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I'm only interested in the first conversion: D to E.

I've been thinking about this for a little bit. My first thought was to use $\ce{TsCl}$ to make a good leaving group then a $\ce{Me}$ nucleophile. This didn't seem like it would work since $\ce{Li-Me/MeMgBr}$ would just displace the leaving group. My second thought was to convert the alcohol to a halide, make it a Grignard reagent, then react it with $\ce{MeBr}$. But that also wouldn't work. Any suggestions?

Question

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    $\begingroup$ Consider going via ketone. $\endgroup$
    – permeakra
    Commented May 15, 2018 at 21:25
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    $\begingroup$ You might also consider an organocuprate as a nucleophile. Ordinarily, I wouldn't suggest a secondary electrophile for SN2, but this is also benzylic so... $\endgroup$
    – Zhe
    Commented May 15, 2018 at 21:30
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    $\begingroup$ I considered ketone, that would add the methyl easily using Me-Li then an OH would be left. Add TsCl then a weak reducing agent, NaB(CN)H3, but would that work to remove the OH? $\endgroup$ Commented May 15, 2018 at 21:48
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    $\begingroup$ Oxidise to ketone, Wittig, then hydrogenate $\endgroup$ Commented May 15, 2018 at 23:38
  • $\begingroup$ @MalcolmHall It should, but there is a more reliable way. The resulting alcohol should dehydrate really easy. In fact, it might dehydrate during acidic treatment of Grignard reagent adduct. $\endgroup$
    – permeakra
    Commented May 16, 2018 at 4:15

1 Answer 1

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As @orthocresol mentions in the comments, the better way to undertake this synthesis would be to first oxide the alcohol to a ketone, perform the Wittig reaction and than hydrogenate the product.

OH to CH3

Another possible reaction sequence might be first convert the alcohol into a ketone, use a Grignard reagent and than reduce the alcohol.

OH to CH3

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