Overview

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Suppose you want fluoroethane, ethane with one fluorine atom in place of a hydrogen. Easy, just pass fluorine gas into ethane, right? Hmm, hold that thought. By the end of this video, you will make fluoroethane properly, starting from ethanol. This chapter maps the main routes to haloalkanes and haloarenes, which come up often in JEE papers. So, let's begin.

0:28The chapter at a glance

So, here is the whole chapter on one page. In the centre sits the haloalkane, R X, where R is the alkyl group, the dash is the bond, and X is the halogen. Okay? On the left, under Preparation, seven arrows point in: seven ways to make it. And they start from just three kinds of compounds: an alcohol, a hydrocarbon, or another halide. Right? On the right, under Reactions, is what a haloalkane turns into, for the coming chapters. Today, we stay on the left. Clear?

1:00From alcohols: replace OH by X

The first three boxes start from an alcohol, ROH, and we replace the O H group by a halogen. ROH plus H X gives R X plus H two O, water; the arrow means gives, and the small two written low counts atoms. With H C l, zinc chloride, Z n C l two, is usually added, as the brackets say. With me? Next, the phosphorus halides. Three ROH plus P X three gives three R X, plus H three P O three. The second box also shows red P plus I two: red phosphorus and iodine make P I three right in the flask. Got it? Phosphorus pentachloride works too. ROH plus P C l five gives R C l, plus POC l three, plus H C l. And now the third box, thionyl chloride, SOC l two. ROH plus SOC l two gives R C l, plus S O two, plus H C l. Look at the two by-products. What do they have in common? Both are gases, yes, exactly! The small upward arrow on screen means a gas that escapes. Nothing is left to separate, so the alkyl chloride comes out pure. That is why this is the best method. Clear?

2:23From hydrocarbons

Now boxes four and five, the hydrocarbons. An alkane plus X two, a halogen, in light or heat, reacts by free radicals. So tell me, will this give you one clean product? No, it gives a mixture! Mono and polyhalides, and position isomers too, so it is of limited use. That is why the map says mixture. Right? Next, an alkene plus H X, which adds by Markovnikov's rule. Keep this in mind for Example one. But H B r with a peroxide adds the other way, by a radical path. The map calls this anti-Markovnikov. With me?

3:05Halogen exchange: one halide from another

Now the last two boxes: one halide from another. On the map, R C l slash R B r means a chloride or a bromide. In the Finkelstein reaction, R X plus sodium iodide, N a I, gives R I plus N a X, so chlorine or bromine is swapped for iodine. Okay? It is done in dry acetone, and here is the clever part: sodium chloride does not dissolve in acetone. So why does that help us? Because it precipitates, correct! Sodium iodide stays dissolved, while sodium chloride or bromide drops out as a solid, and that pulls the exchange to completion. Clear? Next, the Swarts reaction, for fluorides: R X plus silver fluoride, A g F, gives R F plus A g X, the silver halide. Other metal fluorides work too: H g two F two, C o F two, and S b F three, which the map also shows. This is the best way to make alkyl fluorides, R F on screen, and the key to our puzzle. Got it?

4:11Haloarenes: X on a benzene ring

Now the haloarenes, with X directly on a benzene ring. Ethanol and H C l give chloroethane, so phenol and H C l should give chlorobenzene, right? No, no, wait, that reaction does not work at all! In phenol, the carbon oxygen bond, the C dash O on screen, has partial double bond character, so it is too strong to break. Okay? One way is direct: an arene plus chlorine or bromine, with a Lewis acid such as F e C l three, in the dark. The halogen substitutes on the ring. Right? The other way is through a diazonium salt. Ar N H two is an aromatic amine, where Ar is the aryl group. With sodium nitrite, N a N O two, and two H X, it gives Ar N two plus X minus, the diazonium salt; the small raised plus and minus are charges. Along with N a X and water, at two hundred and seventy three kelvin to two hundred and seventy eight kelvin. With me? Then the N two plus group is replaced. In the Sandmeyer reaction, C u C l with H C l gives the aryl chloride, and C u B r with H B r the aryl bromide; the slash means with. Clear? For the iodide, simply warm with potassium iodide, K I, no copper needed. For the fluoride, add HBF four, then heat the dry salt. Got it?

5:36Example 1: HBr with propene, with and without peroxide

Now Example one, an easy one, from the concept From Hydrocarbons in this chapter. Write the product of the addition of H B r to propene, first without a peroxide, and then with one. Pause the video and try it first. Done? Okay, let's check. Propene is C H three, C H, double bond, C H two; the two short lines are the double bond. Without a peroxide, the proton, H plus, adds first. Which carbon does it go to? To the C H two end, the carbon with more hydrogens, yes! That leaves the plus charge on the middle carbon, the more stable carbocation. Bromide adds there: C H three, CHB r, C H three, which is two bromopropane. Right? With a peroxide, H B r adds the other way, so the bromine ends up on the end carbon: C H three, C H two, C H two B r, which is one bromopropane. Careful, this is a common trap: same propene, same H B r, but the peroxide flips the product. Clear?

6:46Example 2: fluoroethane from ethanol via Swarts reaction

And now Example two, a harder one, from the concept Halogen Exchange, and it is our opening puzzle. Starting from ethanol, how would you prepare fluoroethane? Pause the video, look at the map, and try it first. Done? Okay, let's check. Fluorine gas, our first guess, is too violent to control. And on the map, no arrow takes an alcohol straight to a fluoride. So which two arrows can you join? First, turn the alcohol into a bromide. Three C two H five O H, ethanol, plus P B r three gives three C two H five B r, bromoethane, plus H three P O three. Or use N a B r with H two S O four, which makes H B r in the flask. Okay? Then the Swarts reaction: C two H five B r plus silver fluoride gives C two H five F, fluoroethane, plus silver bromide, A g B r. With me? So, ethanol to bromoethane to fluoroethane, and our puzzle is solved! Fluorine goes in indirectly, by exchange. Two arrows from one map; many multi-step questions work like this. Got it?

8:03Quick revision: alcohols, alkenes, exchange, haloarenes

So, quick revision. From alcohols, use H X, the phosphorus halides, or thionyl chloride, the best for chlorides, since its by-products are gases. Alkanes give mixtures, and alkenes add H X by Markovnikov's rule. For exchange, Finkelstein makes iodides and Swarts makes fluorides. And haloarenes come not from phenol, but from ring halogenation or the diazonium salt. Now go and try the practice problems on this page. All the best!