. Write the two resonance hybrids for the carbocation that would be formed by protonation at C-1 of 2-methyl-1,3-pentadiene. Without doing a calculation, would you expect C-2 or C-4 (the two end carbons of the allylic cation) to have the most positive charge on it

Answer is explained in the explanation section below.

Explanation:

This question is requires the diagram, which I have drawn and it is attached in the attachment below. Please for this answer, refer to the diagram attached in the attachment below.

Referring to the diagram, attached. As we know that, Allylic secondary carbocations are more stable than Allylic tertiary carbocations.

Hence,

C2 will have a more positive charge since a tertiary carbocation (C2) is more stable than a secondary carbocation (C4). Therefore, the resonance structure will favor the positive charge at C2.

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the answer is c i took the test