AIEEE and Earlier JEE Main Questions: Alcohols

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. You heat an alcohol with concentrated sulphuric acid, and two alkenes are possible. One has three groups on its double bond, and the other has only two. Saytzeff's rule says the more substituted alkene is the major one, right? But in a question from A I triple E two thousand and ten, the main product is the one with only two groups! How can that be? Keep that puzzle in mind. Today we solve older A I triple E and JEE Main questions on alcohols, which come up often. So, let's begin.

0:38The Lucas test: who is fastest, and how?

First, a favourite of the older papers: the Lucas test. Lucas reagent is concentrated hydrochloric acid, H C l, plus anhydrous zinc chloride, Z n C l two, with the small two written low. The short form on screen means concentrated, and anhydrous means without water. With me? The alcohol dissolves in this reagent, but its alkyl chloride does not, so the solution turns cloudy, which we call turbidity. The small circle written high is read as degree. Tertiary alcohols turn cloudy at once, secondary ones in about five minutes, and primary ones only on heating. Okay? JEE Main two thousand and thirteen asked this. An unknown alcohol is treated with Lucas reagent. Which alcohol reacts fastest, and by what mechanism? The tertiary alcohol, by the S N one mechanism, yes, exactly! S N one, with the small N written low, means substitution, nucleophilic, unimolecular. Right? And why is the tertiary one the fastest? S N one goes through a carbocation, a carbon with a positive charge. The tertiary carbocation is the most stable one, so it forms fastest, and the cloudiness appears at once. Clear?

1:57Dehydration: how does it start?

Next, dehydration. Heat an alcohol with concentrated sulphuric acid, H two S O four, and the arrow says it gives an alkene, plus water, H two O, okay? The picture shows ethanol, C H three, C H two, O H, on the left, ending as ethene, C H two, double bond, C H two, on the right. Right? Now, A I triple E two thousand and three asked this. In this dehydration, what is the initiation step, the very first thing that happens to the alcohol? Protonation of the alcohol molecule, that is correct! In the picture, the two dots on the oxygen are a lone pair, which picks up a proton, H plus, with the plus written high. The pair of half arrows means this fast step is reversible. With me? Next, water leaves, shown as minus H two O under the arrow, and a carbocation remains: C H three, C H two plus, a carbon with a positive charge. This step is slow, so it decides the rate. Right? Then a proton is lost quickly, minus H plus, and we get the alkene. The acid gets its proton back, so it is regenerated. And see the line under the picture: the carbocation may rearrange before it loses a proton. Got it?

3:19Who loses water most easily? And which alkene?

Now, remember, the slow step makes a carbocation. So tell me, primary, secondary or tertiary: which alcohol loses water most easily? The tertiary alcohol, very good! The ease of dehydration is three degree, greater than two degree, greater than one degree, where greater than means easier than. Same reason as the Lucas test: the tertiary carbocation is the most stable. Okay? That is why the conditions get milder. Ethanol needs concentrated sulphuric acid at four hundred and forty three kelvin, and the capital K on screen is kelvin. But a tertiary alcohol needs only twenty percent phosphoric acid, H three P O four, at three hundred and fifty eight kelvin. Clear? And when two alkenes are possible, the major one is the more substituted alkene, with more groups other than hydrogen on its double bond carbons. This is Saytzeff's rule, and our puzzle will test it. Got it?

4:19Example 1: which alcohol dehydrates most easily (AIEEE 2004)

Time for Example one, from A I triple E two thousand and four. Which is dehydrated most easily in acid: three methylpentan three ol, pentan two ol, pentan one ol, or three methylpentan one ol? Pause the video and try it first. Done? Okay, let's check. The class of an alcohol depends only on the carbon that carries the O H group. Now, three methylpentan one ol has a methyl branch. So is it still a primary alcohol? Yes, it is still primary, just like pentan one ol! In both names, the one at the end puts the O H group on carbon one, the end carbon, joined to only one other carbon. The methyl branch on carbon three changes nothing. Clear? Next, pentan two ol: C H three, then C H with the O H group in brackets, then C H two, C H two and C H three, joined by single bonds, the short dashes. The O H carbon is joined to two carbons, so it is secondary. Okay? And three methylpentan three ol: C H three, C H two, then a carbon holding a methyl group, the C H three in brackets, and the O H group, then C H two, C H three. That carbon is joined to three carbons, so it is tertiary. Right? So the answer is three methylpentan three ol, the tertiary alcohol, because it gives the most stable carbocation. Easy marks, if you classify by the O H carbon. With me?

5:58Example 2: ring conjugation beats Saytzeff (AIEEE 2010)

Now the harder one, our puzzle, from A I triple E two thousand and ten. What is the main product when this alcohol is heated with concentrated sulphuric acid? C six H five is the phenyl group, a benzene ring. Then come C H two, C H with the O H group, and C H with two methyl groups, the C H three in brackets with a small two. Pause the video and try it first. Done? Okay, let's check. Water leaves from the O H carbon, then a proton is lost from a carbon next to it, and there are two such neighbours. If the proton goes from the C H with the two methyl groups, we get C six H five, C H two, C H, double bond, C with two methyl groups. Three groups on the double bond, the more substituted alkene. Okay? If it goes instead from the C H two next to the ring, the benzylic carbon, we get C six H five, C H, double bond, C H, then C H with two methyl groups. Only two groups, but this double bond sits right next to the benzene ring, so it is conjugated with the ring. With me? Now, Saytzeff's rule says the more substituted alkene is the major one. So the answer is the first alkene, right? No, no, careful, that is the trap in this question! Loss of water towards the benzylic carbon gives the alkene conjugated with the ring, and that is the main product. Conjugation with the ring makes it the more stable alkene here, so it beats the simple count of groups. And that solves our puzzle from the start. Got it? And now, one last detail. When the phenyl group and the other big group sit on opposite sides of this double bond, we get the trans isomer, the more stable one. Its name begins with a capital italic E in brackets, which marks this trans form: E, three methyl one phenylbut one ene. Clear?

7:57Quick revision: Lucas test, dehydration order, conjugation

So, quick revision. Lucas test: the tertiary alcohol is fastest, by S N one. Dehydration starts with protonation of the alcohol, and the ease is tertiary, then secondary, then primary. Classify an alcohol by its O H carbon, and watch for conjugation with a ring. Now solve the other past questions on this page yourself. All the best!