135B- aromatic chem Flashcards
what is true about all the bonds in benzene
are the same length
why does ethene react with bromine but benzene doesn’t despite both containing double bonds
benzene is much more stable so needs a higher activation energy
what are Hückel’s rules (rules for something to be aromatic)
1/ planar- must be a flat molecule
2/ fully conjugated means alternating single, double bonds
is this cyclic
yes
is this aromatic
yes
describe this diagram of benzene pi bonding
π1 - all lobes are in phase and so maximum overlap
π2/π3- higher in energy because some of the overlap is breaking/ inserting another node into the system across either plane of symmetry
π4/π5- two degenerate levels as both have 2 nodes on 2 lines symmetry
π6*- all orbitals are in opposite directions and no overlap
why are the levels above π3 antibonding
because benzene has 6 electrons and all 6 are used up in π1-π3. Meaning if the higher MO’s are filled they must be antibonding
If electrons go into bonds will start breaking
how many electrons are in this system
8- double bonds don’t count as 2 electrons, just shows that there is conjugation and each C has a filled p orbital
what is the method to drawing a diagram like this
-this can be for any shape aromatic ring not just benzene
anything below the half way point is bonding and anything above is anti-bonding
4n + 2 number of π electrons is true for an aromatic compound. What is the formula for an anti aromatic compound
4n (anti aromatic means its less stable than you’d expect it to be)
why is it rare to find an anti aromatic molecule
anti aromatic molecules are less stable than they are expected to be- and molecules will always try and be the most stable
what Is the frost circle diagram for this anion (is flipped on its head so that the electrons can go where the negative charge is)
there are 8 π electrons which falls into 4n π electron formula - meaning the compound is anti aromatic
what is the frost circle diagram for this cation
-because the number of electrons (6) satisfies 4n + 2 this means it is a aromatic
-meaning its a planar cation
fill in the frost circle diagram
fill in the frost circle diagram
fill in the frost circle diagram
fill in the frost circle diagram
what is the correct way to draw Naphthalene (2 benzene rings together)
which side of the molecule is more stable
left hand side
what are the two resonance forms of Naphthalene
clars sextet theory- what is the other structure of triphenylene
the original structure had 3 π sextets, whereas the other structure only has 1 (in the middle)
- the original one (most sextets) is more stable
which form is most stable
one in purple (has the most sextets)
which way of drawing this compound is most stable/ draw the resonance structures for this (different areas to put the benzene ring)
all 3 are equivalent-no way of drawing it is more stable doesn’t matter where you put the benzene ring
where will phenanthrene react
the orange position is the least aromatic so doing chemistry there is the easiest
what are the 3 resonance structures of this, and therefore when combining them, where is a reaction most likely to take place on the molecule
in the resonance structures, there are more times when the benzene ring is on the outside of the molecule than on the inside
meaning its reactions will mostly happen on the inside of the ring
for linear aromatic structures none of their resonance structures are more preferable to the other. Why do the structures get less stable the more rings you add
only 1 π sextet can exist in a resonance structure at any one time due to it being linear
- the larger the acene, the stable sextet has to be shared between more rings so it is less stable
(e.g naphthalene has one benzene for 2 rings, whereas pentacene has one benzene for 5)
which resonance structure is more stable
the end one where the 2 benzenes are on the outside (the more benzenes the more stable)
why Is the linear formation less stable
has less benzene (sextet) rings
what can you do to reduce Ea
e.g making bromine a better nucleophile is adding a catalyst to make Br+
why is where the positive charge is still sp 2 hybridised like the other carbons and not sp
because the positive charge is just an empty p orbital, that’s still there and can form resonance structures
-still has the potential to form 3 bonds
what is the general mechanism of a SeAr mechanism (electrophile adding to a benzene ring)
why is the first step the slowest
destroying aromaticity which requires a lot of energy (high energy intermediate)
why is the first step the slowest and the last step is the fastest
1/destroying aromaticity which requires a lot of energy (high energy intermediate)
2/ step to regenerate an aromatic compound is very favourable so will happen fast / also fast because its just a loss of a proton
In theory the top reaction will be quicker. However both processes happen at the same speed, so which step is the RDS
theory- Deuterium is twice as heavy as hydrogen, if it is involved in the RDS, the reaction would go slower than the hydrogen reaction because its heavier
answer- because both steps are the same, it means that loosing deuterium isn’t involved in the RDS, meaning step 1 must be the RDS
what are the reactions of the halogens F, Cl, Br and I with benzene
F doesn’t need a catalyst but the reaction is explosive (very rapid reaction)
Iodine doesn’t occur even when you add a catalyst
are the catalysts lewis acid or Lewis bases
Lewis acids
why can sulphuric acid protonate nitric acid
sulphuric acid is a stronger acid, and therefore can protonate another acid
what is the mechanism of nitrating benzene
here is the first step
and this is the product being made
what is a nucleophile and electrophile
nucleophile- thing donating electrons
electrophile- thing accepting electrons
what are two ways to reduce nitrobenzene
1/ tin and HCl
mechanism for sulfonation of benzene
starting with these reactants
to get this product
-generation of electrophile, then normal substitution reaction
alkylation- what is the reaction for the generation of a carbon-carbon bond
starting with these reactants
to get this product
-generation of electrophile, then normal substitution reaction
once made the alkyl electrophile then the reaction is exactly the same as any other benzene substitution
what are the disadvantages of this alkylation reaction
-acylation is better
1. overalkylation- due to excess of cation and the product is more reactive than benzene, so will keep substituting
2. carbocation rearrangment- R group will migrate to make a more stable tertiary carbo cation
what is an acylation reaction mechanism
starting with these reactants
to get the product
-generation of electrophile, then normal substitution reaction
why is the product deactivated (the product is less reactive than the starting material [benzene] so will not further substitute)
the carbonyl groups makes it less reactive
- don’t get rearrangments either because the acylium ion is very stable
how do you replace the =O with 2 hydrogens [fill in the reactants]
-acylate the benzene first (to get this starting material)
-then reduce it either using Wolff–Kishner reduction or Clemmensen
what are the two reactions you need to go through to put an alkyl group on a benzene
- go through a Friedel-Crafts Alkylation
-or Friedel-Crafts Acylation, then reduce using either using Wolff–Kishner reduction or Clemmensen
is this meta, para or ortho
meta
is this meta, para or ortho
para
is this meta, para or ortho
ortho
why do these aromatic nucleophiles get more reactive with bromine
-goes from least to most reactive
-at top of the table you need more harsh conditions
-at bottom reactions can happen in milder conditions
-get better nucleophiles down the group/ electron withdrawing groups make the ring more δ+ which needs to be electron rich
why is a nitro group electron withdrawing shown through resonance structures and what’s this effect called
-how does this explain why you need such strong conditions for nitrobenzene to undergo bromination
mesomeric
- nitrogen draws electrons form benzene ring making it basically non nucleophilic because the ring now has a full on positive charge
-needs harsh conditions to overcome this
how is a CH3 group electron donating
- C-H sp3 bond into C p orbitals
-donating electrons via hyperconjugation
how does nitrogen on the benzene (amine) effect the nucleophilicty of the benzene ring/ is this compound more or less reactive than benzene
makes it more nucleophilic
-this compound is more reactive than benzene as it has more electron density in the ring
how do you decide wether the product is meta, para or ortho
do each substitution at meta, para and ortho and see how many resonance structures there are of each
-one with no resonance structure where the positive charge is next to the nitrogroup is the most stable product
2/ exclamation marks is where positive charge is next to nitro group
would the most stable product be meta, para or ortho
ortho or para [both have the most resonance structures]
-shows why you don’t get reaction at meta
-because is has one resonance structure where the positive charge is next to the NMe2 group, which is good in this case because a 4th resonance structure can be drawn from this
2/ you actually get 2 substitutes both at meta and para
where do the halogens give substitutions [to NO2 for example] at- meta, para or ortho
ortho and para
-the lone paris on the halogens determine the regioselectivity to be ortho or para
why are the halogens all less reactive than benzene
electron withdrawing effect
what is the trend of electron withdrawing effect down the group of halogens
iodine is the best at withdrawing and flourine is the worst
why do iodine and flourine have the rate of reaction with substitution
flourine is more electron withdrawing so you would expect the reaction to be slower than 0.18 [iodine]
-but flourine orbital overlaps benzene ring reasonable well/ so mesmeric stability through resonance is reasonably good
-iodine you would expect to be faster [than 0.18] as its less withdrawing
-but it has worse overlap due to larger orbital so mesmeric stability isn’t as good making it slower
halogen on benzene substitutions favour para and ortho. What happens to the favour of either para or ortho down the group
-and what causes this
para is most likely to be selected at top of group
-this decreases down the group
2/caused by electronegativity
-flourine deactivates benzene the most [most withdrawing] so get more para
-iodine is not as strong electronwithdrawing so you get equal of each
rate flourobenzene, chlorobenzene, bromobenzene and idobenzene in reactivity
goes from most reactive to least
electron withdrawing or donating and what is the effect called
electron donating , mesomeric
what is the difference between mesomeric effect and inductive
mesomeric- interaction of 2 π bonds or between a π bond and a lone pair
inductive- withdrawing or donating of electrons from a σ bond [CH3]
what does this tell you about what compounds do meta and what do para/ortho
ortho/para- activating and react faster than benzene
meta- all of them react slower than benzene and are deactivating
halogens are exception react slower than benzene and are deactivating but are ortho and para
mechanism for converting a phenol to a phenolate to activate it (end product is aspirin)
1/ NaOH and CO2
2/ acid conditions
3/ ketone ester
when reacting bromine with an aniline it substitutes 3 times. How do you deactivate it so bromine only substitutes once
1/ acyl chloride
- amide is less activating than the amide (NH2)
why is an amide less activating than an amine (amine NH2)
[activating meaning making the benzene a better nucleophile and electrons more available]
-in the amide (NH) the lone pair is more around the oxygen which is far away from/can’t donate to the benzene ring
-meaning nitrogen no longer donates to the aromatic ring [resonance structure 2]
the lone pair are less avaliable in the amide (NH) due to nitrogen being sp2 hybridised- it is very much involved in delocalisation
why does the bromine only add In the para position, to the newly deactivated amine (now an amide)
steric block- the extra CH3 group at the top blocks the Br from adding anywhere closer
what can be added as a blocking group for phenol/ blocking groups adds to certain positions to force bromine to be added on a certain carbon
sulphonate group
- as you add each sulphonate group the phenol gets less and less reactive
what can you add to get rid of the sulphonate groups
how do you go from NO2 to NH2
-nirto reduction for meta substituted anilines
what is this effect called
co-operative substituent effect
why is this the case
OH are electron donating (has lone pair)
and aldehyde groups are electron withdrawing ( O is electronegative so draws electrons towards it)
aldehyde groups are electron donating/ withdrawing
withdrawing - O is electronegative so draws electrons towards it
both groups want to push the addition to an ortho position; how is it decided where the group will go
[both groups are ortho,para directing but the para positions are obviously blocked]
electronic effects override steric effects
1. both methyl and NH are electron donating
2. But NH is mesmeric (electron donating through conjugation) whereas methyl is through hyperconjugation so NH takes priority
3/stablisation of positive charge next to NH group is more stabilising than it being by the methyl group
so this wins even though steric hinderance
where would the bromine add to
an activating group can override a deactivating group
1. N-R is activating and CF3 is deactivating
2. While N-R can do ortho and para, an amine is always more inclined to para
3. It will add at the para position
where will the ketone add to
both are electron donating
- OH is mesmeric (electron donating through conjugation) whereas methyl is through hyperconjugation so OH takes priority
- will take OH ortho position
- The circled form is the product
where will the acyl chloride go
both are electron donating
- both are hyper conjugation donating so now streics take precedent
- will add to least hindered position
why could this react with a bad electrophile like iodine but benzene couldn’t
electron rich aromatics can react with poor electrophiles that benzene couldn’t (has two electron donating groups to make it a better nucleophile)
you can’t put iodine on benzene
formylation of electron rich aromatics
i) gatterman reaction [not favoured as using hydrogen cyanide] mechanism
the first step is making the electrophile
Al should be Zn instead
formylation of electron rich aromatics
i) Vilsmeier-Haack reaction [the more favoured and more common] mechanism
the first step is making the electrophile
Azo-coupling of electron rich aromatics
the Azo group is N2+
what are the reactants to go from an Azo group to these products
what is a birch reduction
benzene ring reduced to two double bonds separated by CH2 (cannot be conjugated)
what is the birch reduction mechanism
this is done in an alcoholic solvent so that is why there is an O-H in the reaction
how would a birch reduction play out on a substituted benzene
withdrawing groups- not on a double bond
donating groups- on a double bond
elimination addition reaction mechanism of bromobenzene
using sodamide
you get 50% of this product and then 50% where the NH is In the H position because the bezyne intermediate can be attacked from either side
There are 2 steps to this reaction
1. generating benzyne electrophile with the use of a strong base (sodamide) next to a halide (Br) as shown
2. addition of anything onto the triple bond
There are 2 other ways of getting to the benzyne electrophile. What are these 2 other ways
