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Foundations of Chemistry 1 > Aromatic Chemistry and Organic Carbonyl > Flashcards

Aromatic Chemistry and Organic Carbonyl Flashcards

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1
Q

formula of benzene

A

C6H6

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2
Q

bond angles in benzene

A

120

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3
Q

hybridisation of carbons in benzene

A

sp2

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4
Q

what is the resonance energy?

A

the hydrogenation of benzene (to form cyclohexane) has a less negative enthalpy change than the expected value for the kekule structure

this difference between the enthalpy change for benzene and the calculated value for the kekule structure is called the resonance energy

shows that benzene is more stable than the kekule structure would suggest

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5
Q

what can the stability of benzene be attributed to?

A

the delocalisation of P orbital electrons around the ring

makes it harder for benzene to donate electrons during reactions (acting as a nucleophile)

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6
Q

state huckels rule

A

a molecule is aromatic if it is cyclic, planar, contains uninterrupted
(continuous) conjugation and (4n + 2) π-electrons.

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7
Q

can aromatic compounds have a charge?

A

yes, so long as they satisfy the conditions for being an aromatic compound, i.e. the correct number of pi electrons, planar, monocyclic

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8
Q

can aromatic rings contain heteroatoms?

A

yes
note: lone pairs on the heteroatom may or may not contribute to the pi electron system

note: when the lone pair is not part of the pi ring system, the compound will act as a stronger base - electron pair more easily donated

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9
Q

difference in reaction of alkenes and benzene with bromine?

A

alkenes react with bromine via electrophilic addition (Br2 is added across the C=C bond)

benzene reacts with bromine via electrophilic substitution, requiring presence of a stronger electrophile (e.g. FeBr3)

note: substitution reaction replaces one of the hydrogens on benzene with bromine. The aromaticity of the ring is maintained.

note: benzene does not decolourise bromine water as it does not undergo electrophilic addition

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10
Q

via which mechanism does benzene react with an electrophile?

A

electrophilic substitution
two step reaction.
step 1 (slow): C-E bond formed, aromaticity broken to form nonaromatic carbocation intermediate, stabilised by resonance (wheland intermediate)
step 2 (fast): proton removed from the carbon with C-E bond. aromaticity restored. substituted benzene produced.

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11
Q

Halogenation

A

benzene reacts with bromine or chlorine in the presence of lewis acid (FeBr3 or AlCl3)

step 1: halogen reacts with lewis acid to form coordination complex (strong electrophile) (Br and Cl will not react with benzene alone)

step 2: benzene acts as nucleophile, attacks the partially positive halogen on the coordination complex. nonaromatic carbocation formed.

step 3: nonaromatic carbocation deprotonated by base (FeBr4- or AlCl4-), forming halobenzene + hydrogen halide. lewis acid regenerated.

note: practice drawing on paper

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12
Q

why does monobromination occur in the reaction between benzene and Br2?

A

bromobenzene is a weaker nucleophile than benzene, due to -I effect of bromine

so benzene reacts with bromine in preference to bromobenzene

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13
Q

reaction conditions for halogenation of benzene

A

benzene + halogen + lewis acid catalyst (FeBr3 or AlCl3)

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14
Q

Nitration

A

benzene reacts with nitronium ion (NO2+) to form nitrobenzene

step 1: H2SO4 donates proton to HNO3 to form H2O leaving group on nitric acid

step 2: loss of H2O leaving group forms NO2+ ion, strong electrophile

step 3: benzene acts as nucleophile and attacks NO2+ (nitronium ion). non aromatic carbocation intermediate formed.

step 4: water deprotonates carbocation to form nitrobenzene

draw on paper.
note movement of electrons when benzene attacks nitronium ion.

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15
Q

Reaction conditions for nitration of benzene

A

benzene +
concentrated sulphuric acid and nitric acid
(H2SO4 and HNO3)

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16
Q

Sulfonation

A

benzene reacts with fuming sulphuric acid to form benzenesulfonic acid

note: all three steps in this reaction are reversible - heating benzenesulfonic acid with dilute sulphuric acid converts it back to benzene

step 1: H2SO4 protonates SO3, forming strong electrophile (HSO3+)

step 2: benzene acts as nucleophile and attacks HSO3+, forming nonaromatic carbocation

step 3: HSO4- deprotonates carbocation, forming benzenesulfonic acid. sulfuric acid regenerated

draw this on paper!
note movement of electrons in step 2 to maintain correct number of bonds to sulphur

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17
Q

reaction conditions for sulfonation of benzene

A

fuming sulphuric acid - a mixture of sulphuric acid (H2SO4) and sulphur trioxide (SO3)

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18
Q

friedel-crafts alkylation

A

benzene reacts with chloro or bromoalkanes to form alkylbenzenes

(mechanism for primary haloalkane)

step 1: primary haloalkane reacts with lewis acid to form coordination complex - stronger electrophile

step 2: benzene attacks d+ carbon in coordination complex. nonaromatic carbocation formed.

step 3: lewis acid’s conjugate base deprotonates non-aromatic carbocation, forming alkylbenzene, hydrogen halide, and reforming lewis acid

difference for secondary / tertiary haloalkanes:
coordination breaks down to form carbocation and lewis acid’s conjugate base (secondary and tertiary carbocations are more stable than primary carbocations) carbocation acts as electrophile in step 2

draw this on paper!

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19
Q

reaction conditions for friedel-crafts alkylation

A

benzene + haloalkane +
lewis acid (AlCl3, FeCl3 or FeBr3)

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20
Q

why is friedel-crafts alkylation rarely used in synthetis?

A

a mixture of polyalkylated benzenes is formed as alkylbenzenes are stronger nucleophiles than benzene, due to +I effect of R groups

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21
Q

friedel-crafts acylation

A

benzene reacts with acyl chlorides to form acylbenzenes (ketone group introduced onto the aromatic ring)

step 1: acyl chloride reacts with lewis acid to form coordiantion complex.

step 2: coordination complex breaks down to form an acylium ion (R-C+=O) and lewis acid’s conj. base. acylium ion is stronger electrophile than acyl chloride

step 3: benzene attacks acylium ion to form nonaromatic carbocation

step 4: lewis acid’s conj. base deprotonates nonaromatic carbocation to give acylbenzne, HCl, regenerates lewis acid

step 5: regenerates lewis acid forms coordiation complex with acylbenzene.

step 6: water added to hydrolyse lewis acid, acylbenzene released from the complex

NOTE: LEWIS ACID IS NOT A CATALYST. ONE EQUIVALENT IS REQUIRED

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22
Q

reaction conditions for acylation of benzene

A

benzen + acylchloride + lewis acid (FeCl3 or AlCl3)

23
Q

does friedel crafts acylation form one product or a mixture?

A

one product is formed
acylbenzenes are weaker nucleophiles than benzene due to -I and -M effect of ketone group
forcing conditions are required for further acylation

as one product formed, friedel-crafts acylation often used in synthesis

24
Q

activating groups

A

activating substituents make the aromatic ring more reactive to electrophilic substitution (more nucleophilic)

activating groups stabilise the intermediate carbocation

activating groups push negative charge onto the ring, with +M and +I effects

25
Q

deactivating groups

A

deactivating groups make the ring less reactive to electrophilic substitution (less nucleophilic)

destabilise the carbocation intermediate

withdraw negative charge from the ring with -M and -I effects

26
Q

-NH2, -NHR, -NR2
activating or deactivating

A

+M and -I effect
+M effect is stronger
strongtly activating

27
Q

-OH, -OR
activating or deactivating?

A

+M and -I effect
+M effect is stronger
strongly activating

28
Q

-NHCOR, -OCOR
activating or deactivating

A

+M and -I effect
+M effect is stronger
activating

29
Q

-Ph, -CH=CH2
activating or deactivating

A

+M and +I
activating

30
Q

-R
activating or deactivating?

A

+I
weakly activating

31
Q

-Cl, -Br, I
activating or deactivating?

A

-I and +M effect
-I is stronger
deactivating

32
Q

-CHO, -COR
activating or deactivating?

A

-M, -I
deactivating

33
Q

-COOH, COOR
activating or deactivating?

A

-M, -I
deactivating

34
Q

-SO3H, -NO2
activating or deactivating?

A

-M, -I
strongly deactivating

35
Q

what are the 2, 3 and 4 positions on the benzene ring called?

A

2 - ortho
3 - meta
4 - para

36
Q

regioselectivity of activating groups

A

direct electrophiles to the 2 and 4 position

why? because attack at the 2 and 4 positions produces more stable carbocation intermdiates
carbocation stabilised by +I or +M effect of substituent
substituents with a +M effect create an additional resonance form, so carbocation more stable

37
Q

regioselectivity of deactivating groups

A

direct electrophiles to the 3 position

except halogens - direct electrophiles to the 2 and 4 position due to +M effect

38
Q

for 2,4 directing substituents: which position in favoured, ortho or para?

A

attack at the para (4) position is favoured due to steric hindrance at the ortho (2) position

the degree to which the para position is favoured depends on the size of the substituent and the electrophile

39
Q

are regioisomers seperable?

A

yes, they have different physical properties e.g. boiling point
can be separated by distillation

40
Q

reducing agent used in clemmensen reduction

A

zinc amalgam (zinc dissolved in mercury)

41
Q

on a 1,3 disubstituted benzene, at which position will the next substitution be disfavoured?

A

at the 2 position
due to steric hindrance of substituents on either side

42
Q

NMR peaks for benzene

A

1H -> one peak
13C -> one peak

all hydrogens are equivalent
all carbons are equivalent

43
Q

NMR peaks for monosubstituted benzene

A

1H -> 3 peaks
13C -> 4 peaks

four nonequivalent carbons

44
Q

on paper, decipher the NMR peaks for 1,3 disubstituted benzene
a) when substituents are the same
b) when substituents are different

A

a) 1H = 4, 13C = 4
b) 1H = 4, 13C = 6

strategy : identigy any lines of symmetry!

45
Q

how can regioisomers of electrophilic substitutions of benzene be identified?

A

NMR spectroscopy

46
Q

what transformation can be applied to an NO2 substituent on a benzene ring?

A

reduction to form NH2 substituent

conditions : Sn(tin) and HCl

regioselectivity changes:
NO2 is 3-deactivating, NH2 is 2,4-activating

47
Q

what transformation can be applied to a COR substituent on a benzene ring?

A

clemmensen reduction, forming CH2R substituent

conditions : Zinc amalgam (Zn/Hg) and HCl

regioselectivity changes:
COR is 3-deactivating, CH2R is 2,4-activating

48
Q

clemmensen reduction

A

conversion of acylbenzene to alkylbenzene

acylbenzene is reacted with reducing agent, converting the C=O group into a CH2 group

friedel-crafts acylation -> clemmensen reduction is preferred route in synthesis of alkylbenzenes, as a single product can be formed

49
Q

what transformation can be applied to toluene

A

toluene is benzene with a CH3 substituent

oxidation of CH3 to form COOH substituent

regioselectivty change:
CH3 is 2,4-activating, COOH is 3-deactivating

50
Q

which ion can be formed from aniline?

A

aniline is benzene with NH2 substituent

reaction of aniline with NaNO2 and HCl produced aryl diazonium ion (N2+ substituent on benzene)

51
Q

how can an OH substituent be introduced onto a benzene ring

A

aniline reacted with NaNO2 +HCl to form aryl diazonium ion

aryl diazonium ion reacted with H2O to form phenol

52
Q

how can an aryl diazonium ion be converted to a halobenzene

A

reaction with CuBr or CuCl to from bromo/chlorobenzene

53
Q

structue of an aryl diazonium ion

A

NH2+ group substituted on benzene ring

Foundations of Chemistry 1 (7 decks)

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