Part A: JEE Main Level

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Heat two bromo two methylpentane with alcoholic KOH, and it loses the bromine and one hydrogen to give an alkene. That hydrogen can come from the two methyl groups, with six hydrogens between them, or from carbon three, with only two. Six against two, so six should win, right? Hmm, it is not that simple, and by the end of this video you will see why. This is Part A of the unit problem bank, at JEE Main level. We will sharpen two tools, then solve two of its questions. So, let's begin.

0:39Tool 1: the rate law names the mechanism

So, let us start with tool one. S N two, with the N written low as a subscript, means substitution, nucleophilic, bimolecular. In one step, the nucleophile attacks from the back side as the halide leaves. So rate equals k, times the concentration of R X, times the concentration of N u minus. Square brackets mean concentration, R X is the alkyl halide, N u minus is the nucleophile with its charge written high, and k is the rate constant. So it is second order, okay? Now S N one, where the one means unimolecular. Its slow step involves the halide alone. So let me ask you this: whose concentration sets the rate here? Only the halide, that is correct! The slow step is R, dash, X giving R plus and X minus, a carbocation and a halide ion; the arrow means gives. Then the nucleophile attacks fast. So rate equals k times the concentration of R X, which is first order, with no nucleophile in it. Right? And their substrate orders are opposite. S N two is fastest for C H three X, then one degree, two degree, three degree; the small circle means degree, and greater than means faster than. Bulky groups slow it, and the configuration is inverted. S N one follows carbocation stability, three degree first, and the product is racemised. Clear?

2:12Tool 2: substitution or elimination?

Now, on to tool two, and the picture on the right will help us here. Remember, hydroxide is both a nucleophile and a base, so substitution and elimination compete. With aqueous KOH, in water, we get substitution, and an alcohol. Okay? Now dissolve the KOH in alcohol instead. What do we get now? We get elimination, that is correct, and the product is an alkene. In alcohol, the alkoxide ion is a stronger base, and it pulls off a hydrogen. At the top of the picture, two bromobutane with alcoholic KOH and heat gives the butenes. Right? So which hydrogen goes? In the lower drawing, the carbon holding X is alpha, and its neighbour is beta, both Greek letters. The base, B minus, removes a beta hydrogen as X leaves from the opposite side, the anti side, as the caption says. We get B H, a double bond, and X minus. This is E two, elimination in one step. Clear? Now look at the top again. Two bromobutane has beta hydrogens on both sides of the C Br carbon. Yet the labels say but two ene is eighty percent, and but one ene only twenty. So why does but two ene win? Because it is the more substituted alkene, yes, exactly! This is Saytzeff's rule: when more than one alkene can form, the major one is the more substituted, with more alkyl groups on the doubly bonded carbons. But two ene has two methyls there; but one ene has just one ethyl. Got it? Now, here is a topper's warning. Use a bulky base, such as potassium tert butoxide, and it takes the more exposed hydrogen, giving mainly the less substituted alkene, the Hofmann product. So always check the base. With me?

4:06Example 1: SN1 or SN2 from rate data, and the units of k

Now, example one, question eight of 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. The capital M means molar, and s to the power minus one means per second. 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. Double the concentration of R B r, and the rate doubles; times two means multiplied by two. Double the hydroxide instead, and the rate doubles too. So the rate is first order in each of them. Okay? So both concentrations matter. Which mechanism is it, then? S N two, yes, exactly! Rate equals k, times the concentration of R B r, times the concentration of O H minus. That is second order, just as tool one said. Right? Now let us find the number. k equals the rate over the product of the two concentrations. That is one point two times ten to the power minus five, over zero point zero two times zero point zero two. Clear? Zero point zero two times zero point zero two is four times ten to the power minus four. And one point two times ten to the power minus five, over four times ten to the power minus four, gives zero point zero three. Got it? So the rate constant is zero point zero three per second, right? Oh no, careful, you must check the units here! We divided molar per second by molar squared. So the units are M to the power minus one, s to the power minus one, which is per molar per second. With me? So the mechanism is S N two, and k is zero point zero three per molar per second. The rate law named the mechanism, and gave us the units too. Well done if you caught that!

6:13Example 2: major alkene from 2-bromo-2-methylpentane

And now, example two, question eleven of Part A, our puzzle from the start. Name the major alkene from two bromo two methylpentane and alcoholic KOH, potassium hydroxide in alcohol. Pause the video and try it first. Done? Okay, let's check. Carbon two carries both the bromine and a methyl group. Written out, it is C H three, then C B r with C H three in brackets, then C H two, C H two, C H three; each dash is a single bond. C two, carbon two, holds three other carbons, so this is a tertiary, three degree halide. Okay? Now the beta carbons, the neighbours of carbon two. Carbon one and the methyl group are both C H three, six beta hydrogens together. Carbon three is C H two, with only two. Right? Take a hydrogen from either C H three, and we get C H two, double bond, C with its methyl, then C H two, C H two, C H three; the equals sign is the double bond. That is two methylpent one ene. Clear? Take it from carbon three instead. We get two methyls on one carbon, the C H three in brackets with a subscript two, then a double bond to C H, then C H two, C H three. That is two methylpent two ene. With me? Now, six hydrogens against just two. So the one ene should be the major product, right? No, no, wait, that is exactly the trap! Saytzeff's rule counts the alkyl groups on the doubly bonded carbons, not the hydrogens. The one ene has two, a methyl and a propyl. The two ene has three, two methyls and an ethyl. Got it? So the major alkene is the more substituted one, two methylpent two ene. Puzzle solved: the side with only two hydrogens wins. If you got this one, that is really good work!

8:14Quick revision: rate law, aqueous vs alcoholic KOH, Saytzeff

So, quick revision. First, the rate law names the mechanism: both concentrations means S N two, and the halide alone means S N one. Second, aqueous KOH substitutes, and alcoholic KOH eliminates. Third, for Saytzeff's rule, count the alkyl groups on the double bond, not the hydrogens. Now try the other questions of Part A on this page as a test, before you read the solutions. All the best!