epoxide to ether

I thought that the first step must be using a Grignard reagent to add carbons, but I need a hydroxyl group on the carbon on the other side to make it a diol and produce the product given, but nothing seems to be working.

How do I proceed?

  • $\begingroup$ What if you used Grigniard reagent with chlorine present... $\endgroup$
    – Mithoron
    Commented Aug 1, 2020 at 23:10
  • $\begingroup$ Think about how many carbons you need to introduce. In addition, you need a some kind of functional group handle so that you have something to work with. $\endgroup$
    – Zhe
    Commented Aug 1, 2020 at 23:18
  • $\begingroup$ it would solve it, so basically what can my R chain contain in gringard, i thought only hydrocarbons, it can contain halogens then? $\endgroup$
    – Katia
    Commented Aug 1, 2020 at 23:18
  • $\begingroup$ Yes, halogen (fluorine in particular) may be present, but it may complicate preparation of the reagent. $\endgroup$
    – Mithoron
    Commented Aug 1, 2020 at 23:57
  • 2
    $\begingroup$ Alternatively, you can incorporate some unsaturation into your Grignard reagent, or a different functional group that's protected. $\endgroup$
    – Zhe
    Commented Aug 2, 2020 at 0:59

3 Answers 3


Synthetic scheme

  1. Prepare the tetrahydropyran ether of commercially available 3-bromopropanol using dihydropyran, catalytic PTSA in DCM.

  2. Make the Grignard of the THP-protected bromopropanol, react this with ethylene oxide. Work up and isolate to give the mono-THP-pentane-1,5-diol.

  3. Make the tosylate or mesylate of the mono-THP-pentane-1,5-diol.

  4. Remove the THP group using cat. PTSA in MeOH

  5. Form the alkoxide using LiHMDS in Et2O and cyclisation will occur to give the pyran. Isolate by careful distillation.


For preparation of the Grignard reagent, use allylmagnesium bromide.

$$\ce{CH2=CH-CH2-Br ->[Mg][ether] CH2=CH-CH2-Mg+Br-}$$

Now we react the Grignard reagent formed with ethylene oxide, our starting compound.

$$\ce{C2H4O ->[CH2=CH-CH2-Mg+Br-][Et2O] HO-CH2-CH2-CH2-CH=CH2}$$

This gives us pent-4-en-1-ol. Now we protect the alcoholic group by esterification, using trifluoroacetic anhydride ($\ce{(CF3CO)2O }$).

$$\ce{HO-CH2-CH2-CH2-CH=CH2 ->[(CF3CO)2O][Pyridine] CF3-COO-CH2-CH2-CH2-CH=CH2}$$

Now, we use $\ce{HBr/R2O2}$ to form 5-bromopentyl 2,2,2-trifluoroacetate.

$$\ce{CF3COO-CH2-CH2-CH2-CH=CH2 ->[HBr/R2O2] CF3COO-CH2-CH2-CH2-CH2-CH2-Br}$$

Now we remove the trifluoroacetate protecting group to get 5-bromopentanol.

$$\ce{CF3COO-CH2-CH2-CH2-CH2-CH2-Br ->[K2CO3/MeOH] HO-CH2-CH2-CH2-CH2-CH2-Br}$$

Now, under basic conditions, the compound undergoes intramolecular SN2 to form the six member cyclic ether

$$\ce{HO-CH2-CH2-CH2-CH2-CH2-Br ->[OH-] C6H10O}$$

  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$
    – Waylander
    Commented Aug 2, 2020 at 12:32
  • 1
    $\begingroup$ Actually, I'm not sure if you might actually just get spontaneous cyclisation during the deprotection. Would be a nice bonus. $\endgroup$ Commented Aug 2, 2020 at 13:38
  • $\begingroup$ That should work $\endgroup$
    – Waylander
    Commented Aug 2, 2020 at 15:08

Just summarizing the steps already written in the comments,


One equivalent magnesium will react with iodine preferentially due to it being a better leaving group than chlorine. The bond also has more covalent character.

Subsequent step of reaction will open the oxirane ring easily.

Then hydrolysis produces 5-chloropentan-1-ol.

This is followed by intramolecular SN2 attack by lone pair on oxygen of hydroxide at the chlorine carbon causing cyclized product formation.

  • 1
    $\begingroup$ I would suggest that instead of I, you might want to use Br. It is just a thought since Br would still satisfy the conditions and it seems to be more prevalently used. $\endgroup$ Commented Aug 2, 2020 at 10:03
  • $\begingroup$ How does the magnesium not react with chlorine? Can we use $\ce{SCH3}$ at the far end instead? $\endgroup$ Commented Aug 2, 2020 at 10:04
  • 5
    $\begingroup$ Do you have a reference for the preparation and use of 3-halomagnesium halides? $\endgroup$
    – Waylander
    Commented Aug 2, 2020 at 10:10
  • $\begingroup$ @Waylander pubs.acs.org/doi/pdfplus/10.1021/acs.jchemed.5b00453?src=recsys Not exactly for the above compound, but preferential reaction is observed. $\endgroup$ Commented Aug 2, 2020 at 11:50
  • 1
    $\begingroup$ I think the route falls over at that point. AFAIK you cannot form a Grignard of 3-halo propyl halides and get a standard addition. There is scope for all kinds of alternative reactions - cyclopropane formation possibly. $\endgroup$
    – Waylander
    Commented Aug 2, 2020 at 12:13

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