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Organic Chemistry > Aromatics > Flashcards

Aromatics Flashcards

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1
Q

Kekule model

A

bond lengths were between single and double (+ all same) - 140pm

lower enthalpy of hydrogenation

unreactive in addition reactions - due to aromatic stabilisation energy

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2
Q

Huckel’s rules

A

cyclic

planar

fully conjugated

4n + 2π electrons

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3
Q

aromatic

A

paired e- in bonding/anti-bonding MOs

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4
Q

anti-aromatic

A

unpaired e- in bonding/anti-bonding MOs

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5
Q

PAHs

A

[polycyclic aromatic hydrocarbons]

consist of 2 or more fused aromatics

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6
Q

Clar’s sextet theory

A
  1. resonance structures with the largest number of disjoint aromatic π-sextets = most important
  2. PAHs or resonance structures with more π-sextets in resonance structures = more stable
  3. compounds with resonance structures that have all 6 π-sextets + no double bonds = fully benzenoid (similar to benzene)
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7
Q

π-sextet

A

6-membered ring of 6 π-electrons within a PAH

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8
Q

acenes

A

only 1 π-sextet can exist in a resonance structure at any one time

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9
Q

phenacenes

A

multiple π-sextets can exist in a resonance structure at any one time

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10
Q

why can multiple π-sextets exist in phenacenes but not arenes?

A

arene = linear

phenacene = zig-zag (more stable)

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11
Q

how to lower the activation energy of a reaction?

A
  1. increase energy of HOMO - makes nucleophile more nucleophilic
  2. decrease energy of LUMO - makes electrophile more electrophilic
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12
Q

mechanistic evidence for SEAr

A

[measured via kinetic isotopic effect - swap H for D]

found that the 2nd step (C-H / C-D bond breakage = fast and not rate determining -> measured using relative bond dissociation energies

1st step (breaking aromaticity = slow and rate determining

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13
Q

halogenation of benzene with different halogens

A

F2 - explosion

Cl2 / Br2 - reaction

I2 - no reaction

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14
Q

uses of nitrobenzene

A

explosives

solvents

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15
Q

uses of aniline

A

rubbers and polymers

dyes and pigments

pharmaceuticals

herbicides

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16
Q

which type of carbocation is most reactive in Friedel-Crafts reactions?

A

tertiary

17
Q

disadvantages of Friedel-Crafts reactions?

A
  1. overalkylation
  2. carbocation reactions / Wagner-Meerwein rearrangement

-if C+ is next to methyl group, alkyl group can migrate to make more stable C+
-as it moves, the alkyl group brings pair of e- with it (R-)

18
Q

effect of EWG on RoR of SEAr

A

reduces

withdraw e- density from π system (less reactive)

19
Q

effect of EDG on RoR of SEAr

A

increases

20
Q

direct substitution for EWG

A

meta - provides least destabilised pathway

21
Q

direct substitution for EDG

A

ortho/para - provides most stabilised pathway

22
Q

regioselectivity - halogens

A

despite being EWG, they also direct ortho/para

due to orbital overlap between halogen p-orbital and π cloud

23
Q

regioselectivity trend for halogens

A

selectivity for para over ortho decreases as you go down group due to electronegativity

rate for F > Cl due to improved orbital overlap (2p vs 3p)

rate for I > Br due to decreased electronegativity

24
Q

cooperative effects

A

when multiple sub. direct to same position

25
Q

competitive effects + rules

A

[when multiple sub. direct to different position]

electronic > steric

activating > deactivating

path with most resonance structures = major

when sub. can’t be determined through electronics, steric arguments used

26
Q

Sandmeyer reactions

A

CuCl -> Cl

CuBr -> Br

CuCN -> CN

KI -> I

H2O -> OH

POH3 -> H

27
Q

why is sodium amide used (NaNH2) ?

A

v. strong base => able to deprotonate lots of compounds

28
Q

SNAr - RDS?

A

attack of aromatic ring by Nu (disrupts aromaticity)

29
Q

SNAr - effect of EWG

A

more EWG = more stabilisation of -ve charge of intermediate = faster reaction

30
Q

SNAr - leaving group

A

F = best (opposite to other reactions)

due to high electronegativity

31
Q

SNAr - sterics

A

F&raquo_space; Cl/Br

smaller Nu = easier for nucleophile to attack

bond = v. polarised (Nu attacks more easily due to bigger δ+)

32
Q

SNAr - sub. patterns

A

ortho/para > meta

because negative charge can be stabilised

33
Q

formula for anti-aromatic electrons

A

4n

34
Q

where do EWG direct?

A

meta

with exception of halogens

35
Q

where do EDG direct?

A

ortho/para

36
Q
A

Organic Chemistry (13 decks)

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