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2.5: Rules for Resonance Forms

  • Page ID
    67056
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    Objective

    After completing this section, you should be able to use the concept of resonance to explain structural features of certain species; for example, why all of the carbon-oxygen bonds in the carbonate ion are the same length. This particular compound is discussed in further detail in Section 2.6.

    Rules for estimating stability of resonance structures

    1. The greater the number of covalent bonds, the greater the stability since more atoms will have complete octets
    2. The structure with the least number of formal charges is more stable
    3. The structure with the least separation of formal charge is more stable
    4. A structure with a negative charge on the more electronegative atom will be more stable
    5. Positive charges on the least electronegative atom (most electropositive) is more stable
    6. Resonance forms that are equivalent have no difference in stability and contribute equally. (eg. benzene)

    The above resonance structures show that the electrons are delocalized within the molecule and through this process the molecule gains extra stability. Ozone with both of its opposite charges creates a neutral molecule and through resonance it is a stable molecule. The extra electron that created the negative charge on either terminal oxygen can be delocalized by resonance through the terminal oxygens.

    Benzene is an extremely stable molecule and it is accounted for its geometry and molecular orbital interaction, but most importantly it's due to its resonance structures. The delocalized electrons in the benzene ring make the molecule very stable and with its characteristics of a nucleophile, it will react with a strong electrophile only and after the first reactivity, the substituted benzene will depend on its resonance to direct the next position for the reaction to add a second substituent.

    The next molecule, the Amide, is a very stable molecule that is present in most biological systems, mainly in proteins. By studies of NMR spectroscopy and X-Ray crystallography it is confirmed that the stability of the amide is due to resonance which through molecular orbital interaction creates almost a double bond between the nitrogen and the carbon.

    Example 2.5.1: Multiple Resonance of other Molecules

    Molecules with more than one resonance form

    Some structural resonance conformations are the major contributor or the dominant forms that the molecule exists. For example, if we look at the above rules for estimating the stability of a molecule, we see that for the third molecule the first and second forms are the major contributors for the overall stability of the molecule. The nitrogen is more electronegative than carbon so, it can handle the negative charge more than carbon. A carbon with a negative charge is the least favorable conformation for the molecule to exist, so the last resonance form contributes very little for the stability of the Ion.

    The Hybrid Resonance forms show the different Lewis structures with the electron been delocalized. This is very important for the reactivity of chloro-benzene because in the presence of an electrophile it will react and the formation of another bond will be directed and determine by resonance. The long pair of electrons delocalized in the aromatic substituted ring is where it can potentially form a new bond with an electrophile, as it is shown there are three possible places that reactivity can take place, the first to react will take place at the para position with respect to the chloro substituent and then to either ortho position.

    Exercises

    1) For the following resonance structures please rank them in order of stability. Indicate which would be the major contributor to the resonance hybrid.

    clipboard_e4d50c953548f6f730a984afcb049dd98.png

    2) Draw four additional resonance contributors for the molecule below. Label each one as major or minor (the structure below is of a major contributor).

    E2-18.png

    3) Draw three resonance contributors of methyl acetate (an ester with the structure CH3COOCH3), and order them according to their relative importance to the bonding picture of the molecule. Explain your reasoning.

    4) Below is a minor resonance contributor of a species known as an ‘enamine’, which we will study more in Section 19.8 (formation of enamines) Section 23.12 (reactions of enamines). Draw the major resonance contributor for the enamine, and explain why your contributor is the major one.

    clipboard_e5f1c43f2428856f67f60035a45472d21.png
    5) Draw the major resonance contributor for each of the anions below:

    clipboard_eb6f29cc84455a89c7ee8caa529082d65.png

    Solutions

    1) Structure I would be the most stable because all the non-hydrogen atoms have a full octet and the negative charge is on the more electronegative nitrogen. Structure III would be the next in stability because all of the non-hydrogen atoms have full octets. Structrure II would be the least stable because it has the violated octet of a carbocation.

    2)

    clipboard_eb676d232c751b00958479b3496e955e3.png

     

    3)

    clipboard_e3bb82a6a990867ddeb6199c65af872a3.png

     

    The contributor on the left is the most stable: there are no formal charges.

    The contributor on the right is least stable: there are formal charges, and a carbon has an incomplete octet.

    The contributor in the middle is intermediate stability: there are formal charges, but all atoms have a complete octet.

    4) This contributor is major because there are no formal charges.

    clipboard_ea6176294bc318657ec09a13832ec9648.png

     

    5)

    clipboard_eb02da6558e425431ac6a0ea7f644283b.png

     

    Contributors and Attributions


    2.5: Rules for Resonance Forms is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.

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