Why is pyrene more reactive than benzene?

Answer 1

I would think that it's because pyrene has less resonance stabilization than benzene does (increasing its HOMO-LUMO gap by less), due to its sheer size causing its energy levels to be so close together.

A smaller HOMO-LUMO gap means a more reactive system, despite it having resonance throughout.


DISCLAIMER: THOROUGH/LONG ANSWER!

EXAMINING THE EXTENSIVITY OF RESONANCE STABILIZATION

Consider napthalene, anthracene, and phenanthrene (if you add one benzene ring to the upper-right of phenanthrene, you have pyrene):

The resonance stabilization that one benzene ring gets is #"36 kcal/mol"#. If there were a perfect extensivity with regards to resonance stabilization, we would have expected the amount to be

#"Total Resonance Energy"#

#~~ "Number of Benzene Rings" xx "Resonance Energy"#

But you can see in the above diagram that it isn't:

  • Napthalene has #bb"11 kcal/mol"# less resonance energy than #2xx"benzene rings"#.
  • Anthracene has #bb"25 kcal/mol"# less resonance energy than #3xx"benzene rings"#.
  • Phenanthrene has #bb"17 kcal/mol"# less resonance energy than #3xx"benzene rings"#.

    From this, we could postulate that in general, the more extended the #pi# system, the less resonance stabilization is afforded. The most likely reason for this is probably the volume of the system.

    ENERGY GAPS AS A FUNCTION OF VOLUME (AND ENTROPY)

    The energy gaps (and thus the HOMO-LUMO gap) in any molecule are a function of the system volume and entropy.

    At constant entropy though (which means at a constant distribution of states amongst the energy levels), the trend of volume vs. energy gap can be examined.

    To illustrate this, the following graph was generated and derived from Huckel MO Theory, for which we have:

    #E_k = alpha + 2betacos((2kpi)/n)#,

    where #k# is the energy level index and #n# is the number of fused rings. #alpha# is the nonbonding energy and #beta# is the negative difference in energy from the nonbonding level.

    We can see then that the HOMO-LUMO gap converges as the number of rings increases, i.e. as the system volume increases.

    PARTICIPATION OF HOMO & LUMO IN ELECTROPHILIC ADDITION

    How is this relevant?

    Well, the HOMO and LUMO are both required in electrophilic addition reactions. For example, with adding #"Br"_2#,

    • The HOMO donates #pi# electrons to polarize #"Br"_2# and break the #"Br"-"Br"# bond.
    • The LUMO accepts electrons from the #stackrel(delta^(+))("Br")#.

      And this forms the so-called bromonium complex:

      (Here, the HOMO contained the #pi# electrons in the double bond, and the LUMO accepted the electrons from the bottom #"Br"#.)

      Since the HOMO-LUMO gap gets smaller when the system gets larger, it's very likely that the gap is so small for pyrene that the resonance stabilization (which increases this gap) isn't enough to make it unreactive towards electrophilic addition.

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Answer 2

Pyrene is more reactive than benzene because it has more double bonds, which leads to a higher electron density and greater susceptibility to undergo addition reactions. Additionally, pyrene's structure contains more sites for potential reactions compared to benzene's hexagonal structure.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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