Addition of `HI` on the double bond of propene yield isopropyl iodide and not `n-` propyl iodide as the major product. This is because the addition proceeds through

Options

A. A more stable free radical
B. A more stable carbanion
C. A more stable carbonium ion
D. None of the above being a concerted reaction

Answer

C

Explanation

`HI rarr H^(+)+I^(-)`
`CH_(3)-CH=CH_(2) + H^(+) to CH_(3) - underset({:(("Minor")),(1^(@)"Carboniumion"),(("Less stable")):})(CH_(2))-overset(+)(C )H_(2)+ CH_(3) +underset({:(("Minor")),(2^(@)"Carboniumion"),(("More stable")):})(overset(+)(C)H - CH_(3))`
`CH_(3) underset(2^(@)"Carboniumion")(-overset(+)(C)H-CH_(3)) + I^(-) to underset({:("Isopropyl iodide"),("Major produce"):})underset(I)underset(|)(CH_(3)-CH)-CH_(3)`