The Cannizzaro Reaction

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Take benzaldehyde, add concentrated alkali, and heat it. There is no oxidising agent in the flask, and no reducing agent either. And yet, at the end, half the aldehyde molecules have been oxidised, and the other half reduced. So who did the oxidising, and who did the reducing? That is the Cannizzaro reaction, and it comes up often in JEE questions. So, let's begin.

0:31Self oxidation and reduction of one aldehyde

So, which aldehydes can do this? Only an aldehyde with no alpha hydrogen. The Greek letter alpha you see on screen marks the carbon right next to the aldehyde group, and an alpha hydrogen is a hydrogen on that carbon, okay? Such an aldehyde, heated with concentrated alkali, gives the Cannizzaro reaction. For example, methanal, benzaldehyde, and two, two dimethylpropanal. We will check all three in Example one. Right? Now, in this reaction, one aldehyde molecule gets oxidised. So tell me, when an aldehyde is oxidised, what does it become? An acid, yes, exactly! But in concentrated alkali, we collect the salt of that acid, not the free acid. So the arrow on screen means becomes: one molecule becomes the salt of the acid, and it is oxidised. With me? And another molecule becomes the alcohol, so it is reduced. One aldehyde, oxidising and reducing itself: self oxidation and reduction, as the heading says. Clear?

1:38Writing the equations

Now let us write it, first for methanal, which is HCHO, okay? Two HCHO, plus KOH, potassium hydroxide, and the arrow means gives. We get C H three O H, methanol, where the small three written low means three hydrogens. Plus HCOOK, potassium methanoate, the salt of the acid. Got it? Now benzaldehyde, C six H five CHO, where C six H five is the phenyl group. Two of them, plus KOH, give C six H five C H two O H, benzyl alcohol, plus C six H five COOK, potassium benzoate. One alcohol and one salt, and this is the flask from our puzzle. Right?

2:31How it happens: a hydride ion moves

So now the real question: how does it happen? In step one, O H minus, the hydroxide ion, adds to one aldehyde molecule. The small minus sign up top is its negative charge. Okay? Now this adduct, the aldehyde with the O H added, meets a second aldehyde molecule. To reduce that second molecule, what must it hand over? It hands over a hydride ion, very good! That is step two: H minus, a hydrogen with a negative charge, moves from the adduct to the second molecule. So the molecule that gives away the hydride is oxidised, and the molecule that takes it is reduced. And there is the answer to our puzzle: the aldehyde molecules oxidise and reduce each other. Right? And this hydride transfer is the slow step, so it decides the rate. Questions often ask exactly this. Got it?

3:25Crossed Cannizzaro: methanal with another aldehyde

Now, a clever twist. Take methanal together with another aldehyde which also has no alpha hydrogen. This is the crossed Cannizzaro reaction. Clear? Again, one aldehyde is oxidised, and one is reduced. So which one do you think is oxidised: methanal, or the other aldehyde? Methanal, yes, exactly! Methanal is the more reactive of the two, so it is oxidised to methanoate, HCOO minus. And the other aldehyde is reduced to its alcohol. Okay? So benzaldehyde, heated with methanal and concentrated sodium hydroxide, N a O H, gives benzyl alcohol and sodium methanoate. Methanal becomes the methanoate, as expected. With me?

4:12Two aldehyde groups in one molecule

Now let us look at one more case. Ethanedial, also called glyoxal, is OHC, a short dash for a single bond, and CHO, so it has two aldehyde groups in one molecule. Treat it with concentrated sodium hydroxide. What will its two CHO groups do? That is correct if you said one group is oxidised and the other is reduced, inside the same molecule! The arrow means gives: HOCH two COON a, sodium hydroxyethanoate, or glycolate. One end is an alcohol group, the other end the salt of an acid. See?

4:51Example 1: which aldehydes give the Cannizzaro reaction

Time for Example one, an easy one. Which of these undergo the Cannizzaro reaction: ethanal, methanal, benzaldehyde, propanal, and two, two dimethylpropanal? Pause the video and try it first. Done? Okay, let's check. Ethanal is C H three CHO, and the carbon next to the CHO group is a C H three, with three hydrogens. Propanal is C H three C H two CHO, with two alpha hydrogens on its C H two. So neither gives the Cannizzaro reaction. Right? Now methanal, which is HCHO, has no carbon next to the CHO at all. So no alpha hydrogen is possible, and methanal qualifies. Okay? Next comes benzaldehyde. Its CHO group sits on a ring carbon, which is bonded to two other ring carbons and to the CHO, with no hydrogen left. So benzaldehyde qualifies too. Clear? Now the last one, two, two dimethylpropanal. It is C H three in round brackets, with a small three outside, then C, a short dash, and the aldehyde group, CHO, okay? Look at all those hydrogens, nine of them on the methyl groups! So it surely has alpha hydrogens, and it is out, right? Oh no, careful, because that is exactly the trap! The alpha carbon is the central carbon next to the CHO, and it carries three methyl groups and no hydrogen. All nine hydrogens sit one carbon further away. So it does give the Cannizzaro reaction. With me? So the answer is methanal, benzaldehyde, and two, two dimethylpropanal. Always look at the carbon next to the CHO group, not at the total number of hydrogens. Got it?

6:45Example 2: product masses from 10.6 g benzaldehyde

Now Example two, a little harder, with numbers. Ten point six grams of benzaldehyde undergoes the Cannizzaro reaction completely with N a O H, sodium hydroxide. Find the masses of benzyl alcohol and sodium benzoate formed. Pause the video and try it first. Done? Okay, let's check. Capital M stands for molar mass, and for benzaldehyde, C six H five CHO, it is one hundred and six grams per mole; the small minus one on mol means per mole. So ten point six grams is ten point six over one hundred and six, which gives zero point one mole. Okay? So we get zero point one mole of benzyl alcohol, and zero point one mole of sodium benzoate, right? Oh no, careful, remember the equation! Two molecules of the aldehyde give one alcohol and one salt. So zero point one over two gives zero point zero five mole of each product. With me? Now benzyl alcohol, C six H five C H two O H, has a molar mass of one hundred and eight. Zero point zero five times one hundred and eight gives five point four grams. Clear? And sodium benzoate, C six H five COON a, has a molar mass of one hundred and forty four. Zero point zero five times one hundred and forty four gives seven point two grams. So, five point four grams of benzyl alcohol and seven point two grams of sodium benzoate. Simple, right?

8:23Quick revision: no alpha H, hydride shift, crossed reaction

So, quick revision. An aldehyde with no alpha hydrogen, in concentrated alkali, gives the acid salt and the alcohol. Hydroxide adds first, then a hydride ion moves, and that transfer is the slow step. And in the crossed reaction, methanal becomes the methanoate, the other aldehyde the alcohol. Now go and try the practice problems on this page. All the best!