E1, E2 Reactions & their Comparison | Chemistry Optional Notes for UPSC PDF Download

How Do We Explain What Happens In This Elimination Reaction (Which We Will Call, “E1”)

• Here’s the reaction. First, look at the bonds that are being formed and broken. This is a classic elimination reaction – we’re forming a new C–C(π) bond, and breaking a C–H and C–leaving group (Br here) bond.
  
• But now we want to know more than just “what happens”. We want to understand how it happens. What’s the sequence of bond-forming and bond breaking? To understand HOW it happens, we need to look at what the data tells us.
• That’s because chemistry is an empirical science; we look at the evidence, and then work backwards.

First Clue About The Mechanism of the “E1 Elimination”: The Rate Only Depends On Concentration of Substrate (Not Base)

• Let’s look at the first important clue for this reaction. We can measure reaction rates quite readily. When we vary the concentration of the substrate, the reaction rate increases accordingly. In other words, there is a “first-order” dependence of rate on the concentration of substrate.
• However, if we vary the concentration of the base (here, H₂O) the rate of the reaction doesn’t change at all.
  
• What information can we deduce from this?  The rate determining step for this reaction (whatever it is) therefore does not involve the base. Whatever mechanism we draw will have to account for this fact.
• The reaction is first-order in the substrate (alkyl halide) and zero order in base (i.e. doubling concentration does not double the rate).

The Second Clue About The Mechanism Of The E1 Elimination Reaction – Rate Is Fastest For Tertiary Substrates

• Another interesting line of evidence we can obtain from this reaction is through varying the type of substrate, and measure the rate constant that results. So if we take the simple alkyl halide on the far left (below) where the carbon attached to Br is also attached to 3 carbons (this is called a tertiary alkyl halide), the rate is faster than for the middle alkyl halide (a secondary alkyl halide) which is itself faster than a primary alkyl halide (attached to only one carbon in addition to Br).  (See post: Primary, Secondary, Tertiary)
• So the rate proceeds in the order tertiary (fastest) > secondary >> primary (slowest)
  
• Any mechanism we draw for this reaction would likewise have to account for this fact. What could be going on such that tertiary substrates are faster than primary?

The Third Clue About The Mechanism Of The E1 Elimination Reaction: It Competes with the S_N1 Reaction

• A final interesting clue about the mechanism of this reaction concerns the by-products that are often obtained. For instance, when the alkyl halide below is subjected to these reaction conditions, we do obtain the expected elimination product.
• However, we also get substitution reactions in the product mix as well. Remember – substitution reactions involve breakage of C-(leaving group) and formation of C-(nucleophile).
• What makes this particular starting alkyl halide particularly interesting is that the carbon attached to Br is a stereocenter. And if we start with a single enantiomer of starting material here, we note that the substitution product formed is a mixture of stereoisomers. Note that both inversion and retention of stereochemistry at the stereocenter has occurred.
• We’ve seen this pattern before – it’s an S_N1 reaction!
  

• This last part is a very important clue. If an S_N1 reaction is occurring in the reaction mixture, looking back at the mechanism of the S_N1 could help us think about what type of mechanism might be going on in this case to give us the elimination product.

Putting It Together: The E1 Mechanism Proceeds Through Loss Of A Leaving Group, Then Deprotonation

• Taking all of these clues into account, what’s the best way to explain what happens? This:
  
• The reaction is proposed to occur in two steps: first, the leaving group leaves, forming a carbocation. Second, base removes a proton, forming the alkene. This nicely fits in with the three clues mentioned above. [Also note that the more substituted alkene is formed here, following Zaitsev’s rule].
• Similar to the S_N1 mechanism, this is referred to as the E1 mechanism (elimination, unimolecular).

Example Of An “E2” Reaction: How Do We Explain What Happens In This Reaction?

• Here’s an example of the reaction I’m talking about:
  
• What’s interesting about this reaction is that it doesn’t follow the same rules that we saw for the E1 reaction. We’ll talk about two key differences here.

Clue #1 About The Mechanism Of The E2 Reaction: The Rate Depends  on  Concentration of Both  Substrate and Base

• Remember that the E1 reaction has a “unimolecular” rate determining step (that is, the rate only depends on the concentration of the substrate?)
• Well, when we look at the rate law for this reaction, we find that it depends on two factors. It’s dependent on the concentration of both substrate and the base.
• That means that whatever mechanism we propose for this reaction has to explain this data.
  
• By the way, see how useful chemical kinetics can be? They’re such simple experiments – measure reaction rate versus concentration – and you get these nice graphs out of it.  I can’t even begin to stress how important this data can be in understanding reaction mechanisms. So simple, so elegant, and so useful.
• Another note – you might notice that the base here (CH₃O^–) is a stronger base than we see for the E1 reaction (more on that later).

Clue #2 About The Mechanism Of The E2 Reaction: Stereochemistry Of The C–H Bond And The Leaving Group Is “Anti”

• Here’s the second key piece of information – and we didn’t talk about this for the E1. The reaction below  is very dependent on the stereochemistry of the starting material. 
• When we treat this alkyl halide with the strong base, CH₃ONa, look at this interesting result. What’s weird about this? Well, this seems to fly in the face of Zaitsev’s rule, right? Why don’t we get the tetrasubstituted alkene here?
  
• The mystery gets a little deeper. If, instead of starting with the alkyl halide above, we “label” it with deuterium – that is, we replace one of the hydrogens with its heavy-isotope cousin that has essentially identical chemical properties – we see this interesting pattern:
• Note how the group that is on the opposite face of the cyclohexane ring to the leaving group (Br) is always broken.
• In fact, if we use the molecule above and make just one modification, now we actually do get the Zaitsev product!
• See what’s going on? The hydrogen that is broken is always opposite, or “anti” to the leaving group.
• So how do we explain these two factors? 

Putting It Together: The Mechanism Of The E2 Reaction

• Here’s a hypothesis for how this elimination reaction works. It accounts for all the bonds that form and break, as well as the rate law, and – crucially – the stereochemistry.
  
• In this mechanism, the base removes the proton from the alkyl halide that is oriented anti to the leaving group, and the leaving group leaves – all in one concerted step.
• Since it’s an elimination reaction, and the rate law is “bimolecular”, we call this mechanism the E2.

Comparing The Mechanism Of The E1 and E2 Reactions

Here’s how each of them work:

What Do The E1 and E2 Reactions Have In Common?

Here’s what each of these two reactions has in common:

• in both cases, we form a new C-C π bond, and break a C-H bond and a C–(leaving group) bond
• in both reactions, a species acts as a base to remove a proton, forming the new π bond
• both reactions follow Zaitsev’s rule (where possible)
• both reactions are favored by heat.

How Are The E1 and E2 Reactions Different?

• Now, let’s also look at how these two mechanisms are different. Let’s look at this handy dandy chart:
  
• The rate of the E1 reaction depends only on the substrate, since the rate limiting step is the formation of a carbocation. Hence, the more stable that carbocation is, the faster the reaction will be. Forming the carbocation is the “slow step”; a strong base is not required to form the alkene, since there is no leaving group that will need to be displaced (more on that in a second). Finally there is no requirement for the stereochemistry of the starting material; the hydrogen can be at any orientation to the leaving group in the starting material [although we’ll see in a sec that we do require that the C-H bond be able to rotate so that it’s in the same plane as the empty p orbital on the carbocation when the new π bond is formed].
• The rate of the E2 reaction depends on both substrate and base, since the rate-determining step is bimolecular (concerted). A strong base is generally required, one that will allow for displacement of a polar leaving group. The stereochemistry of the hydrogen to be removed must be anti to that of the leaving group; the pair of electrons from the breaking C-H bond donate into the antibonding orbital of the C-(leaving group) bond, leading to its loss as a leaving group.

E1 vs E2: Why Does One Elimination Give The “Zaitsev” Product, And The Other Elimination Does Not? 

Now we’re in a position to answer a puzzle that came up when we first looked at elimination reactions. Remember this reaction – where one elimination gave the “Zaitsev” product, whereas the other one did not. Can you see why now?

The Key Requirements Of Stereochemistry In The E2 Reaction

So what’s going on here?

• The first case is an E2 reaction. The leaving group must be anti to the hydrogen that is removed.
  
• The second case is an E1 reaction.
  
• In our cyclohexane ring here, the hydrogen has to be axial. That’s the only way we can form a π bond between these two carbons; we need the p orbital of the carbocation to line up with the pair of electrons from the C-H bond that we’re breaking in the deprotonation step. We can always do a ring flip to make this H axial, so we can form the Zaitsev product.
• Here’s that deprotonation step:
  As you can see, cyclohexane rings can cause some interesting complications with elimination reactions! In the next post we’ll take a detour and talk specifically about E2 reactions in cyclohexane rings.