Draw the conjugate base for each compound below (the acidic hydrogen in each case is marked with an *).

(c)

The IUPAC naming of the compound is$\mathbf{5}\mathbf{}\mathbf{-}\mathbf{methyl}\mathbf{}\mathbf{cyclopenta}\mathbf{}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{3}\mathbf{-}\mathbf{diene}$. Naming is done according to the rules of IUPAC nomenclature. Among double bond and methyl groups the priority is given to double bond. It is a cyclic compound having conjugated pi electrons, and it also follows the $\mathbf{(}\mathbf{4}\mathbf{n}\mathbf{+}\mathbf{2}\mathbf{)}$rule; therefore, we can say that the compound is an aromatic compound.

Given compound (c)

In this, the hydrogen that is directly attached to the cyclopentadiene ring is most acidic hydrogen than the hydrogen atom that is present in the ${\mathrm{CH}}_3$

group which is attached to the ring. Double bonds of the ring will withdraw the electron density of the hydrogen atom, making it most acidic among other hydrogen atoms.

Therefore, the conjugate base of the compound $\mathbf{5}\mathbf{}\mathbf{methyl}\mathbf{}\mathbf{cyclopenta}\mathbf{}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{3}\mathbf{-}\mathbf{diene}$is:

Conjugate base of (c)