The Carbonyl Group

Transcript

0:00Introduction

Hello everyone! Let us start with a small lab puzzle. Two bottles have lost their labels. One holds propanal, and the other holds propanone. Both have the carbonyl group, a carbon joined to an oxygen by a double bond. So, with a few reagents and test tubes, how will you find out which bottle is which? We will solve it in Example one. Tests like these come up often in JEE practical chemistry, so let's begin.

0:28Is a carbonyl group present? The 2,4-DNP test

So, the first question in the lab: is there a carbonyl group at all? For that we use two four DNP, short for two four dinitrophenylhydrazine, also called Brady's reagent. With an aldehyde or a ketone, it gives a yellow, orange or red precipitate. Okay? That coloured solid is the two four dinitrophenyl hydrazone. Say the long name slowly once, and write it down in your notes, all right? Now let me ask you this. Ethyl acetate, an ester, also has the C, equals sign, O in the box, a carbon oxygen double bond. So will it give the precipitate? It does not, and many students get this wrong. The carbon oxygen double bond of acids and esters does not react, so only aldehydes and ketones give the precipitate.

1:18Two more tests for aldehydes and ketones

Clear? Next, sodium hydrogensulphite, N a HSO three, where the small three written low counts oxygen atoms. It gives a white crystalline addition compound, but only with aldehydes and methyl ketones, which carry a methyl group on the carbonyl carbon. With me? Now the iodoform test. We add iodine, I two, with sodium hydroxide, N a O H, and the slash means together. The positive result is a yellow precipitate of iodoform, CHI three. Okay? But this test needs one special group: C H three C O, with a dash after it, meaning it is joined to the rest of the molecule. Most aldehydes do not have it. So which aldehyde does? Yes, exactly, it is ethanal. In ethanal, C H three CHO, the methyl group sits right on the carbonyl carbon. So ethanal and methyl ketones give iodoform. Got it?

2:19Aldehydes only: Tollens' reagent

Now the tests that only aldehydes pass. First, Tollens' reagent: silver nitrate, A g N O three, in ammonia, which we call ammoniacal silver nitrate. The square brackets hold one complex ion: silver, A g, with two ammonia molecules, N H three, and the small plus written high is its charge. With an aldehyde, it gives the famous silver mirror. Okay? Here is the reaction. One aldehyde, RCHO, where R is the rest of the molecule, meets two silver complex ions and three hydroxide ions, O H minus. The arrow leads to the carboxylate ion, RCOO minus, two atoms of silver, four ammonia and two water. So the aldehyde is oxidised and silver is reduced. Right? Look at the equation again. One mole of aldehyde gives how many moles of silver? It gives two moles, exactly as the equation says. Each silver goes from plus one to zero by taking one electron, while one aldehyde gives up two electrons as it becomes the carboxylate. So one mole of RCHO deposits two moles of silver. Keep this two in mind for Example two. Clear?

3:36Aldehydes only: Fehling's solution

Next, Fehling's solution, in two parts. Fehling's A is aqueous copper sulphate, C u S O four, and Fehling's B is alkaline sodium potassium tartrate. Okay? With an aldehyde, we get a red precipitate of C u two O, cuprous oxide, as copper goes from plus two to plus one. With me? In the reaction, one aldehyde takes two copper ions, C u with two plus written high, and five hydroxide ions. It gives the carboxylate ion, one cuprous oxide and three water. Got it? Now a classic trap. Benzaldehyde is an aldehyde, right? So surely it reduces Fehling's solution and gives the red precipitate? It does not, and that is the trap. Aromatic aldehydes do not respond to Fehling's solution, so benzaldehyde fails. But it still gives the silver mirror with Tollens' reagent. Remember this pair. Okay?

4:35Schiff's reagent, and a plan for the lab

The last aldehyde test is Schiff's reagent: magenta whose colour has been removed by sulphur dioxide, S O two. Add an aldehyde, and the pink colour comes back, so we say it is restored. Right? And remember this line: ketones give none of these three. No silver mirror, no red precipitate, and no pink colour. Clear? Now a trick toppers use for an unknown compound. First, two four DNP: is there a carbonyl group? Then Tollens': is it an aldehyde? Then iodoform: is the C H three C O group there? In this order, your answer stays organised. Got it?

5:16Example 1: telling propanal from propanone

Now, remember our two bottles? This is Example one. Propanal and propanone are given. State one test positive only for propanal, and one positive only for propanone. Pause the video and try it first. Done? Okay, let's check. Many students first reach for the two four DNP test. Will it tell these two apart? It will not, because both are carbonyl compounds and both give the precipitate. Sodium hydrogensulphite fails too, since both give the white solid: propanal is an aldehyde and propanone is a methyl ketone. Right? So for propanal, C H three C H two CHO, use a test for aldehydes only: the silver mirror with Tollens' reagent, or Fehling's, or Schiff's. Propanone is a ketone, so it gives none of them. Okay? And for propanone, C H three COCH three, use the iodoform test. It has the C H three C O group and gives yellow iodoform, while propanal does not have this group. So our bottle puzzle is solved: the silver mirror picks out propanal, and the yellow precipitate picks out propanone.

6:30Example 2: butanal from the mass of silver deposited

Clear? Now Example two, a harder one, with numbers. Zero point three six grams of a saturated aliphatic aldehyde deposits one point zero eight grams of silver from Tollens' reagent. Find its molar mass, and name the straight chain aldehyde. Silver, A g, has atomic mass one hundred and eight. Pause the video and try it first. Done? Okay, let's check. First, n of A g, the number of moles of silver: one point zero eight over one hundred and eight gives zero point zero one zero mole. Okay? Now, many students write that the aldehyde is also zero point zero one zero mole. Is that correct? It is not, so remember the two from the Tollens' equation. One mole of RCHO gives two moles of silver, so the aldehyde is half of that, which is zero point zero zero five mole. With me? The molar mass, capital M, is mass over moles: zero point three six over zero point zero zero five gives seventy two. The unit, g mol to the power minus one, is read grams per mole. Right? Every saturated aliphatic aldehyde has the formula C n H two n O, where the small n and two n count the carbons and hydrogens. Its mass is twelve times n, plus two times n, plus sixteen, which is fourteen times n, plus sixteen. Set that equal to seventy two, and the double arrow gives n equals four. Clear? That means the molecular formula is C four H eight O, with four carbons. Four carbons in a straight chain, ending in the CHO group, make butanal. So the molar mass is seventy two grams per mole, and the aldehyde is butanal. Very good if you remembered the two!

8:26Quick revision: 2,4-DNP, Tollens', Fehling's, iodoform

So, quick revision. Two four DNP gives a yellow, orange or red precipitate with aldehydes and ketones. Aldehydes only: Tollens' silver mirror, Fehling's red cuprous oxide, and Schiff's pink colour. And the C H three C O group, in ethanal and methyl ketones, gives yellow iodoform. Now try the four practice problems on this page. All the best!