. Write a short on FIVE–MEMBERED HETEROCYCLIC RINGS


Best Answer

The principles that govern the electrophilic substitutions of this group of heterocyclic aromatic compounds will be illustrated by reference to pyrrole.

Pyrrole is highly reactive at both the 2and 3positions. The reason is that the transition state for substitution at each position is strongly stabilized by the accommodation of the positive charge by nitrogen, in just the way that aniline owes its reactivity to the exocyclic nitrogen. 2substitution predominates because the positive charge in the transition state is delocalized over a total of three atoms, compared with two for 3substitution.

The similarity of pyrrole and aniline is particularly apparent in their reactions with aqueous bromine: each reacts at all its activated carbon atoms, pyrrole giving tetrabromopyrrole and aniline giving 2,4,6tribromoaniline. In fact, pyrrole is even more strongly activated than aniline and should perhaps be compared with the phenoxide ion: each undergoes the ReimerTiemann reaction, unlike other benzenoid compounds. In addition, pyrrole undergoes FriedelCrafts acylation in the absence of a catalyst.

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Furan and thiophen are also activated towards electrophiles and react predominantly at the 2position. The underlying theory is similar to that for pyrrole, namely, that the heteroatom is able to delocalize the positive charge on the transition state. Since oxygen accommodates a positive charge less readily than nitrogen, furan is less reactive than pyrrole, just as phenol is less reactive than aniline. The +M effect of sulphur is smaller than that of oxygen because the overlap of the differently sized porbitals of carbon and sulphur is less than in the case of carbon and oxygen so that, understandably, thiophene is less reactive than furan. Thus, the reactivity order of 5membered heterocyclics towards electrophilic substitution would be 

pyrrole > furan > thiophene.   

 

  

 

 

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