IIT-JEE Questions (1997–2009)

Transcript

0:00Introduction

Hello everyone! Here is a puzzle from IITJEE, two thousand five. Two molecules each have a benzene ring, a fluorine atom on that ring, and a nitro group. Treat the first with hydroxide ion, and hydroxide takes the place of the fluorine. But the second does not release fluoride ion at all. Same pieces, so why such different behaviour? We will crack it soon. Today we solve the IITJEE questions from nineteen ninety seven to two thousand nine on nitro compounds, cyanides and amides. Questions like these come up often, so let's begin.

0:38The nitro group on a benzene ring

So, first, the toolkit. N O two is the nitro group, and the small two written low counts the oxygens. It is strongly electron-withdrawing: minus I is the inductive effect, minus R the resonance effect, and the minus sign means pulling electrons away. Okay? On a benzene ring, that pull makes the nitro group strongly deactivating, and it sends a new group to the meta position. Right? Now, IITJEE nineteen ninety eight. Nitrobenzene does not undergo Friedel-Crafts alkylation. Can you tell me why? Yes, exactly, the ring is so strongly deactivated that it does not react with the weak electrophile. In fact, it is even used as a solvent for this reaction. Clear? Now a twist from two thousand five. Nitrosobenzene has the nitroso group, N double bond O; the equals sign is the double bond, and the dash its bond to the ring. It nitrates at ortho and para, while nitrobenzene goes meta. So what does the nitroso nitrogen have that the nitro nitrogen lacks? It has a lone pair, very good. The nitroso nitrogen donates it to stabilise the cation formed by ortho or para attack. The nitro nitrogen has no lone pair, so it only withdraws electrons. Got it?

1:59Halogen on the ring: when does it leave?

Now a statement question from IITJEE nineteen ninety eight. Statement one, on screen, says benzonitrile is made from chlorobenzene, C six H five C l, plus potassium cyanide, KCN, okay? Statement two, in the Think box, says cyanide ion is a strong nucleophile. Cyanide is strong and the chlorine is right there, so statement one is true as well, right? That is exactly the trap, so be careful here. Statement two is true, but statement one is false. With me? That is because aryl halides do not undergo this substitution. The chlorine on a plain benzene ring stays put. Okay? But with help, a ring halogen can be replaced. Here is the rule: a nitro group ortho or para to the halogen, capital X on screen, makes it easy to replace. Right? The figure shows why. Hydroxide ion, O H minus, adds to the carbon holding the chlorine, and the ring takes a negative charge. The double-headed arrow means resonance: in the second form, the para nitro group holds that charge. Then chloride leaves. The note below says a meta nitro group cannot do this. With me? So, back to our puzzle. The first molecule loses its fluorine to hydroxide, but the second releases no fluoride. With this figure in mind, can you explain it now? Yes, exactly, in the first molecule, four fluoro three methylnitrobenzene, the nitro group is on the ring, para to the fluorine. It stabilises by resonance the anion formed when hydroxide adds, so hydroxide replaces the fluorine. Clear? The second molecule is three fluorophenylnitromethane. Its nitro group, O two N, sits on a C H two group, joined by a dash to the ring, C six H four, which carries the fluorine. So the nitro group is on the side chain, and the fluorine is meta: no such stabilisation, and no fluoride ion. That solves our puzzle. Got it?

4:06Cyanides and amides: two more tools

Now two more tools, mapped in the figure. An alkyl halide, R X, with potassium cyanide gives RCN, a nitrile; R is the alkyl group, X the halogen, and the arrow means gives. In the figure, the three lines between C and N are a triple bond. The lower branch, with silver cyanide, gives the isocyanide instead, and the other arrows are reactions you will meet elsewhere in this chapter. Okay? With acid, H three O plus, over the arrow, the nitrile is hydrolysed through the amide to the carboxylic acid, RCOOH, with the same carbons. Right? Here is a quick one from IITJEE two thousand four. What does benzamide give with phosphorus oxychloride, POC l three? It gives benzonitrile, C six H five C N, very good. Phosphorus oxychloride removes water, so the CONH two group becomes a C N group. Clear?

5:07Example 1: 4-fluoronitrobenzene with sodium methoxide (2000)

Time for Example one, from IITJEE two thousand. What is the major product when four fluoronitrobenzene is heated with sodium methoxide? Pause the video and try it first. Done? Okay, let's check. In four fluoronitrobenzene, the nitro group is at position one and the fluorine at four, so they are para. By our rule, the fluorine is easy to replace. Okay? Now, sodium methoxide gives two ions. Which one attacks the ring? It is the methoxide ion, C H three O minus, yes, exactly. It adds to the carbon holding the fluorine, the nitro group takes up the negative charge, and fluoride leaves. Right? So the major product is four nitroanisole, also named one methoxy four nitrobenzene, with a methoxy group where the fluorine was. Clear?

6:03Example 2: benzyl chloride to N-methyl amide, A to D (2004)

Now a harder one, Example two, from IITJEE two thousand four. Benzyl chloride with KCN in DMF gives A, and then A, with sodium ethoxide and benzaldehyde, on heating, gives B, okay? Then B with H three O plus and heat gives C, and C with thionyl chloride, SOC l two, then methylamine, C H three N H two, gives D, the last one. Identify all four. Pause the video and try it first. Done? Okay, let's check. In benzyl chloride, the chlorine is on a C H two group, not on the ring, so cyanide can replace it. So the first compound is phenylethanenitrile: a phenyl group, C six H five, joined to a C H two C N group. Clear? Sodium ethoxide is a base. Which hydrogens in this nitrile can it remove? It is the C H two hydrogens, yes, exactly! The C H two next to the C N group is acidic and condenses with benzaldehyde. Right? Water leaves, made of the two hydrogens of C H two and the aldehyde oxygen, and a carbon carbon double bond forms, the equals sign on screen. Each carbon of it carries a phenyl group, the brackets holding the second, and the C N stays. So B is two, three diphenylpropenenitrile. With me? Now B is heated with H three O plus. Does the hydrolysis of its C N group stop at the amide? It does not stop there: with acid and heat it goes on, through the amide, to the carboxylic acid. So COOH replaces C N, and C is two, three diphenylpropenoic acid. Clear? Then thionyl chloride, SOC l two, makes the acid chloride, and methylamine, C H three N H two, turns that into an amide. Okay? So D is N methyl two, three diphenylpropenamide; the italic N means the methyl sits on the amide nitrogen. And that is all four compounds. Got it?

8:15Quick revision: nitro directing, ring halogen, nitriles

So, quick revision. First, a nitro group on a ring is deactivating and meta-directing, so no Friedel-Crafts reaction. Next, a plain ring halogen does not react with potassium cyanide, but an ortho or para nitro group makes it easy to replace. And in a nitrile, the C H two next to C N is acidic, and acid with heat gives the carboxylic acid. Now try the other past questions on this page. All the best!