Overview

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. In one flask, a secondary bromide is hydrolysed in a solution of hydroxide ions, and some molecules react by S N one while others react by S N two, at the same time. So what fraction goes by each path? The path with the bigger rate constant wins, right? Well, that is a trap, and one simple division gives the real answer. Rate laws for these two mechanisms are a common topic in JEE, so let's begin.

0:32How to use this chapter

So, first, how to use this chapter. This unit problem bank has three parts. Part A is at the JEE Main level, so start there. Okay? Part B is at the JEE Advanced level, so expect harder questions. Right? And Part C is the concept checks: it tests the ideas behind the reactions, not only what happens, but why. Clear? Now, the most important line. Attempt each part as a test, and read the solutions only after that. Our examples today come from Part A and Part B, plus one question from Part C as well. Got it?

1:11Two mechanisms, two rate laws

Let us first set up the two mechanisms. S N two is written as a capital S, a small capital N written low, and a two, for substitution, nucleophilic, bimolecular. N u minus is the nucleophile, and the small minus written high is a negative charge. It attacks the carbon from the back as X minus, the halide ion, leaves, all in one step, with no intermediate. With me? So tell me, in this one step, the rate depends on whose concentration? Both of them, yes, exactly! Rate equals k, times the concentration of R X, times the concentration of N u minus. Here k is the rate constant, R X is the alkyl halide, and the square brackets mean concentration, in moles per litre. So it is second order overall, and the attack from the back inverts the configuration. Okay? Now S N one, with a one, for unimolecular. It has two steps. First, slowly, the bond breaks, as the arrow shows, giving R plus, a carbocation with a positive charge on carbon, and X minus. Then N u minus attacks the carbocation, fast. Right? Now, the slow step sets the speed. So, does the nucleophile appear in the S N one rate law? No, it does not, that is correct! Only R X is in the slow step, so rate equals k, times the concentration of R X: first order, and independent of the nucleophile. And the planar carbocation is attacked from either face, so an optically active halide gives a largely racemised product, clear?

2:53From Part C: the tertiary halide puzzle

Now a question straight from Part C, the concept checks. The small circle written high means degree, so three degree is a tertiary halide. It is the most reactive in S N one, but the least reactive in S N two. So why is that? For S N one, the tertiary halide gives the most stable carbocation, so its slow step is the easiest. Okay? For S N two, its three alkyl groups block the back side attack, so tertiary halides practically do not react this way. Right? And the secondary halides, two degree, can go either way: the nucleophile and the solvent decide. Sometimes both paths run together, as in our puzzle. Clear?

3:37Example 1: find the mechanism and k from rate data

Now Example one, a JEE Main level question from Part A, okay? For a bromide, R B r, reacting with hydroxide ion, the rate is one point two times ten to the power minus five, molar per second, when both concentrations are zero point zero two molar. Capital M means molar, or moles per litre, the small letter s means second, and the minus one written high on the s means per. The rate doubles when either concentration is doubled. Name the mechanism, and find the rate constant. Pause the video and try it first. Done? Okay, let's check. The rate doubles, so it is first order, and first order means S N one, right? Oh no, careful, that is the trap here! Doubling only the bromide doubles the rate, and doubling only the hydroxide also doubles it. So it is first order in each, not first order overall. With me? So rate equals k, times the concentration of R B r, times the concentration of O H minus, the hydroxide ion. It is second order overall, with the nucleophile in the rate law, so the mechanism is S N two. Okay? Now divide the rate by both concentrations; the fraction bar means divided by. Below, it is zero point zero two times zero point zero two. Careful with the zeros: that product is zero point zero zero zero four, which is four times ten to the power minus four. Right? One point two over four gives zero point three, and ten to the power minus five, over ten to the power minus four, gives one tenth. Zero point three times one tenth, so k equals zero point zero three. Now the minus one is written high on the M as well as on the s, so the unit is per molar per second. Got it?

5:34Example 2: what fraction of RBr hydrolysis goes by SN2

Now Example two, from Part B, and it is our opening puzzle. The hydrolysis of a secondary bromide follows this rate law: rate equals k one times the concentration of R B r, plus k two times the concentration of R B r, times the concentration of O H minus. The small one and two written low just label the two rate constants. Here, k one is one point zero times ten to the power minus five per second, and k two is five point zero times ten to the power minus five per molar per second. What fraction goes by the S N two path, when the hydroxide concentration is zero point one zero molar? Pause the video and try it first. Done? Okay, let's check. The first term has no hydroxide in it, so it is the S N one path. The second term has both R B r and O H minus, so it is the S N two path. Okay? Now, k two is five times k one. So S N two wins easily, right? No, no, wait, that is the trap from the start! The two constants have different units, so compare the rates, not the constants. The fraction is the S N two rate over the total rate, and R B r cancels from every term. That leaves k two times O H minus, over k one plus k two times O H minus. With me? First, the top of the fraction: five point zero times ten to the power minus five, times zero point one zero, gives five point zero times ten to the power minus six, per second. Clear? For the bottom of the fraction, note that five point zero times ten to the power minus six is zero point five times ten to the power minus five. Add k one, and one point zero plus zero point five gives one point five times ten to the power minus five, per second. Right? So the fraction is five point zero times ten to the power minus six, over one point five times ten to the power minus five, and that is one third. Only one third goes by S N two, and the other two thirds go by S N one, because the hydroxide concentration is only zero point one zero molar. And that solves our opening puzzle!

7:56Quick revision: SN1 and SN2 rate laws, compare the rates

So, quick revision. Part A is Main level, Part B is Advanced, and Part C checks the ideas; take each one as a test, then read the solutions. For S N two, rate equals k, times R X, times N u minus, and for S N one, rate equals k, times R X alone. And when both paths run at once, compare the rates, not the rate constants. Now open Part A and start. All the best!