Devise syntheses of the following three compounds, starting in each case with any alkene that contains four carbons or fewer. You do not have to write mechanisms, although at this point it may be very helpful for you to do so.

  1. $\ce{(CH3)3COH}$

  2. $\ce{(CH3)2CHOH}$

  3. $\ce{CH3CH2CHOHCH3}$

This is the question I'm trying to tackle. I'm having a bit of trouble with the third part of it, which is the synthesis of butan-2-ol. The text solutions says that I should use but-2-ene, which I can understand but I don't quite understand why my answer is also not plausible. Can someone help me out with this?

enter image description here

  • $\begingroup$ It looks right but using but-2-ene would mean you'd only get the desired product. Your method gives the product as the major product not the exclusive product. Assuming no other, more elaborate, side reactions. $\endgroup$
    – RobChem
    Jul 4, 2016 at 14:10
  • $\begingroup$ @RobChem but for my method, with my product, only a secondary carbocation forms because the primary carbocation is too unstable to form, no? So that means that my method does give the desired product as the exclusive product, not just the major product, no? $\endgroup$ Jul 4, 2016 at 14:12
  • $\begingroup$ Because from what i see, there are only two carbocations that could form and the primary carbocation itself is too unstable, in this reaction anyway, to form? $\endgroup$ Jul 4, 2016 at 14:13
  • $\begingroup$ The main point here I think here is that using but-2-ene eliminates any arguments about selectivity of carbocation formation because of the symmetry of the alkene. Yes the secondary carbocation is more stable in your proposed synthesis but I suspect there may be small amounts of the undesired terminal alcohol or perhaps rearrangement products. Using but-2-ene is just better because there is no possibility of these complications (no matter how small) so why would you use your method. Your reasoning is sound but the other way is just better $\endgroup$
    – RobChem
    Jul 4, 2016 at 14:21
  • $\begingroup$ Oh okay so in general, for synthesis questions, the thing that i have to worry about is whether i get my desired product exclusively and which starting compound would help me achieve that? $\endgroup$ Jul 4, 2016 at 14:22

1 Answer 1


You propose to use but-1-ene rather than but-2-ene as a starting material to synthesise butan-2-ol via addition of water to the double bond. There is no reason why it would not happen the way you drew.

However, using but-2-ene allows only one possible reaction pathway. With the relatively high symmetry of the molecule ($C_\mathrm{2h}$, assuming trans), attacking either side of the double bond will lead to an identical product: The molecule’s intrinsic $C_2$ axis will transform one into the other. Thus, it is only possible to generate butan-2-ol.

Starting from but-1-ene, you have two non-equivalent first steps, generating either the primary or the secondary carbocation. These obviously cannot be transformed into each other by symmetry (the alkene’s symmetry is now $C_\mathrm{s}$). Thus, you will get a side product: butan-1-ol. These products will be difficult to separate. If anything, careful distillation would do the job.

As RobChem already noted in the comments, (organic) chemists typically attempt to avoid side reactions and different products, hence why the book suggests using the more symmetric alkene.

  • $\begingroup$ Of course, i can understand what the book was thinking when it gave that compound as the starting material. Like, i could understand why the book gave that as the answer and that's because i wrote out the mechanism for but-2-ene. $\endgroup$ Jul 5, 2016 at 0:31
  • $\begingroup$ I was just wondering if there was anything wrong with the starting material i used haha. I didn't realize that the product from the primary carbocation would form in the ever so slightest amounts, despite how unstable it is. $\endgroup$ Jul 5, 2016 at 0:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.