Directive Influence of Substituents

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. When we nitrate toluene, a benzene ring carrying a methyl group, C H three, the nitro group, N O two, goes to carbon two or to carbon four. But when we nitrate benzoic acid, which carries COOH, the major product has the nitro group on carbon three. Same reaction, different position. So who tells the nitro group where to go? Interesting, right? Soon you will predict it in seconds. This comes up often in JEE, so let's begin.

0:34Where is the ring rich in electrons?

So, first, the names of the positions. The course figure shows three rings, each with one group on top. The two carbons right next to the group are the ortho positions, written as a small o. Okay? The two carbons marked with a small m are the meta positions, one carbon further away. And the carbon opposite the group is para, a small p. So, if the group sits on carbon one, then carbon two is ortho, carbon three is meta, and carbon four is para. Right? Now, the first ring is phenol, with O H on top. Its oxygen has a lone pair and pushes electrons into the ring; the figure calls this plus R, the resonance effect. See the delta minus signs? They mean a small negative charge, at ortho and para, but not at meta. Electrophiles love electrons. So which positions will they attack? Yes, exactly, ortho and para, the electron-rich positions! So an electron donor, a plus R group, is ortho, para-directing; the double arrow means therefore. And since it makes the ring richer, the ring reacts faster than benzene. We call it activating. Clear? Now the middle ring, nitrobenzene, with N O two on top; the small two written low is a subscript. The nitro group pulls electrons out of the ring, minus R, so ortho and para now carry delta plus, a small positive charge. So where will the electrophile go now? It goes to meta, that is correct! Ortho and para are electron-poor, so the electrophile avoids them. So a minus R group, an electron withdrawer, is meta-directing, and since it drains the whole ring, it is deactivating too. Got it? Now the third ring, chlorobenzene, with C l, chlorine, on top. Chlorine is very electronegative and pulls electrons out of the ring. So it should be meta-directing, right? Oh no, careful, this is the classic trap! The figure shows delta minus at ortho and para again. Chlorine pulls electrons by the inductive effect, minus I, but its lone pair still donates, plus R, by resonance. Minus I is greater than plus R, the greater than sign on screen. So the ring is deactivated, yet halogens still direct ortho and para. With me?

2:54Three classes of substituent

Now let us sort the groups into three classes. Class one is ortho, para-directing and activating. O H; O R, where R is an alkyl group; N H two; NHR; NHCOCH three; OCOCH three; alkyl groups; and C six H five, phenyl. The short dash is the bond to the ring. Okay? Now look at the atom joining these groups to the ring: oxygen in O H, nitrogen in N H two. What do these atoms carry? Yes, exactly, a lone pair! The atom joined to the ring has a lone pair, or it is an alkyl carbon, and it releases electrons into the ring. Phenyl, C six H five, is the odd one out: it gives electrons through its own ring, so just remember it. Right? Class two is small: the halogens. They direct ortho and para, but deactivate the ring, just like our chlorine trap. Clear? Class three is meta-directing and deactivating. N O two; C N; CHO, the aldehyde group; COR; COOH, the acid; and COOR, the ester. Then S O three H; N R three, with a plus sign above the N for its positive charge; C C l three; and C F three. With me? So, here is the trick to spot class three. In most of these, the ring atom carries a multiple bond to a more electronegative atom, like oxygen, or a positive charge, like that nitrogen. C C l three and C F three are on the list too, so just remember those two. Okay?

4:30Using the rules

Now, four rules for using all this. Rule one: an ortho, para director gives a mixture, and para usually predominates over ortho, because para is less crowded. Right? Rule two: with two groups on the ring, the more strongly activating group decides the position. Rule three: a position between two groups that are meta to each other is too crowded to be attacked. Clear? Rule four, for synthesis: the order of steps decides the pattern, so introduce first the group whose directing effect you need. Okay? Finally, the rate of reaction with electrophiles. N H two and O H come first, then OCH three, NHCOCH three, and C H three. Then H, plain benzene, then X, any halogen, then COOH, and N O two last. Each greater than sign means reacts faster than. With me? So tell me, which one is nitrated faster: chlorobenzene or benzene? Yes, benzene is faster, because X, the halogen, sits below H in the order. So chlorobenzene, C six H five C l, is nitrated more slowly than benzene.

5:45Example 1: classify CH3, NO2, OH, COOH, Br and CHO

Right? Now example one, an easy one. Classify each group as ortho, para-directing or meta-directing. The groups are C H three, N O two, O H, COOH, B r and CHO, six of them. Pause the video and try it first. Done? Okay, let's check. C H three is an alkyl group, and O H has a lone pair, so both are ortho, para. B r, bromine, is a halogen, so it is ortho, para too, even though it deactivates. If you wrote meta, remember the chlorine trap. Okay? Now N O two, COOH and CHO: in each, the ring atom has a multiple bond to oxygen, so all three are meta. And see, our puzzle is solved! Toluene carries C H three, so nitration goes ortho and para, carbons two and four. Benzoic acid carries COOH, so it goes meta, carbon three. Clear?

6:41Example 2: make 1-bromo-3- and 1-bromo-4-nitrobenzene

Now a harder one, about the order of steps. From benzene, how would you prepare these two? Part A is one bromo three nitrobenzene, and part B is one bromo four nitrobenzene; the letters in brackets mark the parts. Hint: think about which group should go in first. Now pause the video and try it. Part A first: bromine on carbon one, the nitro group on carbon three, meta to each other. So we brominate first, then nitrate, right? Oh no, careful, that is the trap! Bromine is ortho, para-directing, so it would send the nitro group to carbons two and four, not to three. Positions one and three mean meta, so nitrate first, then brominate: the meta-directing nitro group sends bromine to carbon three. Got it? Part B: positions one and four, para to each other. So the ortho, para director, bromine, must go in first: brominate, then nitrate. With me? But nitrating bromobenzene gives both ortho and para products. Para usually predominates, that was rule one, and we separate the para isomer. Remember, the first group introduced directs the second. Clear?

7:57Quick revision: activating, halogen and meta directors

So, quick revision. Lone pair or alkyl carbon on the atom joined to the ring: ortho, para and activating. Halogens: minus I beats plus R, so they deactivate, yet direct ortho and para. Multiple bond to a more electronegative atom, or a plus charge: meta and deactivating. Now go and try the practice problems on this page. All the best!