A reaction involving two different reactants can never be a

  1. unimolecular reaction
  2. bimolecular reaction
  3. second order reaction
  4. first order reaction

The answer as per my book is 1. unimolecular.
But the nucleophilic substitution SN1 is unimolecular but it involves two different reactants. Let us take an example, $$\ce{(CH3)3C-Cl + Br- -> (CH3)3C-Br + Cl-}$$ This follows SN1 mechanism. Yet, it involves two different reactants even though the R.D.S depends only on the formation of carbocation and the rate law expression being $$\text{rate} = k[\ce{(CH3)3C-Cl}]$$ where $k$ is the rate constant. So, is the answer that unimolecular reactions can never involve two different reactants wrong? Or am I missing something trivial?
Also it would be great if other options are discussed.

  • 1
    $\begingroup$ I think the confusion is primarily caused due to the lack in distinguishing between overall reaction and elementary reaction. $\endgroup$ – Eashaan Godbole Mar 27 '18 at 19:19

First of all,

Molecularity isn't defined for a complex reaction.

From Physical Chemistry by Ira. N Levine: (Page 531, Rate Laws And Equilibrium Constants For Elementary Reactions ):

The number of molecules that react in an elementary step is the molecularity of the elementary reaction. Molecularity is defined only for elementary reactions and should not be used to describe overall reactions that consist of more than one elementary step. The elementary reaction $\ce{ A → products}$ is unimolecular. The elementary reactions $\ce{A + B → products}$ and $\ce{ 2A → products}$ are bimolecular. The elementary reactions $\ce{A + B + C → products}$, $\ce{2A + B → products}$, and $\ce{3A → products}$ are trimolecular (or termolecular).

Atkins' mentions: (page 810, Elementary Reactions):

The molecularity of an elementary reaction is the number of molecules coming together to react in an elementary reaction.It is most important to distinguish molecularity from order: reaction order is an empirical quantity, and obtained from the experimental rate law; molecularity refers to an elementary reaction proposed as an individual step in a mechanism.

Now, from Wikipedia:

This difference can be illustrated on the reaction between nitric oxide and hydrogen:

${\displaystyle {\ce {2NO + 2H2 -> N2 + 2H2O}}}$

The observed rate law is ${\displaystyle v=k{\ce {[NO]^2[H2]}}}$, so that the reaction is third order. Since the order does not equal the sum of reactant stoichiometric coefficients, the reaction must involve more than one step. The proposed two-step mechanism is

$${\displaystyle {\ce {2 NO + H2 -> N2 + H2O2}}}$$

$${\displaystyle {\ce {H2O2 + H2 -> 2H2O}}}$$

On the other hand, the molecularity of this reaction is undefined, because it involves a mechanism of more than one step. However, we can consider the molecularity of the individual elementary reactions that make up this mechanism:

The first step is termolecular because it involves three reactant molecules, while the second step is bimolecular because it involves two reactant molecules.

As far as your example of SN1 reactions is concerned, Peter Skyes' acknowledges while discussing the kinetics of SN1:

The molecularity refers to the number of species that are undergoing bond-breaking or making in one step of the reaction, usually, in the rate limiting step.

Thus, when we say SN1 (S= substituition, N= nucleophilic, 1= unimolecular) we are essentially talking with respect to the RDS in which the carbocation is formed. Note that though most books identify 1 for unimolecular, Clayden's book refers to it as 1st order, which, in my honest opinion, is much more clearer.

The slope of the first graph is simply the first-order rate constant because rate = $k_1[t\text{-}\ce{BuBr]}$. But the slope of the second graph is zero. The rate determining step does not involve $\ce{NaOH}$ so adding more of it does not speed up the reaction. The reaction shows first-order kinetics (the rate is proportional to one concentration only) and the mechanism is called SN1, that is, Substitution, Nucleophilic, 1st order.


The question is poorly framed. However, since molecularity is defined only for an elementary reaction, maybe after looking at the options, the question setters expect you to assume that they are talking about an elementary reaction. If this is the case, only then option 1 (but along with option 4) can be marked correct. However, because they haven't mentioned anything, the candidate attempting the question is bound to get confused. If they are talking about the RDS step, then your argument is valid. An SN1 reaction is indeed unimolecular wrt the RDS step (according to all standard books like Solomon Frhyle, Peter Skyes', etc.) which would deem all of the options incorrect.

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A reaction involving two reactants can be:

  • a first order reaction, for instance when the rate solely depends on the concentration of one of the reactants. An example is the acid hydrolysis of sucrose, water being used in large excess (which makes the variation of its concentration negligible)
  • a second order reaction, for instance in basic hydrolisis of an ester (both the concentrations of base and ester affect the reaction rate)
  • a bimolecular reaction, for instance the classic SN2 reaction: both the reactants are involved in the transition state

It cannot be a unimolecular reaction: there is no way that two reactants could be involved in a transition state in which only one of them participates, after which you can expect a reaction between both of them (it would be bimolecular, if that happened).

There are indeed unimolecular reactions, for instance the thermal rearrangement of azulene to naphthalene, but they require a single reactant.

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  • $\begingroup$ Why is my example/explanation not correct? Also can you explain “transition state”? Isn’t hydrolysis of ester-pseudo first order? $\endgroup$ – MollyCooL Mar 27 '18 at 17:26
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    $\begingroup$ The example of the Sn1 is not correct because you are not asked about the molecularity of the rate determining step, but of the whole reaction! Yes, my example is pseudo-first order. Radioactive decay is an example of a pure first order reaction $\endgroup$ – The_Vinz Mar 27 '18 at 17:37
  • $\begingroup$ Oh! So they are called unimolecular but actually molecularity of entire step is not 1? $\endgroup$ – MollyCooL Mar 27 '18 at 17:42
  • $\begingroup$ This may be a good read: !en.wikipedia.org/wiki/Molecularity $\endgroup$ – Mathew Mahindaratne Mar 27 '18 at 17:47
  • $\begingroup$ Yep: in a Sn1 reaction you have two reactants and get two products, right? Bimolecular in the whole reaction, unimolecular at the intermediate step. After all, since only one of the reactants partecipates in the formation of the carbocation, you see that until that step the reaction is unimolecular. But the whole Sn1 proceeds beyond that step. $\endgroup$ – The_Vinz Mar 27 '18 at 17:48

A unimolecular reaction is an elementary reaction of a single molecule, which produces one or more molecules of product.

e.g., thermal decomposition, cis-trans isomerization (by light or heat), radioactive chemical decomposition (or simply, decay), etc.

For great example, see: Unimolecular Reactions in the $\ce {CF3CH2Cl ↔ CF2ClCH2F}$ System: Isomerization by Interchange of $\ce {Cl}$ and $\ce {F}$ Atoms: J. Phys. Chem. A, 2011, 115(6), 1054–1062 (DOI: 10.1021/jp108955m).

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Reaction between t-butyl chloride and a bromide ion is not unimolecular. There are two reactants, $\ce{Br-}$ and $\ce{(CH3)3Cl}$. Also, unimolecular and first-order are not synonymous. Molecularity refers to the amount of distinct molecules involved in a reaction while order refers to the exponent to which the concentration terms are raised to obtain the rate of reaction.

A unimolecular reaction isn't exactly a "reaction" if you look through common sense, as a single reactant cant actually "react" with anything. However unimolecular reactions are usually decomposition reactions, where a molecule simply dissociates into other molecules upon heat or certain other factors. Decomposition of calcium carbonate is a famous example for a unimolecular reaction.

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  • $\begingroup$ Why is it not unimolecular? Molecularity = 1 and Being unimolecular are equivalent right? Also, I never involved order here. $\endgroup$ – MollyCooL Mar 27 '18 at 17:25
  • $\begingroup$ No, $\pu{S_{N}}1$ reactions (while the name suggests) are not exactly unimolecular. They are pseudo-unimolecular. They appear to follow unimolecular rate statistics due to the rate determining step being unimolecular. $\endgroup$ – Pritt says Reinstate Monica Mar 27 '18 at 17:33
  • $\begingroup$ Can you elaborate on that please? $\endgroup$ – MollyCooL Mar 27 '18 at 17:35
  • $\begingroup$ See my edit to my earlier comment. And what @The_Vinz mentioned in the comment to his answer is correct. Molecularity of a single step (rate determining step) and of the whole reaction are two different things. $\endgroup$ – Pritt says Reinstate Monica Mar 27 '18 at 17:41
  • $\begingroup$ Yup, just saw, so are there no reactions that involve different reactants but also unimolecular? I get what a unimolecular reaction is and why my example is incorrect, but not exactly how to explain it taking a general case. Any help? Thanks. $\endgroup$ – MollyCooL Mar 27 '18 at 17:46

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