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organic sem 2 > lecture 5 > Flashcards

lecture 5 Flashcards

1
Q

benzene + normal bromine

A

no reaction

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2
Q

benzene and mcPBA

A

no reaction

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3
Q

why can’t bromine and mCPBA react with benzene

A

bc benzene is aromatic, more stable
- cyclic
- conjugated
- flat
- huckels rule

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4
Q

what happens to e- when they are conjugated

A

they are shared between the double bonds that are conjugated

they are shared between the system

this can be shown in resonance forms

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5
Q

what must happen to bromine in order to react with benzene

A

it must be activated

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6
Q

explain how bromine is activated

A

Br2 must react with a lewis acid

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7
Q

what is a lewis acid

A

electron deficient
accepts e-

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8
Q

what lewis acid is used to activate a bromine

A

AlCl3

aluminium tri chloride

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9
Q

how is bromine activated using AlCl3

A

Bromine lone pair attacks the Al in AlCl3

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10
Q

product of bromine activation with AlCl3 (lewis acid)

A

Br-Br- Al-Cl3
+ -

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11
Q

how does AlCl3 activate bromine

A

the lewis acid weakens the Br-Br bond by removing some e- density

allowing it to react with aromatics

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12
Q

draw the reaction between aromatics and an activated bromine

A
  • aromatic arrow to bromine
  • bromine arrow to the + Bromine
  • the 2 products are a CC+ cyclohexene with a bromine attached and a Br-Al-Cl3 molecule
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13
Q

what can occur in the product when aromatics react with an activated bromine

A

the CC+ cyclohexene bromine can have resonance forms in the molecule. (it is no longer aromatic)
the Br-AlCl3 has a - charge on the Al

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14
Q

what is needed in order for the cc+ cyclohexene with bromine attached to regain its aromaticity

A

it needs to have a proton removed

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15
Q

how does cc+ cyclohexene with bromine regain its aromaticity

A

it loses a Hydrogen

the e- in the Br-Al bond in Br-Al-Cl3 attack the H which gives its electrons to the ring, regaining its aromaticity

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16
Q

okay so steps of the reaction of bromine with an aromatic

A
  • bromine + Al-Cl3 gives Br-Br(+)-Al(-) -Cl3
  • the double bond then attacks one br giving Br-Al(-)-Cl3
  • the Br-Al then attacks the H giving HBr, AlCl3 (-) and an aromatic
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17
Q

what is needed for a normal nitration reaction

A

HNO3 (nitric acid)

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18
Q

can nitric acid HNO3 react with an aromatic

A

nope!!
it must first be activated

19
Q

how is nitric acid, HNO3 activated

A

it uses H2SO4 to protonate itself
lone pair on one O attacks the H of H2SO4
H e- are moved onto the O
- protonated HNO3 now has a H2O group (good LG)
- O- electrons kick off the H2O

20
Q

draw a nitric acid molecule

A

HNO3
+
-:O: - N = O
|
:O:H
(this one attacks the H in H2SO4)
and is later forced to leave

21
Q

when HNO3 reacts with H2SO4 it is activated,, what does this give

A

a very reactive electrophile
nitronium ion
+
O=N=O

22
Q

what is used to nitrate aromatics

A

activated HNO3 (nitric acid)
aka a nitronium ion ( O=+N=O)

23
Q

reaction between an aromatic and a nitronium ion step 1

A

double bond attacks the + charged N
e- are taken to the double bonded O

24
Q

reaction between an aromatic and a nitronium ion product of step 1

A

aromatic is bonded to the NO2
the N is +
the O is -

25
Q

step 3 of the reaction between an aromatic and a nitronium ion

A

the aromatic wants to regain the aromaticity lost when bonded to the NO2 so it loses a H and it’s e- are donated to the ring system, regenerating it’s aromaticity

26
Q

what is used to remove the H in the reaction of an aromatic and a nitronium ion in order for it to regain its aromaticity

A

a HSO4 (-) charged

27
Q

what is the final product of nitric acid // nitronium and a benzene aromatic

A

a nitrobenzene product

benzene with NO2 bound to it

28
Q

benzene and nitric acid is called aaaaa

A

nitration reaction

29
Q

what is needed in a sulfonation reaction

A

2 H2SO4 in order to activate themselves

30
Q

how is H2SO4 activated in a sulfonation reaction

A

it uses another H2SO4 in order for one of the single bonded Os to gain a H by attacking it, this will give a H2O which can then be lost as a good LG (one step of attacking the H and the H2O leaving) the lone pairs on the other single O donate : to form a double bond with the S and neutralise it as it was +

31
Q

what is the final product when H2SO4 is activated

A

a protonated SO3 // HSO3
O(+) charged, S is neutral, other Os are neutral

a strong electrophile which can react with aromatic

32
Q

how to remember the products

A

remember they have to be electrophiles so will probably have a + charge

33
Q

what does the aromatic attack in a sulfonation

A

it attacks the S in the middle
the double bond e- then go to the newly made O(+)

this O will have a + charge

34
Q

what is used to remove a H from the sulfonation product

A

the HSO4 (-)
this helps it regain its aromaticity

35
Q

what is the product of the sulfonation of benzene

A

benzene
sulfonic
acid

36
Q

what is the RDS when the aromatic and the electrophile react

A

the rds is the breaking of aromaticity

when the aromatic attacks the activated electrophile (SO3, NO2, Br-AlCl3)

37
Q

what is the fast step when an aromatic attacks an activated electrophile

A

the fast step is the regeneration of an aromatic

when the H is removed

rearomatisation

38
Q

electrophilic aromatic substitution reactions

A

bromination
nitration
sulfonation

39
Q

what is needed in a bromination reaction with an aromatic

A

Br2
AlCl3
combine

40
Q

what is needed in a nitration reaction with an aromatic

A

HNO3
H2SO4

nitronium ion

41
Q

what is needed in a sulfonation reaction

A

H2SO4
H2SO4

42
Q

when an aromatic attacks an electrophile,, what is made

A

a cyclohexene attached to the electrophile

it has resonance forms as it is a CC+ intermediate

43
Q

what then occurs to the benzene x electrophile x CC+ intermediate

A

a hydrogen is removed in order to regain its aromaticity

44
Q

steps in electrophilic aromatic substitution

A
  • activation of electrophile
  • attack of electrophile + CC+ intermediate formation
  • removal of H to regain aromaticity

organic sem 2 (14 decks)

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