Halogen Exchange

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Some reactions can run both ways, forward and backward. Now suppose you want such a reaction to go all the way forward, and your main trick is the choice of solvent. Can a solvent really push a reaction to completion? Sounds strange, right? Today we will see exactly how. We will meet three reactions: one that swaps a halogen for iodine, one that makes alkyl fluorides, and one where the product comes out with one carbon fewer than you started with. These reactions are part of the JEE syllabus, under the preparation of haloalkanes. So, let's begin.

0:40Finkelstein reaction: Cl or Br swapped for I

So, reaction number one is the Finkelstein 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. Here X is chlorine or bromine, written C l or B r on screen. Add N a I, sodium iodide, and the arrow means gives: R, dash, I, an alkyl iodide, plus N a X, sodium chloride or sodium bromide. So iodine simply takes the place of the chlorine or bromine. Okay? Now, this exchange can also run backwards, with the iodide swapping back. So what pushes it all the way forward? Remember our puzzle. The solvent, exactly as our puzzle said! The reaction is done in dry acetone. Acetone is the solvent, and dry means it has no water in it. Right? Now look at solubility. N a I, sodium iodide, dissolves in acetone. But N a C l, sodium chloride, and N a B r, sodium bromide, do not dissolve in it. With me? So, as soon as sodium chloride or sodium bromide forms, it comes out of the solution as a solid; it precipitates. Since a product keeps leaving the solution, by Le Chatelier's principle the exchange is driven forward, all the way to completion. And there is our puzzle solved: the solvent drives the reaction. Clear?

2:09Swarts reaction: making alkyl fluorides

Now reaction number two, the Swarts reaction. Take an alkyl halide, R, dash, X, and add A g F, silver fluoride. We get R, dash, F, an alkyl fluoride, plus A g X, a silver halide. Got it? Silver fluoride is not the only choice. Other metal fluorides also work. H g two F two is a fluoride of mercury, C o F two is cobalt fluoride, and S b F three is antimony fluoride. The small numbers written low tell you how many atoms of each element there are. Okay? And remember this line: the Swarts reaction is the best way of making alkyl fluorides. So, Finkelstein brings in iodine, and Swarts brings in fluorine. Clear? Now a quick question from the course. Take one mole of S b F three, with excess alkyl chloride. How many moles of alkyl fluoride, R, dash, F, can it give? Three moles, yes, exactly! Each S b F three carries three fluorine atoms, so it can replace the chlorine in three molecules of R C l, an alkyl chloride. Three R C l plus S b F three gives three R F, plus S b C l three, antimony trichloride.

3:27Hunsdiecker reaction: from a silver salt

Right? Now reaction number three, the Hunsdiecker reaction. RCOOA g is the silver salt of a carboxylic acid. Add B r two, bromine; the small two written low means two bromine atoms. The products are R, dash, B r, an alkyl bromide, then C O two, carbon dioxide, and A g B r, silver bromide. With me? The solvent here is C C l four, that is, carbon tetrachloride. Okay? Now look carefully at the equation. The silver salt has the carbons of R, plus one more carbon in the COO group. The alkyl bromide has only the carbons of R, nothing more. So where did the carbon of the COO group go? Into carbon dioxide, very good! So the bromide has one carbon fewer than the acid. Remember this; it is the key to our second example. Got it?

4:24Example 1: 1-chlorobutane with NaI in dry acetone

Time for Example one. What is formed when one chlorobutane is heated with sodium iodide in dry acetone? And name the reaction. Pause the video and try it first. Done? Okay, let's check. One chlorobutane is an alkyl halide, R, dash, X, where the halogen is chlorine, C l, as the first line shows. And the reagent is N a I, sodium iodide, in dry acetone. These are exactly the conditions of reaction number one. Right? So, write the swap. On screen, one chlorobutane is C H three, then three C H two groups, then C l: a straight chain of four carbons with chlorine at the end. Add N a I, and it gives the same chain with iodine at the end, plus N a C l, sodium chloride. Okay? Now, the exchange can run both ways. So will it stop halfway, at an equilibrium? No, it goes to completion! Sodium chloride does not dissolve in acetone, so it precipitates, and that drives the reaction forward. Clear? So the product is one iodobutane, and the reaction is the Finkelstein reaction. Same chain, same position, and only the halogen has changed, okay?

5:43Example 2: silver butanoate with bromine in CCl4

Now a harder one, Example two. Silver butanoate is refluxed with bromine in C C l four, carbon tetrachloride. Name the organic product. Pause the video and try it first. Done? Okay, let's check. Silver butanoate is the silver salt of butanoic acid, written C three H seven COOH on screen. Count the carbons: three in C three H seven, and one in COOH, so four carbons in all. Okay? Four carbons in the acid, so the product must be one bromobutane, right? No, no, wait, that is exactly the trap! Write the Hunsdiecker equation. C three H seven COOA g, plus B r two, gives C three H seven B r, plus C O two, plus A g B r, silver bromide. With me? See, the carbon of the COO group leaves as carbon dioxide. So only three carbons are left in the bromide, one carbon fewer than the acid. Right? The chain of butanoic acid is straight, and bromine takes the place of the COOA g group, at the end of the chain. So the product is one bromopropane. Four carbons went in, three came out, and the missing carbon left as carbon dioxide. Clear?

7:05Quick revision: Finkelstein, Swarts and Hunsdiecker

So, quick revision. Finkelstein: an alkyl chloride or bromide with sodium iodide in dry acetone gives the alkyl iodide, because sodium chloride or sodium bromide precipitates. Okay? Swarts: an alkyl halide with silver fluoride, or another metal fluoride like mercury, cobalt or antimony fluoride, gives the alkyl fluoride, the best way to make one. Right? And Hunsdiecker: the silver salt of an acid with bromine in carbon tetrachloride gives an alkyl bromide with one carbon fewer. You have done really well today. Now go and try the practice problems on this page. All the best!