Overview

Transcript

0:00Introduction

Hello everyone! Let us start with a one-minute question, in the JEE Main style. Suppose you want pentane, the alkane with five carbons. So you take iodoethane, with two carbons, and one iodopropane, with three carbons, and you join them with sodium in dry ether. Two plus three is five, so the flask gives you pure pentane, right? Hold that thought, because we will come back to it in a few minutes. This chapter gives you three timed sets of mixed problems on hydrocarbons, so let us see how to use them, with a few quick tools on the way. So, let's begin.

0:37How this chapter works

So, first, what is in this chapter? Three sets, with fifteen original problems in each, written in the style of recent JEE Main papers. Okay? Each set runs through the whole unit: alkanes, alkenes, alkynes, and arenes, the aromatic hydrocarbons like benzene. So for each problem, you first decide which family and which idea it needs. Right? Now, the timing. Give yourself twenty five minutes per set, and only then check the solutions. So tell me, twenty five minutes for fifteen problems: how much time is that for each problem? Twenty five times sixty, over fifteen; the cross means times, and the long line is a fraction bar. That is one thousand five hundred seconds over fifteen, which gives one hundred seconds per problem. So keep one eye on the clock.

1:28Alkanes: the Wurtz reaction and combustion

Clear? Now our first tools, for alkanes, starting with the Wurtz reaction. R dash X is an alkyl halide: R is the alkyl group, and X is the halogen. Two of them react with two sodium atoms, written N a, and the words dry ether over the arrow give the solvent. The arrow means gives: R dash R, plus two N a X, sodium halide. With me? So here is a quick question. Can the Wurtz reaction ever give methane, the alkane with one carbon? No, it can never give methane. Two alkyl groups join, so the chain is doubled, and the product has at least two carbons. Keep this in mind; we will need it for our opening question. Okay? Next, the combustion of alkanes. C n H two n plus two is the formula of any alkane; the small letters and numbers written low are subscripts, counting the atoms. One mole of it burns in three times n, plus one, over two moles of O two, oxygen, giving n moles of C O two, and n plus one, in brackets, moles of H two O, water. Right? Let us test it on methane, where n is one. How many moles of oxygen does one mole of methane need? Three times one, plus one, is four, and four over two gives two. So one mole of methane needs two moles of oxygen; the mol on screen is short for mole. Remember this formula; Example one uses it. Got it?

3:02Alkenes: where does HBr go?

Now let us move on to alkenes. Remember Markovnikov's rule: when an unsymmetrical reagent H X adds to an unsymmetrical alkene, the hydrogen goes to the doubly bonded carbon that already has more hydrogens, and X goes to the other carbon. The brackets give the reason: this way, the more stable carbocation is formed. Clear? Now here is a twist. Add a little peroxide to the flask with H B r, hydrogen bromide. So what changes in the product? The addition turns around! With a peroxide, H B r adds by free radicals: the bromine atom adds first, to the carbon with more hydrogens. So we get the anti-Markovnikov product. This is the peroxide effect. Okay? But careful, this is a classic trap. The peroxide effect works for H B r only. H C l does not show it, because the hydrogen chlorine bond is too strong, and H I, hydrogen iodide, does not either, because iodine atoms combine with each other instead. With me?

4:05Alkynes and arenes: quick checks

Next come the alkynes. In a terminal alkyne, the triple bond, the three lines on screen, sits at the end of the chain, so one hydrogen is on a triply bonded carbon. That carbon is s p hybridised, with fifty percent s character, which holds an electron pair tightly, so this hydrogen is weakly acidic. Got it? That gives two quick tests. First, silver nitrate, A g N O three, with ammonia, N H three; the slash means with. This ammoniacal silver nitrate gives a white precipitate, and ammoniacal cuprous chloride, C u two C l two with N H three, gives a red one. Only terminal alkynes give these precipitates. Right? Now arenes, and the directing groups. The small o and p stand for ortho and para. O H, O R, N H two and alkyl groups send the next group to the ortho and para positions. The halogens also direct ortho and para, but they deactivate the ring. Clear? And these groups are meta directing: N O two, C N, CHO, COR, COOH, and S O three H, all of them. In each, the atom joined to the ring has a multiple bond to a more electronegative atom. Okay?

5:26Example 1: the alkane that burns in 11 moles of oxygen

Now Example one, an easy one from Set one. One mole of an alkane needs eleven moles of oxygen for complete combustion. Identify the alkane. Pause the video and try it first. Done? Okay, let's check. Three times n, plus one, over two, equals eleven; the double arrow means implies. Multiply both sides by two: three times n, plus one, equals twenty two. So three times n is twenty one, and n is seven. Right? So the alkane has seven carbons. Now tell me, how many hydrogens does it carry? Two times seven, plus two, gives sixteen hydrogens. So it is C seven H sixteen, heptane. Clear? And a quick check, the way toppers do it. Seven carbons give seven C O two, and n plus one gives eight water molecules. They hold fourteen plus eight, that is twenty two oxygen atoms, and twenty two oxygen atoms make eleven O two molecules, exactly as given. Got it?

6:34Example 2: Wurtz with iodoethane and 1-iodopropane

And now Example two, a medium one from Set one, and it is our opening question. Here it is: a mixture of iodoethane and one iodopropane is treated with sodium in dry ether. Which alkanes are formed? Pause the video and try it first. Done? Okay, let's check. Iodoethane gives ethyl groups, C two H five, with two carbons; the capital C on screen counts carbons. One iodopropane gives propyl groups, C H three, C H two, C H two, with three carbons in a straight chain. With me? Sodium joins two alkyl groups. So ethyl joins propyl, two plus three is five carbons, and we get only pentane, just as we guessed at the start, right? Oh no, careful, sodium does not choose partners! Two ethyl groups can join each other too, and two plus two gives four carbons, butane. Okay? Ethyl with propyl does give five carbons, pentane. And two propyl groups give three plus three, six carbons, hexane. Clear? So the flask holds three alkanes, butane, pentane and hexane, and they are hard to separate. That is why the Wurtz reaction is useful only for symmetrical alkanes, R dash R, made from a single halide, and these always have an even number of carbons, as the brackets say. That settles our opening question. Got it?

8:01Quick revision: Wurtz, combustion, peroxide effect, alkynes

So, quick revision. Three sets of fifteen, from alkanes to arenes: twenty five minutes each, then check the solutions. The Wurtz reaction doubles the chain, and two different halides give three alkanes. An alkane burns in three times n, plus one, over two moles of oxygen, and the peroxide effect is for H B r only. And only terminal alkynes give the silver and copper precipitates. Now open Set one, start your twenty five minute timer, and then check the solutions. All the best!