# Favoured reaction product in acid catalysed addition of methanol to alkene [closed]

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

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

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.

• Thank you insipidintegrator! My original idea is explained here: chadsprep.com/chads-organic-chemistry-videos/… How can I modify this to maximise RCH2CH2OMe ? Sep 8 at 11:24
• 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 $\ce{RCH_2CH_2OH}$. Sep 8 at 12:34
• Thanks again, Insipidintegrator. What would adding H2O2 do to the reaction mechanism? Would we get a Kharasch type anti-Markovnikov addition to favour RCH2CH2OMe? Sep 8 at 13:20
• 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. Sep 8 at 13:27
• Pure alcohols will be protonated by organic acids altho the reaction is promoted by traces of water or by bases such as pyridine. Sep 10 at 0:10