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Year 2 Chem > RSA > Flashcards

RSA Flashcards

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1
Q

What factors are there to consider when designing a synthetic route?

A
  • A pure sample of the desired product must be produced
  • The synthesis is efficient and short
  • The synthesis is convenient, ideally at room temperature and pressure, using non-toxic reagents
  • A flexible synthesis to allow for the preparation of analogues (for biological evaluation)
  • An inexpensive synthesis
  • High level of atom economy (maximise the number of atoms in the starting materials that end up in the products
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2
Q

What is chemoselectivity?

A

A chemoselective reaction is a reaction in which one functional group reacts in preference to another functional group or groups

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3
Q

What is regioselectivity?

A

A regioselective reaction is a reaction that leads to the selective formation of one structural isomer (regioisomer)

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4
Q

What is stereoselectivity?

A

A stereoselective reaction is a reaction in which one enantiomer, one diastereomer or one double bond isomer is formed selectively over others

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5
Q

What is a protecting group?

A

A protecting group converts a reactive functional group into an unreactive form such that after the desired transformation of the molecule, the original functional group can be regenerated

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6
Q

What is the most common protecting group for an aldehyde/ketone?

A

Acetals, aldehydes and ketones are susceptible to attack by nucleophiles, whereas acetals don’t react with nucleophiles

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7
Q

What is the most common protecting group for alcohols?

A

Silyl ethers (ROSiR3), for example the TBDMS group

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8
Q

How does the protection of an alcohol group work?

A
  • ClSiMe2tBu (TBDMSCl) reacts with R-OH as Si forms a stronger bond to O than Cl, this forms R-OSiMe2tBu (ROTBDMS) causes the loss of a HCl molecule
  • The reaction is now carried out with the part of the molecule that you want, the alcohol group is protected so doesn’t react
  • Bu4NF (TBAF) is then used to deprotect the alcohol group to form RO-, this works because Si forms a stronger bond to F than O
  • RO- is converted to ROH on workup with H+ and FSiMe2tBu is formed when TBAF reacts with the protected alcohol group
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9
Q

What is the most common protecting group for amines?

A

Carbamates, these are formed when an amine reacts with a BOC anhydride, the lone pair on the N reacts with one of the carbonyls, this forms R-NHCO2tBu, CO2 and OtBu-

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10
Q

How are carbamates deprotected?

A

CF3CO2H is reacted with carbamates to convert them back to amines

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11
Q

Why are carbamates weaker nucleophiles than amines?

A

Carbamates have the lone pair on N conjugated with the C=O

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12
Q

What is used as the protecting group for carboxylic acids?

A

Esters, they are not deprotonated by weak bases whereas carboxylic acids are

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13
Q

When drawing a retrosynthetic arrow which side are the products of the forward reaction on?

A

The products come before the arrow and the reactants come after the arrow

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14
Q

What is the general rule for which materials are readily available?

A
  • Compounds with up to around 6 carbon atoms and with one functional group are usually readily available
  • Straight chain compounds with one functional group are usually available up to 8 carbons
  • Cyclic compounds with one functional group from 5 to 8 membered are also available
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15
Q

What is the synthetic equivalent for a Benzene synthon when it is reacting with a relatively weak electrophile?

A
  • C6H5MgX for reactions with weak nucleophiles

- For reactions with strong nucleophiles, benzene is the synthetic equivalent

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16
Q

For benzenes with multiple substituents how do you choose which bond to disconnect first?

A
  • When disconnecting a bond from a benzene you need to keep the directing effects of the substituent that you disconnect second
  • For example if you have a benzene with a Br and an NO2 attached to the benzene and you want the Br and the NO2 1,4 to each other then you would have to disconnect the NO2 as the Br is 2,4 directing so some of the products you form will be your desired products
  • If you disconnect Br first then you will be attaching a Br while NO2 is directing so your Br will end up in the 3 position and some products you form will have it in the 5 position, this means you form no desired products
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17
Q

When you have two substituents that need to be 1,4 to each other but neither of them is ortho-para directing, what is the solution?

A

Perform a FGI to convert one of the substituents to a group that is ortho-para directing. In the forward synthesis this means you start with a substituent that is orhto-para directing and this is converted to the desired substituent in the final step of the forward synthesis

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18
Q

If you have a group that is ortho-para directing and you want to add Br at the 4 position, how can you improve regioselectivity?

A

You can perform a FGI to form a ortho-para director that is bulkier meaning steric hinderance prevents the Br from being added to the 2 position. If you pick a group to convert to that is less activating than the original ortho-para director then you don’t need to worry about polybromination

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19
Q

How do Grignard reagents react with electrophiles to form C-C bonds?

A

Electrons from the C-Mg bond attack the electrophile to form a new C-C bond and this eliminates MgX+ from the molecule

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20
Q

What do you get if you react an aryl Grignard with an aldehyde?

A

A secondary alcohol, workup is required to protonate O-

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21
Q

What do you get if you react an aryl Grignard with methanal?

A

Primary alcohol, workup is required to protonate O-

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22
Q

What do you get if you react an aryl Grignard with an epoxide?

A

A primary alcohol with a long chain, workup is required to protonate O-

23
Q

What do you get if you react 2 equivalents of an aryl Grignard with an ester?

A

A tertiary alcohol, the aryl Grignard reacts with the ester to form a ketone intermediate. Another aryl Grignard molecule reacts with the ketone to form a tertiary alcohol. Workup is required to protonate O-

24
Q

What do you get if you react an aryl Grignard with CO2?

A

A carboxylic acid, workup is required to protonate O-

25
Q

What do you get if you react an aryl Grignard with a ketone?

A

A tertiary alcohol, workup is required to protonate O-

26
Q

When performing a disconnection on a benzene ring with a substituent attached, where do you disconnect?

A

The bond between the benzene ring and the first atom of the substituent

27
Q

What is a stabilised ylide?

A
  • An ylide with an R group that can delocalise the negative charge on the carbon of the ylide
  • R= COR, CO2R, Ar, CN
28
Q

What is an unstabilised ylide?

A
  • An ylide where the R group can’t stabilise the negative charge
  • R = Alkyl group
29
Q

How are stabilised ylides prepared?

A

By deprotonation of a phosphonium salt, Ph3P+CH2R where R is an electron withdrawing group. The deprotonation is done with a weak base

30
Q

Which isomer of alkenes do stabilised ylides produce when reacting with aldehydes?

A

E alkenes because stabilised ylides often have large R groups, they favour the production of the E isomer as they are less sterically hindered and therefore more thermodynamically stable

31
Q

Do stabilised ylides react with ketones?

A

No, they only react with aldehydes as ketones aren’t reactive enough. This is because a ketone has 2 +I groups which stabilise the delta positive charge making the ketone less reactive

32
Q

Which isomer of alkenes do stabilised ylides produce when reacting with aldehydes?

A

Z alkenes

33
Q

Why can we use a weak base to deprotonate a phosphonium salt to form a stabilised ylide?

A

Because the anion is stable it means it readily loses H+

34
Q

How does selectivity of alkenes from stabilised ylides arise?

A
  • From reversible formation of two possible betaines
  • Erythro betaine is an intermediate where the bulky groups are on the same side hence it is less stable than threo betaine where the bulky groups are on opposite sides
  • The intermediate threo betaine has less energy hence the formation of the alkene from this intermediate is faster, the threo betaine produces the E isomer hence stabilised ylides favour formation of E alkenes
35
Q

How are destabilised ylides prepared?

A

Non-stabilised ylides are prepared by deprotonation of a a phosphonium salt, Ph3P+CH2R where R is an electron donating group. The deprotonation is done with a strong base as the anion isn’t stable meaning that H+ isn’t readily lost

36
Q

Do destabilised ylides react with ketones as well as aldehydes?

A

Yes, because destabilised ylides are more reactive than stabilised ylides meaning they can react with ketones which are less reactive than aldehydes

37
Q

When determining the synthons for a Wittig RSA, how do you determine which R group came from the ylide and which came from the aldehyde/ketone?

A
  • We know which isomer the starting material is so we know whether a stabilised or unstabilised ylide was used to produce it
  • If it is a stabilised ylide then we know that the R group with an electron withdrawing group came from the ylide
  • If it is an unstabilised ylide then we know that the R group that has an electron donating group (eg. an alkyl chain) came from the ylide and the electron withdrawing group was part of the molecule containing the aldehyde/ketone
38
Q

What is the Horner-Wadsworth-Emmons reaction?

A
  • A variation of the Wittig reaction that allows the formation of an E alkene from a ketone, this is significant because unstabilised ylides are the only ylides that can react with ketones and these form Z alkenes
  • Horner-Wadsworth-Emmons reaction uses a phosphonate ester that can stabilise the negative charge on the carbon, this means that it has a bulky R group and favours the formation of an E alkene
  • The negatively charge enolate type ion is far more reactive to C=O bonds than a neutral ylide
39
Q

How can alkynyl anions be used to form C-C bonds?

A

RC≡C- reacts with a range of electrophiles to form C-C bonds

40
Q

If you’ve got an alkyne with an OH group in the molecule and you want to perform an RSA on the molecule, why must you disconnect the carbon on the side of the triple bond with the OH?

A

Because if you try to deprotonate the alkyne and it has an OH attached then you will deprotonate the OH instead of one of the carbons in the triple bond

41
Q

Which base is used to deprotonate an alkyne?

A

NaNH2

42
Q

Which reagents are used to reduce and alkyne to an alkane?

A

Na, NH3

43
Q

When reacting an alkynyl anion with an alcohol what must you do before carrying out the reaction?

A

Protect the alcohol group with TBDMSCl as the alkynyl anion will act as a base and deprotonate the OH group

44
Q

What is the common synthetic equivalent for an O- attached to an alkyl chain?

A

An alcohol

45
Q

When forming an ester from an alkyne molecule containing an alcohol, why does the alkyne have to be reduced before converting the alcohol to an ester?

A

Because Na/NH3 that is used to reduce the alkyne bond can also be used to reduce C=O bonds

46
Q

What is the Heck reaction?

A
  • A disubstituted alkene is formed from a terminal alkene and an aryl or vinyl halide (the carbon bonded to X must be sp2 hybridised)
  • R-X + H2=CHR1 → RHC=CHR1, where R= aryl or vinyl
  • This reaction requires a palladium(0) catalyst
47
Q

What selectivity occurs in the Heck reaction?

A
  • Regioselectivity, the new R group is introduced at the least hindered carbon, because the Heck reaction starts with a mono-substituted alkene, the R group is added to the carbon that R1 isn’t attached to
  • Stereoselectivity, an E alkene is formed as this isomer has less steric hinderance between bulky R groups
48
Q

How are α,β-unsaturated carbon compounds produced?

A

Via an aldol condensation reaction

49
Q

Why are α,β-unsaturated carbon compounds susceptible to attack at both the 2 and 4 positions?

A

α,β-unsaturated carbon compounds have resonance forms where there the positive charge can be at both the 2 position and the 4 position

50
Q

Do Grignard reagents mainly attack Michael acceptors by 1,2 or 1,4 addition?

A
  • 79% 1,2 , 21% 1,4
  • The reaction is described as being kinetically controlled as the stronger C=O bond is broken and the weaker C=C bond is retained
51
Q

Do organocopper reagents and cuprates mainly attack Michael acceptors by 1,2 or 1,4 addition?

A
  • 98% 1,4 , 2% 1,2
  • Reaction described as being under thermodynamic control because the stronger C=O bond is retained and the weaker C=C bond is broken eg. more stable product is formed
52
Q

How are organocoppers prepared?

A

From organolithium compounds

53
Q

In a 1,4 Michael reaction, how does a cuprate attach to Michael acceptor?

A
  • The bond between the Cu and an R group attacks the 4 position, the double bond moves to between carbon 2 and 3, and the C=O bond becomes C-O-
  • This movement of electrons forms an enolate intermediate that reacts with an H+ ion and the C=O bond is reformed
54
Q

Why do some nucleophiles have a preference for 1,2 and others have a preference for 1,4 addition?

A
  • The carbon in the 2 position that is directly attached to oxygen is very delta positive and is described as a hard electrophile, this preferably reacts with a hard nucleophile which is highly delta negative. The Grignard reagent is a hard nucleophile because the R group has a high δ- density as it next to a highly δ+ Mg atom and therefore the Grignard reagent prefers to react by 1,2 addition
  • The carbon in the 4 position is not very delta positive and is described as a soft electrophile, this preferably reacts with a soft nucleophile which has low δ- density (delocalised nucleophile - charge spread out). The R groups in R2CuLi have a relatively low δ- density as they’re next to a weakly δ+ Cu atom, hence R2CuLi is a soft nucleophile and prefers 1,4 addition

Year 2 Chem (5 decks)

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