IIT-JEE Questions (1996–2008): Reactions

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle from IITJEE two thousand and eight. Two carbonyl compounds are each treated with the same reagent, potassium hydroxide. The first one gives an alcohol and one more compound. The second one joins with a copy of itself into a bigger molecule. Same reagent, two completely different reactions. So what decides? Keep that question in mind. Today we solve older IITJEE questions on aldehydes and ketones, and you will crack this puzzle yourself. So, let's begin.

0:36One hydrogen decides: the alpha hydrogen

First, let us learn one name. In the carbonyl group, C double bond O, the equals sign is the double bond. The carbon right next to it is the alpha carbon, marked by the Greek letter alpha on screen, and a hydrogen on it is an alpha hydrogen. Okay? Now let me ask you this. Usually, a hydrogen on carbon is not acidic at all. So can a base really pull off this alpha hydrogen? Yes, it can, that is correct! It is weakly acidic, with a p K a of about nineteen to twenty. That is a small p, a capital K and a small a written low. Right? When a base removes it, we get the enolate ion, stabilised by resonance. Got it? So with an alpha hydrogen and dilute alkali, the enolate forms and we get the aldol reaction, shown in the top row of the picture. The short form d i l means dilute. With me? Now look at benzaldehyde, C six H five CHO, in the bottom row. Its alpha carbon is a ring carbon with no hydrogen on it, so no enolate can form. So what happens to it in alkali? With concentrated alkali, it undergoes the Cannizzaro reaction, the bottom row of the picture. Here c o n c is short for concentrated. So this one hydrogen is the switch. Clear?

2:02Aldol: join two molecules, then lose water

The enolate of one molecule attacks the carbonyl carbon of another molecule. The new carbon carbon bond joins the alpha carbon of one to the carbonyl carbon of the other. Okay? Take ethanal, C H three CHO, where the small three written low counts the hydrogens. The arrow means gives, and two ethanal molecules give three hydroxy butanal. Its O H sits on the beta carbon, marked by the Greek letter beta, so this is a beta hydroxy aldehyde, the aldol. Right? Now heat the aldol. It loses water, H two O, and gives but two enal, with a new carbon carbon double bond. Join, then lose water: that is aldol condensation. Got it? Now a topper's trick for working backwards from a product. Cut the carbon carbon double bond. The beta carbon was a carbonyl carbon, so put its oxygen back, and the alpha carbon gets back the two hydrogens it lost. Example two needs this. Clear?

3:03Cannizzaro: one oxidised, one reduced

Now the Cannizzaro reaction. Two molecules of benzaldehyde, which is what the two in front means, react with potassium hydroxide, KOH, and they give benzyl alcohol, C six H five C H two O H, plus potassium benzoate, C six H five COOK, a salt. Okay? One molecule was reduced to the alcohol, and the other was oxidised to the salt of the acid. With me? And how does one molecule reduce the other? It hands over a hydride ion, and this transfer is the slow step. IITJEE nineteen ninety six asked about this slow step. Right? Now a crossed Cannizzaro, from IITJEE two thousand and one. Benzaldehyde and methanal are heated with aqueous sodium hydroxide, and neither has an alpha hydrogen. So which one is oxidised? It is methanal, very good! Methanal, HCHO, is the more reactive one, so methanal is oxidised, and benzaldehyde, C six H five CHO, is reduced. Clear? So the products are sodium methanoate, HCOON a, also called sodium formate, and benzyl alcohol, C six H five C H two O H, the same alcohol as before. Got it?

4:26Example 1: which aldehydes give aldol condensation (1998)

Time for Example one, from IITJEE nineteen ninety eight. Which of these undergo aldol condensation: ethanal, propanal, benzaldehyde, and trideutero ethanal, C D three CHO? Pause the video and try it first. Done? Okay, let's hunt for alpha hydrogens. Ethanal, C H three CHO, has three, on its methyl carbon. Propanal, C H three C H two CHO, has two, on the C H two group. So both of them react. Okay? Benzaldehyde's alpha carbon is a ring carbon with no hydrogen. So no enolate, and no aldol. With me? Now the last one, C D three CHO, okay? Its alpha carbon carries only D atoms, and no H at all. So no aldol here either, right? Oh no, careful, that is the trap in this question! D is deuterium, which is just heavy hydrogen, an isotope of hydrogen. The base removes an alpha deuterium in the same way, the enolate forms, and the aldol goes ahead. Got it? So the answer is all of them except benzaldehyde. Clear?

5:43Example 2: identify H to L from Cannizzaro and aldol clues

And now, our puzzle from the start. There is a tertiary alcohol, called H, which acid dehydrates to give I, okay? Ozonolysis of I gives J and K, the two carbonyl compounds of our puzzle. J with KOH gives benzyl alcohol and a compound called L, and K with KOH gives only the product on screen. In that formula, the dashes are single bonds and the C H three in brackets is a branch. Your job is to identify all five of them. Pause the video and try it first. Done? Okay, let's check, starting with the easiest clue, compound J and its benzyl alcohol. Which reaction turns an aldehyde into an alcohol and one more product? It is the Cannizzaro reaction, yes, exactly! The alcohol is benzyl alcohol, so J must be benzaldehyde, C six H five CHO, with no alpha hydrogen. Okay? Then L is the other product, potassium benzoate, C six H five COOK, the salt of the acid. Right? Now compound K, okay? Its product has a carbon carbon double bond right next to the C O group, the mark of an aldol condensation. It is the only product, so K reacted with itself. With me? So, what is K? Use the topper's trick: cut the double bond and work backwards. So, let us cut the double bond. The beta carbon carries the C H three and a C six H five, so it gets back its oxygen, and that half is C six H five COCH three. The alpha carbon gets back its hydrogens, and that half is the very same compound! So K is acetophenone, which has alpha hydrogens. Clear? Next comes the alkene I: ozonolysis cut its double bond into J and K, so join them back at their carbonyl carbons. That gives C six H five C H, double bond, C with C H three in brackets, then C six H five: one two diphenyl propene. Got it? And finally H, the tertiary alcohol. Add water back to I so that the O H sits on the carbon joined to three other carbons: one two diphenyl propan two ol. And our puzzle is solved: J had no alpha hydrogen, so it split by Cannizzaro, while K had alpha hydrogens, so it joined up by aldol. Okay?

8:22Quick revision: alpha H, aldol, Cannizzaro, work backwards

So, quick revision. With an alpha hydrogen and dilute alkali, we get the aldol reaction, and heating removes water. With no alpha hydrogen and concentrated alkali, we get the Cannizzaro reaction, an alcohol and a salt. And to work backwards, cut the carbon carbon double bond: the beta carbon was a carbonyl carbon. Now try the other past questions on this page yourself, before you read the solutions. All the best!