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I was wondering how the reaction between neopentyl alcohol and triphenylphosphite/methyl iodide looks like from a mechanistic point of view.
It should involve the formation of a phosphonium iodide salt, which can then use the phosphorus' oxophilicity; however, for that to work, the alcohol needs to act as a nucleophile, which in my opinion is not feasable, as alcohols in general are weak nucleophiles.
As evident by the Ref.1, the first step is the reaction of triphenylphosphite and methyl iodide to give methyltriphenoxyphosphonium iodide (See path B of Ref.1). This reaction is conditional and some details of conditions are described in Ref.2. Then, one of the phenolate ion in methyltriphenoxyphosphonium iodide is replaced by neo-pentyloxy group giving away a phenol molecule. The resulting compound, suggested by the original authors of the method (Ref.2), undergoes $\mathrm{S_N}2$ mechanism to give the resulting neo-pentyl iodide in good yield:
However, the suggested $\mathrm{S_N}2$ mechanism leans heavily on the claim that rearrangement does not occur, but in light of the finding of Ref.3 (the presence of more than 6% of t-amyl iodide in the reaction mixture that can only achieved by the rearrangement of neo-pentyl group, loses much of its credibility. Accordingly, to explain the rearrangement, Ref.3 suggests following two paths for the mechanism:
According to Ref.2, $\bf{A}$ and $\bf{B}$ are in equilibrium. The covalent form $\bf{A}$ would undergo “cis elimination” thus giving rise to a cyclic transition state, $\bf{a}$ leading to give pure neo-pentyl iodide (See Path 1). Alternatively, the ionic form $\bf{B}$ would break up to give an ion pair $\bf{b}$, which collapses to neo-pentyl iodide before rearrangement occurs (Path 2). However, when combination of neo-pentyl and iodide ions is delayed, the rearrangement would occur and t-amyl iodide would be formed (Also, according to Ref.3, the material balance has never exceeded 80% and the deficiency may well be due to olefin formation).
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