Set 1

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Treat two bromopentane, a five carbon chain with bromine on carbon two, with sodium cyanide. The cyanide simply replaces the bromine; nothing else is added. Yet the product is called two methyl pentanenitrile. So where did a methyl group come from? We will find out today. This is set one of our mixed problems on halogen compounds. Questions like these come up often in JEE Main, so let's begin.

0:30Idea 1: preparations you must know by name

So, idea one: preparations by name. On the map, haloalkanes, R X, are made on the left and react on the right. Problem five: which reagent turns an alcohol, ROH, into a chloride with only gaseous by-products? Thionyl chloride, SOC l two, that's it! ROH plus SOC l two gives R C l, plus S O two and H C l, and the small up arrows mean both escape as gases. So the chloride is left behind pure. Okay? Problem six uses the Finkelstein reaction. R X, a chloride or a bromide, with sodium iodide, N a I, in dry acetone, gives the iodide, R I, okay? Sodium chloride and bromide do not dissolve in acetone, so they precipitate, and that drives the reaction. Clear?

1:22Idea 2: SN2, one step from the back

Now idea two, S N two: a capital S, a small capital N written low, and a two. The nucleophile, N u minus, attacks as the halide leaves, in one step. So rate equals k, times the concentration of R X, times the concentration of N u minus; the square brackets mean concentration, and k is the rate constant. Second order overall. Okay? In the picture, hydroxide, H O minus, attacks from the side opposite to the bromine. On the left, the two hydrogens and the methyl group lean towards it. In the product on the right, where do they point? To the other side, yes, exactly! The three bonds turn inside out, like an umbrella in the wind, so the configuration is inverted. This is Walden inversion. Right? And S N two prefers an uncrowded carbon. The greater than sign means faster than, and the small circle is read as degree: so methyl halide, C H three X, first, then primary, secondary and tertiary. It also prefers a polar aprotic solvent, like DMSO, in brackets. With me?

2:30Idea 3: SN1 goes through a carbocation

Now idea three, S N one, with a one in place of the two. In the slow step, R X splits into R plus, a flat carbocation, and X minus; the nucleophile attacks it in the fast step. So rate equals k, times the concentration of R X only. Clear? Silver ion, A g plus, helps this path by pulling off the halide. So in problem ten, which gives a precipitate fastest with alcoholic silver nitrate: one chloropentane, two chloropentane, or two chloro two methylbutane? The tertiary one, two chloro two methylbutane, very good! The rate follows carbocation stability: tertiary, then secondary, then primary. Okay? S N one prefers a polar protic solvent, like water. And on the right of the picture, the nucleophile attacks the flat carbocation from either face, so an optically active halide gives a largely racemised product. S N two inverts, and S N one racemises. Got it?

3:32Idea 4: alcoholic KOH gives the Saytzeff alkene

Now idea four: when heated with alcoholic KOH, a haloalkane loses the halogen and a beta hydrogen, giving an alkene. In the lower picture, the Greek letter alpha marks the carbon holding X, and beta marks the next carbon, whose hydrogen the base, B minus, pulls off. In the figure, two bromobutane gives eighty percent but two ene, the more substituted alkene, and twenty percent but one ene. This is Saytzeff's rule. Okay? Problem eleven heats the compound on screen, three chloro two methylpentane, with alcoholic KOH as the base. Chlorine is on carbon three. So which beta hydrogen goes, from carbon two, or from carbon four? From carbon two, yes! Then the double bond, between carbons two and three, carries three alkyl groups. So the major product is two methylpent two ene. Right?

4:27Idea 5: an acidic hydrogen destroys a Grignard reagent

Now idea five: R X with magnesium, M g, in dry ether, written over the arrow, gives R Mg X, the Grignard reagent. Its carbon behaves like a carbanion, a strong base. Clear? Problem twelve adds ethylmagnesium bromide to propyne. Many students expect a new carbon carbon bond. Is that what happens? It is not, and this is the trap! Any compound with an acidic hydrogen, written H A, destroys the reagent: R Mg X plus H A gives R H, plus Mg X with the A in round brackets, okay? Propyne, C H three C triple bond C H, is a terminal alkyne, and its end hydrogen is acidic. So C two H five Mg B r, with propyne, gives ethane, C two H six, and C H three C triple bond C Mg B r. Got it?

5:21Example 1: SN2 rate, halide halved and nucleophile times 4

Alright, example one, problem eight. 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. Some students say the nucleophile is not in the rate law, so the rate just halves. Is that right? That is not right here, because that rule belongs to S N one. In S N two, rate equals k, times R X, times N u minus, so both concentrations count. Okay? So the new rate is k, times half of R X, times four times N u minus. Each change is its own factor. Right? Half times four gives two. So the rate doubles. That's it!

6:11Example 2: (S)-2-bromopentane + NaCN, product name and R/S

Now example two, problem nine: our puzzle from the start. The S form of two bromopentane reacts with sodium cyanide in DMSO; the S in round brackets labels its configuration. Name the product, with its configuration. Pause the video and try it first. Done? Let's check. Cyanide, C N minus, is a strong nucleophile, and DMSO is polar aprotic. So this secondary halide reacts by S N two: attack opposite to the bromine, with inversion. Okay? So C N takes the place of B r on carbon two, shown in round brackets on the right. Right? Now for the name, many students write two cyanopentane. Is that right? No, and here is our methyl! The nitrile is the principal group, and its carbon is counted in the chain as carbon one. So the chain runs from N triple bond C to the far end: five carbons, pentanenitrile. The old carbon one, the C H three in brackets, is now a methyl on carbon two. Clear? Now the R or S label. It depends on the priorities, so inversion does not always flip it, and we must check. Here the greater than sign means higher priority. Before, the order is bromine, propyl, methyl, hydrogen. After, it is C N, propyl, methyl, hydrogen. With me? Why is C N first? For the triple bond, its carbon counts as bonded to N, N and N, which beats C, H and H for propyl. So C N takes the top place bromine had, and the order is unchanged. The arrangement is inverted, so the S form becomes the R form, as the arrow shows. Got it? So the answer is the R form of two methylpentanenitrile. Puzzle solved: the chain is simply counted from the new end. That's it!

8:04Quick revision: SOCl2, SN1 vs SN2, Saytzeff, Grignard

So, quick revision. Thionyl chloride gives only gaseous by-products, and Finkelstein uses sodium iodide in dry acetone. S N two has rate k, times R X, times N u minus, with inversion; S N one goes through a carbocation, with racemisation. Alcoholic KOH gives the Saytzeff alkene, and an acidic hydrogen destroys a Grignard reagent. And count the nitrile carbon as carbon one, then check R or S with the priorities. Now give yourself twenty five minutes for the fifteen problems on this page. All the best!