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Organic Chemistry > Heteroaromatics > Flashcards

Heteroaromatics Flashcards

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1
Q

pyridine

A

nitrogen l.p. 90 to ring system (maintains aromaticity)

less e- rich than benzene (EW effect of N)

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2
Q

which compound does pyridine have similar chemistry to?

A

C=N looks like an imine

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3
Q

pyrrole

A

cyclopentadienyl ring with nitrogen

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4
Q

how does benzene form pyrrole?

A

replace HC=CH with N (ring contraction)

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5
Q

how does pyrrole still count as an aromatic compound?

A

to maintain Huckel’s rule, e- lone pair on nitrogen is within system (6 pi electrons)

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6
Q

furan

A

cyclopentadienyl ring with oxygen

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7
Q

thiophene

A

cyclopentadienyl ring with sulfur

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8
Q

which compound does pyrrole have similar chemistry to?

A

C=C-NH looks like an enamine

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9
Q

are pyrroles electron rich or deficient

A

[rich]

N lone pair adds electron density to pi system

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10
Q

are the protons more or less deshielded in pyrrole compared to benzene

A

less - more upfield (more e- density = more shielding)

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11
Q

properties of pyridine lone pair?

A

nucleophilic + can act as a base (lone pair on N = available)

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12
Q

how can the e- density of pyridine change?

A

adding/removing substituents

e.g. EDG would add e- density and increase nucleophilicity

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13
Q

why is unsub. pyridine so resistant to SEAr?

A

HOMO is too low in energy (addition of nitrogen)

N has given 2 e- to make bond ∴ e- deficient + pulls e- density from aromatic ring + makes intermediate v. low in energy

+ N lone pair is v. basic -> react with acid to form conjugate base, which drops HOMO/LUMO even further in energy

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14
Q

how do you make SEAr possible on pyridine?

A

add EDG group to ring / activating group

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15
Q

examples of strong EDG

A

OH, OR, NH2, NHR, NR2

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16
Q

why isn’t an EWG required for SNAr of pyridine?

A

-ve charge stabilised by nitrogen

if LG is @2/4 position, stable Meisenheimer intermediate is formed

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17
Q

quinoline

A

10 pi electron system

N lone pair perpendicular to pi system

additional benzene ring reduces aromaticity

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18
Q

quinoline sub. reactions

A

SEAr on benzene side - not regioselective (mixture at both 5 and 8 position)

SNAr on pyridine side - needs activating like in pyridine (form N-Oxide)

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19
Q

which protons are acidic in quinoline?

A

C2/C4 protons

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20
Q

conditions for hydrogenation of pyridine ring side of quinoline

A

H2, Pt, MeOH, rtp

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21
Q

conditions for hydrogenation of benzene side of quinoline

A

H2, Pt, 12M HCl, rtp

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22
Q

what compound is pyrrole compared to?

A

C=C-N-H resembles an enamine

pyrrole =. e- rich (N lone pair = e- donating)

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23
Q

pyrrole protons compared to benzene protons

A

pyrrole protons = more upfield

-> more e- density on them
-> more shielded

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24
Q

pyrrole SEAr

A

easy as it’s e- rich

most stable @C2 as there’s more resonance structures

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25
Q

what happens if pyrrole is placed in acidic conditions

A

just polarise to pyrrole

makes nitration/sulfonation difficult

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26
Q

oxygen equivalent of pyrrole

A

furan

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27
Q

sulfur equivalent of pyrrole

A

thiophene

28
Q

pyrrole + equivalents - aromaticity

A

furan -> pyrrole -> thiophene

thiophene -> most aromatic as charge is most delocalised due to difference in orbital size
furan -> least aromatic as O holds onto lone pairs more tightly

29
Q

pyrrole + equivalents - reactivity

A

thiophene -> furan -> pyrrole (most)

^ due to different orbital overlap between lone pair and 2p carbon orbitals

30
Q

indole

A

10 pi electron system

e- rich -> N lone pair = ED

31
Q

indole protons

A

C3 = more shielded than benzene

C2 = more similar to benzene (unlike in pyrrole)

^ only 1 area of increased e- density (@C3)

32
Q

where does SEAr occur in indole + reason

A

C3

despite more resonance for C2 -> resonance = destabilising due to breaking aromaticity in benzene ring

33
Q

how does a group migrate to C2 position during indole SEAr

A

sigma bond on group overlaps with pi* in iminium ion

= highly electrophilic

34
Q

how to access C-H acidic proton in indole

A

NH must be protected

e.g. Boc group or transmetallation (Li group -> Cu to form cuprate)

35
Q

advantages of Bartoli reaction

A

v. quick (10mins)

36
Q

disadvantages of Bartoli reaction

A

involves Grignard -> chemical incompatibility

37
Q

azole

A

> 1 heteroatom in ring structure

38
Q

imidazole

A

[2 nitrogens in ring system] - 1,3 position

6 pi electron system

N = sp2 hybridised to maintain aromaticity (1 lone pair in pi system and other at right angle)

39
Q

what are the effects of adding sp2 N?

A
  1. ring less electron rich (N pulls e- density)
  2. N l.p. = v. basic (resonance)
  3. N-H proton = more acidic
40
Q

types of 1,3-azoles

A

pyrrole -> imidazole

furan -> oxazole

thiophene -> thiazole

41
Q

effect of that adding O/S on basicity of N

A

reduces basicity - addition of adjacent heteroatom

oxygen = v. EW but can’t push as much e- density back into system
-> lone pair not as available => less reactive

42
Q

order of basicity of 1,3-azole derivatives

A

oxazole < thiazole < imidazole

43
Q

SEAr of 1,3-azole

A

C5 = preferred

44
Q

C2 deprotonation of 1,3-azole

A

C2 = most acidic AFTER NH (must be protected first)

45
Q

pyrazole

A

[2 nitrogens in ring system] - 1,2 position

6 pi electron system

lone pair perpendicular and donates 1 e- from unbonded p-orbital

46
Q

1,2-azole derivatives

A

furan -> isoxazole

thiophene -> isothiazole

47
Q

which is the least basic 1,2-azole derivative?

A

[isoxazole]

when forming conjugate base, N next to EW heteroatom (O)

got no e- density to give away -> v. unstable

48
Q

why is pyrazole less basic than imidazole?

A

cation adjacent to EW atom -> pulls e- density = destabilisation

lone pair is less available

49
Q

SEAr in 1,2-azoles

A

C4 = preferred

50
Q

SNAr in 1,2-azoles

A

doesn’t occur -> LUMO = too high in energy

51
Q

C5 deprotonation in 1,2-azoles

A

C5 proton = most acidic after NH (must be protected)

52
Q

what are cross-coupling reactions used for?

A

form new bonds between sp2-sp2 hybridised carbons

53
Q

what catalyses cross-coupling reactions?

A

transition metals

mainly palladium -> due to its ability to jump between 16e- to 14e- complexes

54
Q

Pd(0) properties

A

less stable

55
Q

Pd(II) properties

A

more stable

can be converted to active Pd(0) in situ

although it WON’T cross-couple -> higher cost; higher chance of side reactions

56
Q

what is the intermediate for organometallic synthesis?

A

aryl lithium species

57
Q

why does direct-lithiation / halogen-metal exchange work?

A

sp2 protons = more acidic than sp3 (has more s character)

need base with higher pKa e.g. tBuLi @ LOW TEMP.

58
Q

what limits direct deprotonation of unsub. aromatic rings?

A

pKa and anion stability

-> can use protecting groups to overcome this (direct metalating groups / DMG)

59
Q

when does lithium halogen exchange work?

A

when halide is large (bond = weak)

60
Q

mechanism for lithium-halogen exchange

A

E2

61
Q

Suzuki-Miyaura reaction advantage

A

greener alternative -> waste = boron salt (not v. toxic + easy to dispose)

62
Q

oxidative addition - KINETICS

A

weaker carbon-halogen bond = faster process

oxidative addition = RDS

occurs at most e- deficient carbon

63
Q

transmetallation - KINETICS

A

more nucleophilic organometallic = easier reaction

boronic acids need to be activated (not nucleophilic)
-> converted to boronate in situ

boronate nucleophilic @ carbon

64
Q

Negishi reaction advantage

A

faster than Suzuki

65
Q

Stille reaction disadvantage

A

waste = toxic (tin)

66
Q

N3 in synthesis

A

= protected ammonia equivalent

can be deprotonated to NH3 by Pd/C and H2

Organic Chemistry (13 decks)

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