Which of the following resonating structures of 1-methoxy-1, 3-butadiene is least stable
A$\mathrm{CH}_2 - \mathrm{CH} = \mathrm{CH} - \mathrm{CH} = \mathrm{O} - \mathrm{CH}_3$
B$\mathrm{CH}_2 = \mathrm{CH}_2 - \mathrm{CH} - \mathrm{CH} = \mathrm{O} - \mathrm{CH}_3$
C$\mathrm{CH}_2 - \mathrm{CH} - \mathrm{CH} = \mathrm{CH} - \mathrm{O} - \mathrm{CH}_3$
D$\mathrm{CH}_2 = \mathrm{CH} - \mathrm{CH} - \mathrm{O} - \mathrm{CH}_3$
Answer & Solution
Correct answer: C. $\mathrm{CH}_2 - \mathrm{CH} - \mathrm{CH} = \mathrm{CH} - \mathrm{O} - \mathrm{CH}_3$
To compare resonance structures, the least stable contributor is the one with the greatest charge separation, incomplete octets, or an unfavorable placement of charges.
For $1$-methoxy-$1,3$-butadiene, the methoxy group can donate electron density into the conjugated system. The more stable resonance forms are those in which all atoms have complete octets and the negative charge, if produced, is preferably on oxygen.
Among the given forms, option $C$ corresponds to a structure with extensive charge separation in which carbon and oxygen bear unfavorable formal charges relative to the other contributors. Such a contributor is much less stable than neutral forms or forms with negative charge on oxygen.
Thus, after comparing the listed options, the least stable resonance structure is $C$.
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