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1. What the question is asking (in plain language)

The textbook says that when toluene (methylbenzene) is treated

  • with HCl alone → a σ‑complex (the usual Wheland intermediate) is formed and the mixture stays colourless.

  • with HCl + AlCl₃ (a Lewis‑acid catalyst) → a π‑complex of toluene with AlCl₃ is formed and the mixture turns green.

We have to explain why the two reactions give different colours.
In particular, does the green colour come from the π‑complex, and if so, what electronic process is responsible for it?

2. Step‑by‑step explanation

Step 1 – Identify the two species that are actually present

Reaction condition Species that is generated (major) Structural picture
HCl only (no Lewis acid) σ‑complex (also called the Wheland intermediate) – a cyclohexadienyl cation where the aromatic ring has temporarily lost aromaticity because H⁺ has added to one carbon. ![σ‑complex] (a benzene ring with one sp³‑hybridised carbon bearing a H⁺)
HCl + AlCl₃ π‑complex – AlCl₃ coordinates to the π‑electron cloud of the aromatic ring, giving a “Lewis‑acid/π‑base” adduct. The aromatic system stays planar; AlCl₃ sits above the ring. ![π‑complex] (toluene with AlCl₃ above the ring)

The σ‑complex is a carbocation that is strongly localized and does not involve any new metal‑ligand orbitals. The π‑complex, on the other hand, involves a metal‑to‑π charge‑transfer interaction.

Step 2 – What colours do we normally expect from these species?

Species Typical electronic transitions Position of absorption Observed colour
Uncomplexed benzene/toluene π → π* (UV, ~260 nm) Far‑UV, outside the visible range Colourless
σ‑complex (cyclohexadienyl cation) Mostly σ → σ* and π → π* but still high‑energy (UV) because the conjugation is broken UV, < 350 nm Colourless
π‑complex (aryl‑AlCl₃) Charge‑transfer (CT) transition: donation of electron density from the aromatic π system to the empty Al 3p orbital (π → Al) Visible region (≈ 500–560 nm) Green (the complementary colour of the absorbed red‑orange light)

Why does a CT band appear only for the π‑complex?

AlCl₃ is a strong Lewis acid; it possesses an empty 3p orbital that can accept electron density. When the aromatic ring sits over AlCl₃, the π electrons are partially transferred into this orbital, creating a π → Al charge‑transfer excited state. This transition costs less energy than the ordinary π → π* transition, so its absorption is shifted from the UV into the visible part of the spectrum. The band lies roughly at 550 nm, which removes red/orange light and makes the solution appear green.

Step 3 – Sketch of the electronic picture

   π electrons of toluene   →   empty Al 3p orbital (Lewis‑acid)
   ------------------------------------------------------------
   Ground state:   π (filled)   |   Al 3p (empty)
   Excited state:  π (partially empty) | Al 3p (partially filled)

The energy gap (ΔE) for this CT transition is:

[ \Delta E = h\nu \approx \frac{hc}{\lambda} ]

Taking λ ≈ 540 nm (green‑absorbing red/orange),

[ \Delta E \approx \frac{(6.626\times10^{-34}\,\text{J·s})(3.00\times10^{8}\,\text{m·s}^{-1})}{5.40\times10^{-7}\,\text{m}} \approx 3.68\times10^{-19}\,\text{J} \approx 2.3\,\text{eV} ]

This is much lower than the ≈ 5 eV required for a typical π → π* transition in benzene, confirming that the AlCl₃‑π interaction creates a low‑energy, visible‑region transition.

Step 4 – Why the σ‑complex stays colourless

The σ‑complex does not involve a metal centre with low‑lying empty orbitals. Its electronic spectrum is dominated by localized σ‑bond transitions and the broken aromatic π system. The lowest‑energy allowed transition remains in the UV, so no visible light is absorbed → the solution looks colourless.

Step 5 – Summarise the answer to the original question

Yes, the green colour is directly caused by the formation of the π‑complex between toluene and AlCl₃.
The colour originates from a π → Al charge‑transfer transition that absorbs red/orange light, leaving the transmitted light green. The σ‑complex formed with HCl alone lacks such a metal‑centred acceptor and therefore does not absorb visible light, remaining colourless.

3. Final answer

  • The σ‑complex (Wheland intermediate) formed with HCl alone absorbs only in the UV; it is colourless.
  • The π‑complex formed when AlCl₃ is present coordinates to the aromatic π‑system, creating a metal‑to‑π charge‑transfer interaction. This interaction introduces an electronic transition in the visible region (≈ 500–560 nm), which removes red/orange light and makes the solution appear green.

Thus, the observed colour difference is indeed due to the formation of the π‑complex, specifically the π → Al charge‑transfer band it introduces.

4. Common mistakes for this type of problem

Mistake Why it’s wrong How to avoid it
Assuming the colour comes from the HCl itself HCl is a colourless gas; it does not generate visible‑range absorptions. Focus on the complex formed, not on the reagent.
Attributing the green colour to a simple “chlorination” product Electrophilic chlorination of toluene gives p‑/o‑chloro‑toluene, which are also colourless. Remember the question is about complexes (intermediates), not the final substitution product.
Confusing σ‑complex with a σ‑bonded AlCl₃ adduct σ‑complex refers to the carbocation intermediate, not a metal‑bound species. Keep the definitions clear: σ‑complex = Wheland (C–H σ bond formed), π‑complex = Lewis‑acid/π‑base adduct.
Neglecting the role of charge‑transfer transitions Colour in many organometallic/π‑complexes arises from CT bands, not from π → π* alone. Explicitly consider whether an empty orbital on a metal (AlCl₃) can accept electron density from the aromatic π system.
Thinking the colour must be “green because of AlCl₃” Not every AlCl₃ complex is green; the colour depends on the energy of the CT transition. Relate colour to the wavelength of the absorbed light, not just the identity of the metal.

By keeping these points in mind, you can correctly diagnose colour changes in aromatic Lewis‑acid complexes.

Original question: What explains the color in clear sigma versus green pi complexes formed by methylbenzene? on Chemistry Stack Exchange, licensed CC BY-SA.