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I do understand that H2; Lindlar will give a cis double bond and Na;NH3(l) will give trans double bond. However, I am having difficulty understand how to determine which reagent will work in this question between H2; lindlar catalyst and Na;NH3(l). The answer is given as H2;Lindlar. Second step is to add Br2 to the alkene. Then a strong base such as NaNH2 will deprotonate OH and the compound will undergo a Sn2 reaction to form the ring. The part I am having difficulty is choosing between H2; lindlar catalyst vs Na;NH3(l).

Thanks for any help.


This is a really interesting question. I'm not sure I agree with the end product, but I will get to that in a second. There does appear to be a piece missing from this equation, which is a second step (something has to be responsible for the new bromine in the product).

First thing is to identify how to get to the final product from an intermediate. You've correctly ruled out $\ce{NaNH2}$ which is just a strong base. It's frequently used to deprotonate a terminal alkyne (which you do not have), and in this example, all it would do would be to remove the hydrogen on the alcohol.

You've also ruled out $\ce{H2, Pt}$. This is a standard hydrogenation and will reduce a triple bond fully to the alkane. If that happens, you won't have the functional groups to do anything like a ring closure.

That gives two possibilities, hydrogenation with Lindlar's catalyst or a sodium ammonia reduction. Both of these reactions reduce the alkyne to an alkene (and then stop). The difference is that concerted hydrogenation will add hydrogen in a syn fashion to afford the cis-alkene, whereas the sodium/ammonia reduction is not concerted and will give the more stable trans-alkene.

The cyclization happens via addition of bromine. This is a standard reaction: halohydrin formation. Except that the intramolecular opening via the attached alcohol will be fast, so the alcohol attacks instead of water to provide an ether. Assuming this is the mechanism, you have to think about how the relative stereochemistry of the alkene is carried forward.

The key intermediate is the bromonium ion (the three member ring with a bromine). The intermediate retains the stereochemistry of the alkene. You might need a model to identify this, but the correct bromonium ion that gives the indicated product has the hydrogens on the same side. That means that they were originally syn across the double bond, so the double bond was cis. Then Lindlar hydrogenation was the appropriate reaction for the first step.

Here's my issue with this question: I don't actually think you'll get this product. The technical point is that the opening can proceed via a 5-exo-tet attack or a 6-endo-tet attack. Generally, in these types of reactions, the exo attack is favored for stereoelectronic reasons. So my money is on obtaining the 5-member ring instead of the 6-member ring.


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