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If we add HI to a substrate Ph-C-C-OH what will be the major product and what reaction mechn. will be followed?

My attempt: as -OH is bad Leaving Group, we cannot just simply proceed with Sn2 and remove the OH. So first H2O (due to the H+ and lone pair on O) will form and then it will leave the substrate and then 1-2 hydride shift in Ph-C-C+ --> Ph-(C+)-C and we will add I- at the carbocation.

My dilemma is how can we even form a 1* carbocation (even if it can be stabilised by rearrangement )

Is my attempt valid? If not what would be the real mechanism and why?

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This reaction has been studied and yields 2-phenylethyliodide (Ph-CH2-CH2-I) in around 90% yield. The authors postulate a symmetrical non-classical carbocation with involvement of the aromatic system.

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  • $\begingroup$ Yes thank you.....(SNGP will take place) $\endgroup$
    – Ash_Tag
    Commented Jun 12, 2023 at 13:39
  • $\begingroup$ Another possibility: a hydride ion half-shifts to float over the carbon-carbon bond of what is originally the $\ce{CH2CH2OH}$ group. This would also displace the water molecule and distribute the positive charge over both carbons. $\endgroup$
    – Oscar Lanzi
    Commented Jun 12, 2023 at 18:36

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