AIEEE and Earlier JEE Main Questions: Substitution

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Take a tertiary chloride, with three methyl groups on the carbon that holds the chlorine. By one substitution mechanism, it is the fastest of the simple alkyl chlorides. By the other, it practically does not react at all! Two opposite answers for the same molecule? Keep that puzzle in mind, because the secret is the mechanism. Today we solve older A I triple E and JEE Main questions on substitution, and ordering questions like these come up often. So, let's begin.

0:36SN2: crowding decides the speed

So, first the S N two path. In the heading, S N two is a capital S followed by a small N and a small two, both written low: substitution, nucleophilic, bimolecular. N u minus is the nucleophile; the small minus written high is its charge. In the picture, H O minus, the hydroxide ion, comes in along the orange arrow, from the back, opposite to the bromine, and B r minus leaves, all in one step. The solid wedge points towards you, and the dashed bond points away from you. With me? Both the halide and the nucleophile take part in that one step. So rate equals k, times the concentration of R X, times the concentration of N u minus. R X is the alkyl halide, k is the rate constant, and the square brackets mean concentration. Okay? Now look inside the big square brackets of the picture. The double dagger marks a transition state, the dashed lines to H O and B r are bonds half made and half broken, while the dashed line to C H three still just points away from you. Delta minus is a partial negative charge. Five groups crowd round that one carbon. So tell me, by this path, which reacts faster, a methyl halide or a tertiary halide? The methyl halide, yes, exactly! The order is C H three X, the methyl halide, where X is the halogen, then one degree, two degree and three degree. The small circle is read as degree, for primary, secondary and tertiary, and the greater than sign means faster than. Each extra alkyl group on that carbon hinders the attack from the back. Clear? And the tertiary halide is practically inert to S N two. The reason is steric hindrance: its three alkyl groups block the back of the carbon. In fact, A I triple E two thousand and five asked for exactly this reason. Got it?

2:32SN1: the carbocation decides the speed

Now the other path, S N one: substitution, nucleophilic, unimolecular. Step one is slow: R X breaks by itself into R plus and X minus. The arrow means gives, and R plus is a carbocation. In the picture, R one, R two and R three, with small numbers written high, are any three groups, and C plus in the middle is the flat carbocation. Then, in the fast step, N u minus attacks it, along the orange arrows, from either face. That gives the two products on the right, labelled inversion and retention. With me? Only R X takes part in the slow step, and the slow step decides the speed. So rate equals k times the concentration of R X, nothing else, so it is first order. Right? Now, back to our puzzle. On this path, the speed depends on how stable the carbocation is. So, is a tertiary halide fast or slow by S N one? Fast, the fastest of these four types, exactly! Reactivity follows carbocation stability: three degree, then two degree, then one degree, then C H three. It is the S N two order, completely reversed. Puzzle solved: the tertiary chloride is last by S N two, and first by S N one. Okay? And two special families. In allylic halides, the halogen is on a carbon next to a carbon carbon double bond; in benzylic halides, next to a benzene ring. Their cations are stabilised by resonance, so they are very reactive by S N one. Example two needs this. Clear?

4:08Which path? The deciding factors

Now, the conditions. S N two is favoured by a strong nucleophile and a polar aprotic solvent. And since the attack comes from the back, the configuration of the carbon is inverted, like an umbrella turning inside out in the wind. Okay? S N one is favoured by a weak nucleophile and a polar protic solvent, which solvates both ions. And because the carbocation is flat, the nucleophile can attack either face, so an optically active halide gives a largely racemised product. Right? And now a simple trap that catches many students: the wording. Some questions ask for decreasing order, and some for increasing. So here is a topper's habit: underline that word first, and write your answer in that direction. Got it?

4:56Example 1: SN2 order of methyl to tertiary chlorides

Time for Example one, from JEE Main two thousand and fourteen. Arrange these four chlorides in order of reactivity in S N two reactions: C H three C l, C H three C H two C l, and two more with methyl groups in brackets. Pause the video and try it first. Done? Okay, let's check. First, classify each one by the carbon that holds the chlorine. In C H three C l, that carbon has no other carbon on it: methyl. In C H three C H two, dash, C l, the dash is the bond to chlorine, and that carbon is joined to one carbon, so it is primary. With me? Next, C H three in brackets with a small two outside means two methyl groups on that carbon, so it is secondary. And with a small three outside, there are three methyl groups, so it is tertiary. Okay? Now, a student writes the tertiary chloride first, because it is the most substituted, and the methyl chloride last. Is that order right? No, no, careful, that is the S N one order! S N two hates crowding: the more alkyl groups, the harder the attack from the back. So C H three C l is the fastest, then the primary, then the secondary, and the tertiary is the slowest. Clear?

6:12Example 2: increasing SN1 order, p-methoxybenzyl chloride

Now Example two, the harder one, from JEE Main two thousand and seventeen. Arrange three chlorides in order of increasing reactivity in S N one reactions. The Roman numerals one, two and three label them: two simple chain chlorides, and para methoxybenzyl chloride. Pause the video and try it first. Done? Okay, let's check. It is S N one, so we compare the carbocations. In compound one, C H three, CHC l, C H two, C H three, the chlorine sits on a carbon joined to two carbons. So losing chloride gives a secondary carbocation. Okay? In compound two, the chlorine is on the end carbon, joined to only one carbon. So it gives a primary carbocation, the least stable here. Right? Now compound three. The small italic p means para, and C six H four is a benzene ring carrying two groups opposite each other: C H three O, the methoxy group, and C H two C l. Losing chloride gives a benzylic cation, and this para methoxybenzyl cation is strongly stabilised by resonance. With me? So the order of cation stability is three, then one, then two. So the answer is three, greater than one, greater than two, right? Oh no, careful, read the question again! It asks for increasing reactivity. So we write it from the slowest: two, less than one, less than three, where the less than sign means slower than. Same chemistry, written in the direction the question asks. Got it?

7:50Quick revision: SN2 and SN1 orders, increasing vs decreasing

So, quick revision. S N two: the less crowded carbon is faster, so methyl, then primary, secondary and tertiary. S N one: the more stable carbocation is faster, so tertiary, secondary, primary, then methyl. Allylic and benzylic halides are very reactive by S N one, and always check whether the question wants increasing or decreasing order. Now solve the other past questions on this page yourself. You can do this. All the best!