Why can't an enolate anion successfully displace a halogen from a benzene?

Question

Why can't an enolate anion successfully displace a halogen from a benzene? I am told that the geometry is incorrect; that such S_N2 substitutions take place on $\mathrm{sp^3}$ hybridized (i.e. tetrahedral) substrates.

This reasoning is circular. So what's a better reason?

I can think of one reason but it is not satisfactory; it's still a bit circular. S_N2 = backside attack; how would one attack a planar bromobenzene? From within the ring? Seems like there would be a lot of possibly unfavorable interactions with the p-orbitals for something to get within the ring. Also this doesn't explain why substitution reactions with enolate anions don't take place at other $\mathrm{sp^2}$ carbon atoms - i.e. bromoethene.

Answer 1

explain why substitution reactions with enolate anions don't take place at other sp2 carbon atoms

As explained in this earlier answer, $\ce{S_{N}2}$ reactions do occur at $\ce{sp^2}$ carbons, they're just higher energy and consequently, less common. So the question becomes, "why is this pathway higher in energy?"

Since the $\ce{S_{N}2}$ reaction involves "backside" approach to the carbon undergoing substitution, steric considerations play a role. However, the attacking reagent could alter its approach by 15 degrees or so lessening steric interactions (e.g. in benzene the attacking nucleophile could approach at an angle of 15 degrees above the plane of the benzene) and not increasing the activation energy too much.

One way to explain the high energy associated with $\ce{S_{N}2}$ attack at a vinylic carbon is as follows. Consider the ethylenic double bond as a bent bond system, that is, we have a two-membered ring (instead of a traditional sigma-pi double bond) and each carbon in the ring is $\ce{sp^{4.3}}$ hybridized. This bent bond description is equivalent to the alternate pi bond description in all regards including electronic (btw, it makes the reason [strain] for the high heat of combustion of olefins more palpable). If possible, build a model to help visualize the situation. As the attacking nucleophile approaches and begins to interact with one of the ring carbons, that carbon will attempt to rehybridize to an $\ce{sp^2}$ carbon (just like in the traditional case of $\ce{S_{N}2}$ attack on an $\ce{sp^3}$ carbon). In this case we're forcing the already highly strained bond angle in our two-membered ring to try and open further - towards 120 degrees. Hence, this is a high-energy pathway that is not often observed.

Alternately, this same concept can be rephrased in terms of traditional $\ce{sp^2}$ sigma-pi hybridization, by noting that we start with an $\ce{sp^2}$ hybridized carbon which becomes $\ce{sp}$ hybridized in the transition state. It is known that $\ce{sp}$ hybridized carbons are roughly 5 kcal/m less stable than $\ce{sp^2}$ hybridized carbons (more electrons in higher energy p orbitals, not offset by the energy lowering of the electrons shifting from $\ce{sp^2}$ bonds to $\ce{sp}$ bonds) thereby increasing the energy required to reach the transition state.

Another point to consider is that since the $\ce{S_{N}2}$ reaction proceeds with inversion of configuration, such a reaction in a ring system would lead to the formation of a trans-double bond. This would be an extremely high energy process in the case of a small ring system such as benzene.

Finally, $\ce{sp^2}$ hybridized carbon is more electronegative than $\ce{sp^3}$ hybridized carbon due to the increased s-character. This will increase the electron density around an $\ce{sp^2}$ carbon and, as a nucleophile approaches, the increased electron repulsion will make this a higher energy pathway that nucleophilic attack at a less electron rich $\ce{sp^3}$ carbon.

Answer 2

When attacking a benzene ring for an SN2 reaction, I think your explanation is good enough. The $\ce{C-X}$ antibonding orbital lies in the plane and inside the ring. The nucleophile would be forced to approach through the rest of the atoms of the ring.

As for other $\mathrm{sp^2}$ hybridized organohalides, there was a good explanation recently on this site: Why do SN1 and SN2 reaction not occur at sp2 centres?

Answer 3

Why can't an enolate anion successfully displace a halogen from a benzene?

As you know enolate is the conjugate base of enol which is a tautomeric form of ketone. Thus being an electron rich molecule it basically acts like a nucleophile.

Aryl halides are extremely less reactive towards nucleophilic substitution reactions due to the following reasons:

• Resonance effect : In haloarenes, the electron pairs on halogen atom are in conjugation with π-electrons of the ring and the following resonating structures are possible. $\ce{C—Cl}$ bond acquires a partial double bond character due to resonance. As a result, the bond cleavage in haloarene is difficult than haloalkane and therefore, they are less reactive towards nucleophilic substitution reaction.

• Difference in hybridisation of carbon atom in $\ce{C—X}$ bond: In haloalkane, the carbon atom attached to halogen is $\mathrm{sp^3}$ hybridised while in case of haloarene, the carbon atom attached to halogen is $\mathrm{sp^2}$-hybridised. The $\mathrm{sp^2}$ hybridised carbon with a greater s-character is more electronegative and can hold the electron pair of $\ce{C—X}$ bond more tightly than $\mathrm{sp^3}$-hybridised carbon in haloalkane with less s-chararcter. Thus, $\ce{C—Cl}$ bond length in haloalkane is 177pm while in haloarene is 169 pm. Since it is difficult to break a shorter bond than a longer bond, therefore, haloarenes are less reactive than haloalkanes towards nucleophilic substitution reaction.

• Instability of phenyl cation: In case of haloarenes, the phenyl cation formed as a result of self-ionisation will not be stabilised by resonance and therefore, SN1 mechanism is ruled out.
• Because of the possible repulsion, it is less likely for the electron rich nucleophile to approach electron rich arenes.