Part C: Concept Checks

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. You want propylbenzene, a benzene ring carrying a straight chain of three carbons. So you take benzene, add one chloropropane, a straight three carbon chain with chlorine at the end, and anhydrous aluminium chloride. So the straight chain goes onto the ring, right? But the flask contains mainly isopropylbenzene, where the chain is joined by its middle carbon. So where did the straight chain go? Hold that thought, because Example two answers it. This is Part C of the unit problem bank: short concept checks, where every answer needs a reason. Reasoning questions of this kind are a familiar JEE pattern, so let's begin.

0:43Check 1: HBr, with and without peroxide

Let us begin with check number one: propene and H B r, hydrogen bromide, with no peroxide. In the figure, propene is on the left: C H three, C H, two short lines for the double bond, and C H two; the small numbers written low count hydrogen atoms. H plus adds first, and the small raised plus is its positive charge. So which carbocation forms? Yes, exactly, the secondary one, on the upper path! H plus goes to the C H two end, and the plus sign sits on the middle carbon. Two with a small circle written high is read as secondary. It is more stable than the primary cation on the lower path, so it forms faster, and the figure marks it as the fast path. Okay? Then B r minus attacks the cation, and the major product is two bromopropane. That is Markovnikov's rule: the hydrogen goes to the carbon that already has more hydrogens. One bromopropane, at the end of the lower path, is only the minor product. Right? Now add a peroxide to the flask. Suddenly one bromopropane becomes the major product, the anti-Markovnikov product. So what has changed? The whole mechanism has changed! The peroxide starts a free-radical chain, and now a bromine atom adds first, not H plus. On screen, B r with a dot is that bromine atom, and the dot is its unpaired electron. It adds to the C H two end, the carbon with more hydrogens. With me? Why that end? Because then the unpaired electron sits on the middle carbon, the more stable secondary radical; the double-lined arrow means which leads to. That radical takes a hydrogen from another H B r, and we get one bromopropane. The more stable intermediate wins again, only now it is a radical. Clear? And here is the trap: H C l and H I do not show this effect. With H C l, the hydrogen chlorine bond is too strong to be broken by the carbon radical, and with H I, the iodine atoms combine with each other instead of adding. So the peroxide effect is for H B r only. Got it?

2:58Check 2: why is OH ortho, para-directing?

Now check number two, on benzene rings. The figure shows three of them: phenol with O H on the left, nitrobenzene with N O two in the middle, and chlorobenzene on the right; the small letter m marks a meta position. An electrophile attacks phenol mainly at the ortho and para positions. So why does the O H group send it there? Look at the oxygen. The short dash in front of O H is its bond to the ring. The oxygen carries lone pairs, and one lone pair is delocalised into the ring. That is the plus R effect, R for resonance, and you can see it written under phenol. Okay? And where does this extra electron density go? To the ortho and para carbons, the places marked delta minus, the Greek letter delta with a minus sign for a small negative charge. So those carbons are electron-rich, and the electrophile goes straight to them. Right? There is a second reason as well. When the electrophile attacks ortho or para, the lone pair of oxygen stabilises the arenium ion formed there, through an extra resonance structure. Meta attack gets no such help. On screen, the small o and p are short for ortho and para. With me? So O H is ortho, para directing, and activating too, because the whole ring becomes richer in electrons. Compare N O two in the middle: it pulls electrons out of the ring, the minus R effect, so its ortho and para carbons are marked delta plus. That makes it meta directing and deactivating. Clear? Now chlorine, on the right, also has lone pairs. So it should activate the ring just like O H, right? No, no, careful, because this is the trap! For the halogens, the minus I effect, I for inductive, the pull of electrons through the sigma bond, is stronger than the plus R effect; the pointed sign means greater than. So the ring is deactivated overall, but the lone pair still helps ortho and para attack. Halogens are deactivating, yet ortho, para directing. Got it?

5:01Example 1: why alkanes substitute but alkenes add

Now the two examples. Example one asks: why do alkanes react by substitution, but alkenes by addition? Pause the video and try it first. Here is a hint: look at the kinds of bonds. Done? Okay, let's check. In an alkane, every carbon is s p three, with the small three written high. Each carbon is joined to four atoms by strong sigma bonds, sigma being the Greek letter on screen. Okay? So suppose a chlorine atom wants to join an alkane. Every carbon already has four partners. Where can it fit in? It cannot just add on, because there is no room for more atoms. So a hydrogen must be replaced, and that is substitution. Right? Now the alkene. The two carbons of the double bond are joined by a sigma bond and a weaker pi bond, pi being this other Greek letter. With me? So the reagent breaks only the weak pi bond, and two new sigma bonds form in its place, one to each carbon. Nothing has to leave, and that is addition. Clear?

6:11Example 2: 1-chloropropane gives isopropylbenzene

And now Example two, a step harder, and it is our opening puzzle. Why can propylbenzene not be made in good yield by direct Friedel-Crafts alkylation with one chloropropane? Pause the video and try it first. Here is a hint: ask what the electrophile is. Done? Okay, let's check. One chloropropane is C H three, C H two, C H two, C l. Aluminium chloride, written A l C l three, takes the chloride, and we get a primary carbocation, or its complex. The number one with a small circle written high is read as primary. Okay? Now, a primary carbocation is the least stable kind. Will it just sit there and wait for benzene? No, it rearranges first, by a hydride shift! On the left, the plus written above the end carbon is the positive charge. The middle carbon passes a hydrogen, with its electron pair, to the end carbon, and the plus moves to the middle; the arrow means changes into. That is the isopropyl cation, secondary, and more stable. Right? So benzene meets mainly the isopropyl cation, and the major product is isopropylbenzene, not propylbenzene. Puzzle solved: the chain did join the ring, but by its middle carbon. With me? So how would you make propylbenzene, with the chain kept straight? Here is the topper's trick. Acylate instead, with propanoyl chloride and A l C l three, because the acylium ion does not rearrange. Then reduce the ketone with zinc amalgam, written Z n dash H g, and H C l. The result is propylbenzene, with a straight chain. Clear?

7:55Quick revision: peroxide, OH group, pi bond, rearrangement

So, quick revision. With a peroxide, the bromine atom adds first, and that works for H B r only. The O H group pushes a lone pair into the ring, so it is ortho, para directing. Alkanes substitute, while alkenes add across the weak pi bond. And in Friedel-Crafts alkylation, the carbocation may rearrange. Now go and try the practice problems on this page. All the best!