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Chemistry > Reactions And Mechanisms In Organic Chemistry: Part 2 > Flashcards

Reactions And Mechanisms In Organic Chemistry: Part 2 Flashcards

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1
Q

What is the mildest reducing agent that can be used as. A source of hydride?

A

NaBH4

For reduction of ketones, acid chlorides/anhydrides and esters (very SLOW) to alcohols

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2
Q

What possible hydride reducing agents may be used, from mildest to strongest ?

A

NaBH4, LiBH4, LiAlH4, BH3 (borane)

Use the mildest possible!!

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3
Q

Make a list of carbonyl derivatives, from most reactive to least reactive to incoming nucleophiles.

A 
Acid chlorides
Acid anhydrides
Ketones/aldehydes
Esters
Amides (so unreactive will not react with NaBH4)
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4
Q

What can LiBH4 be used to reduce?

A

Acid chlorides/anhydrides, ketones/aldehydes and esters.

Therefore chemoselectivity for esters over amides

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5
Q

What can LiAlH4 be used to reduce?

A

All carbonyl derivatives (although v slow with carboxylic acid- use borane)

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6
Q

Describe the reaction of borohydride with acid chloride / anhydrides

A

Attacks carbonyl group, forming tetrahedral intermediate with good leaving group (Cl or RCOO) so aldehyde formed. This is reduced further by the borohydride to the alkoxide which can be protonated by the solvent/workup.

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7
Q

Why aren’t reactions of LiAlH4 (e.g. reduction of Ester in analogous mechanism to reduction of acid chloride) done in alcoholic solvents?

A

Very reactive and incompatible with water.

Therefore protonation of the alkoxide is done in a workup step.

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8
Q

Why does LiAlH4 reduce carboxylic acids very slowly?

A

The first step of this reaction is removal of the proton to leave a carboxylate anion and H2 gas, which is less reactive to attack by the hydride anion (both negative).

The reaction with borane requires 3 RCOOH equivalents to produce triacylborate, which reacts with borane once more to produce the alcohol, RCH2OH.

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9
Q

What is different about the reduction of amides compared to all other carbonyl compounds using LiAlH4?

A

Produce AMINES, instead of alcohols.

The tetrahedral intermediate does not have a good leaving group and so the oxyanion coordinates the remaining AlH3 (into vacant p orbital), following which the nitrogen lone pair can expel oxygen, rather than the other way around.
The resultant iminium ion is analogous to the aldehyde in previous reductions- the C=N bond can be attacked by AlH4- and an anime formed.

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10
Q

How do organometallic reagents react with acid chlorides/anhydrides?

A

1 EQUIVALENT of reagent added to acid chloride/anhydride to produce a KETONE.

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11
Q

How do esters react with organometallic reagents?

A

2 EQUIVALENTS of reagent added to Ester (but also acid chlorides/anhydrides, though these are less easy to handle) to produce an alcohol.

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12
Q

Why do the products vary when organometallic reagents react with acid chlorides/anhydrides and esters?

A

In the case of acid chlorides/anhydrides: starting material is more reactive than the ketone product, so will react with the organometallic species in preference (only very small amount of alcohol produced).

In the case of esters: intermediate ketone is more reactive than the starting material so reacts in preference (note: if only 1 equiv present, 50% of starting material will remain).

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13
Q

How will organometallic species react with CO2?

A

To form a carboxylic acid!

After protonation

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14
Q

How do the reaction conditions for hydrolysis of each of the carbonyl derivatives to form carboxylic acids vary?

A

Acid chlorides - fast at 20oC
Acid anhydrides- slow at 20oC
Esters- only on heating with acid/base catalyst
Amides- only on prolonged heating with strong acid/base

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15
Q

How does the addition of acid increase the rate of hydrolysis of esters/amides?

A

Protonates the carbonyl oxygen, increasing its electrophilicity.
Protonates the leaving group

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16
Q

How do the mechanisms for acid-mediated hydrolysis of esters and amides vary?

A

Ester: truly catalytic in acid (catalyst regenerated)
Amide: the amine that leaves is protonated, so overall one equivalent of acid is used up .

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17
Q

How does the addition of base increase the rate of hydrolysis of Esters/amides?

A

Creating a negatively charged nucleophile (OH-)
Deprotonating the carboxylic acid product, pulling the equilibrium over irreversibly towards the product.

NEITHER are catalytic in base.

(Note: mechanism same for both, except equilibrium in first step of amide hydrolysis is further towards the starting materials due to relative pKa values)

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18
Q

How does acid-mediated esterification of carboxylic acids occur?

BASES MEDIATED ESTERIFICATION DOES NOT WORK

A

By the opposite mechanism of Ester hydrolysis; it is an equilibrium process, which can be pushed towards the formation of Ester by having an excess of alcohol present.

(Pushed towards carboxylic acid by an excess of water).

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19
Q

What is transesterifation?

A

The process of exchanging the R’ group in an Ester for the R’’ group of an alcohol.

This can be acid or base mediated

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20
Q

What is a soft nucleophile?

E.g. bromide

A
  • Large
  • High energy lone pairs (i.e. high HOMO)
  • Often uncharged or have charge spread diffusely over large orbitals.
  • Tend to react quickly at saturated carbon but slowly at C=O groups.

This means:
Reactions are dominated by FMO interactions and electrostatics are not important.

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21
Q

What are hard nucleophiles?

E.g. OH-

A
  • Small and high charge density
  • Attack C=O groups rapidly
  • Electrons concentrated close to the nucleus (low energy HOMO)
  • Basic

This means:
Large HOMO-LUMO gap means reactions are driven by electrostatics before orbital interactions can occur.

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22
Q

What is the difference between an SN1 and SN2 nucleophilic substitution reaction?

A

SN1- first order (one molecule in RDS) with formation of reactive sp2 intermediate
SN2- second order (two molecules in RDS) with formation of transition state

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23
Q

What are the 4 main reasons for differences in the mechanism of nucleophilic substitution at saturated carbon?

A

Substrate structure
Nucleophile
Leaving group
Solvent

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24
Q

How does substrate structure determine whether or not a reaction will go via SN1 or SN2?

A

STERIC HINDRANCE: SN2 more likely with fewer substituents on C centre (approach of Nu more difficult/higher energy transition state due to steric clashes, especially with 90 bond angles), whilst SN1 more likely with more substituents (stabilisation of carbocation).
CARBOCATION STABILITY IN SN1: hyperconjugation, pi-bond and lone pair donation increase carbocation stability.
TRANSITION STATE STABILITY IN SN2: increased by adjacent double bonds and carbonyl groups.
HYBRIDISATION STATE: neither SN1 nor SN2 will occur at sp2 centres.

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25
Q

What is hyperconjugation?

A

(Sigma-conjugation)

Carbocation stabilised by WEAK/PARTIAL DONATION of adjacent SIGMA bond electrons into the empty P-ORBITAL of carbocation due to slight overlap of orbitals (imperfect alignment).
This can come from C-H or C-C bonds (almost equiv in stabilising abilities), with a max of 3 donations occurring simultaneously in a 3 carbocation (only one sigma bond from each group will be parallel to empty p orbital).
This means no donation of this sort is possible in a CH3+ carbocation.

26
Q

How does pi bond donation increase carbocation stability?

A

Positive charge stabilised by delocalisation.

More resonance forms=more stabilised.

27
Q

How does lone pair donation increase carbocation stability?

A

Lone pairs may delocalised into empty p orbital. However it is also likely that the lone pair assists in departure of the leaving group in an intramolecular process.

28
Q

Order the stabilising factors in terms of their relative strength at stabilising the carbocation intermediate.

A 
Weakest: C-X sigma bond donation (X= N>O>S>Hal)
C-H/C-O
Ar/C=C
O lone pair
N lone pair
29
Q

Why is it that BOTH electron-donating and withdrawing substituents are able to stabilise the SN2 transition state?

A

The two partial bonds formed have some character of both filled and vacant orbitals (filled HOMO and vacant LUMO mixed).
This means other orbitals offering stabilising delocalisation may themselves be vacant or filled.

30
Q

Which major groups stabilise the SN2 transition state?

A

Double bonds (e.g. C=C, phenyl)- adjacent pi and pi* stabilisation via withdrawal AND donation.

C=O adjacent- only stabilised by pi* (vacant) because of strong pull by O in pi, meaning only electron withdrawal.
NOTE: C=O also lowers E of starting material LUMO for a faster reaction though overlap of C=O pi* and C-LG sigma*.

31
Q

What are the stereochemical outcomes of SN1 and SN2 reaction processes?

A

SN1- racemic mixture

SN2- single enantiomer

32
Q

Why won’t SN1 nucleophilic substitution reactions occur at sp2 centres?

A

Positive charge of carbocation cannot be conjugated/stabilised because all occupied p orbitals are perpendicular to the empty sp2 HAO.

33
Q

Why won’t SN2 nucleophilic substitution reactions occur at sp2 centres?

A
  1. Approach of Nu repelled by electron density of double bonds.
  2. Sigma* orbital points into the aromatic ring so approach of Nu blocked.
  3. Inversion impossible.
34
Q

In which nucleophilic substitution mechanism is the nucleophile important?

A

SN2- plays apart in the RDS

35
Q

What makes a good nucleophile for NS at the Carbonyl group?

A

A base

i.e. the higher the pKa of HY, the better the nucleophile Y

36
Q

What makes a good nucleophile for NS at the saturated carbon (SN2 reactions)?

A

If the atoms forming the bond to carbon are the same (e.g. OH-, TsO-), then a strong base (weakest acids).

If the atoms forming the bond to carbon are different (e.g. PhO-, PhS-), the better HOMO-LUMO interactions (soft nucleophiles- small electronegativity difference at carbon so not under charge control)

37
Q

In which NS mechanism is the leaving group important?

A

Both- involved in RDS.

38
Q

Which halides act as the best leaving groups?

A

Those with the LOWEST BOND STRENGTH and LOWEST pKa (of HX)

So I- is the best.

39
Q

Why is OH- not usually a good leaving group?

A

PKa of H2O is 15.7.

Nucleophile often just deprotonates OH group.

40
Q

How can the OH group be made into a better leaving group?

A

By lowering its pKa:

PROTONATION by acid- formation of ROH2+, with pKa=-1.7 (loss of water is RDS).

SULFONATE ESTER FORMATION- using TsCl and pyridine to form ROTs (pKa TsOH=-0.6). This requires OH group to act as a nucleophile and attack the sulfonyl chloride, followed by deprotonation by pyridine to give Py-Cl as a byproduct. This is the better option in a reaction where acidic conditions cannot be used (e.g. organometallic species).

41
Q

Why are ethers so unreactive in NS reactions but epoxides are ?

A

Ethers- little bond angle strain, strong C-O bond, alkoxide poor LG (better if protonated).

Epoxide- although alkoxide LG, there is severe bond angle strain (60, 49.5 lower than desired) and C-O weaker due to poorer overlap.

42
Q

What is the optimum solvent to use in an SN1/SN2 reaction?

A

SN1- polar, protic (stabilises carbocation and sequesters LG. Solvent may also be nucleophile- solvolysis).

SN2- polar, aprotic (e.g. acetone, DMSO- no solvation of nucleophile but solvation of its countercation, freeing it up).

43
Q

What are the two types of mechanism for elimination?

A

E1- formation of carbocation intermediate, and if it has a proton present, elimination may occur.

E2- base removes proton and leaving group leaves simultaneously (transition state forms).

44
Q

Describe the orbital interactions in an E1 elimination.

A

For an alkene to form from the carbocation (and not NS), the C-H sigma bind of the proton being removed must be parallel to the empty p orbital.

Two HOMO-LUMO interactions:

  1. Lone pair on base with C-H sigma*
  2. C-H sigma with empty p orbital
45
Q

Describe the orbital interactions in an E2 elimination.

A

Most efficient overlap when C-H sigma bond and C-LG sigma* are antiperiplanar to one another (180 degrees, same plane), allowing for the “flow of electrons” from Nu to LG.

Two HOMO-LUMO interactions:

  1. Lone pair on B, C-H sigma*
  2. C-H sigma bond and C-LG sigma*
46
Q

What are the stereochemical consequences of these elimination mechanisms?

A

E1- trans isomer is major product (least steric clash, rotation possible)
E2- dependent on arrangement of substituents in SM

47
Q

How does substrate structure determine whether E1/E2 will occur?

A

Steric factors which disfavour SN2 reactions do not affect elimination reactions.
Otherwise many of the same rules apply to E1/E2 substrate structure as SN1/SN2 (though more often both mechanisms are possible).

48
Q

Give examples of substrates which will not eliminate.

A

Me-X

ArCH2-X

(CH3)3CH2-X

49
Q

How does nucleophile basicity determine whether a substitution or elimination reaction will occur?

A

Strong base - elimination (better at removing proton)

Weak base- NS

50
Q

How does size of the nucleophile determine whether a substitution or elimination reaction will occur?

A

Easier access to exposed H atom as opposed to C centre means that:
Bulky Nu favour elimination.
Non-bulky Nu favour substitution

51
Q

How does temperature determine whether a substitution or elimination reaction will occur?

A

High T- elimination

(Due to positive entropy change in elimination and associated free Gibbs energy change).

52
Q

What can make a double bond nucleophulic/electrophilic?

A

Nucleophilic (electron rich)- lone pair donation or sigma donation.
Electrophilic (electron poor)- electron withdrawing groups nearby

53
Q

With the addition of H-X across an unsymmetrical double bond, what determines the major product?

A

The most stable carbocation intermediate- note that Markovnikovs rule (H adds into atom already with most H’s) does not work for all electrophilic addition reactions.

54
Q

What is the mechanism for the bromination/iodination of a double bonding inert solvent?

A

Initial REVERSIBLE formation of a bromonium/iodonium ion, a three membered ring, which is possible because bromine and iodine are such large atoms (HOMO-LUMO: C=C pi bond and X-X sigma*).
The bromide/iodide ion then adds onto the bromonium/iodonium via an SN2 mechanism.

55
Q

What is the stereochemical outcome of bromination/iodination?

A

Because of the nucleophilic attack on the fused ring system, which is a ‘back-side attack’, the product is TRANS.

56
Q

What happens if bromination occurs in water?

A

Nucleophile is water which attacks the bromonium ion, forming a BROMOHYDRIN product.

Same would occur if an alcohol was the solvent.

57
Q

How may epoxides be formed from trans-halohydrins?

A

Treat with base to deprotonate the alcohol. This leads to a rapid SN2 reaction, expelling the halide.

58
Q

How else can epoxides be formed?

A

Synthesised in one step from an alkene precursor and the peroxy-acid m-CPBA.
Overall one O atom is transferred from the m-CPBA to the alkene, in a single step mechanism:
1. HOMO-LUMO: C=C pi bond and O-O sigma*
2. Push rest of arrows round, forming carboxylic acid and other epoxide bond.

Here cis-alkenes give cis-epoxide and trans-alkenes give trans-epoxides.

59
Q

How is it possible to obtain the alcohol product of opposite regioselectivity from the addition of water to an alkene?

A

Addition of borane followed by hydrogen peroxide, NaOH and H2O.

60
Q

What is the first step of this hydroboration?

A

Addition of borane across the double bond:

  • Pi bond e’s donated into empty B p orbital, with the partial positive charge going onto the carbon best able to stabilise it.
  • BH2 than H so ends up where there is least steric hindrance.

This can occur 3 times to form a trialkylated boron.

61
Q

What is the second step of this hydroboration?

A

Breakdown of the trialkylborane with hydroperoxide anion, during which the C-B bonds are oxidised to C-O bonds:

  1. HO-O- adds into empty boron p orbital, generating unstable intermediate.
  2. O-O bond is weak and breaks to lose OH. At the same time a C-O bond is formed by MIGRATION of one of the alkyl groups from B to O (into O-O sigma*).
  3. Hydroxide anion assists departure of boron species.

Chemistry (7 decks)

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