Types of Organic Reactions

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Mix ethene with bromine, and two molecules go in, but only one comes out. Now take ethanol, and one molecule goes in, but two come out, ethene and water. So why does the count go down in one reaction, and up in the other? Keep that question in mind. Organic chemistry has thousands of reactions, but most of them fit into just four basic types, and questions on them come up often in JEE, so let's begin.

0:31Substitution: one group replaces another

So, type number one is substitution. One atom or group leaves, and another takes its place, a simple swap. Okay? R X is an alkyl halide, where R is the alkyl group and X is the halogen. It reacts with O H minus, the hydroxide ion, and the small minus written high is its charge. The arrow means gives, so we get ROH plus X minus. The attacker is a nucleophile, so this is nucleophilic substitution, S N, with the N written low. Right? When the attacker is an electrophile, we get electrophilic substitution, S E, with the E written low. The example to remember is the nitration of benzene. With me? And the third kind is free-radical substitution, like the chlorination of methane in light. Light is your clue for radicals. Clear?

1:21Addition: across a multiple bond

Now, type number two is addition: two species add across a multiple bond, like the double bond of an alkene. Got it? One pi bond, that is the Greek letter pi, is replaced by two sigma bonds, the Greek letter sigma, one to each new group. So the double bond becomes a single bond. Okay? Again there are three kinds. Adding HBR, hydrogen bromide, to an alkene is electrophilic addition. Right? Adding HCN, hydrogen cyanide, to a carbonyl group is nucleophilic addition. Now tell me, with the same alkene and the same HBR, is that addition always electrophilic? No, no, careful, that is a common trap! With a peroxide, the same HBR adds to the alkene by free radicals. So always read the conditions. With me?

2:10Elimination and rearrangement

Type number three is elimination, where the molecule loses two atoms or groups. Addition used up a pi bond, remember? So what do you think forms when the two groups leave? Yes, exactly, a new pi bond forms! The two groups usually leave from adjacent carbons, so this is called beta elimination, and beta is the Greek letter on screen. Clear? Two common examples are the dehydration of alcohols, where water is lost, and dehydrohalogenation, where a hydrogen halide is lost. Right? And type four is rearrangement, where the skeleton, or the position of a group, changes within the molecule. So the full set is substitution, addition, elimination and rearrangement. Got it?

2:55Transition state and intermediate

Now let's look inside a reaction. In the left graph, energy goes up the side, and reaction progress runs along the bottom. The reactants climb a hill and come down as products, and the top of the hill is the transition state, an energy maximum. The climb, marked E a, that is E with a small a written low, is the activation energy. Okay? At the transition state, bonds are partly made and partly broken. So tell me, could you collect it in a bottle and study it? No, you cannot, because it cannot be isolated! It is only the top of the hill, not a substance you can keep. Clear? The right graph has two hills, T S one and T S two, so two transition states. The species in the valley between them is an intermediate, an energy minimum between two transition states. With me? Unlike a transition state, an intermediate is a real species, though short-lived: a carbocation, a carbanion or a free radical. Here the graph shows a carbocation. Right?

3:56Two paths: SN1 and SN2

Now, the two paths of nucleophilic substitution. S N one is the two-step path on the right, through a carbocation. The label under the right graph says the first, slow step decides the rate, and in that step the alkyl halide alone ionises to form the carbocation; the nucleophile comes in only after that. So, does the rate depend on the nucleophile? No, the rate does not depend on the nucleophile at all. Rate equals k times R X in square brackets; the square brackets mean concentration, and k is the rate constant. Reactivity follows carbocation stability: three degree, greater than two degree, greater than one degree, that is, tertiary, secondary, primary, where the small circle written high is the degree sign. Okay? Benzyl and allyl halides are also fast. And if the carbon is a chiral centre, it is racemised in S N one. Got it? Now S N two, the one-step path on the left, with no intermediate. The nucleophile attacks from the back side, opposite the leaving group. Clear? So both species appear in the rate: rate equals k, times R X, times N u minus, the nucleophile, each concentration in square brackets. And the configuration is inverted. Now here is a trap for you. Tertiary was fastest in S N one, so it is fastest in S N two as well, right? No, no, careful, in S N two it is the exact reverse! In back-side attack the nucleophile must reach the carbon, and bulky groups block the way; this is steric hindrance. So methyl, C H three, with the small three written low, is fastest, then primary, then secondary, and tertiary is slowest.

5:33Example 1: classify substitution, addition or elimination

Okay? Alright, example one: classify these three reactions. In part a, C H three B R reacts with O H minus, giving C H three O H and B R minus. In part b, ethene reacts with bromine, B R two, and the equals sign between the carbons is a double bond. In part c, ethanol gives ethene and water. Pause the video and try it first. Done? Okay, let's check. In part a, O H replaces the bromine, and the attacker, O H minus, is a nucleophile. So this is nucleophilic substitution. Easy, right? Part b is the first half of our puzzle. Bromine adds across C double bond C, and the pi bond is used up. Two molecules become one: this is addition, and addition to an alkene like this is electrophilic. Okay? And part c is the other half. Water is lost and a C double bond C forms, so this is elimination, the dehydration of an alcohol. One molecule became two, and our puzzle is solved. See how quickly you can classify now?

6:40Example 2: SN1 or SN2 from rate data, and favoured halides

Now example two, a harder one. The rate of a substitution doubles when the concentration of the alkyl halide is doubled, and is unchanged when the concentration of the nucleophile is doubled. Which mechanism operates, and which alkyl halides favour it? Pause the video and try it first. Done? Okay, let's check. Doubling the halide doubles the rate, but doubling the nucleophile changes nothing. So the rate equals k times the concentration of R X, and the slow step involves only the halide. Now, which mechanism has exactly this rate law? It is S N one, yes, the two-step path through a carbocation! Now for the second part, which alkyl halides favour it, primary halides maybe? No, no, careful, primary halides are the fast ones in S N two! S N one follows carbocation stability, so it is favoured by tertiary, benzylic and allylic halides. Clear?

7:41Quick revision: four reaction types, SN1 vs SN2

So, quick revision. Four types: substitution, addition, elimination and rearrangement. The transition state is a maximum that cannot be isolated, while an intermediate is a real species at a minimum. S N one goes with k times R X, tertiary fastest, and racemisation. S N two goes with k times R X times N u minus, methyl fastest, and inversion. Now go and try the practice problems on this page. All the best for your preparation!