Question
Write three resonance structures for $CH3 – CH = CH – CHO.$ Indicate their relative stabilities and explain.
✓
Answer
Three resonance structures are:
Stability order$: I > II > III$
$I:$ Contains more number of covalent bonds, each carbon atom and oxygen atom has complete octet, and involves no separation of opposite charges. Therefore, the most stable resonance structure.
$II:$ Contains one covalent bond less than in $I,$ one carbon $(C^+)$ has only $6$ valence electrons, involves separation of opposite charges; the resonance structure
$II$ has $-ve$ charge on more electronegative $‘O \ ’$ and $+ve$ charge on more electropositive $‘C \ ’.$ It has intermediate stability.
$III:$ Contains one covalent bond less than in $I,$ oxygen has only $6$ valence electrons, involves separation of opposite charge, has $-ve$ charge on the more electropositive $‘C \ ’$ and $+ve$ charge on more electronegative $‘O \ ’.$ All these factors are unfavourable for stability. Therefore, it is the least stable.
Stability order$: I > II > III$
$I:$ Contains more number of covalent bonds, each carbon atom and oxygen atom has complete octet, and involves no separation of opposite charges. Therefore, the most stable resonance structure.
$II:$ Contains one covalent bond less than in $I,$ one carbon $(C^+)$ has only $6$ valence electrons, involves separation of opposite charges; the resonance structure
$II$ has $-ve$ charge on more electronegative $‘O \ ’$ and $+ve$ charge on more electropositive $‘C \ ’.$ It has intermediate stability.
$III:$ Contains one covalent bond less than in $I,$ oxygen has only $6$ valence electrons, involves separation of opposite charge, has $-ve$ charge on the more electropositive $‘C \ ’$ and $+ve$ charge on more electronegative $‘O \ ’.$ All these factors are unfavourable for stability. Therefore, it is the least stable.
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