How can one get 3,4-dimethylhex-3-ene from but-2-ene ?

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Using $\ce{Br_2}$ then two equivalents of $\ce{CH_3CH_2MgBr}$, I can get the corresponding alkane (3,4-dimethylhexane) but I don't see how to do it while preserving the double bond...

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    $\begingroup$ Are you looking for a synthetic route that produces the trans isomer? That's the isomer pictured in your diagram. $\endgroup$ – ron Jul 15 '15 at 14:17
  • $\begingroup$ Convert 2-butene to 2-butanone via hydroboration and oxidation. Couple 2-butanone using the McMurry reaction (TiCl3; Zn/Cu). See this --> chemistry.stackexchange.com/questions/108388/… $\endgroup$ – user55119 Mar 9 at 23:52

Your approach, which uses $\ce{Br2}$ to convert 2-butene into 2,3-dibromobutane and react that with ethylmagnesium bromide $\ce{CH3CH2MgBr}$ has two flaws:

  1. This approach removes the alkene and there is no obvious way to get it back (there is a way)
  2. More seriously, Grignard reagents are pretty terrible for $\text{S}_N 2$ reactions. They are very strong bases, so elimination can compete, but more importantly, they can undergo metathesis:

$$\ce{R-Br + R'-MgBr <=> R-MgBr + R'Br}$$

Assuming your approach worked to generate 3,4-dimethylhexane, you could carefully radically brominate it to produce 3,-bromo-3,4-dimethylhexane, and I am sure you can get back to the alkene from there.

A better approach uses two molecules of 2-butene. Do you see how you can break your target molecule up into two 4-carbon pieces? Convert one into an electrophile and the other into a nucleophile. It might take two reactions in each case. If you choose your electrophile correctly (think about what makes a great combo with a Grignard reagents), you will form the $\ce{C-C}$ bond between the two pieces and you will still have a functional group left that you can use as a handle to direct the elimination reaction to regenerate the alkene.


Ben is right that your product is simple dimer of substrate. However, there is much easier approach to conduct dimersisation of butene - it happens easily if you use strongly acidic catalyst, like conc. sulfuric acid or ion-exchange resin, such reaction is used for example in production of isooctane. It's similar to cationic polymerization of alkenes, so parameters have to be carefully adjusted to reduce creation of heavier oligomers.


To follow up on Mithoron's answer and to answer this question, a journal article covers alkylation of isobutane. Its graphical abstract seems to suggest that isobutene is acidified to a tertiary carbocation. Another isobutene adds to that carbocation and combine themselves to be isooctane carbocation at C-4 location. A hydride transfer from a third isobutane molecule to the isooctane carbocation gives the isooctane.

If this analogy is possible imaginarily, then but-2-ene could be acidified to a butane secondary carbocation. Another but-2-ene adds to that carbocation and combine themselves to be 3,4-dimethylhexane secondary carbocation. Internal hydride shifts from the secondary to a tertiary carbocation. Elimination of or returning that acidic hydrogen to the stronger acidic catalyst creates that desired double bond at that location.

Comparing this imaginary case to that of the journal article, hydride shift is one of numerous pathways to product in that abstract but it must occur in this case.


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