Consider the organic molecule 2-methylbutane (...

The structure of 2-methylbutane is $CH_3-CH(CH_3)-CH_2-CH_3$.

Removing a hydride ion ($H^-$) from different carbon atoms generates different carbocations:

Removal from C1 gives a primary carbocation: $CH_2^{+}-CH(CH_3)-CH_2-CH_3$.

Removal from C2 gives a tertiary carbocation: $CH_3-C^{+}(CH_3)-CH_2-CH_3$.

Removal from C3 gives a secondary carbocation: $CH_3-CH(CH_3)-CH^{+}-CH_3$.

Removal from C4 gives a primary carbocation: $CH_3-CH(CH_3)-CH_2-CH_2^{+}$.

The stability of carbocations follows the order: tertiary $>$ secondary $>$ primary. The tertiary carbocation formed at C2 is the most stable due to the maximum number of $\alpha$-hydrogens (8 $\alpha$-hydrogens) providing hyperconjugation, along with the +I effect of three alkyl groups.

Therefore, removal of a hydride ion from Carbon C2 forms the most stable carbocation.

Answer: Carbon C2