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PT IA Chemistry > Reactions and Mechanisms in Organic Chemistry Pt. 1 > Flashcards

Reactions and Mechanisms in Organic Chemistry Pt. 1 Flashcards

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1
Q

Define chirality

A

Chiral molecules are molecules which are non-superimposable mirror images of each other

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2
Q

what is required for chirality in molecules and why is it important

A
  • to have chirality a molecule must have a chiral centre, this is often a carbon attached to 4 different groups
  • Enantiomers (pairs of chiral molecules) are mirror images but do not necessarily have the same properties
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3
Q

how are bonds out of plane, bonds into plane and partial bonds represented

A

bonds out of plane of paper = wedge

bond into plane of paper = hashed

partial bond = dashed

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4
Q

why are C-C or C-H bonds not very reactive

A
  • they have a high bond enthalpy and are not strongly polarised
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5
Q

why are C-X bonds more reactive and in what way are they generally reactive

A
  • usually polarised and reactive towards nucleophilic attack
  • generally the carbon is attacked as it is delta +ve
  • they are much easier to break in a chemical reaction
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6
Q

what is the difference between primary, secondary and tertiary alcohols, give the names for the propane type alcohols for each

A

primary = -OH attached to a carbon that’s attached to 1 carbon e.g. propan-1-ol

secondary = -OH attached to a carbon that’s attached to 2 carbons e.g., propan-2-ol

tertiary = -OH attached to a carbon that’s attached to 3 carbons e.g. 2 methyl, propan-2-ol, or tert-butanol

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7
Q

what is an ether and what is something to note about them

A

the functional group of
R1-O-R2

R1 does not have to be the same as R2
e.g. tert-butyl methyl ether

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8
Q

what is an amine and what the difference between primary, secondary, tertiary

A

amine is the R(x+1)-NH(2-x)
x < 3

  • the terms primary/secondary/tertiary refer to the number of hydrogens on the ammonia which have been replaced
  • a primary amine is where one hydrogen on the ammonia has been replaced
  • a secondary amine is where 2 hydrogens have been replaced
  • a tertiary amine is where 3 hydrogens have been replaced
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9
Q

what is an amine and what the difference between primary, secondary, tertiary

A

amine is the R(x+1)-NH(2-x)
x < 3

  • the terms primary/secondary/tertiary refer to the number of hydrogens on the ammonia which have been replaced
  • a primary amine is where one hydrogen on the ammonia has been replaced
  • a secondary amine is where 2 hydrogens have been replaced
  • a tertiary amine is where 3 hydrogens have been replaced
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10
Q

what is the functional group of halides

A

molecules with the functional group of

R-X
where X is any one of the halogens, F,Cl,Br,I

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11
Q

what are generally considered carbonyl group derivatives

A
  • aldehydes, ketones, carboxylic acids, esters, amides and acid halides
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12
Q

difference between alkene and alkyne

A

alkene = C=C double bond

alkyne = C=_C triple bond

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13
Q

what is an aldehyde and what is a ketone

A
  • an aldehyde is where a hydrogen is attached to the carbon of the carbonyl bond i.e. R-CHO
  • a Ketone is where the carbon of a carbonyl bond is not terminal i.e. RCOR
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14
Q

what is a carboxylic acid and what is an acid halide

A
  • a molecule containing the functional group RCOOH
  • an acid halide is a molecule containing the functional group RCOX where X is one of the halides (C is terminal)
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15
Q

what is an ester, how do we name them

A

a molecule containing the functional group R(1)COOR(2)
Note R(1) and R(2) do not have to be the same

to name do alkyl alkyloate
e.g. methyl ethanoate
this would be where R(1) is ethane based and R(2) is methane based

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16
Q

what is an acid anhydride

A

R(1)COOCOR(2)

Note: R(1) and R(2) do not have to be the same but usually are

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17
Q

what is an amide

A

a molecule containing the group

RCONH2 = primary amide

RCONHR(1) = secondary amide

RCONR(1)R(2) = tertiary amide

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18
Q

what is a nitrile

A

a molecule containing the functional group R-C=_N (triple bond)

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19
Q

what is retrosynthetic analysis and what type of arrow do we used to represent ‘can be made from’

A
  • it’s a process where we look at the target molecule and ask ‘How can I make this’
  • we use a fat arrow
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20
Q

what must two or more orbitals be in order to interact

A
  • close in energy
  • overlap efficiently
  • be of suitable symmetry
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21
Q

what things must we consider when analysing molecular interactions (usually with HAOs)

A

primarily:
- HOMO/LUMO interactions
- electrostatic interactions

related:
- hardness and softness and hence, whether a reaction is FMO or electrostatics driven
- orbital coefficients
- symmetry

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22
Q

what is the important thing to consider when deciding which HOMO/LUMO interaction is most effective

A
  • relative, not absolute energy levels are important
  • this means the interaction with the lowest difference in energy between the HOMO of the nucleophile and LUMO of electrophile is the most effective
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23
Q

what is a curly arrow and where is it used, what are the different types

A
  • curly arrows represent the movement of a single or pair of electrons from a filled orbital (HOMO) to an empty orbital (LUMO)
  • the arrow starts with its tail on the source of the moving electrons representing the HOMO
  • it finishes with the head of the arrow pointing close to the empty orbital (LUMO)
  • a double headed arrow represents a pair of electrons
  • a single headed arrow represents a single electron
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24
Q

what is a nucleophile

A
  • an electron pair donor
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25
Q

what is an electrophile

A
  • an electron pair acceptor
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26
Q

what is the order of relative nucleophilicities

A

-ve charge > lone pair > pi-bond > sigma bond

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27
Q

what is the order of relative electrophilicities

A

carbocation > empty bonding orbital > pi* orbital > sigma* orbital

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28
Q

What are the two main steps in the mechanism of the nucleophilic addition of a carbonyl

A

1) nucleophilic addition to the carbonyl group

2) protonation of the anion that results

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29
Q

what are the key differences between the initial molecule and final molecule in the nucleophilic addition of a carbonyl

A
  • the initial molecule has a trigonal planar sp2 carbon
  • the final molecule has a tetrahedral sp3 carbon
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30
Q

what causes the nucleophile to attack the carbon of the carbonyl (4 main points)

A
  • the C=O bond has a large dipole, this means on electrostatic attraction alone we would expect the electron rich nucleophile to attack the electron poor carbon
  • both the C and O in the carbonyl group are sp2 hybridized, therefore the lone pairs on the oxygen are perpendicular to the pi system and are in sp2 HAOs
  • the LUMO is the pi* C=O orbital
  • the largest coefficient in the pi* orbital is the carbon, this makes it the most susceptible to attack
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31
Q

why does the nucleophile attack at 107 degrees

A
  • the ideal angle of attack would be >90 degrees to the C=O to give maximum orbital overlap with ‘splayed out pi*’
  • and repulsion from the filled pi bonding MO forces the nucleophile to attack at an even more obtuse angle
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32
Q

what are the three types of reaction we consider in the nucleophilic addition of carbonyls

A
  • hydride addition = NaBH4
  • organometallic reagents = compounds with C-Metal bonds
  • water and alcohols = formation of hydrates, hemiacetals, acetals
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33
Q

what is the thing we need to note about the way hydrides react with a carbonyl

A
  • the nucleophilic attack is NOT by the hydride ion itself
  • this is because it ‘prefers’ to act as a base to another molecule forming H2 and X-
  • this is because the filled 1s is of a preferred size to interact with the hydrogen atom’s contribution to the sigma* orbital of an H-X bond than the carbon’s contirbution to the C=O pi*
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34
Q

what reagent do we actually use for hydride addition reactions with carbonyls and why

A

NaBH4, it is a mild source of hydride for the reduction of an aldehyde

it’s a reducing agent

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35
Q

what is the mechanism for the hydride addition to a carbonyl group, give details of where electrons come from / go to etc.

A
  • a electrons from the B-H sigma bond attack the delta +ve carbon, this is the nucleophilic addition to the carbonyl group step
  • the pi bond of the C=O group breaks by heterolytic fission giving O a -ve charge
  • the O(-) attacks the H of the H-X group e.g. an alcohol or water. And the H-X bond breaks by heterolytic fission, this is the protonation step

NOTE: the boron does have a negative charge but the arrow does not start from the negative charge because there’s no lone pair on the boron so it must start from the B-H bond

H-X is a solvent molecule

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36
Q

what are the HOMO and LUMO in the borohydride/carbonyl addition reaction

A

HOMO = B-H sigma bonding MO

LUMO = C=O pi* antibonding MO

  • these orbitals are well matched so borohydride acts more as a reducing agent and not as a base
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37
Q

what happens to the BH3 generated after the loss of the hydride ion (in the addition to a carbonyl group)

A
  • BH3 is sp2 hybridized and electron deficient
  • it has an empty p orbital and therefore acts as a Lewis acid
  • it reacts very quickly with either the oxyanion or a molecule of the solvent
  • this can produce another tetravalent boron anion which can transfer a second hydride ion to another molecule of the starting carbonyl
  • in theory this can continue until all four hydrogen atoms have been used

SEE pg.30 for a detailed mechanism

38
Q

what is an organometallic reagent and what are the two types we consider

A

‘an organometallic reagent is any compound that contains a carbon bonded to a metal’

the two most discussed (in this section) are:
- Organomagnesium compounds (Grignard reagents)
- Organolithium compounds

39
Q

what is an important property of organometallic reagents and what does this allow to form

A
  • the electronegativity of the carbon is greater than any of the metals they bond to
  • this gives a delta(-ve) carbon which is unusual
  • it allows the C-C bonds to form
40
Q

How are organomagnesium compounds made

A

Grignard reagents/organomagnesium compounds are made by reacting alkyl, aryl or vinyl halides with magnesium ‘turnings’ and ether (diethyl ether/ ethyl ethanoate)

e.g.
CH3CH2CH2Br –(Mg, Et2O)—> CH3CH2CH2MgBr

41
Q

How are alkyl/aryl/vinyl organomagnesium compounds made

A

Grignard reagents/organomagnesium compounds are made by reacting alkyl, aryl or vinyl halides with magnesium ‘turnings’ and ether (diethyl ether/ ethyl ethanoate)

e.g.
CH3CH2CH2Br –(Mg, Et2O)—> CH3CH2CH2MgBr

This is a process called magnesium insertion
Don’t worry about mechanism in NSTIA

42
Q

How are alkyl/aryl/vinyl organolithium compounds made

A

A similar process to organomagnesium compounds, alkyl, aryl or vinyl halides are reacted with lithium, but this time the stoichiometry is 2:1 not 1:1
halide salts are also formed

e.g.

CH3Cl —(2Li, Et2O) —> CH3Li + LiCl

this process is called lithium halogen exchange

43
Q

how are alkynyl orgametallics made

A
  • a normal alkyl/aryl/vinyl organometallic is reacted with an alkyne
  • the C-Metal bond in the organometallic attacks the hydrogen on the end of the alkyne
  • the C-H bond of the alkyne then breaks by heterolytic fission where arrow starts on bond, ends on carbon
  • this formed an alkyne with the metal from the organometallic compound on the end and the rest of the initial organometallic compound with an H added on

e.g.

CH3C=_CH + H3CLi —–> CH3C=_CLi + CH4

or

RC=_CH + RCH2MgBr —–> RC=_CMgBr + RCH3

the alkyne can also be deprotonated using a strong nitrogen base such as Sodium Amide (NaNH2), here the negative Nitrogen attacks the hydrogen instead forming ammonia and the +ve sodium attaches to the Alkyne

44
Q

what are the stages the nucleophilic addition reactions of organometallic compounds with carbonyls and what is an important point to note

A

1) nucleophilic attack of the organometallic carbon into the carbonyl pi* (carbon end), the arrow comes from the middle of the C-Metal bond, reagent = organometallic compound, including alkynyl organometallics

2) protonation from alcohol or water (or whatever solvent is being used), if water + acid then do H3O+, the oxygen lone pair/-ve charge attacks the hydrogen on the water or H3O+ or alcohol and the O-H bond breaks by heterolytic fission, reagent = water, water+acid, alcohol

NOTE: DO NOT DO STAGES 1 AND 2 SIMULTANEOUSLY, stage 1 MUST be completely finished first before water or another protic solvent is added

45
Q

why should the two stages of the nucleophilic addition of organometallics and carbonyls not be done simultaneously

A
  • organometallic compounds R-Li and R-MgBr are incompatible with water or other protic solvents (such as alcohols, Note: if an alcohol group is on the reactant molecule you will need 2eq not 1)
  • this is because if they were then the C-metal electrons would attack the hydrogen on the protic solvent and the protic solvent would break down into a negative ion, the +ve metal ion would be removed from the organometallic and you end up with a normal organic molecule and a metal hydroxide

e.g.
CH3CH2=CH2CH2Li + H2O ——> CH3CH2=CH2CH3 + LiOH

46
Q

what is formed in the reactions of carbonyls with water and alcohols

A
  • Hydrates are formed when water reacts with aldehydes or ketones
  • Hemiacetals and acetals form when alcohols react with alcohols and ketones
47
Q

what is the overall reaction of aldehydes/ketones with water, what is the HOMO and LUMO

A
  • when water reacts with aldehydes/ketones, it forms 1,1 diols which are called hydrates

R(1) CO R(2) —(reversible)—> R(1)C(OH)2 R(2)

  • overall water is added across the C=O double bond

HOMO = l.p on oxygen
LUMO = C=O pi*

48
Q

describe each step of the reaction mechanism of an aldehyde/ketone with water to form a hydrate

A
  • arrow from lone pair on water’s oxygen to carbon of C=O bond (pi*), the C=O pi bond breaks by heterolytic fission (arrow to oxygen), forming O(-)
  • O-H bond on the molecule breaks as another water molecule protonates from the lost OH
  • the R(1)CO(-)OH R(2) protonates from another water
  • this forms a hydrate with an sp3 hybridized carbon
49
Q

Do aldehydes or ketones favour hydrate formation more and why

A
  • Aldehydes form more significant amounts of hydrate than ketones
  • this is because in the initial aldehyde/ketone the carbon is sp2 hybridized so has a bond angle of 120 degrees
  • when it undergoes the reaction with water it becomes tetrahedral/sp3 hybridized and has a bond angle of 109.5 degrees
  • this means it is harder to form the hydrate with larger ‘R’ groups because there will be a greater steric clash in the product than the initial carbonyl
50
Q

What molecules (other than aldehydes) also favour hydrate formation more

A
  • strained ring structures e.g. 3 or 4 member rings
  • this is because when the carbon is sp2 hybridized in the reactant the bond angle is 120 degrees, this is much higher than the ideal 60 degrees and forces other bond angles to be very small
  • when the reaction occurs and the carbon becomes sp3 hybridized with a 109.5 degree bond angle the bond angles are closer to their ideal size, this releases ring strain
51
Q

what is the mechanism for the reactions of aldehydes and ketones with alcohols and what are the products, HOMO/LUMO?

A
  • the (initial) product is a hemiacetal
  • the mechanism is exactly the same as the mechanism for water and an aldehyde/ketone
  • overall R-OH has been added across the carbonyl group

HOMO = l.p. on oxygen
LUMO = C=O pi* (on carbon)

52
Q

what is a catalyst

A

“A catalyst is a substance which increases the rate of a reaction by providing an alternative reaction pathway with a lower activation energy without being used up itself”

53
Q

what are the two ways to catalyse hydrate/hemiacetal formation and how do they work, do these work for acetal formation?

A
  • acid catalysis, this works by making the carbonyl group more electrophilic
  • base catalysis, this works by making the nucleophile more nucleophilic
54
Q

what is the mechanism for the formation of a hemiacetal (or the same for hydrate) using an acid catalyst

A
  • the alcohol (or water) is protonated to be R-OH2(+)
  • the lone pair on the oxygen of the carbonyl attacks the proton and becomes an activated C=OH(+) group
  • the carbon is then attacked by the lone pair of the oxygen on the alcohol (or water) as usual and the C=O pi bond breaks by heterolytic fission to form C-OH
  • the spare proton on the ester group is then attacked by another alcohol molecule to form our hemiacetal and then the proton will leave the alcohol group to be reformed and hence act as a catalyst

NOTE: ALL STEPS ARE STILL REVERSIBLE

55
Q

what is the mechanism for the formation of a hemiacetal (or the same for hydrate) using an base catalyst

A
  • the base will attack the hydrogen on the alcohol group (or water), this leaves an R-O(-) molecule
  • the lone pair on this group attacks the pi* LUMO of the carbon on the carbonyl as usual and the C=O pi bond breaks by heterolytic fission as usual
  • the O(-) on the C-O(-) then attacks the Base-H molecule formed in step 1 to protonate and form the hemiacetal

NOTE: ALL STEPS AR STILL REVERSIBLE

56
Q

what conditions are required for acetal formation

A
  • all of the steps of acetal formation are reversible
  • to push the acetal formation through to completion, we must use an excess of alcohol and/or remove the water from the reaction mixture as it forms
  • although hemiacetal formation can be catalysed by acid or base, acetal formation can only be catalysed by acid because the OH group must be made to be a good leaving group.
57
Q

what is the full mechanism for the formation of an acetal from a hemiacetal

A
  • the whole mechanism occurs under acidic conditions
  • the reactant alcohol is protonated, this protonated alcohol is attacked by the lone pair of the OH group of the lone pair on the hemiacetal
  • the lone pair of the oxygen on the ester attacks back into the structure and the the C-O bond to the OH2(+) group breaks by heterolytic fission
  • this causes the water to leave and forms a C=O(+)R group on the ester part
  • the carbon of this carbonyl is attacked further by another lone pair on an alcohol forming an ester and a C-O(+)HR group
  • this proton is removed by another alcohol and then the alcohol breaks down to reform the proton as the catalyst
  • an acetal is formed
  • for details see page 38
58
Q

what must we consider if we want to think about the reactions of nucleophiles with molecules other than aldehydes/ketones

A
  • leaving group ability
  • these molecules will undergo nucleophilic substitution which is best understood by looking at leaving groups
59
Q

what is a general mechanism for nucleophilic substitution (including intermediate)

A
  • the lone pair on the nucleophile (HOMO) attacks the carbon pi* of the C=O bond and the C=O pi bond breaks by heterolytic fission forming a C-O(-)
  • this forms a negatively charges intermediate, this intermediate breaks down by the lone pair on the C-O(-) being contributed back into the structure to reform the C=O double bond and another group leaves as its bond will break by heterolytic fission
  • this forms a carbonyl with a Nu on and the X that was on it originally is now a free negative ion
60
Q

what is the difference between the nucleophilic substitution situation and the nucleophilic addition situation

A
  • in nucleophilic addition the only groups present are R groups as it’s aldehydes and ketones
  • in nucleophilic substitution there are other groups present labelled ‘X’, these could be Cl- or RO-
  • the negatively charged tetrahedral intermediate formed in nucleophilic substitution is much less stable than that formed in nucleophilic addition so it encourages another group to leave
61
Q

what is the difference between the nucleophilic substitution situation and the nucleophilic addition situation

A
  • in nucleophilic addition the only groups present are R groups as it’s aldehydes and ketones
  • in nucleophilic substitution there are other groups present labelled ‘X’, these could be Cl- or RO-
  • the negatively charged tetrahedral intermediate formed in nucleophilic substitution is much less stable than that formed in nucleophilic addition so it encourages another group to leave
62
Q

in the nucleophilic substitution reaction of an acyl chloride and an alcohol ion such as EtO(-) how can we determine which group will leave the negative intermediate

A
  • we can consider how acidic the conjugate acids of each leaving group will be

for example if we had

MeCOCl + EtO(-) —MeC(O(-))(OEt)Cl–> MeCOOEt + Cl-

we know the Cl- will leave because the conjugate acids of the groups on the intermediate are

HCl = good acid
EtOH = not very acidic
CH4 = not acidic

  • this shows Cl- will be the most stable anion and is hence likely to be the best leaving group
63
Q

if we apply the same leaving group conjugate acid logic to the negative intermediate formed in the nucleophilic addition of an aldehyde or ketone what does it show

A
  • we will obtain an intermediate something like

PhC(O(-))(Me)Bu

considering the conjugate acids gives

Benzene = not acidic
Methane = not acidic
Butane = not acidic

hence the most stable solution is for the O(-) to protonate and form our alcohol (or similar product) as expected

64
Q

what are the Bronsted definitions of acids and bases

A
  • an acid is a species with a tendency to lose an proton
  • a base is a species with a tendency to gain a proton
65
Q

what are the key points about the pH scale and what are its issues

A
  • strongly acidic = 0
  • neutral =7
  • strongly basic = 14
  • it’s only a measure of the acidity of a particular solution and says nothing about the relative acidities of different acids in themselves, so tells us nothing about leaving group ability
66
Q

define pH

A

pH = -log([H3O+])

67
Q

for two solutions of equal concentration, what does a lower pH show

A
  • the solution of lower pH contains more acidic molecules
  • it is more dissociated and produces more H3O+ ions
68
Q

what is a better way to show tendency to lose a proton than pH and how is it calculated

A
  • we can consider the position of equilibrium in the dissociation of an acid
  • from this we can calculate a Ka value

for the reaction
HA + H2O —-(REVERSIBLE)—> H3O(+) + A(-)
Keq = [A-][H3O+] / [H2O][HA]

this can be simplified to Ka because the conc. of water remains virtually constant (55.56moldm^-3)

Ka = [A-][H3O+] / [HA]

69
Q

define pKa

A

pKa = -log(Ka)

70
Q

How does the stability of the conjugate base of an acid affect its pKa value and why is it just this factor

A
  • in all cases of acid dissociation water acts as a base and accepts a proton from the acid so H3O+ is always formed
  • this means the only difference from acid to acid is the stability of the anion formed when the acid is deprotonated.
  • the ease with which the anion is (re)protonated by H3O+ determines where the equilibrium sits
    e.g. Cl- is not easily protonated as its stable so the equilibrium position of HCl is far to the right, but CHCO2(-) is easily protonated so the equilibrium position of CH3COOH is further left
71
Q

make a statement about the strength of an acid compared to its conjugate base and its pKa

A
  • the stronger the acid, the weaker its conjugate base
  • the weaker the acid, the stronger its conjugate base
  • the stronger the acid, the more dissociated it is, so the higher the Ka value so the lower the pKa value
72
Q

How can we link pKa value to leaving group ability

A

the lower the pKa of the conjugate acid of a leaving group the better the leaving group because it implies a stronger acid so a more stable anion

73
Q

what are the 3 main things that a higher pKa value implies

A
  • implies weaker acid
  • anions decrease in stability
  • leaving group ability decreases
74
Q

How can we link pKa back to the difference between nucleophilic addition/substitution of a carbonyl

A
  • nucleophilic addition occurs where there is no good leaving group on the carbonyl carbon, this is where every group has a high pKa value
  • nucleophilic substitution occurs where there is a good leaving group on the carbonyl carbon, this is where there is (at least) one group with a low(er) pKa value
75
Q

what are the three key factors when determining anion stability

A
  • presence of electronegative atoms
  • delocalization of negative charge
  • strength of the A-H bond
76
Q

what is the general principle when considering how the presence of electronegative atoms links to anion stability, give an example of anions that show this

A

General principle: as we increase the electronegativity of the atom upon which a negative charge sits, we increase the stability of the anion

e.g.
C(-)H3 < N(-)H2 < O(-)H < F-
—-> increasing anion stability/decreasing pKa

this is because as you move across these elements the atom with the negative charge becomes more electronegative so the anion becomes more stable and the leaving group ability increases

77
Q

what is the general principle when considering how delocalization of negative charge affects anion stability, give an example of a series of anions that show this

A

General principle: the more resonance forms we can draw, the greater the stability of the anion

an example of this is considering the anions
ClO(-), ClO2(-), ClO3(-), ClO4(-)

as you move across this series you can draw more resonance forms, this means the anion stability increases, as is shown by the decreasing pKa value across the series

  • the ClO(-) anion has no delocalization, just inductive stabilisation on the negative oxygen
  • the ClO4(-) has 4 resonance forms so more delocalization and hence a more stable anion

increase delocalization —> decrease pKa —> increase anion stability —> better leaving group

78
Q

what is the general principle when considering how the strength of an A-H bond affects anion stability/acid strength, give an example of a series of acids that show this

A

General principle: the weaker the A-H bond, the stronger the acid generally

(not technically linking to anion stability but more how easily proton is lost)

HF > HCl > HBr > HI
—–> decreasing bond strength, and decreasing pKa

  • this means
    decrease bond strength —> decrease pKa —> increase acidity —> better leaving group
79
Q

derive and explain the order of decreasing leaving group ability/ decreasing acidity for
RSO2OH, RCOOH, ArOH, ROH

A
  • the acids become less acidic in that order/ become worse leaving groups
  • to explain this, consider their anions formed after deprotonation
  • look at resonance structures
    RSO2OH has 3 resonance forms with the negative charge on the electronegative oxygen
    RCOOH has 2 resonance forms with the negative charge on the electronegative oxygen
    ArOH has 1 resonance form with the negative charge on the electronegative oxygen and 3 resonance forms with the negative charge on a not very electronegative carbon
    ROH has no resonance forms
80
Q

what can be done to convert acetals back into hemiacetals and further back to ketones

A
  • treat with aqueous acid
  • the reverse reaction of acetal formation occurs and the backwards mechanism can also be derived from the forwards mechanism
81
Q

what is a carbon acid and why are they generally weaker than oxygen acids

A
  • a carbon acid is one where a proton is removed from carbon instead of oxygen
  • they are generally weaker than oxygen acids because carbon is less electronegative than oxygen
  • again, their strength/pKa value can be dependent on which atoms are nearby to the carbon which has the negative charge and whether they can delocalise
82
Q

explain the differences of pKa of the carbon acids of alkanes, alkenes, ketones, nitroalkanes, diketones

A
  • pKa decreases across this series i.e. acids become stronger

considering electronegativity and resonance structures:
- alkanes have no possible resonance and the charge is localised on the not very electronegative carbon so poor stabilisation
- alkenes can (if the correct proton is removed) delocalize to form different resonance structures e.g. propene has two resonance structures. still only carbon (low E-ve) so poor stabilisation (but better than alkanes)
- Ketones also allow for delocalization and different resonance structures but this time it can delocalize onto the oxygen which is more electronegative so the anion is more stable
- nitroalkanes can delocalize onto the nitro group (NO2) which is even more stable
- diketones can delocalize a lot onto two different oxygens so have good stability

83
Q

what is the effect of hybridization on pKa and why

A
  • the more s-character an HAO which contains the negative lone pair has the greater the stability of the anion
  • this is because s-orbitals are held closer to the nucleus than p-orbitals
  • so the electrons are lower in energy, and hence more stable
  • hence the anion is more stable so stronger acid etc.
84
Q

what are the three main factors to consider when looking at the nucleophilic substitution of different carbonyl groups

A
  • strength of the incoming nucleophile (identity of Nu: and pKa of conjugate acid)
  • reactivity of the carbonyl group (strength of delta+ on the carbonyl C)
  • leaving group ability (identity of X- and pKa of its conjugate acid)
85
Q

what are the main things to consider when looking at the strength of the incoming nucleophile (relating to the feasibility of nucleophilic substitution of carbonyls), give some examples

A
  • generally, the higher the pKa of HNu, the better the nucleophile Nu:
  • good nucleophiles are poor leaving groups because generally species which readily form bonds to hydrogen (high pKa) also readily form bonds to carbons

e.g.
R- is a good nucleophile but poor leaving group

Cl- is a poor nucleophile but good leaving group

86
Q

what are the main things to consider when looking at the reactivity of the carbonyl group (relating to the feasibility of nucleophilic substitution of carbonyls), give some examples

A

The inductive effect
- relates to the relative electronegativity of the group adjacent to the carbonyl carbon and withdrawal of electron density through the sigma framework
- the more electronegative the atom adjacent to the carbonyl carbon, the more delta+ the carbonyl carbon is so the more reactive it is towards attack

however we must also consider

The conjugative effect
- the conjugative effect involves delocalisation of the lone pair from the attached group X into the pi* system
- e.g. an amide group will have the lone pair of the nitrogen attacking the pi* on the delta+ carbon, this reduces the delta+ on the carbon and makes it less reactive towards nucleophilic attack
- the order of lone pair donation is N>O>Cl

taking the overall effect of these two gives Cl as the group attached to the carbonyl carbon which will give the greatest reactivity

87
Q

give the order of most reactive to least reactive carbonyls, when considering nucleophilic attack

A
  • Acid Chloride, inductive effect dominant
  • Anhydride, inductive effect dominant, central O delocalised into two C=O so each C is more delta+ than a standard ester
  • Ester, inductive and conjugative effects balance
  • Amide, conjugative effect dominant
  • Carboxylate anion, conjugative effect dominant
  • this order also corresponds to the C=O stretching frequencies (IR) for these molecules
88
Q

what are the main things to consider when looking at the ability of the leaving group (relating to the feasibility of nucleophilic substitution of carbonyls)

what are the leaving groups for the main types of carbonyl and how does this link to reactivity

A
  • acid chloride, leaving group = Cl(-) = good leaving group, low pKa (-7)
  • acid anhydride, leaving group = RCO2(-) = ok leaving group, low pKa (5)
  • ester, leaving group = RO(-) = poor leaving group, medium pKa (16)
  • amide, leaving group = R1R2N(-) = very poor leaving group, high pKa (33)
89
Q

explain why converting an acid chloride to an ester is likely to work using the three main principles for the nucleophilic substitution of a carbonyl

A

1) reactivity of the carbonyl, Cl is very electron withdrawing (high inductive effect) and has a low conjugative effect, making the carbon very delta+ and very reactive

2) incoming nucleophile = ROH, relatively poor

3) leaving group = R1,R2 or Cl(-)
Cl(-) is a very good leaving group

  • so combining these factors means overall this reaction is likely to work
90
Q

explain why the reaction of an ester to an amide may be able to work (under harsher conditions) using the three main principles for the nucleophilic substitution of a carbonyl

A

1) reactivity of carbonyl, OMe is quite electron withdrawing and not a very strong pi-donator so ester carbonyl is quite delta+ and quite reactive

2) strength of nucleophile, H3N: is an ok nucleophile (relatively high pKa of 33)

3) leaving group ability, OMe has a pKa of 16 so medium, this makes it an ok leaving group

overall this means we have a quite reactive carbonyl, an ok nucleophile and an ok leaving group

91
Q

explain why the reaction of an amide to an ester will not work using the three main principles for the nucleophilic substitution of a carbonyl

A

1) reactivity of carbonyl group, amides are the least reactive carbonyl molecule

2) strength of nucleophile, MeOH is a reasonably poor nucleophile

3) NH2(-) is not a good leaving group

overall this means the reaction will not take place

PT IA Chemistry (9 decks)

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