Question

Calculate the λ_max of the following molecule:

• A. 283 nm
• B. 273 nm
• C. 234 nm
• D. 244 nm

Answer 1

The Correct Option is B

The correct option is (B): 273 nm.

Answer 2

The structure shown is a bicyclic ring system carrying two double bonds that sit next to each other inside the same six-membered ring, forming a locked, homoannular (ring-held) diene. To find its \(\lambda_{max}\) we apply the Woodward-Fieser rules for dienes. Let's check what each answer choice would mean.

1. 283 nm: This value would only come out if we mistakenly added an extra +30 nm increment for a further conjugated double bond that this molecule does not have. The diene here has just two double bonds in conjugation, so this overshoots the true value.
2. 273 nm: Because the two double bonds are held in one ring, the base value used is that of a homoannular diene, 253 nm. The four sp2 carbons of the diene each carry a ring-residue (the rest of the fused bicyclic skeleton acts as an alkyl substituent at each end), and each ring residue adds +5 nm. With four such ring residues, we add \(4 \times 5 = 20\) nm to the base, giving \(253 + 20 = 273\) nm. There is no exocyclic double bond in this system, so no further increment applies.
3. 234 nm: This lower figure would follow only if the heteroannular (open-chain type) base value of 217 nm were used instead of the correct homoannular base of 253 nm. Since the diene here is locked inside one ring, not spread across two separate rings, the heteroannular base does not apply.
4. 244 nm: This would result from combining the heteroannular base with extra increments that do not belong to this structure, or from missing one of the four ring-residue contributions. Either way it does not reflect the actual substitution pattern.

Using the correct homoannular base of 253 nm with four ring-residue increments of 5 nm each gives \(253 + 20 = 273\) nm, matching the true diene system in the molecule.

So the correct answer is 273 nm.