1
$\begingroup$

Refluxing, say, one mole of 1,5-dibromo-pentane with excess ammonia in ethanol can result in the formation of rings. (e.g. $1\lambda^2$-piperidine)

Picture

However, it is also possible for the 1,5-dibromo-pentane molecules to be connected together, forming longer chains without forming rings. The longer chains can be reacted again to form larger rings. So in theory, we can have rings that are arbitrarily large.

This is about the rings' size distribution

Question: What is the expected number of rings (large and small) after the reaction finishes? How many of the rings are $1\lambda^2$-piperidine? Generally, is there a mathematical distribution that describes the expected number of rings of any given size?

$\endgroup$
  • 2
    $\begingroup$ Common teaching is that the formation of rings greater than, say, 8 members is difficult. It usually requires templating (internal or external to the molecule). You are basically asking about ring polymers and their size distribution. $\endgroup$ – TAR86 Feb 25 '19 at 7:45
  • $\begingroup$ @TAR86 Could you explain more about it or tell me where can i learn more? $\endgroup$ – Ma Joad Feb 25 '19 at 9:22
  • 1
    $\begingroup$ Do you know what $1\lambda^2$-piperidine means and why you're getting that? ChemDraw circled the nitrogen atom in red for a reason.. $\endgroup$ – orthocresol Feb 25 '19 at 23:47
1
$\begingroup$

There is no simple answer here because the outcome will be heavily dependent on the concentration and on the rate of addition.

The most salient point in your question to me is: "excess ammonia."

You can ask what the competing reactions are.

  1. 1,5-dibromopentane reacts with ammonia to form 1-amino-5-bromopentane
  2. 1-amino-5-bromopentane reacts with itself to form piperidine
  3. 1-amino-5-bromopentane reacts with ammonia to form 1,5-diaminopentane
  4. 1-amino-5-bromopentane reacts with another straight chain alkyl amine to elongate the chain
  5. 1,5-dibromopentane reacts with piperidine to form N-(5-bromopentyl)piperidine
  6. and others...

The main problem I see is that in excess ammonia, unless you perform a slow addition of ammonia, the concentration may be such that the ammonia will over add via (3). Then you're going to get any product. The intermediates are also reactive, so the reactant concentration has to be relatively low as well, otherwise, you may see (4) or (5) or other side reactions. You may also need to run the reaction at lower temperature to disfavor reaction of the product with any alkyl bromide intermediate or reactant.

These issues need to resolved before you even begin to think about distribution of ring sizes because you might not even form any rings.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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