Discussion
A butadiene with a base value of \( 217 \ \mathrm{nm} \) carries three alkyl substituents worth \( 5 \ \mathrm{nm} \) each. Its predicted maximum is
Answer: D
Adding each increment to the base value:
\[ 217 + 3 \times 5 = 232 \ \mathrm{nm} \]
giving 232 nanometres. The scheme is additive by construction, so the arithmetic is trivial and the skill lies in identifying which increments apply - an exocyclic double bond, a ring residue, an auxochrome each carrying its own value.
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