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








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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!




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](https://en.wikipedia.org/wiki/Markovnikov%27s_rule)**:   
Use either [Acid Catalysed Hydration](https://en.wikipedia.org/wiki/Hydration_reaction#Alkenes) or [Oxymercuration-Demercuration](https://en.wikipedia.org/wiki/Oxymercuration_reaction). $\ce{RCH=CH\_2\longrightarrow RCH(OH)CH\_3}$
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](https://en.wikipedia.org/wiki/SN2_reaction). (Williamson’s Ether Synthesis)


### For $\ce{RCH\_2CH\_2OMe}$ from $\ce{RCH=CH\_2}$:


1. **Anti-Markovnikov Water Addition**: Use [Hydroboration-Oxidation](https://en.wikipedia.org/wiki/Hydroboration#Reactions_involving_substituted_alkenes).
2. **Add a small amount of sodium or sodium hydride**
3. **Add iodomethane** ([Williamson’s Ether Synthesis](https://en.wikipedia.org/wiki/Williamson_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](https://en.wikipedia.org/wiki/Ether#Synthesis) for more details.


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