Iodoform and the Haloform Reaction

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Picture two test tubes, one with methanol and one with ethanol, two simple alcohols. We warm each one with iodine and sodium hydroxide. In one tube, a yellow solid appears, with a characteristic smell. The other shows nothing. So which alcohol gave the yellow solid, and why only that one? Interesting, isn't it? That yellow solid is iodoform, and today we learn the haloform reaction behind it. This test comes up often in JEE papers. So, let's begin.

0:36The group the test looks for

So, what does this test look for? On screen is C H three, then C O, and each dash is a bond. The small three written low counts the hydrogens. So it is a methyl group joined to a C O, a carbonyl group, and on its other side there must be a carbon or a hydrogen, okay? Now, an alcohol has no C O group at all. So can an alcohol ever pass this test? Yes, some alcohols can, and here is how. Look at this group: C H three, then C H, with O H in brackets as a side group. It can be oxidised to C H three C O, so a compound with this group passes too, got it? Now the reagent: a halogen, X two, where X stands for a halogen atom, and sodium hydroxide. Or simply sodium hypohalite, written as sodium, then O X, clear?

1:31The haloform reaction

Now the full equation, where R stands for the rest of the molecule. RCOCH three, plus three X two, plus four sodium hydroxide, with me? The arrow means gives. We get RCOO sodium, the sodium salt of an acid, plus CHX three, the haloform, plus three sodium X, the sodium halide, and three molecules of water. Now look at both sides. The C H three carbon on the left, where did it go? Yes, exactly, it left as the haloform, CHX three. So the salt has one carbon fewer than the starting compound. That helps you work backwards in questions, right? With iodine, I two, the haloform is CHI three, iodoform, a yellow solid with a characteristic smell. That is the iodoform test, got it?

2:26Who passes the iodoform test?

So, who passes? First, ethanal, C H three CHO, where our group is joined to a hydrogen. It is the only aldehyde that passes, okay? Next, all methyl ketones, RCOCH three, with R a carbon group. Every one of them passes, right? Then ethanol, the only primary alcohol that passes. That is part of our puzzle, and the first example shows why, with me? And secondary alcohols of the type C H three CHOHR, where R is a carbon group. They carry the C H three CHOH group, which is oxidised to a methyl ketone, so they pass. Clear?

3:09Who fails? Watch the traps

Now, who fails? Methanol, propan one ol, the other aldehydes, and tertiary alcohols, okay? Next, ketones without a C H three C O group. Pentan three one is C two H five, C O, C two H five, with an ethyl group on each side. Benzophenone has a phenyl group, C six H five, on each side. No methyl on the C O, so both fail, got it? Now the classic trap. Ethanoic acid also has a C H three C O group. So, does it pass the test? No, it fails, and many students get this wrong. Its C O is joined to an oxygen, not to a carbon or a hydrogen. Its esters and amides fail too. So always check what the C O is joined to, right?

3:59Iodoform as an antiseptic

Now one fact about iodoform itself. It was used as an antiseptic. So tell me, is it iodoform that does the work? No, the action is due to the iodine that iodoform liberates, not to iodoform itself. This point comes up often in questions, clear? And because of its objectionable smell, it has been replaced by other iodine preparations, okay?

4:24Example 1: which alcohols give the iodoform test

Now let us use all this. The first example is an easy one, and it settles our puzzle. Which of these give the iodoform test: ethanol, methanol, propan one ol and propan two ol? Pause the video and try it first. Done? Okay, let's check. Ethanol is C H three C H two O H, so its O H carbon carries a C H three and two hydrogens. That is our C H three CHOH group, with a hydrogen on the free bond. It is oxidised to C H three CHO, ethanal, so ethanol passes, right? Next, propan two ol, C H three CHOHCH three. Its O H is on the middle carbon, next to a C H three, so it is oxidised to C H three COCH three, propanone, and it passes too. Now, methanol and propan one ol. Do they have the C H three CHOH group? No, they do not. Methanol, C H three O H, has only one carbon. In propan one ol, the O H carbon is joined to a C H two, not to a C H three. So both fail, clear? So the answer is ethanol and propan two ol. And our puzzle is solved: the yellow solid came from the ethanol tube. See how simple that was?

5:53Example 2: identify C8H8O and the mass of iodoform formed

Now let us try a harder one. Here is a compound, C eight H eight O, which gives iodoform and sodium benzoate with iodine and alkali. Identify it, and find the mass of iodoform from one point two grams of it. Pause the video and try it first. Done? Okay, let's check. Sodium benzoate is C six H five COO sodium, where C six H five is the phenyl group. So the salt keeps C six H five C O, okay? Now, the salt has one carbon fewer. Which group was lost to make the iodoform? Yes, exactly, it was the C H three group. Put it back, and we get C six H five COCH three, acetophenone, a methyl ketone. Eight carbons, eight hydrogens and one oxygen, so it matches C eight H eight O, right? Now let us find the mass. M, the molar mass, is eight times twelve, plus eight times one, plus sixteen, which gives one hundred and twenty. So one point two grams, over one hundred and twenty, gives zero point zero one mole, with me? Now be careful here. The equation uses three molecules of iodine. So do we get three moles of iodoform from each mole of ketone? No, and that is the trap. One mole of the ketone gives just one mole of the haloform. So we get zero point zero one mole of CHI three, and its molar mass is twelve, plus one, plus three times one hundred and twenty seven for the three iodine atoms, which gives three hundred and ninety four, okay? So the mass of iodoform is zero point zero one times three hundred and ninety four, which gives three point nine four grams. And the compound is acetophenone. Very good if you got both parts!

7:53Quick revision: groups needed, who passes, who fails

So, quick revision. The test needs C H three C O on a carbon or a hydrogen, or C H three CHOH, which is oxidised to it. It gives the haloform, CHX three, and an acid salt with one carbon fewer. Ethanal, methyl ketones, ethanol and C H three CHOHR alcohols pass. Methanol, propan one ol, pentan three one and ethanoic acid fail. Now go and try the practice problems on this page. All the best!