The compound that produces aromatic species upon deprotonation of the indicated hydrogen is:
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The cyclopentadienyl anion is one of the classic examples of non-benzenoid aromatic systems. Its conjugate acid (cyclopentadiene) has an unusually low $\text{pK}_a$ ($\approx 16$) for a hydrocarbon because of this aromatic stabilization of the conjugate base.
Step 1: Recall what makes a ring aromatic. A ring is aromatic when it is flat, fully conjugated all the way around, and holds a special number of pi electrons given by $4n+2$ (so 2, 6, 10, and so on). If after losing a proton the ring fits this count, it becomes a stable aromatic ion.
Step 2: Understand what deprotonation does. When we pull off the marked hydrogen, the carbon left behind keeps a lone pair. If that carbon then becomes part of the conjugated ring, its electrons join the pi system and change the electron count. So we just check the new electron count after the proton leaves.
Step 3: Test the cyclopentadiene case. 5-methylcyclopentadiene has a $\text{CH}_2$ type carbon. Remove one proton from it and that carbon now donates a lone pair into the ring. The ring then has five carbons sharing six pi electrons, which matches $4n+2$ with $n=1$. That is the famous aromatic cyclopentadienyl anion.
Step 4: Test the cyclopropene case. Deprotonating 3-methylcyclopropene gives a carbanion with four pi electrons in the ring. Four electrons is $4n$, which is anti aromatic, not aromatic. So this one is rejected.
Step 5: Test the cycloheptatriene case. Pulling a proton off cycloheptatriene gives a carbanion with eight pi electrons. Eight is again $4n$, anti aromatic, so this is unstable, not aromatic. Rejected.
Step 6: Pick the aromatic one. Only the cyclopentadienyl anion reaches the magic six electron count after losing its proton. So the compound that becomes aromatic on deprotonation is 5-methylcyclopentadiene.
\[ \boxed{\text{5-methylcyclopentadiene gives an aromatic anion (6 pi electrons)}} \]
