Things I've learnt from failed past papers - nucleophilic substitution

Apologies for the terrible puns in this post, but I wanted to cheer myself up after what's been an inefficient study session.

Don't put all your electrons in one basket

• If you want to maximise the odds of a fast SN2 reaction, you better have a good leaving group with as little steric hindrance as possible. Those two factors go hand in hand, however. 
• Resonance is often a bigger contributing factor to the rate of an SN2 reaction, so compounds with existing C=C or C=O bonds that can engage in conjugation will often react faster than typical primary alkyl halides. The reason for this is that this conjugation will stabilise the transition state in an SN2 reaction - the more stable the transition state, the higher the rate of reaction.
• The best of these conjugated compounds is the ɑ-carbonyl halide, where you have a halide on the ɑ position of the carbonyl - ɑ just means "next to the carbonyl". Here, the relative rate of reaction can be several times faster with a given nucleophile, compared to the same reaction but with a regular alkyl halide. This is due to the conjugation present within the carbonyl, which allows for a more stable transition state:

Left to right: ɑ-carbonyl halide, benzyl halide, allyl halide. All have conjugation which will increase the stability of the transition state, hence enabling a faster rate of reaction.

• For the same reason, more resonance means a worse nucleophile. That's because for, say, a negatively charged nucleophile, that charge is spread out across multiple atoms in a conjugated nucleophile, whereas you can have that charge concentrated at a specific atom without conjugation.
• So in a nutshell:
• SN2 favours sterics, but conjugation can play an even bigger role in making a reaction go faster;
• Nucleophiles would rather not be conjugated, so as to concentrate their charge in a specific position.

Let's solv some problems!

• I mentioned last time that SN1 reactions prefer polar, protic solvents, whereas SN2 reactions favour polar, aprotic solvents. But why is that?
• Both times, polar solvents help ensure a more stable transition state which is lower in energy. In fact, that's key - more stable transition states = transition states that are lower in energy. 
• But the reason why that is is because polar solvents solvate the nucleophile. Essentially, in an SN1 reaction, we form a carbocation and an anion, and the solvent will basically surround the ions, stabilising them. For SN2 reactions, it's similar, where polar solvents in general will lower the transition state sufficiently so that a reaction can take place at a fast enough rate. 
• But crucially, the more polar a solvent is in an SN2 reaction, the lower the rate. 
• Similar reasons apply to why we use protic and aprotic solvents respectively:
• Protic solvents form strong hydrogen bonds between their acidic proton and the solvated ions. The anion is trapped, and the nucleophile's strength - its nucleophilicity - is weakened. This stabilises the anion, and thus using an aprotic solvent favours an SN1 reaction.
• The opposite applies to an SN2 reaction, as the weaker the nucleophile, the worse the rate of reaction. Aprotic solvents will still solvate the anion, but weakly - therefore, we get a stronger nucleophile in the process.
• So in a nutshell:
• Polar: lowers energy of transition state
• Protic: solvates anions, stabilising them
• Aprotic: solvates but not too much; stronger nucleophile.

Could an SN1 reaction in theory occur along a primary carbon?

Yes it can, but only under certain conditions.

For one, you need to be sure of the reaction conditions facilitate an SN1 reaction in the first place. For example:

• Do you have a polar protic solvent?
• Do you have what would be a tertiary carbon in your molecule?

Those are two good indications that you'll probably have an SN1 reaction, even if your leaving group is bonded to a primary carbon. Here's an example of what I mean:

Ethanol is a clear indicator this reaction will be SN1 - I mean come on, you don't get much more polar (O is highly electronegative in that C-O bond) and protic (that hydrogen is great at being lost) than that. 

Most notably, though, we also have a very large tosylate group bonded at the end there. This is a very good leaving group, and in this reaction, we'll have a substitution where the tosylate group is lost, and we replace it with an ethoxy group.

And yet...it's not as clear-cut as it may appear. You see, we also have an allyl group lurking there, which will result in additional resonance structures once we eliminate the tosylate group and form an intermediate, as we would in an SN1 reaction:

From this, it becomes clear that the ethanol's best place to attack isn't at the terminal carbon, as it initially seemed, but rather at the tertiary carbocation that's in the resonance structure on the right. And as this resonance structure is more stable, it follows that this is where the ethanol will attack. 

This is very familiar to the concept of a hydride shift, where in an electrophilic substitution reaction, a proton might shift to a carbocation so as to provide a more stable structure overall. And even though the example I've given isn't quite the same, this following example is closer to the hydride shift idea:

Here, we initially have a primary carbocation, but by shifting a methyl group across, we end up getting a tertiary carbocation. Much more stable, thanks to all the hyperconjugation, and therefore a much better place for a nucleophile to attack. Again, worth noting this is exclusive to SN1, because in SN2 reactions we have no intermediate - we just speed past via a transition state. And again, it's worth noting this is only going to happen if there's a reason for there to be a shift; that's if we can get a more stable carbocation out of it. 

So in a nutshell...

• Read the question and don't be a fool - sometimes tertiary carbocations could come dyed in the wool! 

Elimination reactions 

• I haven't really touched on alkenes in these blogposts recently, mainly because I feel rather secure in my alkene addition reactions. But it is worth mentioning briefly how we form alkenes from alkanes:
• Elimination reactions will occur when our nucleophile is a base. A very strong base will typically undergo an E2 mechanism, which is analogous to SN2; a weaker base might follow an E1 mechanism, which is similar to SN1.
• It's also worth noting, however, that both E1 and E2 mechanisms will favour tertiary carbons over secondary carbons over primary carbons. 
• The reason for this is different for both E1 and E2 mechanisms. For E1, it's down to greater hyperconjugation between the C-C and C-H σ orbitals overlapping, which provides more stable carbocations. This will also make allylic and benzyllic compounds great due to resonance, as I mentioned earlier.
• But in E2 mechanisms, it's actually because you have way more hydrogens bonded to the substituent carbons, making it far more likely you'll end up with an elimination reaction.
• There's another crucial factor swaying the balance in favour of elimination over substitution - temperature. 
• High temperature favours elimination because of entropic effects. In general, increase the temperature, and you'll increase entropy; elimination reactions result in three products from two reactants, meaning you get a positive change in entropy from a reaction. Substitution reactions involve two reactants becoming two products, so there is no change in entropy.
• So in a nutshell:
• If a base acts as a nucleophile;
• and it's a strong base;
• and the reaction is occurring at a high temperature;
• it's an elimination.

Why this alkene and not that?

• A common trope I've fallen into is saying an alkene will form somewhere because it's more stable. But that's far from helpful. In fact, it means absolutely nothing.

• This is a dialkene. Imagine we had an elimination reaction alongside either of those two carbons, like in the image below. Which C=C bond is more likely to form?

• The answer is the bromine on the left will be eliminated at a greater rate than the one on the right, because it's at a tertiary carbon. Therefore, the preferred product we'd get, if we could only have one elimination, would be for the C=C bond across the two tertiary carbons (the one on the left in the dialkene).
• And the reason for that? Favourable orbital overlap between the C-C orbitals, and σ conjugation as a whole.
• So in a nutshell:
• The preferred alkene product will be along a more stable carbocation: tertiary > secondary > primary;
• due to greater σ conjugation between the substituent groups, along with more favourable C-C orbital overlap.

And...

• Not all primary alkyl halides are great for SN2. Take neopentyl bromide, for example. You might think, because the bromide is on a primary carbon, that it would react superbly.
• But it won't. SN2 reactions involve a backside attack where the nucleophile will attack from the opposite side of the leaving group. And, well...neopentyl bromide is very bulky. This is a visualisation I made at home:

• Yeah...it's going to be difficult for the nucleophile to make its way past those bulky methyl groups.
• In fact, this effect does persist as you increase the length of the carbon chain, but it does weaken over time. A really bulky tertiary carbon won't have that much of an effect on the rate of reaction if you end up with an infinitely long alkyl halide chain afterwards.
• So in a nutshell:
• Think before you answer because I had no idea this was in the lecture slides, and yet this exact example was lurking there. If only I properly read them, I'd have got the question right.