Solutionsortho-para Directors

1. Ortho-para groups

Predict the result of the following reactions:

Eight reactions: bromination of toluene, sulfonation of anisole, nitration of tert-butylbenzene, alkylation of ethylbenzene with cyclohexene, styrene with H₂SO₄, 1-phenylcyclohexene with Br₂ without and with FeBr₃, allylbenzene with H₂SO₄ and 4-phenylbutan-1-ol with BF₃

Products: p-bromotoluene, p-methoxybenzenesulfonic acid, 1-tert-butyl-4-nitrobenzene, 1-cyclohexyl-4-ethylbenzene, a styrene polymer, the trans dibromide of 1-phenylcyclohexene (plus para bromination with FeBr₃), indane and tetralin

With bulky groups such as tert-butyl, ortho substitution is hindered and almost only the para product is obtained.
Without a catalyst Br₂ adds to the double bond of 1-phenylcyclohexene (anti addition); with FeBr₃ the ring is brominated as well.
Allylbenzene and 4-phenylbutanol form carbocations that attack their own ring (intramolecular alkylation) to give indane and tetralin.

2. Ortho-para deactivators

Why do halogens deactivate the ring and yet direct to ortho and para?

Because of their electronegativity they withdraw electron density from the ring (inductive effect) and deactivate it. But in the ortho and para benzenonium ions a halogen lone pair can stabilize the positive charge by resonance; in the meta ion it cannot.

3. Aniline in acidic medium

Nitration of aniline with HNO₃/H₂SO₄ gives substantial amounts of m-nitroaniline. Why? How is p-nitroaniline obtained?

In acidic medium aniline is protonated to the anilinium ion; the –NH₃⁺ group is deactivating and meta-directing. To obtain the para isomer the nitrogen is protected as acetanilide (with acetic anhydride), the compound is nitrated (–NHCOCH₃ is an ortho-para director and, being bulky, favors para) and the amide is hydrolyzed.

← Back to the exercises