Overview

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Someone hands you a bottle labelled C four H ten O, and it is an alcohol. But four different alcohols share this formula. So which one is in the bottle? By the end of this video, you will know. Today we see the whole chapter on alcohols on one map. It comes up often in JEE, so let's begin!

0:25The whole chapter on one map

So, here is the whole chapter on one page. In the centre sits the alcohol, ROH, where R is an alkyl group, the dash is the bond, and O H is the hydroxyl group. On the left, seven arrows point in, seven ways to make it; on the right, ten arrows point out, its reactions. Okay? The map keeps writing one, two or three with a small raised circle, the degree sign, for primary, secondary and tertiary. So look at C H three, C H with O H in brackets, C H three. Which class is it? It is secondary, yes, exactly! Count the carbons attached to the carbon bearing the O H, here, two. One makes it primary, two secondary, three tertiary. Right?

1:13Making alcohols from alkenes

Let's start on the left, with the alkenes. Water with an acid, H two O slash H plus; the slash means with, and the raised plus is a proton. It adds the Markovnikov way, through a carbocation that may rearrange. So in propene, C H three, C H, double bond, C H two, which carbon gets the O H? The middle carbon, exactly! The proton goes to the end C H two, so the arrow gives C H three, CHOH, C H three, propan two ol. Okay? Next, H g, O A c in brackets, two, with water, then N a B H four: oxymercuration demercuration. It is also Markovnikov, but with no rearrangement. With me? Then hydroboration oxidation: B two H six, then H two O two with O H minus, the raised minus being a charge. This one is anti-Markovnikov, so the O H goes to the less substituted carbon, here the end carbon. Keep this for Example one. Clear?

2:22Making alcohols from carbonyl compounds

Next, the carbonyl compounds. Reduce an aldehyde to get a primary alcohol, or a ketone to get a secondary one. Use N a B H four, sodium borohydride, or L i A l H four, lithium aluminium hydride; the map adds hydrogen over nickel. Got it? Now here is a classic trap. Sodium borohydride reduces aldehydes and ketones, so surely it reduces an ester too, right? What do you think, does it? No, no, careful, it does not! Sodium borohydride does not reduce acids or esters. Of these two, only lithium aluminium hydride does, giving a primary alcohol. Don't fall for this one, okay? Then the Grignard reagent, R M g X, followed by H three O plus, that is, dilute acid. Methanal, HCHO, gives a primary alcohol; any other aldehyde, RCHO, a secondary one; and a ketone, a tertiary one. Right?

3:23Reactions of the O–H and C–O bonds

Now the right side. With sodium, two ROH plus two N a gives two RON a, the alkoxide, plus hydrogen gas, H two. With me? With H X, where X is a halogen, P X three or thionyl chloride, SOC l two, we get the alkyl halide R X, okay? With an acid, R prime COOH, and H plus, or an acid chloride, R prime COC l, we get an ester; the small raised dash is called prime. Clear? Now heat ethanol with concentrated sulphuric acid, H two S O four. The map gives two temperatures, four hundred and thirteen kelvin and four hundred and forty three kelvin. Which one gives the alkene? The higher one, four hundred and forty three kelvin, yes! Water is removed, giving an alkene. At four hundred and thirteen kelvin, with the alcohol in excess, we get an ether; this route suits primary alcohols only. Don't swap these two! Got it?

4:28Oxidation: the class decides

Next, oxidation, where the class decides everything. PCC, pyridinium chlorochromate, takes a primary alcohol to an aldehyde, and a secondary one to a ketone. Okay? Potassium permanganate, K M n O four, or potassium dichromate, K two C r two O seven, with acid, takes a primary alcohol all the way to a carboxylic acid. Right? And a tertiary alcohol? Under these conditions it resists oxidation. Why do you think that is? Because the carbon bearing the O H has no hydrogen to lose, exactly! In a tertiary alcohol it carries three carbons instead. With me?

5:09Tests: telling alcohols apart

Finally, the tests our puzzle needs. Lucas reagent is concentrated H C l with anhydrous zinc chloride, Z n C l two. The alcohol dissolves in it, but the alkyl chloride formed does not, so the liquid turns turbid, or cloudy. Okay? Tertiary alcohols turn turbid at once, secondary ones in about five minutes, and primary ones only on heating. Clear? The iodoform test: iodine with sodium hydroxide gives yellow iodoform, CHI three, but only if the alcohol has the group C H three, CHOH, joined to the rest. Right? And after oxidation, only the aldehyde from a primary alcohol reduces Tollens' reagent; the ketone from a secondary alcohol does not. Got it?

5:56Example 1: but-1-ene by acid hydration vs hydroboration

Now Example one, an easy one, from the concept Preparation of Alcohols. Name the alcohol formed from but one ene by part a, acid catalysed hydration, and part b, hydroboration oxidation. Pause the video and try it first. Done? Okay, let's check. But one ene is C H two, double bond, C H, C H two, C H three. Part a follows Markovnikov, so which carbon gets the O H? Carbon number two, yes! The proton goes to the end C H two. So we get C H three, CHOH, C H two, C H three: butan two ol, a secondary alcohol. Okay? Part b, hydroboration oxidation, is anti-Markovnikov, so the O H goes to carbon number one. H O, C H two, C H two, C H two, C H three: butan one ol, a primary alcohol. Same alkene, two reagents, two different alcohols, a common trap. Clear?

7:10Example 2: identify C4H10O from Lucas and iodoform tests

And now Example two, a harder one, from the concept Oxidation and Tests for Alcohols. It is our opening puzzle! An alcohol A, C four H ten O, gives turbidity with Lucas reagent in about five minutes, and a yellow precipitate with iodine and alkali. Oxidation gives B, C four H eight O, which does not reduce Tollens' reagent. Identify both A and B, okay? Pause the video and try it first. Done? Okay, let's check. Turbidity with Lucas reagent in about five minutes: which class of alcohol is that? It is secondary, exactly! So A is a secondary alcohol. Right? The yellow precipitate with iodine and alkali is iodoform, so A must carry the group C H three, CHOH, okay? Of the four alcohols C four H ten O, only one is secondary: C H three, CHOH, C H two, C H three, butan two ol. And it has the iodoform group too. Look, it is the alcohol from Example one, and our puzzle is solved! Oxidising a secondary alcohol gives a ketone, which does not reduce Tollens' reagent, as the question says. So B is C H three, C O, C H two, C H three, butanone, with two hydrogens fewer. Clear?

8:42Quick revision: making, reacting and testing alcohols

So, quick revision. Make alcohols from alkenes, minding Markovnikov or anti, by reducing aldehydes and ketones, or with a Grignard reagent. They react with sodium, with H X, and with acids to give esters; ethanol with concentrated sulphuric acid gives an alkene at four hundred and forty three kelvin, and an ether at four hundred and thirteen kelvin. The class decides oxidation and the Lucas test, and iodoform needs the C H three, CHOH group. Now go and try the practice problems on this page. All the best!