10b Flashcards
what is benzyne
benzene with a triple bond
how is benzyne formed
the H adjacent to the substituent is removed by a super strong base giving an anion in its place.
this is used to kick off the substituent and give a triple bond.
this triple bond is odd,, the original pi bond is due to the overlap of p orbitals in the aromatic system. the new pi bond is due to the overlap of 2 sp2 orbitals outside of the aromatic system. meaning its very weak and unstable
what can happen to benzyne next,,
a nucleophile can attack the triple bond,, push the triple bond e- to the place the H was at,, to give an anion here.
this anion can then attack another H.
so weve basically removed a substituent and added another one. weve swapped the substituents from one we didnt want to one we wanted using a strong base.
new triple bond is due to
overlap of the two sp2 orbitals (the C-X) and the anion (lone pair in the sp2 orbital)
is benzyne stable
nope
benzyne isnt stable,, the new triple bond is very weak. §
the nucleophile that removes the H to form benzyne must be what
it must be basic
why cant the anion be stabilised by resonance around the aromatic ring
bc the p orbitals and the sp2 orbital with the lone pair are orthogonal meaning they dont overlap.
the sp2 C-X and sp2 lone pair do overlap tho!! which cases the CX bond to break and a triple bond to form.
why is the triple bond formed by the overlap of 2 sp2 orbitals weak
bc they have poor overlap
is the new triple bond inside or out of the ring
its out of the ring,, bc the sp2 orbtials face out
3 ways u can make benzynes
- Halobenzenes + strong base (halogen as a substituent,, remove the adjacent H, to give an anion using a strong basic nuc)
- Aryl lithium-halogen exchange (2 halogens as substituents,, Li replaces one of the halogens, the e- from the C-HALOGEN form the triple bond) to give benzene.
- Diazotisation of ortho-aminobenzoic acid under basic conditions (cooh and nn as substituents,, base deprotonates cooh,, this kicks off the n2 and forms a triple bond) benzyne
when benzynes are formed,, are any of the previous substituents still on it
nope!!
when the triple bond is formed,, theres no substituents near it,, bc we use the one of the substituents e- in the bond to form the triple bond and then we kick the other one off, so there are no susbtituents left.
benzyne formation
deprotonate using a basic nucleophile to give an anion // use the e- in lithiums bond to form a triple bond and kick off the other substituent.
Diazotisation of ortho-aminobenzoic acid under basic conditions driving forceeeee
gives out lots of gas so this reaction is entropically favoured and is entropically driven.
releases co2 and n2
when nucs attack the triple bond in benzyne,, where do the e- get accepted into
the empty sp2 pi* orbital due to the sp2 overlap
describe how a group is added to benzyne
the nuc attacks one side of the triple bond
the other side of the triple bond gets a negative charge
which then attacks a H
what determines what side the nuc attacks from
sterics
electronic factors
electronic factors + the side of the benzyne the nuc attacks from
think: where would the anion // negative charge like to be near : is the other susbitiuent ewg or edg
if its an ewg,, the anion would want to be near it
if its edg,, the anion would want to be further from it.
sterics + the side of benzyne the nuc attacks from
steric hinderance if the nuc is close to the substituent as theyre in the same plane.
ewg + edg + the side of the benzyne the nuc attacks from,, what do we mean by this
we think of induction!!!
does it inductively attract e- towards it or no. does this stabilise a negative charge or no.
ignore resonance stabilisation bc there is no resonance bc theyre not in the same plane.
do two anions like being close
nope
so u attack closer to the first anion so the other one could be further away
if theres para substitution aka the 2 groups are paa and one of them is an alkyl group,, does it rlly matter what side the nuc attacks from
nopeeeee
the group doesnt rlly have a steric effect as its too far away
if the substituent is an OH // and electron repelling anion as its deprotonated to give O-,, AND THERES A BR PARA TO IT,, where would the NH2 attack
the nh2 will attack meta position.
bc u remove the H next to Br,, either meta positions to form a triple bond and kick off Br.
then u can attack where Br was and put the anion in meta. or attack meta and put the anion in para.
at the end u want the anion to be para as the O- is very negative and will repel the anion. so its best to just put it as far as possible in the first place.
