Question
Question: Which of the following compounds do not exhibit hyperconjugation? (a)\(C{{H}_{3}}CH_{2}^{\bullet }...
Which of the following compounds do not exhibit hyperconjugation?
(a)CH3CH2∙
(b) (CH3)2CH∙
(c) CH3CH=CH2
(d) (CH3)2C−CH2+
Solution
The hyperconjugation is the overlapping g between the p-orbital or a pi-orbital and a sigma -orbital and in this the carbon next to sp2 hybridized carbon should have the hydrogen bond attached to it. Now you can easily answer the given statement accordingly.
Complete step-by-step answer: First of let’s discuss what is hyperconjugation. Hyperconjugation is another kind of resonance or delocalization of the electrons which takes place through the overlap between the p-orbital or a pi-orbital and a sigma -orbital of an adjacent C-H bond.
Condition for the compounds to undergo hyperconjugation is that the carbon which is attached to the sp2 hybridized carbon should have hydrogen attached to it.
Now considering the statement as;
In option (a) CH3CH2∙ , the carbon attached to the sp2 hybridized carbon consists of the hydrogen atom and thus, it exhibits hyperconjugation.
In option (b) (CH3)2CH∙, the carbon attached to the sp2 hybridized carbon consists of the hydrogen atom and thus, it also exhibits hyperconjugation.
In option (c) CH3CH=CH2, the carbon attached to the sp2 hybridized carbon consists of the hydrogen atom and thus, it also exhibits hyperconjugation.
In option (d) (CH3)2C−CH2+, the carbon attached to the sp2 hybridized carbon does not consist of the hydrogen atom and thus, it does not exhibit hyperconjugation.
Hence, option (d) is correct.
Note: Hyperconjugation helps to explain the relative stabilities of alkenes( i.e. which consists of the C=C bond), carbocations (i.e. which consists of the positive charge) and the free radicals(i.e. which consists of a single electron).