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I would like to add methanol to an alkene RCH=CH2, catalysed by conc. sulphuric acid. I would like to know which of these endproducts will be preferentially produced:

  1. RCH(OMe)-CH3
  2. RCH2-CH2OMe

Many thanks in advance!

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    $\begingroup$ Ethers are hardly made like that. $\endgroup$
    – Mithoron
    Sep 8 at 0:02
  • $\begingroup$ Then kindly suggest an alternative method to produce the endproducts from the starting materials. $\endgroup$ Sep 8 at 0:09

1 Answer 1

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As the OP asks in the comments, I shall provide a pathway for producing the end products.

For $\ce{RCH(OMe)CH_3}$ from $\ce{RCH=CH_2}$:

  1. Electrophilic addition of water across double bond following Markovnikov’s Rule:
    Use either Acid Catalysed Hydration or Oxymercuration-Demercuration. $\ce{RCH=CH_2\longrightarrow RCH(OH)CH_3}$
  2. Add a small amount of sodium or sodium hydride: There is formation of hydrogen gas as the alcohol acts like an acid in presence of $\ce{Na}$ or $\ce{NaH}$: $\ce{RCH(OH)CH_3\longrightarrow RCH(O^-Na^+)CH_3}$
  3. Add iodomethane: The nucleophilic oxide ion attacks the iodomethane to form $\ce{RCH(OCH_3)CH_3}$ via SN2 reaction mechanism. (Williamson’s Ether Synthesis)

For $\ce{RCH_2CH_2OMe}$ from $\ce{RCH=CH_2}$:

  1. Anti-Markovnikov Water Addition: Use Hydroboration-Oxidation.
  2. Add a small amount of sodium or sodium hydride
  3. Add iodomethane (Williamson’s Ether Synthesis)

Alternative Idea:

You may want to explore adding aq. acidified methanol solution into $\ce{RCH(OH)CH_3}$ and $\ce{RCH_2CH_2OH}$. This will give you $\ce{RCH(OMe)CH_3}$ as the major product. See here for more details.

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  • $\begingroup$ Thank you insipidintegrator! My original idea is explained here: chadsprep.com/chads-organic-chemistry-videos/… How can I modify this to maximise RCH2CH2OMe ? $\endgroup$ Sep 8 at 11:24
  • $\begingroup$ Unfortunately, $\ce{RCH_2CH_2OMe}$ always remains the minor product as the mechanism is an Electrophilic addition involving formation of a carbocation intermediate which is much more stable on a secondary carbon than a primary one. Also, you can get a bit of $\ce{RCH(OH)CH_3}$ because of water addition as a side reaction. (Note that you cannot operate in $100$% pure alcohol(as a way of side-stepping the reaction with water) as organic acids do not ionise in non-polar solvents) A pathway to produce the ether you want involves use of diazomethane and $\ce{BF_3}$ with $\ce{RCH_2CH_2OH}$. $\endgroup$ Sep 8 at 12:34
  • $\begingroup$ Thanks again, Insipidintegrator. What would adding H2O2 do to the reaction mechanism? Would we get a Kharasch type anti-Markovnikov addition to favour RCH2CH2OMe? $\endgroup$ Sep 8 at 13:20
  • $\begingroup$ If you have studied the mechanism of the Kharasch effect, you’ll notice that the first step involves the homolytic cleavage of the radical initiator and subsequently the reagent, i.e. $\ce{HBr}$ (not $\ce{HCl}$ or $\ce{HI}$). I’m not very sure about the viability of the formation of $\ce{CH_3O•}$ radical from methanol. However, I found this: pubs.acs.org/doi/10.1021/ja00328a011 It involves photolysis of methanol. $\endgroup$ Sep 8 at 13:27
  • $\begingroup$ Pure alcohols will be protonated by organic acids altho the reaction is promoted by traces of water or by bases such as pyridine. $\endgroup$
    – jimchmst
    Sep 10 at 0:10

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