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Q8.11 (c):
Draw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation. (c) $CH_3CH=CHCHO$

Solution :

Step 1: Analyzing the Conjugated System

The molecule is but-2-enal, $CH_3CH=CH-CH=O$. It possesses a conjugated system consisting of alternating double and single bonds ($C=C-C=O$). The $\pi$ electrons are delocalized over the four atoms: $C_2, C_3, C_4,$ and the oxygen atom of the carbonyl group. [Per the principle of resonance involving conjugated systems].

Step 2: Identifying Electron Displacement

Due to the high electronegativity of oxygen ($EN = 3.5$) compared to carbon ($EN = 2.5$), the carbonyl group is polar. The $\pi$ electrons of the $C=O$ bond shift toward the oxygen atom, creating a carbocation character at the carbonyl carbon ($C_4$). This induces a shift of the $C_2=C_3$ $\pi$ bond toward the $C_3-C_4$ bond to stabilize the positive charge, resulting in a positive charge at $C_2$.

Step 3: Visual Representation of Resonance

CH3-CH=CH-CH=O CH3-CH(+)-CH=CH-O(-)

Step 4: Formal Resonance Structure Derivation

The electronic displacement is defined by the following resonance contributor sets:

Resonance Structure Description of Charge Distribution
Structure I: $CH_3-CH=CH-CH=O$ Neutral form; all octets complete.
Structure II: $CH_3-CH^+-CH=CH-O^-$ Charge-separated form; positive charge on $C_2$, negative charge on oxygen.

Step 5: Theoretical Justification

  • Curved Arrow Notation: Arrows originate from the $\pi$ bond (donor) and terminate at the electronegative atom or into the space between atoms to form a new $\pi$ bond [By IUPAC convention for electron movement].
  • Stability Analysis: Structure I is the major contributor because it has no formal charges and all atoms satisfy the octet rule [Per the criteria for resonance stability]. Structure II is a minor contributor due to charge separation.

Final Solution: The resonance hybrid of but-2-enal is represented by the delocalization of $\pi$ electrons from the $C=C$ and $C=O$ bonds, resulting in the canonical forms: $CH_3CH=CHCHO \leftrightarrow CH_3CH^+CH=CHO^-$.


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