Substitution or Elimination?

Transcript

0:00Introduction

Hello everyone! Let us begin with a small puzzle. Take an alkyl halide and potassium hydroxide, KOH, okay? With the KOH dissolved in water, you get an alcohol. With the same KOH dissolved in alcohol, the same halide gives an alkene! Only the solvent has changed. So why does the product change completely? That is interesting, isn't it? Choosing between substitution and elimination comes up often in JEE questions, so today we learn to predict the winner. Let's go.

0:33One reagent, two jobs

So, start with the reagent itself: the hydroxide ion, written O H minus, with the minus sign written high. Now tell me, is it a nucleophile, or is it a base? It is both, yes, exactly! Any nucleophile is also a base, because the same lone pair that attacks a carbon can also pick up a hydrogen ion. Okay? If it attacks the carbon carrying the halogen, the halide ion is pushed out, and we get substitution, by S N one or S N two, with the N written low. The double arrow means gives. Right? But it can also pull off a beta hydrogen, from the next carbon; the curly letter is the Greek letter beta. The halide leaves too, a double bond forms, and we get an alkene. That is elimination, by E one or E two. So the two pathways always compete. With me?

1:27Aqueous KOH or alcoholic KOH?

Now, back to our puzzle. Aqueous KOH, in water, gives substitution, an alcohol. Alcoholic KOH, in an alcohol, gives elimination, an alkene. Okay? So let me ask you this. The reagent is the same, so why should the solvent change the winner? Look at water first. In water, the hydroxide ion is heavily solvated, meaning tightly surrounded by water molecules, and it acts as a nucleophile. Right? In alcohol, the base present is the alkoxide ion, R O minus, where R stands for an alkyl group. It is a stronger base, so it removes a beta hydrogen. Got it? Also, alcohol is less polar than water, so it does not stabilise the ionic substitution pathway as well. Both effects push towards elimination, and that solves our puzzle.

2:20Predicting the outcome

Clear? Now the table that predicts the outcome. First, sort the reagent. The strong bases are hydroxide, O H minus, and alkoxide, R O minus. Okay? The second group are good nucleophiles but weak bases: iodide, I minus; cyanide, C N minus; R S minus, which has sulphur in place of oxygen; and azide, N three minus, with the small three written low. Right? Now the halide. The small circle written high is read as degree, so one degree means primary. Methyl and primary halides have an open, easy to reach carbon, so they go by S N two with both groups. Only a bulky base switches a primary halide to E two. But a methyl halide has no beta hydrogen at all, so it can never eliminate. With me? Next, the secondary halides, two degree. With a strong base like ethoxide, which pathway wins? It is mainly E two, very good! The arrow here means leads to. But with a good nucleophile that is a weak base, like iodide or cyanide, it goes by S N two. Okay? Now the tertiary halides, three degree. Earlier we saw that tertiary halides favour S N one. So with a strong base too, it is S N one, right? No, no, careful, that is exactly the trap, because with a strong base a tertiary halide gives E two. Only with a weak base does it ionise, giving S N one plus E one together. Clear?

3:56What pushes the balance towards elimination

Now, what tips the balance? If you heat the reaction more strongly, which pathway gains, substitution or elimination? It is elimination, yes, exactly! Elimination produces more particles and has the higher activation energy, so a higher temperature increases its share. Okay? Elimination is also favoured by a stronger or bulkier base, and by a more branched halide, whose crowded carbon is harder to attack. Right? See this case from the course. One bromobutane is primary, so you might expect S N two. But potassium tert-butoxide is too bulky to reach the carbon, so it removes a beta hydrogen instead, giving but one ene. The plus sign means reacts with. Got it? And finally, solvolysis, where a tertiary halide reacts with the solvent itself, water or an alcohol. Here both S N one and E one products form, from the same carbocation. Okay?

4:54Example 1: 1-bromobutane vs tert-butyl bromide with ethoxide

Now an example from the course. Predict the major product in part a, one bromobutane with sodium ethoxide in ethanol, and in part b, two bromo two methylpropane with sodium ethoxide in ethanol. Pause the video and try it first. Done? Okay, let's check. In part a, there are four carbons in a chain with the bromine on the end carbon; the small numbers written low count the hydrogens. So it is primary. Ethoxide is strong but unhindered, not bulky, so S N two wins, and the product is one ethoxybutane, an ether. Compare that with tert-butoxide a minute ago, which gave an alkene from the same halide. Right? Now for part b, look at the formula. C H three in brackets with a small three outside means three methyl groups on the carbon carrying the bromine, and the dash is the bond. So it is tertiary. Which pathway wins with ethoxide? It is E two, correct! The carbon is too crowded for S N two, and ethoxide is a strong base, so it removes a beta hydrogen from one of the methyl groups. Clear? So the product is two methylpropene. In its formula, the equals sign inside is the carbon to carbon double bond. Got it?

6:12Example 2: cyclohexyl bromides with ethoxide, cyanide, water

Now a harder one, in three parts. Predict the main product and mechanism. Part a, bromocyclohexane with sodium ethoxide in ethanol. Part b, bromocyclohexane with sodium cyanide in DMSO, a polar aprotic solvent. Part c, one bromo one methylcyclohexane in warm water. Pause the video and try it first. Done? Okay, let's check. In bromocyclohexane, the carbon carrying the bromine, B r, is joined to two ring carbons, so it is secondary, and the reagent decides. Right? In part a, ethoxide is a strong base, and a secondary halide with a strong base gives mainly E two. So we get cyclohexene. Okay? Now part b, the same secondary halide, but with cyanide. Which pathway wins now? It goes by S N two, very good! Cyanide, C N minus, is a good nucleophile but a weak base, so it replaces the bromine, giving cyclohexanecarbonitrile. With me? Part c is the tricky one. Here the bromine and a methyl group sit on the same ring carbon, so that carbon is tertiary. There is no strong base, only warm water, which is both the solvent and the reagent. So this is solvolysis. Clear? So both S N one and E one products form. The main product is one methylcyclohexanol, by S N one, with a smaller amount of one methylcyclohexene, by E one. Three parts, and one table answered them all! Got it?

7:50Quick revision: aqueous vs alcoholic KOH, heat, bulky bases

So, quick revision. Remember, a nucleophile is also a base, so substitution and elimination compete. Aqueous KOH gives an alcohol, and alcoholic KOH gives an alkene. And heat, a stronger or bulkier base, or a more branched halide, all push towards elimination. Now go and try the practice problems on this page. You will find them easy. All the best!