Nitro Compounds

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Treat an alkyl halide with silver nitrite, and you mainly get a nitroalkane. Now treat the same halide with potassium nitrite. Same nitrite ion, so the same product, right? But this time you mainly get a different compound, an alkyl nitrite. So how can one ion give two different products? Today we study nitro compounds: their structure, preparation, acidity and reduction, and what they do to a benzene ring. Questions on these come up often in JEE exams. So, let's begin.

0:36The nitro group: a resonance hybrid

So, first the structure. Here are two structures of the nitro group, and the dash on the far left is its bond to carbon. Nitrogen has a double bond, the equals sign inside the brackets, to one oxygen, and a single bond to the other oxygen, which carries a minus charge. The small plus above N is a positive charge on nitrogen, and the double-headed arrow between the structures means resonance. Okay? The real nitro group is a resonance hybrid of these two, so both nitrogen oxygen bonds are equal, not one long and one short. And nitrogen carries a positive charge. Clear? So the nitro group is strongly electron-withdrawing. Minus I is the inductive effect and minus R is the resonance effect, and the minus sign means the group pulls electrons away. Remember this pull, because it comes back again and again today. Got it?

1:29Nitroalkanes and alkyl nitrites

Now, two look-alikes. In RNO two, where R is an alkyl group and the small two written low counts the oxygens, the carbon is joined to nitrogen, and this is a nitroalkane. In RONO, the carbon is joined to oxygen, and this is an alkyl nitrite. Same formula, different functional groups, so they are functional isomers. Right? The nitrite ion, N O two minus, is an ambident nucleophile. Ambident means it can attack through two different atoms, the nitrogen or an oxygen. With me? With silver nitrite, written A g N O two, an alkyl halide R X gives mainly the nitroalkane. Here, X stands for a halogen, and the arrow means gives. Okay? So now, with potassium nitrite, KNO two, what does the same alkyl halide give mainly? Yes, exactly, it gives mainly RONO, the alkyl nitrite. That solves our puzzle: the ion can attack through two different atoms, and the salt we use decides the main product. So remember it like this: silver gives nitro, and potassium gives nitrite.

2:40Acidity: the α-hydrogens

Clear? Now, acidity. The Greek letter alpha marks the carbon that carries the nitro group, and the hydrogens on it are the alpha hydrogens. In primary and secondary nitroalkanes, these hydrogens are acidic, which is surprising for a carbon hydrogen bond, right? So tell me, why? What keeps the anion stable when this hydrogen leaves as a proton? Yes, the anion is stabilised by resonance with the nitro group, because the negative charge spreads onto its oxygen atoms. So primary and secondary nitroalkanes dissolve in N a O H, sodium hydroxide. Clear? But in a tertiary nitroalkane, the carbon bearing the nitro group has no hydrogen at all. No alpha hydrogen, so it does not dissolve. Keep this in mind for Example two. Got it?

3:29Reduction of nitrobenzene: the medium decides

Now, reduction. In the figure, nitrobenzene, C six H five N O two, is on the left, and four arrows lead to four products: the medium decides which. In acid, tin or iron with hydrochloric acid, written S n slash H C l and F e slash H C l, gives aniline, C six H five N H two. So does hydrogen gas over palladium, written H two slash P d. Okay? In neutral medium, zinc with ammonium chloride, N H four C l, we get N phenylhydroxylamine. The italic N tells you the phenyl group is joined to the nitrogen, and the figure gives its formula, C six H five NHOH, right? In alkaline medium, zinc with sodium hydroxide, we get hydrazobenzene, where two phenyl N H groups are joined by a nitrogen nitrogen single bond. With me? And the last arrow is lithium aluminium hydride, L i A l H four. Many students expect aniline here too. Is that right? It does not give aniline, and this is a common trap. It gives azobenzene, where the two rings are joined through an N double bond N link. So in this figure, only acid or hydrogen over palladium gives aniline. Got it?

4:51The nitro group and the benzene ring

Now, aromatic nitro compounds are made by nitration with nitric acid and sulphuric acid, HNO three and H two S O four, and the electrophile is N O two plus, the nitronium ion. Okay? Once it is on the ring, the nitro group pulls electrons out of the ring. So it is strongly deactivating, and it directs a new group to the meta position. Clear? So here is a question for you. Will nitrobenzene itself undergo the Friedel Crafts reaction? The answer is no, nitrobenzene does not undergo it. In fact, nitrobenzene is used as a solvent for the Friedel Crafts reaction, which is a nice point to remember. Right? And one effect works the other way round. When a nitro group sits ortho or para to a halogen, the halogen becomes easy to replace by nucleophiles. With me?

5:44Example 1: 1-iodopropane with AgNO2 and with KNO2

Now Example one, an easy one. One iodopropane is treated separately with silver nitrite and with potassium nitrite. Name the main product in each case. Pause the video and try it first. Done? Okay, let's check. One iodopropane is a chain of three carbons with iodine at the end. Silver nitrite gives mainly the nitroalkane, with the nitro group in place of iodine, so we get one nitropropane. Right? Now potassium nitrite. It is the same nitrite ion, so is it one nitropropane again? No, and this is our opening puzzle again. Potassium nitrite gives mainly the alkyl nitrite, with the propyl group joined to oxygen, so we get propyl nitrite. Both products are C three H seven N O two, so they are functional isomers.

6:34Example 2: C4H9NO2 nitroalkanes, how many dissolve in NaOH

Okay? And now Example two, a little harder. How many nitroalkanes have the formula C four H nine N O two? And how many of them dissolve in aqueous sodium hydroxide? Pause the video and try it first. Done? Okay, let's check. Take a straight chain of four carbons and put the nitro group on the end carbon, and that gives one nitrobutane. Right? On the second carbon, we get two nitrobutane; the brackets show the nitro group hanging on that carbon. The third carbon just gives the same compound again. Clear? Now the branched chain, where the bracket with a small two means two methyl groups on one carbon. With the nitro group on an end carbon, we get two methyl one nitropropane. With me? And on the central carbon, joined to three methyl groups, we get two methyl two nitropropane. So there are four nitroalkanes in total. Okay? Now the second part. All four are nitroalkanes, so all four dissolve in sodium hydroxide, right? Not all four, and this is the trap. In two methyl two nitropropane, the carbon bearing the nitro group has no hydrogen, so there is no alpha hydrogen and it does not dissolve. So the answer is four nitroalkanes, and three of them dissolve. Got it?

7:54Quick revision: AgNO2 vs KNO2, acidity, reduction, ring

So, quick revision. Silver nitrite gives the nitroalkane, and potassium nitrite gives the alkyl nitrite. Primary and secondary nitroalkanes have acidic alpha hydrogens and dissolve in sodium hydroxide, but tertiary ones do not. In the reduction of nitrobenzene, acid gives aniline, and the medium decides the product. And on a ring, the nitro group is deactivating and meta-directing. Now go and try the practice problems on this page. All the best!