic OH
-same with the C=O. MgBr and C=O would just react together within the same molecule
what are the products what all of these carbonyl groups are reacted with a gringard [R]
should be able to do this so far
how do you go from an acid to an acyl chloride
how do you go from an acid to an acid chloride
SCL2 / PCL5 / COCl2 and DMF
why are carboxylic acids strong acids
stability of conjugate base
what H’s are acidic in carbonyl groups
and the H on OH,
also true for the H on any other alcohol OH
3 ways to make an acid
-need to know and remember
what is the point of turning an acid into an acid chloride
to turn the carbonyl carbon more reactive
-Cl is a much better leaving group than OH
-Therefore the chiral carbon is much more reactive
what is the mechanism for this
the O-Ph bond and O=Ph bond are very strong and the driving force for most organic reactions
what is the functional group on an anhydride
what is the reactant to go from an acid chloride to an acid anhydride
what is the mechanism for this reaction (the last step)
the most important thing acid chlorides are good for is making esters and amides
-when acyl chloride is left in air, it will just hydrolyse back into its acid [reacts H2O in air]
ester and amides are key to remember
what is the mechanism for this
what does this form
different ways to make esters
mechanism for going from a ketone to an ester
this is a key reaction
-what are the reagents to go from an ester back to its acid
you can also do it with a base like NaOH
-doesnt have to be under acid conditions [i.e. with H2SO4]
must know the mechanism for this
with this reaction ester –> acid
not all esters will react as quick
-what makes an ester react slower/quicker
if the ester has a bulky group it will react slower
-the last one is completely inert to based hydrolysis
the last one is inert to the top reaction so this is the bottom reaction for it
mechanism going from an ester to a tertiary alcohol with gingard reagent
how do we stop this reaction at the aldehyde
use only one eq. and at very low temperature
-ignore how reactants are different they do the same job
these are ways to reduce an amide [get rid of the C=O]
-which reaction will use BH3, and which will used LiAlH4
the BH3 is selective
-but the LiAlH4 is not
must know the mechanism for the reduction of amides
reagents for acid to amide
this is amide reduction
top reaction: the amide is completely reduced to an alkane
bottom reaction: the amide is stopped halfway through reduction at an aldehyde
how is this?
-1 eq in the halfway reaction
amide —> aldehyde
-what other reagent can you use
amide —> ketone?
stable tetrahedral intermediate
for a reduction reaction, to stop it going all the way [to an alkane] you can stop it halfway at an aldehyde or ketone
-for these half reactions, the mechanism must have a stable tetrahedral intermediate
which groups do addition reactions, and which do substitution
hydration of an aldehyde with water and alcohol [R-OH]
Acetyls are easy to isolate
the formation of all of this is carefully controlled by your choice of conditions
-ROR is quite slow so you need to add an acid or a base
going from a semi-acetal to an acetal only works under what conditions
acidic
is ethanol a good or bad nucleophile
bad nucleophile
this is a reaction from an aldehyde to a hemi acetal
-ethanol is the nucleophile but it is not very good
-what conditions is this done under to speed it up
acid or base
are chained or cyclic hemi-acetals more stable
-and why is one of them more stable than the other
cyclic
-entropy favours the cyclisation process [is positive in this direction]
aldehyde [or ketone] –> acetal conditions and reactant
-acid must be in water
-EtOH can be any alcohol cyclic or an-cyclic or diol
-pTSA can be any acid (e.g HCl)
-acid increases the rate of reaction
what is the mechanism for this reaction
-acid protonates the reactant to make them more reactive
all these arrows are reversible; how do we force the reaction to go to completion and get a high yield
we need to remove water we can use
-molecular sieves
-MgSO4
-dean stark
what is an acetal funtional group
C bonded to two oxygens which are OR groups
are cyclic or an-cylic acetals more stable
cyclic ones are more stable/ need less heat to break down
why won’t this reaction work
[reduction reaction]
-and what would the product look like instead
the ester group is less reactive than the ketone, therefore the gringard will attack the ketone first [1 eq]
-1.5eq of the gringard will be left to attack the ester to reduce it back to its tertiary alcohol
how would we stop the blue highlighted product being formed instead of the desired product [in red]
- reactant + diol under toluene dean stark conditions
-this makes the next product inert to nucleophiles, so the top bit will not be attacked
in this example, making the gringard straight from the reactant wouldn’t work, why?
It would just react with itself because it still has a carbonyl group [ketone] on it
you need to add an extra step before you make the reactant a gringard. You need to make the reactant less reactive
-what step would you add in to do this
- reactant + diol under toluene dean stark conditions
-THEN turn that into a gringard [that won’t react with itself]
-then follow through rest of reaction
-ignore fact it doesn’t come out to final product in top equation you couldn’t make that anyway
instead of diols, you can also use the sulphur equivalent
doils can protect aldhyde and ketone functional groups, you react it with the ketone, let a reaction run its course, then turn it back to a ketone with acid
[explained in previous cards]
What are the reactants to go from primary aldehyde —> imine
then from imine to amine
the second step can be any acid in water
what is the mechanism for this
what’s usually the first step in any carbonyl mechanism
protonate the carbonyl group
this imine can also exist as another compound called an enamine
-draw this structure
how would this be different if the aldehyde was treated with a secondary amine, not a primary one
you will form the enamine, instead of the imine because the equilibrium lies to the enamine
-forming the enamine is more stable
aldehyde + primary amine –>
aldehyde + secondary amine –>
primary= imine
secondary= enamine
what does this have similar reactivity to
-will it react with a nucleophile or electrophile
electron rich alkene
-because N lone pair can also align itself with the double bond
-react with electrophiles
mechanism for this reaction
-nucleophillic sub
-protonate O-
what pH should this be done at
5
why is this done under dean stark conditions
to remove the water from the reaction
-to stop the reaction just going in a cycle
inmine/enamine chem is important to know
inmine and examines are tautomers of each other,
what are the different reactivities of each
imine-reacts like aldehyde or ketone [δ+ carbon]
-acidic H so can be deprotonated
enamines are nucleophiles
inmines can react with nucleophiles and enamines react with electrophiles
What would form if an inmine reacts with a
- gringard
-hydride source (H-)
gringard= corresponding tertiary amine
hydride = secondary amine
[same as if its a aldehyde or ketone]
which one is the inmine and which is the enamine
what products do they form
enamine just acts as an alkene
what is the major product formed
-addition
-get rid of charges with H+
-make and get rid of the water
-get rid of charges
aldehyde/ ketone + secondary amine –> enamine
-so why does C go to the iminium ion instead of the enamine
-what do you have to have to form enamine
an acidic proton/ proton in alpha postion
-1 has an alpha proton, but 2 doesn’t, so 2 will stop halfway whereas 1 will go to completion [to the enamine]
where is the alpha vs beta hydrogen
where is the alpha vs beta hydrogen
is an amine [NH2]
a nucleophile/electrophile/acid/base
nucleophile and a base
what carbonyl compounds are electrophiles
halogenoalkanes- list of most to least electrophilic compounds
if the product you’ve made is more reactive than your reactant, it will probably just react with your other reactant and you will not get the product you want
-you either want your product to be less reactive
-or you need to isolate it before the reaction can continue
mechanism for this
what are the tautomers pairs
what are the other tautomer forms of this ketone
what are the other tautomer forms of this ketone
what are the tatuomer forms of this and what useful reactant
what are the tautomer forms of this and what useful cyclic reactant does this form
enols are very useful intermediates, but usually the equiblirum doesn’t lie in their favour
-how do we force the equilibrium to form enols
aqueous acid
how does an enol react
similar to an alkene
which is the thermodynamic enolate, and which is the kinetic one
-which is more stable
thermodynamic one Is more stable
what conditions could you put the reaction under to make sure the thermodynamic proudut is formed
-use non bulky bases [small base]
-nucleophillic bases
-room temp or above
what conditions would you put the reaction under to get the kinetic product
-bulky base
-low temperatures
what is the difference between a kinetic and thermodynamic enolate
kinetic - less stable and less substituted
thermodynamic- more stable and more substituted
(substitution on the double bond)
no temperature information usually means its being done at room temp
react the product the MeI
consequence on enolisation = loose any stereochemistry, the product becomes racemic
how would a hard/ soft nucleophile react
what is the difference between a hard and soft electrophile
hard- more concentrated negative charge
soft- more diffuse negative charge
what are some examples of hard electrophiles
how would this molecule react with bromine Br2
which solvents encourage reaction at the oxygen and which promote reaction at the carbon
polar protic- oxygen
ethereal- ether and alcohol in its make up
draw an enol and anolate
which would form an enol and which the anolate
what is an aldol reaction
-what is the electrophile and what is the nucleophile
reaction between two carbonyl species to make a new C-C
-have to turn one into a nucleophile by making an anion, and leave the other to be the electrophile
aldol reaction- you need to turn one carbonyl into a nucleophile
-what does the molecule need to have to be the nucleophile
an acidic proton that can be removed
what is the problem of using an acid or base to make the carbonyl nucleophile for an aldol reaction as shown in this reaction
the acid/base are just regenerated at the end, which means the reaction just continues and doesn’t stop at the aldol product
aldol reaction
why could this reaction format not very good and forms many different product
both protons are as acidic as each other so it can make 2 enolates instead of 1
what does a 2eq usually indicate
its going to add to the ketone twice
what are 2 (3) different ways to stop the acid/base regenerating
-what other reagents can you use
lithium enolates-good for ketones
silyl enols - can be made two different ways [Me3 SiCl with either lithium enolate or a weak base]
notes- for second reaction need to make the aldehyde more reactive because the SiMe compound isn’t as reactive as the lithium enolate compound
what is the equation for how many diastereoisomers there are in a molecule
2 ^n
n=number of stereo centres in the molecule
are entatiomers similar to each other
-are disteroisomers?
enantiomers- are all the same (mp, bp, IR spec) except for the way they rotate plane polarised light
disteroisomers- completely different from each other
what is an intramolecular aldol reaction
nucleophillic and electrophilic areas of the same molecule which can react with each other
-usually forming (6 membered) rings
what are the different sites where enolisation could take place on this molecule
enolisation- an acidic proton {the proton next to a carbonyl group} reacts with an acid or base
-arrows are where hydrogens are next to the carbonyl group
with intramolecular aldol reactions, what is the preferred size of the ring formed
(5 and) 6 membered rings
why is enolate 2 favoured over enolate 1
-first step is forming both enolates that’s why there’s too many arrows
enolate 2 will form a 6 membered ring, whereas the first one only forms a 4 membered ring
-6 is move favourable
what are the steps to an (intramolecular) aldol reaction
base used in all
-deprotonation of the starting material to make an enolate (nucleophile) by base / acid
- reaction with another carbonyl to make product
-deprotonation 2 to get intramolecular aldol [make nucleophillic and electrophilic end] to make a ring
-addition of H to O-
-elimination of acidic proton and alcohol group to make a double bond and a C=O
product of this aldol reaction
product of this aldol reaction
rank these in term of reactivity (electrophiles)
what would be the mechanism for this aldol reaction
steps
-base used in all
-deprotonation of the starting material to make an enolate (nucleophile) by base / acid
- reaction with another carbonyl to make product
-make alcohol
-deprotonation 2 to get intramolecular aldol [make nucleophillic and electrophilic end] to make a ring
-addition of H to O-
-elimination of acidic proton and alcohol group to make a double bond and a C=O
