Which Mechanism? The Deciding Factors

Transcript

0:00Introduction

Hello everyone! Let us start with a small mystery. Take one bottle of two bromobutane. Pour some into sodium iodide in acetone, and the rest into water containing silver nitrate. Same compound, same carbon to bromine bond. So the same mechanism in both flasks, right? No, not at all, the two flasks follow two completely different mechanisms. So who decides? The halide, the nucleophile, or the solvent? Choosing between S N one and S N two comes up often in JEE questions, so let's find out.

0:38Two paths, side by side

So, first the two paths side by side, because every deciding factor comes from here. In S N one, with the N written low, there are two steps: the halide first ionises to a carbocation, a carbon carrying a plus charge, and then the nucleophile attacks it. With me? S N two is just one step. The nucleophile comes in from the back as the halide leaves, at the same time, so there is no intermediate at all. Okay? Now tell me this. In which of the two mechanisms does the concentration of the nucleophile appear in the rate law? In S N two, yes, exactly! For S N one, rate equals k times the concentration of R X, the alkyl halide; the square brackets mean concentration, and k is the rate constant. For S N two, rate equals k, times the concentration of R X, times the concentration of N u minus, the nucleophile, with the minus sign written high. Clear? Next, the stereochemistry at a chiral carbon. S N one goes through a flat carbocation, so the nucleophile can attack from either face, and we get racemisation. S N two always attacks from the back, so we get inversion. Got it? And the last line is about rearrangement. Think about it: in which mechanism can the carbon skeleton rearrange, and why? In S N one, that is correct, because its carbocation can shift a hydride or an alkyl group to become more stable. S N two has no intermediate at all, so there is nothing to rearrange. Right?

2:14Factor 1: the carbon that carries the halogen

Now the first deciding factor: the substrate, meaning the carbon that carries the halogen. S N one depends on how stable the carbocation is. So which halides react fastest by S N one? Yes, the tertiary halides are the fastest! The small circle written high is read as degree, so three degree means tertiary. The order is tertiary, greater than secondary, greater than primary, greater than methyl, C H three. Okay? And for S N two, it is exactly the reverse. The nucleophile must reach the back of the carbon, and bulky groups crowd it out. So methyl is the fastest, then primary, then secondary, and tertiary halides practically do not react this way. With me? Now here is the trap. Secondary sits in the middle of both lists. So which path does a secondary halide take? See, it can go either way! The carbon alone cannot decide for a secondary halide. Here, the nucleophile and the solvent decide. Clear?

3:16Factors 2 and 3: the nucleophile and the solvent

Next, the nucleophile. Remember the rate laws we just saw. Does S N one need a strong nucleophile? No, it does not! In S N one, the nucleophile only attacks the carbocation in the fast step, so a weak one is enough. But S N two needs a strong nucleophile, in high concentration, because the nucleophile is right there in its rate law. Right? Now let us look at the solvent. S N one likes a polar protic solvent, such as water or an alcohol. They stabilise both ions as they form, so ionisation becomes easier. Okay? And S N two likes a polar aprotic solvent, such as DMSO, which is dimethyl sulfoxide, or acetone. This helps because such a solvent leaves the negative nucleophile free to attack.

4:04The leaving group: the same order for both

Got it? Now, what about the leaving group, the halide ion that goes away? Does a better leaving group push the reaction towards S N one, or towards S N two? Careful, it does neither of these! The order is iodide, greater than bromide, greater than chloride, greater than fluoride, and it is the same for both mechanisms. So it changes the speed, not the path. With me? And iodide is the best for a simple reason. The rule is that the weaker the base, the better it leaves, because a weak base is stable on its own. Iodide is the weakest base of the four, so it leaves most easily. Clear?

4:43Topper's trick: read the question in this order

Now, here is a trick that toppers use. Read the question in a fixed order. First, the carbon. Methyl and primary halides usually go by S N two, and tertiary halides by S N one. The arrow here means leads to. Okay? Second, only if the halide is secondary, look at the reagent and the solvent. With a strong nucleophile and an aprotic solvent, you get S N two. Right? With a weak nucleophile, a protic solvent, or the silver ion, written A g plus, you get S N one. Two quick checks, and the answer is ready. Got it?

5:16Example 1: CH3Br with NaCN in DMSO, t-butyl bromide in water

Now let's put this to work. Predict the mechanism in part a, bromomethane with sodium cyanide in DMSO, and in part b, two bromo two methylpropane in water. Pause the video and try it first. Done? Okay, let's check. In part a, bromomethane is C H three B r, a methyl halide, the best substrate for S N two. The cyanide ion, C N minus, is a strong nucleophile, and DMSO is a polar aprotic solvent that leaves it free. The double arrow means so, and the answer is S N two. Clear? Now for part b, C H three in brackets with a small three outside means three methyl groups, all on the carbon that carries the bromine. So it is tertiary, and S N two is practically ruled out. Water is a weak nucleophile and a polar protic solvent, perfect for ionisation. So the answer is S N one. With me?

6:12Example 2: 2-bromobutane, NaI in acetone vs AgNO3 in water

Now the harder one, and it is our mystery from the start. Two bromobutane reacts, in part a, with sodium iodide in acetone, and in part b, with water containing silver nitrate. Predict the mechanism in each case. Pause the video and try it first. Done? Okay, let's check. In the formula, B r in brackets means the bromine is attached to the C H just before it. That carbon is joined to two other carbons, so this is a secondary halide. And you remember the trap: a secondary halide can go either way, so the conditions decide. Right? So look at flask a, with iodide ions in acetone. Which path does it favour? It favours S N two, exactly! The iodide ion, I minus, is a strong nucleophile, and acetone is a polar aprotic solvent. That is the S N two combination. Okay? Now look at flask b, with water and silver nitrate. Here, water is a weak nucleophile and a polar protic solvent, and the silver ion, A g plus, pulls the bromide off and assists ionisation. So this flask goes by S N one. Same compound, two mechanisms, and now you know why! Clear?

7:27Quick revision: when to choose SN1 and when SN2

So, quick revision. Methyl and primary halides favour S N two, and tertiary halides favour S N one. Secondary halides follow the conditions: a strong nucleophile in an aprotic solvent gives S N two, and a weak nucleophile in a protic solvent gives S N one. And the leaving group order, iodide, bromide, chloride, fluoride, is the same for both. Now go and try the practice problems on this page. You will find them easy. All the best!