30. Arrange the protons Ha,,Hb,Hc, and Hd in decreasing order of bond energy?
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The decreasing order is Hb>Hc>Ha>Hd.
Option (B) is correct.
Explanation:
- First look the bond energies. Breaking of the bonds creates any unstable compound.
- For example in the above question, Hb-C bond breaks.
- It results to the formation of a carbocation on the double bond, which makes the compound unstable. So, C-Hb has highest bond energy.
- Then check if breaking of the bond forms any stable carbocation.
- They will easily break the bond to attain that stable structure.
- So, their bond energy will be less. For ex in the above compound Ha-C, Hc-C, Hd-C on breaking form stable carbocation.
- Between Ha and Hc, Ha involves in conjugation where as Hc doesn't. Thus C-Ha bond energy is less than that of C-Hc bond energy.
Hence The decreasing order is Hb>Hc>Ha>Hd.
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