Compound in which all C-N bond lengths are no | vedclass

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(D) In the given options,we analyze the $C-N$ bond lengths:$A$. Guanidinium ion $[(H_2N)_3C]^+$: Due to resonance,all three $C-N$ bonds are equivalent

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$HN=CH-C(=CH_2)-CH=NH$

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(D) In the given options,we analyze the $C-N$ bond lengths:$A$. Guanidinium ion $[(H_2N)_3C]^+$: Due to resonance,all three $C-N$ bonds are equivalent.$B$. Imidazolium ion: The two $C-N$ bonds in the ring are equivalent due to resonance.$C$. $1,4$-Dinitrobenzene: The $C-N$ bonds are not involved in resonance that would make them equivalent to each other in a way that averages their lengths to be identical.$D$. $HN=CH-C(=CH_2)-CH=NH$: In this structure,the $C-N$ bonds are clearly distinct as some are single and some are double bonds,and they do not undergo resonance to become equivalent.However,looking at standard chemistry problems of this type,the question typically asks for the compound where resonance does not make all $C-N$ bonds equivalent. Among the choices,$1,4$-Dinitrobenzene $(C)$ and the structure in $(D)$ have non-equivalent $C-N$ bonds. Given the context of resonance-stabilized ions,the structure in $(D)$ is the most obvious example where the $C-N$ bonds are chemically distinct (single vs double) and do not participate in a symmetric resonance system.

Compound in which all $C-N$ bond lengths are not same:

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