AIEEE and Earlier JEE Main Questions: Preparation

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Fluorine is a halogen, just like chlorine and bromine. So, to make an alkyl fluoride, just take an alkane and pass fluorine gas through it. Easy, right? Well, it does not work like that. With an alkane, fluorine reacts explosively, and it even breaks the carbon to carbon bonds of the chain. So how do chemists get a fluorine atom into a molecule at all? Today we solve two older past-paper questions from this chapter, and both of them ask exactly this, one for alkyl fluorides and one for fluorobenzene. Questions on these named preparation reactions do come up in JEE Main. So, let's begin.

0:43Alkyl fluorides: the Swarts reaction

So, the first back door is for alkyl fluorides, and it is called the Swarts reaction. On screen, R, dash, X is an alkyl halide: R is the alkyl group, the dash is the single bond, and X is the halogen already there, such as chlorine or bromine. Add silver fluoride, written A g F, and the arrow means gives. We get R, dash, F, the alkyl fluoride, plus A g X, the silver halide. Okay? Silver fluoride is not the only choice. H g two F two, a fluoride of mercury, C o F two, cobalt fluoride, and S b F three, antimony fluoride, all work too. The small numbers written low, the subscripts, count the atoms of each element. Right? And underline this line: the Swarts reaction is the best way of making alkyl fluorides. Our first example uses almost exactly these words. Clear? Now, this chapter has a second halogen exchange, and students often mix the two up. Tell me, which reaction swaps chlorine or bromine for iodine? The Finkelstein reaction, very good! R, dash, X with sodium iodide, N a I, in dry acetone, gives R, dash, I, the alkyl iodide. So keep them apart: Finkelstein brings in iodine, and Swarts brings in fluorine. Got it?

2:13Aryl fluorides: through the diazonium salt

Now the second back door, for aryl fluorides, where the fluorine sits on a benzene ring. We start from an aromatic amine, written A r N H two, with A r for the aryl group. Add sodium nitrite, N a N O two, and two H X, a mineral acid. We get A r N two plus X minus, the diazonium salt; the small raised plus and minus are charges. The rest is N a X and two molecules of water. With me? Now, this step is always done ice cold. Why do you think? Because diazonium salts are unstable. Above about two hundred and seventy eight kelvin, they decompose in water to phenols, with loss of nitrogen. So we work at two hundred and seventy three to two hundred and seventy eight kelvin; the capital K on screen stands for kelvin. Okay? Now the N two plus group can be replaced, and nitrogen gas leaves. In the Sandmeyer reaction, C u C l, cuprous chloride, with H C l gives A r C l, an aryl chloride; the slash just means with. And C u B r, cuprous bromide, with H B r gives A r B r, an aryl bromide. Right? But for fluorine, we use the Balz Schiemann route, named on screen. We add HBF four, fluoroboric acid, to the diazonium salt. The new diazonium salt comes out as a solid, and then we heat the dry salt. Clear? It breaks up into A r F, the aryl fluoride, plus B F three, boron trifluoride, plus N two, nitrogen gas. This is the usual way to make aryl fluorides. Got it?

3:56Example 1: best way to make alkyl fluorides (JEE Main 2015)

Time for Example one, from the JEE Main two thousand fifteen paper. By which reaction is the synthesis of alkyl fluorides best accomplished: Finkelstein, Swarts, free-radical fluorination, or Sandmeyer? Pause the video and try it first. Done? Okay, let's check the options one by one. Finkelstein uses sodium iodide in dry acetone, so it brings in iodine and gives an alkyl iodide, not a fluoride. So that option is out. Okay? Next, Sandmeyer. It uses cuprous chloride or cuprous bromide on a diazonium salt, and gives an aryl chloride or an aryl bromide. So it brings in the wrong halogen, and on a benzene ring, not on an alkyl group. With me? Now, free-radical fluorination. We want fluorine, and this puts fluorine in directly. So it must be the best method, right? Oh no, careful, that is the trap! Remember our puzzle: fluorine reacts with an alkane explosively, and breaks even the carbon to carbon bonds. It is too violent to control. Clear? So the answer is the Swarts reaction: an alkyl halide with silver fluoride, or another metal fluoride, gives the alkyl fluoride. It is exactly the line we underlined, the best way of making alkyl fluorides. Got it?

5:18Example 2: making fluorobenzene from aniline (AIEEE 2006)

Now Example two, the harder one, from the AIEEE two thousand six paper. How can fluorobenzene be made in the laboratory? Option one, direct fluorination of benzene with F two. Option two, from bromobenzene and a solution of N a F, sodium fluoride. Option three, by heating phenol with H F and K F, hydrogen fluoride and potassium fluoride. And option four, from aniline, by diazotisation, then heating the diazonium salt with HBF four. Pause the video and try it first. Done? Okay, let's check. Option one is out at once. Fluorine gas is too violent to control, just as in our puzzle. Okay? Now option two looks very tempting. Swarts swaps a halogen for fluorine, so bromobenzene with sodium fluoride should give fluorobenzene, right? No, no, wait, this is the trap! Bromobenzene is an aryl halide, and one lone pair of its bromine spreads into the ring. So the C dash B r bond on screen, carbon to bromine, has partial double bond character, and it is hard to break. That is why aryl halides resist nucleophilic substitution, where an ion like fluoride would replace the halogen. Got it? But note the condition. The halogen is replaced only when electron-withdrawing groups sit at the ortho or para positions, or under harsh conditions. Plain bromobenzene in sodium fluoride solution has neither, so option two fails. With me? Next, option three, heating phenol with H F and K F, hydrogen fluoride and potassium fluoride. In phenol too, a lone pair, this time of the oxygen, spreads into the ring, so the C dash O bond has partial double bond character, and the ring carbon is s p two. So the O H group of phenol cannot be replaced by a halogen, the way the O H of an alcohol can. Option three fails too. Clear? That leaves option four, our back door. C six H five N H two is aniline. Over the first arrow are the reagents, sodium nitrite and H C l, and below it is two hundred and seventy three to two hundred and seventy eight kelvin, the cold condition. That gives the diazonium chloride, C six H five N two plus, C l minus. Then HBF four, over the second arrow, turns it into C six H five N two plus, B F four minus. Okay? Now heat it. It gives C six H five F, which is fluorobenzene, plus B F three and N two, nitrogen gas. So the answer is option four: from aniline, by diazotisation, then heating with HBF four. And see the common lesson of both questions: fluorine never goes in directly; it always comes in through a back door. Right?

8:19Quick revision: Swarts, diazonium salt with HBF4, never F2

So, quick revision. Alkyl fluorides come from the Swarts reaction: an alkyl halide with silver fluoride, or the fluoride of mercury, cobalt or antimony. Aryl fluorides come from the diazonium salt, with HBF four, and then heat. And never use fluorine gas directly; Finkelstein gives iodides, and Sandmeyer gives aryl chlorides and bromides. Now go back to these two past-paper questions on this page and solve them yourself, without looking at the answers. You can do this. All the best!