SN2 mechanism: What the Substitution Reaction looks like

Do you struggle to comprehend the SN2 mechanism, or the difference between SN2 vs SN1? You are not alone! All of us need models and practice to understand what the molecules look like in their 3D structure.

SN2 mechanism: Watch the video on Youtube!

I suggest to watch the video above – the point of animations is to see them, duh! If you can’t, I still wanted to share some explanations here in written form as well.

SN2 Mechanism: it takes two to tango

Our first model is the nucleophilic substitution of 2-bromobutane with the phenolate anion, also called a Williamson ether synthesis.

chiral electrophile for SN2 (bimolecular nucleophilic substitution)

2-Bromobutane is chiral as one of the carbons has four different substituents. We are looking at the (R)-enantiomer here – this will be important for the stereospecificity of the reaction. Electrophiles provide the LUMO for reactions, in this case the antibonding sigma star orbital between carbon and our leaving group. Note that bromide and iodide are particularly potent leaving groups due to high acidity of conjugate acids but also weak bonds with carbon. This is due to weak overlap of atomic orbitals, resulting in a low-energy sigma star that is accessible to our nucleophile, phenolate.

SN2 highest occupied molecular orbital

This electron-rich anion is completely planar due to conjugation of one of the oxygen electron pairs with the aromatic ring. Its HOMO is localized on the oxygen as you would expect – but we can also nicely see resonance with delocalization across the pi-system.

SN2 Transition state

To ensure decent orbital overlap, the substitution proceeds via back-side attack. Because the nucleophile needs to get pretty close to the already tetrahedral carbon, steric factors are more important for the SN2 reaction compared to SN1.

SN2 transition state

The SN2 mechanism proceeds in one concerted step with both electrophile and nucleophile present in the transition state – that’s why we call it 2, for bimolecular. The carbon-bromine bond is partially broken, and the carbon-oxygen bond partially formed. Remember that is just a transient energy maximum and not a real intermediate, carbons are never actually five-coordinate!

After the transition state, the product moves to a more comfortable conformation but importantly, features the inverted stereochemistry due to back side attack. This changes the (S) enantiomer in the starting material to the (R) enantiomer product. As a good leaving group, the bromide anion enjoys its solitude and buzzes off.

Steric effects in SN2 substitutions

steric hindrance in SN2 reactions

Due to the five-coordinate transition state, more sterically hindered substrates react much slower or not at all. While it’s not intuitive on paper, the model nicely visualizes that surrounding substituents can block the nucleophiles back side attack. There’s simply too much unwanted repulsion. Instead, depending on reaction conditions like solvent polarity, we would see more step-wise SN1.

Intramolecular SN2 reaction Mechanism

Let’s look at a slightly more advanced example. I’m TotalSynthesis, so I just had to take a cute natural product that was isolated from random tropical algae in Brazil. As fate wanted it, this also has a secondary alkyl bromide, so it fits perfectly.

aldingenin C, a natural product

We’re interested in this epoxide opening step as it showcases a common question on diastereoselectivity. The reaction is intramolecular but is pretty similar to a SN2-type reaction. Given our fixed starting configuration, the side chain and the nucleophilic alkoxide hover on the bottom side of the ring.

intramolecular SN2 epoxide opening

As you can see, the nucleophile has a perfect position for the backside attack, leading to the 1,2-anti product. The leaving group is now much worse than bromide, but relieving the strain energy present in the epoxide drives the reaction forward.

two potential products

Is there another potential substitution? Indeed, the other epoxide carbon is also an electrophile. However, the methyl group at this position adds some steric hindrance. Given the quaternary center, this substitution could also proceed stepwise or “asynchronous”, with C-O bond breaking being more advanced prior to addition.

Taking the longer approach forms a 7-membered ring. Compared to the six-member ring on the right, it’s not as rapidly formed or as stable – but the pathway is still significant with 19% yield.

After six additional reactions, a surprising twist showed that the original proposal was incorrect. It turned out this unique natural product never existed to begin with! Instead, it was a mis-assigned, already known molecule, which is even a bit cooler given it includes two bromides and even a chloride. Well, it happens to the best of us.

I’m looking forward to explaining simple, beginner-level content in addition to my other educational videos. Let me know if this helped you!






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