Set 1

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. When an alkane is chlorinated in light, a secondary hydrogen reacts almost four times as fast as a primary one. So the secondary chloride should be the biggest product, right? Well, not always, and in one problem from today's set, a primary chloride comes out on top. This is set one of our mixed hydrocarbon problems, with alkanes, alkenes, alkynes and arenes together. Questions of this kind are common in JEE Main, so let's begin.

0:33Idea 1: alkanes, count the carbons

So, the first idea: with alkanes, count the carbons. The figure is a map of how alkanes are made and how they react, and the Wurtz reaction is the third box on the left. In the Wurtz reaction, two molecules of an alkyl halide, R dash X, react with two atoms of sodium, N a, with dry ether written over the arrow; R is an alkyl group, and X is a halogen. R dash R joins two alkyl groups, so the chain is doubled, and sodium halide, N a X, is left over. Okay? Now, what if we put two different alkyl halides in the same flask? How many alkanes do we get? Three alkanes, yes, not one! And that is the trap. Call the two halides R X and R prime X; the small dash after the second R is called prime, and it marks the second group. The groups join as R with R, R with R prime, and R prime with R prime, a mixture that is hard to separate. Got it? In the first problem of this set, iodoethane, C two H five I, with one iodopropane, the straight chain C three H seven I, gives butane, pentane and hexane. With two iodopropane, the branched one, the products would be different. Clear? Next, soda lime, which is sodium hydroxide with calcium oxide. Heat the sodium salt of an acid, RCOON a, with soda lime, and the arrow gives R H, an alkane with one carbon less. With me? And to burn one mole of an alkane, C n H two n plus two, you need three n plus one, over two, moles of O two. Right?

2:13Idea 2: halogenation, count times reactivity

Now the second idea, chlorination in light. The small circle up high is read as degree: one degree means primary, two degree secondary, and three degree tertiary. At two hundred and ninety eight kelvin, capital K on screen, the reactivity per hydrogen is one, to three point eight, to five. Okay? So tell me, does the most reactive kind of hydrogen always give the major product? No, not always, and this is the trap! The amount of a product is proportional to the number of equivalent hydrogens, times the reactivity per hydrogen; the sign like a stretched alpha means proportional to. So many slow hydrogens can beat a few fast ones. Remember this for example two. Got it?

3:02Idea 3: adding HBr, with and without peroxide

Now let us move to alkenes. When H X adds to an unsymmetrical alkene, the hydrogen goes to the doubly bonded carbon that already has more hydrogens. That is Markovnikov's rule, and it gives the more stable carbocation. Clear? In the figure, propene takes H plus on its end carbon, giving the secondary carbocation, marked two degree, with the plus sign up top. Bromide attacks it, so two bromopropane is the major product. Got it? But with a peroxide, H Br adds the other way round, by free radicals. The bromine atom adds first, to the carbon with more hydrogens, so we get one bromopropane. This is anti-Markovnikov addition. Okay? Now, with a peroxide present, does H Cl also add the anti-Markovnikov way? No, it does not; the peroxide effect is for H Br only. The H Cl bond is too strong for the carbon radical to break, and with H I, the iodine atoms combine with each other instead. Right?

4:05Idea 4: alkynes and arenes

The fourth idea starts with alkynes. In a terminal alkyne, the triple bond is at the end of the chain, so a hydrogen sits on the triple bond carbon, and that hydrogen is weakly acidic. So only a terminal alkyne gives a white precipitate, p p t on screen, with ammoniacal silver nitrate, A g N O three; the slash means silver nitrate together with N H three, ammonia. Clear? Now for arenes, look at chlorobenzene, the third ring in the figure. Halogens deactivate the ring, but they still direct ortho and para; the small delta minus signs mark the positions rich in electrons. With me? So, when bromobenzene is nitrated, which product is major, the ortho one or the para one? The para one, yes, because the para position is less crowded. So the answer is one bromo four nitrobenzene. Got it? And groups like N O two and COOH direct to meta; the dash in front shows the bond to the ring. In nitrobenzene, the middle ring, the delta plus signs show that ortho and para are poor in electrons. Okay?

5:16Example 1: alkane that burns in 11 mol O2 is heptane

Alright, example one, which is problem four of this set, an easy 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. The general equation: C n H two n plus two, plus three n plus one, over two, O two, gives n C O two, plus n plus one, H two O, which is water. Right? So three n plus one, over two, equals eleven; the double arrow means it follows. Then three n plus one equals twenty two, and n equals seven. If you forget the over two, you get ten over three, not even a whole number, so be careful. Clear? Seven carbons and sixteen hydrogens, so the alkane is heptane, C seven H sixteen. Got it? And check by balancing. C seven H sixteen, plus eleven O two, gives seven C O two, plus eight H two O, the water. Oxygen atoms: twenty two on the left, and fourteen plus eight, which is twenty two, on the right. That's it!

6:28Example 2: chlorination percentages of 2,2-dimethylbutane

Now example two, which is problem three of this set, the harder one, and it is our puzzle from the start. Two, two dimethyl butane is monochlorinated, and the reactivities of primary, secondary and tertiary hydrogens are one, three point eight and five. Find the percentage of each structural isomer. Pause the video and try it first. Done? Let's check. Draw it: C H three, then a carbon carrying two C H three groups, in brackets, then C H two, then C H three. That second carbon has no hydrogen, so no tertiary hydrogen at all, okay? Now, four methyl groups means twelve primary hydrogens. But are all twelve of them the same? No, and this is where many students slip! The three methyls on carbon two are equivalent, nine hydrogens. Chlorine there gives one chloro two, two dimethyl butane, with a share of nine times one, which is nine. Right? The C H two group has two secondary hydrogens, and two times three point eight gives seven point six, for three chloro two, two dimethyl butane. With me? And the methyl at the far end has three primary hydrogens of its own, so three times one is three. Numbered from the other end, this product is one chloro three, three dimethyl butane. Got it? Add the shares: nine, plus seven point six, plus three, gives nineteen point six. Nine over nineteen point six gives forty five point nine percent. Seven point six over nineteen point six gives thirty eight point eight percent. And three over nineteen point six gives fifteen point three percent. So our puzzle is solved! The secondary hydrogens are more reactive, but there are only two of them, and the nine slow primary hydrogens win.

8:18Quick revision: Wurtz, soda lime, chlorination, peroxide

So, quick revision. Wurtz doubles the chain, and two different halides give three alkanes. Soda lime removes one carbon, and burning needs three n plus one, over two, moles of oxygen per mole. In halogenation, multiply the number of equivalent hydrogens by the reactivity per hydrogen. And the peroxide effect is for H Br only, halogens direct ortho and para, and COOH directs meta. Now give yourself twenty five minutes for all fifteen problems of this set on this page. All the best!