A2 Chemistry 9701: Orbital Explanation for Bromobenzene's Unreactivity

Bromoethane and bromobenzene are separately heated under reflux with aqueous sodium hydroxide. Bromoethane is readily hydrolysed to ethanol; bromobenzene shows no detectable reaction even after prolonged heating.

(a) Describe, in terms of orbitals, how a lone pair on the bromine atom in bromobenzene interacts with the aromatic ring's pi system, and state the resulting effect on the length and strength of the C–Br bond compared with the C–Br bond in bromoethane. [3]

(b) Explain how this interaction reduces the electron deficiency (partial positive charge) at the ring carbon bonded to bromine, and use this to explain why $\text{OH}^-$ cannot attack that carbon effectively. [3]

(c) Suggest why a mechanism analogous to SN1, proceeding via a carbocation, is also not a feasible alternative pathway for bromobenzene. [2]

Part (a): Orbital overlap and its effect on the C–Br bond

In bromobenzene, the carbon bonded to bromine is part of the aromatic ring’s continuous, delocalised $\pi$ system (formed by overlap of the p-orbitals on each ring carbon, each contributing one electron). One of bromine’s lone pairs occupies a p-orbital of similar orientation and energy, which can overlap with this ring $\pi$ system. This lone pair partially delocalises into the ring, giving the C–Br bond some double-bond character.

Because of this partial double-bond character, the C–Br bond in bromobenzene is shorter and stronger than the C–Br bond in bromoethane, where bromine’s lone pairs have no adjacent $\pi$ system to delocalise into and the bond remains a normal single bond.

Part (b): Reduced electrophilicity at the ring carbon

The electron density that delocalises from bromine’s lone pair into the ring increases the electron density around (and partially “shields”) the ring carbon bonded to bromine. In bromoethane, that carbon carries a significant partial positive charge, $\delta^+$ , because bromine (more electronegative) pulls electron density away through the $\sigma$ bond with little compensation. In bromobenzene, the delocalisation effect pushes electron density back towards that same carbon, substantially reducing its partial positive charge.

Nucleophilic attack by $\text{OH}^-$ depends on being electrostatically and orbital-attracted to an electron-deficient carbon. With much less electron deficiency at the ring carbon in bromobenzene, $\text{OH}^-$ has far less driving force to attack there, so nucleophilic substitution does not proceed under conditions that readily hydrolyse bromoethane.

Part (c): Why an SN1-type pathway is also not feasible

An SN1-type pathway would require the C–Br bond to break heterolytically first, generating a carbocation on the ring carbon. However, this cation would have to sit in an orbital on an $sp^2$ ring carbon that is not part of the delocalised $\pi$ system (a so-called phenyl-type cation). It cannot be stabilised by the ring’s aromatic delocalisation in the way an alkyl carbocation is stabilised by neighbouring alkyl groups. Forming such a cation would also force that carbon out of the planar arrangement the ring needs to maintain its aromatic stability, which is highly energetically unfavourable. Because neither the direct-attack (SN2-type) route nor the carbocation (SN1-type) route is accessible, bromobenzene resists nucleophilic substitution by both mechanisms.

Final answers

(a) Bromine’s lone pair delocalises into the ring’s $\pi$ system, giving the C–Br bond partial double-bond character; this makes it shorter and stronger than in bromoethane.
(b) Delocalisation increases electron density at the ring carbon bonded to bromine, reducing its $\delta^+$ character, so $\text{OH}^-$ has little electrostatic attraction to attack it.
(c) A carbocation on the ring carbon would not be stabilised by the aromatic system and would disrupt the ring’s planarity/aromaticity, so this pathway is also not feasible.