1
$\begingroup$

In my textbook, I have read about unimolecular elimination and nucleophilic unimolecular substitution reaction of organic compounds. In those reactions, firstly one bond breaks automatically then as soon as a carbocation forms then a nucleophile attacks the carbocation to complete the reaction.

Now, my question is: Why doesn't the same happen in case of addition reaction of molecules having double bond? One bond breaking may cause a charge separation and then nucleophile/electrophile can attack the positive or negative charged side? I know molecules with charge separation aren't supposed to be much stable. But even in the case of a carbocation the same happens.

I have Googled for unimolecular addition reaction but all the results were about unimolecular substitution/elimination.

|improve this question
$\endgroup$
  • 2
    $\begingroup$ Search "intramolecular cycloadditions". $\endgroup$ – RBW Oct 19 '16 at 18:42
1
$\begingroup$

The problem I see with your question is that you mix up two completely different concepts. The first concept is the reaction order (i.e. order of the kinetics, for example unimolecular = first order) which is dependent on the number of components that take part in the rate determining step of the reaction. The other concept is the type of the reaction, e.g. addition, substitution, functional group interconversion.

Naturally, most addition reactions will be of the type $\mathrm{A + B \rightarrow C}$ or something like that, but that neither means that they will be of second order kinetically, nor that all addition reactions are of that type. Intramolecular additions would be one example for a unimolecular addition.

Please be careful though not to mix up the order of the reaction and the stoichiometry of the written down reaction.

Imagine a reaction were you have $\mathrm{A+B \rightarrow C}$ and $\mathrm{B}$ is used as the solvent, so that $\mathrm{B}$ is available in large excess. While there are clearly two components needed for the addition to proceed it will appear to be unimolecular (first order reaction).

Strictly speaking, the stoichiometry of the reaction has nothing to do with the reaction order.

However, most addition reactions can be conceptualized to be of the form $\mathrm{A+B \longrightarrow C}$ whether that means that $\mathrm{A}$ and $\mathrm{B}$ are part of the same molecule or different chemical entities or if the rate law will actually appear to be of second order as well then depends on the specific case.

In that regard, I don't really see the point in asking whether an addition reaction might be unimolecular, as the kinetics will depend on the specifics of the reaction while the stoichiometry can be inferred just by looking at the equation. I hope that clearifies it.

|improve this answer
$\endgroup$
  • $\begingroup$ Can Organic addition reactions be unimolecular too? we see separate mechanisms for unimolecular and second order elimination and substitution. But no such is discussed in general textbooks for addition reaction. $\endgroup$ – Tahmid Hossain Oct 20 '16 at 18:13
  • $\begingroup$ The idea of a unimolecular addition as a basic mechanism doesn't make sense. You need at least two things to add together. Even in the case of intramolecular additions, you still need the two components for the addition to take place. $\endgroup$ – Zhe Oct 20 '16 at 19:48
  • $\begingroup$ @Zhe That is not correct! The molecularity of a reaction depends only on the number of molecules, and in an intramolecular reaction there is only one molecule, thus the reaction is said to be unimolecular -- it is of first order as only one component engages in the rate determining step of the reaction. $\endgroup$ – logical x 2 Oct 20 '16 at 19:56
  • $\begingroup$ @Zhe Your use of the word "component" seems not clearly defined to me in that context. The way you say it here it implicates that if two groups in the same molecule have to come together the reaction is of second order but that is not correct! $\endgroup$ – logical x 2 Oct 20 '16 at 20:04
  • $\begingroup$ @ketbra I disagree. An $\mathrm{S_{N}2}$ reaction is still still a bimolecular process from a mechanistic standpoint even if it is intramolecular. As you say, the rate may not reflect the fact that we still consider the basic mechanism to be bimolecular. I think we're taking about the same thing, but "bimolecular" and "unimolecular" are overloaded terms, w.r.t. to kinetics and nomenclature of basic mechanisms. $\endgroup$ – Zhe Oct 20 '16 at 20:05

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.