Identifying an Unknown Compound

Transcript

0:00Introduction

Hello everyone! Let us start with a small lab puzzle. On your bench there is one bottle with no label, and a whole rack of reagents: litmus, ferric chloride, Tollens' reagent, and many more. The question says: identify the compound. So which reagent do you pick up first? It is tempting to start anywhere, and then the results soon get confusing. Today we learn a fixed order of tests, five stations, and then how the molecular formula takes us from the tests to the actual structure. Questions that give test results and a molecular formula come up often in JEE practical chemistry, so let's begin.

0:41Station 1: is it a carboxylic acid?

So, here is the plan. We use a fixed order of tests, and we will walk through it station by station. Station one asks a simple question: is it an acid? Touch blue litmus paper with the compound. If the blue litmus turns red, the compound is acidic. Okay? So the litmus turned red. That means we have a carboxylic acid, right? That is not quite right, and it is the first trap of today. Phenols also turn blue litmus red. So litmus alone only tells us that the compound is acidic. It cannot tell an acid from a phenol. Clear? That is why station one also uses sodium hydrogencarbonate, N a HCO three on screen, where the small three written low counts the oxygen atoms. Brisk effervescence of carbon dioxide, C O two, means a carboxylic acid group, COOH, and the little dash in front of it is its bond to the rest of the molecule. Phenols do not fizz, so this test separates the two. Got it?

1:43Station 2: is it a phenol?

Station two asks: is it a phenol? Add neutral ferric chloride, F e C l three, where the small three counts the chlorine atoms. With a phenol you get a violet colour, and, as the brackets on screen say, sometimes blue, green or red instead. Right? So that colour means a phenolic O H group. But a compound with an enolic O H group gives it too, so keep that condition in mind. With me?

2:11Station 3: 2,4-DNP, then Tollens' reagent

Station three is for the carbonyl group, C double bond O, and the two little lines on screen are the double bond. The reagent is two four DNP, short for two four dinitrophenylhydrazine. With an aldehyde or a ketone, it gives a yellow, orange or red precipitate. Okay? Only after that comes Tollens' reagent, the second half of station three. With an aldehyde, Tollens' reagent gives a silver mirror. Got it? Now suppose two four DNP gave a precipitate, but Tollens' reagent gave no silver mirror. So there is no carbonyl group after all, right? That is the trap, so be careful here. The precipitate with two four DNP has already proved the carbonyl group. No silver mirror only tells us that it is not an aldehyde, so the compound is a ketone. Clear?

3:01Stations 4 and 5: alcohol, then amine

Station four is the alcohol test, with ceric ammonium nitrate. This yellow reagent turns red, wine red, with an alcohol. Right? And station five is for amines. An amine that is insoluble in water dissolves in dilute hydrochloric acid, H C l, because it forms a salt. With me? Then comes the carbylamine test. Warm the compound with chloroform, CHC l three, and with alcoholic KOH, potassium hydroxide. The foul smell of an isocyanide means a primary amine. Only primary amines give this test, so that is its condition. Okay?

3:39Topper's trick: the same order every time

So here is the topper's trick, the whole order in one line, and each arrow on screen means next. Acid, then phenol, then carbonyl, aldehyde or ketone, then alcohol, then amine. The same order, every time. Got it? And here is the key idea. One positive test fixes the group. Then the molecular formula fixes the structure: count the carbons, hydrogens and oxygens, and find the structure that fits. Let us see it in action now.

4:09Example 1: C2H4O2 as acetic acid and methyl formate

Time for Example one. Here, a liquid with formula C two H four O two gives effervescence with sodium hydrogencarbonate. Its isomer, with the same molecular formula, has a fruity smell and is neutral. Name both. Pause the video and try it first. Done? Okay, let's check. Effervescence with sodium hydrogencarbonate means station one fires straight away. So the liquid is a carboxylic acid. Okay? Now the formula fixes the structure. The COOH group uses one carbon, two oxygens and one hydrogen. That leaves one carbon and three hydrogens, which is a methyl group, C H three. So the liquid is C H three COOH, acetic acid. Right? Now the isomer. It has the same formula, but it is neutral, and it has a fruity smell. Where have we met a fruity smell before? Yes, exactly, it is the smell of an ester. Esters are neutral, and they have a fruity smell, so the isomer is an ester. Clear? With two carbons and two oxygens, only one ester is possible: HCOOCH three, methyl formate. So the answers are acetic acid and methyl formate. Got it?

5:30Example 2: C8H8O as acetophenone and its aldehyde isomer

Now a harder one. This time, a compound C eight H eight O gives a precipitate with two four DNP, and a yellow precipitate with iodine and alkali, but no silver mirror. Its isomer gives a silver mirror and no iodoform. Identify both. Pause the video and try it first. Here is a hint: go station by station. Done? Okay, let's check. There is a precipitate with two four DNP, so station three says a carbonyl group. There is no silver mirror with Tollens' reagent, so it is not an aldehyde, and the compound is a ketone. Okay? Now the extra clue: a yellow precipitate with iodine and alkali. What does that tell you about this ketone? This is the iodoform test: iodine, I two, with sodium hydroxide, and the slash on screen just means with. Aldehydes and ketones with the C H three C O group give yellow iodoform, CHI three, and the dash at the end is the bond to the rest of the molecule. Acetic acid has that group too, but it gives no iodoform. So we have a methyl ketone. Right? Now the formula fixes the structure. Take the C H three C O group away from C eight H eight O, and what is left is C six H five, a phenyl group, which is a benzene ring. So the compound is a phenyl group joined to the COCH three group, and that is acetophenone. Clear? Now the isomer. It has the same formula, but it gives a silver mirror, and no iodoform. What kind of compound is it? Yes, exactly, a silver mirror means an aldehyde. And no iodoform means there is no C H three C O group. With me? So one answer is phenylethanal: a benzene ring, then a C H two group, then the aldehyde group, CHO, at the end. Count it, and you get eight carbons, eight hydrogens and one oxygen. The bracket on screen adds that a methylbenzaldehyde, with a methyl group on the ring, also fits. Neither of them has the C H three C O group, so neither gives iodoform. See how the stations and the formula did all the work?

7:51Quick revision: test order, group, then structure

So, quick revision. Work in a fixed order: acid, phenol, carbonyl, with Tollens' reagent right after two four DNP, then alcohol, then amine. One positive test fixes the group. And the molecular formula then fixes the structure. Now go and try the practice problems on this page. You have all the tools now. All the best!