Skip to main content
Chemistry LibreTexts

20.3: Starting With an Alkane, Provide a Synthesis For an Alkane With the Same Number of Carbons

  • Page ID
    216853
  • First question to ask in a lot of synthetic problems is, given the functional group I’m supposed to prepare, how many ways do I know right now that I can use to form that particular functional group? In this case I’m supposed to prepare an alkene. At this point the only way I know to prepare alkenes is by means of elimination reactions. According to the lecture notes, the most efficient approaches are the E2 reaction (dehydrohalogenation of bulky halides in the presence of strong bases), and the acid-catalyzed E1 dehydration of secondary and tertiary alcohols.

    clipboard_e2dd2fee497c8ed2e99ac53ec5f21a25c.png

    The E2 reaction requires a halogen as the leaving group, so I can prepare an alkyl halide from the required alkane by the free radical halogenation reaction, just as I did before. The E1 reaction on the other hand requires the preparation of an alcohol as the starting material. However, I find that at this point the only preparations of alcohols that I know start with alkenes (acid catalyzed addition of water, oxymercuration reaction, and hydroboration sequence). It would be redundant to start with an alkene to prepare an alcohol, only to dehydrate it back to alkene. Therefore, at this point, I will stick with the E2 reaction because it is more compatible with the requirements of the problem.

    EXAMPLE 3. Starting with cyclohexane, provide a synthesis for cyclohexene.

    clipboard_e7c2097f82f693491628e11e4665a67ad.png

    No problemo. By now I’m a whiz of organic synthesis. so here is the quick solution.

    clipboard_e76a28756c39d929f0e1e777e0412ef35.png

    EXAMPLE 4. From n-butane, prepare 2-butene

    clipboard_ee320845d93ec97082021def9e61d948f.png

    Again, no problem. A retrosynthetic analysis yields the following solution.

    clipboard_e8b08e2d77030c74278da46049f284c2a.png

    The only thing to note in this approach is that the last step above would yield a mixture of cis and trans isomers, since in this particular case, that step would not involve a stereospecific reaction.