Overview

Transcript

0:00Introduction

Hello everyone! Let us start with a one-minute question, in the JEE Main style. Take two flasks, each with the same halide, two bromopropane, and the same base, potassium hydroxide. Yet one flask gives an alcohol, and the other gives an alkene. So how is that possible? Hold that thought. This chapter gives you three sets of mixed problems from the whole unit, so let us see how to use them, and sharpen the tools they need. Let's begin.

0:31How this chapter works

So, first, what is inside this chapter? Three sets, each with fifteen original problems, in the style of recent JEE Main papers. Okay? And each set mixes all the chapters of the unit: naming, preparation, substitution, elimination, metals, haloarenes and polyhalogen compounds. Right? Now, the timing. Give yourself twenty five minutes for each set, and only then check the solutions. So, twenty five minutes for fifteen problems: how much time is that for each one? Yes, one hundred seconds, exactly! Twenty five minutes is fifteen hundred seconds, with s for seconds, and fifteen hundred over fifteen gives one hundred; the division sign means over. So the key rules must be ready in your head. With me?

1:23Tool 1: SN1 or SN2?

Now, tool one, needed in every set: S N one or S N two? On screen, that is a capital S, then a capital N written low, as a subscript, and then the number. S N two is one step, and its rate equals k, times the concentration of R X, times the concentration of N u minus. R X is the alkyl halide, N u minus is the nucleophile, with its minus written high, k is the rate constant, and the square brackets mean concentration. Okay? S N one goes in two steps, through the carbocation, R plus, with the plus written high, and its rate equals k times R X only. The nucleophile is not in it. Right? Now tell me, which reacts faster by S N two, a methyl halide or a tertiary halide? The methyl halide, yes, exactly! The nucleophile attacks from the back, and alkyl groups crowd it. So the order is C H three, with the small three written low, then one degree, two degree and three degree; the small circle means degree, and the greater than sign, faster than. And at a chiral carbon, S N two gives inversion. Clear? S N one runs the other way, because it follows carbocation stability, so tertiary comes first. And an optically active halide gives a largely racemised product. Got it? Polar aprotic solvents, like acetone, DMSO and DMF, favour S N two; polar protic solvents, like water and alcohols, favour S N one. With me? And secondary halides can go either way; the nucleophile and the solvent decide, so always look at both.

3:00Tool 2: substitution or elimination?

Now, back to our two flasks, a problem from Set two. See, a nucleophile is also a base, so substitution and elimination compete. In aqueous KOH, potassium hydroxide in water, two bromopropane is substituted and gives the alcohol, propan two ol; the arrow means gives. Okay? So what happens in alcoholic KOH, potassium hydroxide dissolved in alcohol? Yes, we get elimination, that is correct! The halide loses a hydrogen and the bromine, and gives the alkene, propene. So the one word that differs, aqueous or alcoholic, settles our opening question! Right? Now, when two alkenes can form, look at the top of the figure. Two bromobutane heated with alcoholic KOH gives but two ene, eighty percent, and but one ene, just twenty percent. That is Saytzeff's rule: the more substituted alkene, with more alkyl groups on the double bond, is major. Below is the E two step: the base, B minus, removes a hydrogen from the beta carbon, as X minus leaves the alpha carbon from the opposite side; beta and alpha are the Greek letters you see. The result is B H, the alkene, and X minus. Clear? Now a trap from Set three. Two bromobutane is treated with potassium tert butoxide, a bulky base. So but two ene is the major alkene again, right? No, no, careful, that is the trap! The bulky base cannot easily reach the inner hydrogen, so it takes a more exposed one, and gives mainly but one ene. This less substituted alkene is called the Hofmann product. Got it?

4:42Tool 3: metals and haloarenes

Now, tool three, for metals and haloarenes. An alkyl or aryl halide, R X, with magnesium, M g, in dry ether, written over the arrow, gives R M g X, a Grignard reagent; the short dashes are bonds. Okay? Its carbon behaves as a carbanion, a strong base. So any acidic hydrogen, H A on screen, from water or an alcohol, destroys it, giving R H, and M g, A in brackets, X, and that is why the ether must be perfectly dry. Right? Haloarenes resist nucleophilic substitution, mainly because the carbon halogen bond has partial double bond character, from resonance, and the ring carbon is s p two. They react only under harsh conditions, or with a nitro group, N O two, at the ortho or para position. Clear? And with electrophiles, the halogen deactivates the ring, yet it directs to ortho and para, with para as the major product. With me?

5:39Example 1: SN2 rate, halide halved and nucleophile times 4

Now, Example one, a medium problem from Set one. In an S N two reaction, the concentration of the halide is halved, and that of the nucleophile is made four times. By what factor does the rate change? Pause the video and try it first. Done? Okay, let's check. The halide is halved, so the rate is halved too, right? Oh no, careful, that would be the S N one rate law! In S N two, rate equals k, times R X, times N u minus, so both concentrations count. Okay? The halide gives a factor of one half, the nucleophile a factor of four, and the cross means times. One half times four gives two, so the rate doubles. Clear?

6:26Example 2: SN1 enantiomeric excess, inversion vs retention

Now, Example two, a hard one from Set two. An optically pure halide is hydrolysed by the S N one mechanism, and the alcohol formed has an enantiomeric excess of ten percent of the inverted form; that is the percent sign. What percentage of the molecules reacted with inversion, and what percentage with retention? Pause the video and try it first. Done? Okay, let's check with the figure. On the left, the halide, with groups R one, R two and R three, the small numbers written high, loses bromide in the slow step. The flat carbocation in the middle, C plus, has an s p two carbon, with the small two written high. In the fast step, N u minus attacks either face, as the red arrows show, giving the two products on the right, inversion and retention. Okay? Now, enantiomeric excess is how much more of one mirror image form there is than of the other. So a ten percent excess of the inverted form means ten percent inversion and ninety percent retention, right? No, no, wait, the excess is not the whole amount! Ninety percent of the product is racemic, a fifty fifty mixture, so forty five percent of each form. Right? The remaining ten percent, the excess, is all inverted. Clear? So inversion is forty five plus ten, which gives fifty five percent, and retention is forty five percent. Okay? Let us check. Fifty five minus forty five is ten, the excess, and fifty five plus forty five is one hundred. Got it?

7:57Quick revision: SN1 vs SN2, KOH, Grignard, haloarenes

So, quick revision. S N two: rate equals k times R X times N u minus, with inversion. S N one: rate equals k times R X, largely racemised. Aqueous KOH substitutes, and alcoholic KOH eliminates, Saytzeff's way, unless the base is bulky. Grignard reagents need dry ether, and haloarenes need a nitro group at ortho or para, or harsh conditions. Now take Set one, start a twenty five minute timer, and check the solutions only at the end. All the best!