In a heterolytic bond fission of $2 -$ methyl $1 -$ phenylpropane, which of the following stable carbocation will form?

C_{6} H_{5} - C H_{2}^{+}

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(C H_{3})_{2} C H^{+}

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C H_{3}^{+}

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C_{6} H_{5}^{+}

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Solution

The correct option is A $C_{6} H_{5} - C H_{2}^{+}$
A benzylic carbocation is a resonance-stabilized carbocation. In each of the two equally stable major resonance forms of which the formal charge of +1 is on benzylic carbon.

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Similar questions

C_{6} H_{5} \leftarrow C H_{2}

C_{6} H_{5} \leftarrow C H_{2} C H_{2}

C_{6} H_{5} \leftarrow C H \leftarrow C H_{3}

C_{6} H_{5} \leftarrow C (C H_{3})_{2}

The correct order of stability of the following carbanions is:

$(C H_{3})_{3} C^{-},^{-} C C l_{3}, (C H_{3})_{2} C H^{-}, C_{6} H_{5} - C H_{2}^{-}$

I_{2} / N a O H

C_{6} H_{5} \oplus C H_{2}

C_{6} H_{5} C H_{2} \oplus C H_{2}

C_{6} H_{5} \oplus C H C H_{3}

C_{6} H_{5} \oplus C (C H_{3})_{2}

C H_{2} = C H - ⨁ C H_{2}

C_{6} H_{5} - ⨁ C H_{2}

C H_{3} - ⨁ C H - C H_{3}

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