Aniline Yellow and Iodoform

Transcript

0:00Introduction

Hello everyone! Let us start with a small lab puzzle. We add a diazonium salt to aniline, hoping for a dye called aniline yellow. Sure enough, a yellow compound appears. But is that yellow compound really our dye? Keep that question in mind, because the answer is a lovely trap. Today we prepare two yellow compounds, aniline yellow, a dye, and iodoform, an antiseptic, with their conditions, equations and yields. Both preparations are part of JEE practical chemistry, so let's begin.

0:35Aniline yellow: diazotise, then couple

So, aniline yellow first, and step one is diazotisation. We treat aniline with sodium nitrite, N a N O two, where the small two written low counts the oxygen atoms, and with hydrochloric acid, H C l, as well. The temperature stays at zero to five degrees Celsius, and the little circle means degrees. Okay? This gives the diazonium salt, on the left. The small plus written high is a positive charge, and the small minus after C l is the chloride ion. It reacts with a second aniline, and the arrow means gives. On the right is aniline yellow: a benzene ring, then two nitrogens joined by a double bond, the double line, then a second ring carrying an amino group, N H two. The single dashes are single bonds, and the other product is hydrogen chloride, H C l, as you see. Right? Its other name is para aminoazobenzene, and the small italic p on screen stands for para. The N double bond N group is the azo group, so aniline yellow is an azo dye. Its molar mass, M, is one hundred and ninety seven. Got it? The coupling step is an electrophilic substitution, and the diazonium ion is the electrophile, but only a weak one. Keep that word, weak, in mind. With me?

1:52Where does the coupling happen?

Now, back to our puzzle. Aniline has an electron-rich ring, and also a lone pair on the nitrogen of its amino group. So where does the diazonium ion attack first: the para carbon of the ring, or the amino nitrogen? Most students say the ring, but the first attack is on the nitrogen. In the weakly acidic solution, the most available electron pair is the lone pair on the amino group, N H two, where the dash in front is its bond to the ring. So this nitrogen attacks the diazonium ion faster than the ring carbon does. Clear? So the first product is diazoaminobenzene. Look at its formula: the azo group is joined to the second ring through an N H group, so through a nitrogen, not a carbon. And the bracket says it is yellow. So the yellow compound from our puzzle is not aniline yellow at all! Okay? To get the dye, we warm diazoaminobenzene at forty to forty five degrees Celsius, with aniline and a little aniline hydrochloride. Right? Now it rearranges to para aminoazobenzene, our aniline yellow, so the coupling is now on the para carbon. Both compounds have the same formula, C twelve H eleven N three, so they are isomers. Got it? One more condition. We said the solution is weakly acidic. So why must the coupling medium not be strongly acidic? Think about what strong acid does to aniline. In strong acid, aniline is completely protonated, and it becomes the anilinium ion. The anilinium ion has no lone pair, and its ring is deactivated. So it does not couple at all. With me?

3:29Iodoform, CHI₃, from acetone and ethanol

Now our second compound, iodoform, CHI three. We start from acetone. Acetone, plus three iodine molecules, I two, plus four sodium hydroxide, gives iodoform, sodium acetate, three sodium iodide and three water. Okay? Here is what happens inside. Acetone has the C H three C O group, and the dash at the end is its bond to the rest of the molecule. The three hydrogens of this group are replaced by iodine. Then the alkali splits off the C I three group, and that group becomes iodoform. With me? Now take ethanol, ordinary alcohol. It has no carbonyl group at all, no C double bond O, as the badge says. So it cannot give iodoform, right? No, no, careful, that is the trap, because ethanol does give iodoform, and here is its equation. Ethanol, plus four iodine, plus six sodium hydroxide, gives iodoform, sodium formate, five sodium iodide and five water. Four iodine and six alkali, not three and four as with acetone, so do not copy the acetone numbers. Clear? Here is how ethanol manages it. Hypoiodite, made from iodine and alkali, first oxidises ethanol to ethanal, which has the C H three C O group. Then the same steps follow as for acetone, and the extra iodine and alkali go into that first oxidation. Got it?

4:52Iodoform in the lab: method and properties

Now the lab method. We add iodine to acetone, or to ethanol, with aqueous alkali, and we warm the mixture. Okay? We keep warming until the yellow solid separates. Then we filter it off, and recrystallise it from ethanol. Right? Iodoform forms yellow hexagonal crystals with a characteristic smell. Its melting point, written m p on screen, is one hundred and nineteen degrees Celsius, and its molar mass is three hundred and ninety four. Clear? And iodoform is used as an antiseptic, because it slowly releases iodine. Got it?

5:29Example 1: mass of aniline yellow from 4.65 g of aniline

Time for Example one. Find the theoretical mass of aniline yellow from four point six five grams of aniline that is diazotised, and then coupled with excess aniline. Pause the video and try it first. Done? Okay, let's check. Aniline, C six H five N H two, has molar mass ninety three. So four point six five over ninety three, where the slash means divided by, gives zero point zero five moles. Right? Now careful here. Each molecule of aniline yellow is made from two aniline units. So should we halve the moles, and use zero point zero two five? No, we should not halve them. The four point six five grams is the aniline that gets diazotised, and the second aniline is in excess. So one mole of diazotised aniline gives one mole of dye. Okay? So the mass is zero point zero five moles times one hundred and ninety seven grams per mole, and the minus one written high means per mole. That gives nine point eight five grams of aniline yellow. Clear?

6:36Example 2: iodoform yield and iodine mass from 2.9 g acetone

Now a slightly harder one. Two point nine grams of acetone is converted into iodoform. Find the theoretical yield, and the mass of iodine needed. Pause the video and try it first. Here is a hint: use the balanced equation for acetone. Done? Okay, let's check. Acetone has molar mass fifty eight, so two point nine over fifty eight gives zero point zero five moles. Got it? From the equation, one mole of acetone gives one mole of iodoform. So zero point zero five times three hundred and ninety four gives nineteen point seven grams. That is the theoretical yield. Okay? Now the iodine. Many students write zero point zero five moles here too. Is that right? Look at the acetone equation once more. That is not right, because the equation has three I two for every acetone. So we need three times zero point zero five, which gives zero point one five moles of iodine. Right? And iodine, I two, has molar mass two hundred and fifty four. So zero point one five times two hundred and fifty four gives thirty eight point one grams of iodine. Clear?

7:49Quick revision: diazotise, couple on N, iodoform, yields

So, quick revision. Aniline yellow: diazotise at zero to five degrees Celsius. Coupling happens on the nitrogen first, and the product rearranges at forty to forty five degrees. Iodoform: the C H three C O group with iodine and alkali, and ethanol is first oxidised to ethanal. And for yields, take the moles from the balanced equation. Now go and try the practice problems on this page. All the best!