Set 1

Transcript

0:00Introduction

Hello everyone! Let us start with a little detective story. We have an unknown compound, C nine H ten O, okay? Clue number one: with two four DNP, it gives a yellow, orange or red precipitate. Clue number two: with Tollens' reagent, there is no silver mirror. Clue number three: there is no iodoform either. And the last clue: a reduction called the Clemmensen reduction turns it into propylbenzene. Can four small clues really pin down one exact structure? They can, and by the end you will name it yourself. This is Set one of the mixed problems on organic compounds containing oxygen. Questions like these come up often in JEE Main, and the secret is to know which test catches which group. So, let's begin.

0:50Tool 1: which test catches which group?

So, tool one: which test catches which group. First, two four DNP, short for two four dinitrophenylhydrazine; the comma just separates the positions two and four. It gives a yellow, orange or red precipitate with any aldehyde or ketone. So it tells you aldehyde or ketone, but not which one. Okay? Next, the silver mirror with Tollens' reagent. It is given by all aldehydes, aliphatic and aromatic. And here is a surprise: methanoic acid, HCOOH, gives it too, even though it is an acid. Ordinary ketones do not give the mirror. Right? Now the iodoform test, with iodine, I two, plus sodium hydroxide, N a O H; the small two written low counts the atoms. It needs a C H three C O group, or a C H three CHOH group, where the O H in round brackets hangs on that carbon. The dash at the end is the bond to the rest of the molecule, and it must lead to a carbon or a hydrogen, the C or H on screen. Then you see yellow iodoform, CHI three. Clear? Now question seven of this set. Methanal, benzaldehyde, propanone, methanoic acid, ethanoic acid and two methylpropanal. How many of them give a silver mirror with Tollens' reagent? If you said three, you missed one, so be careful! Methanal, benzaldehyde and two methylpropanal are the three aldehydes; aromatic benzaldehyde counts too. But methanoic acid also gives the mirror, remember? So three aldehydes plus methanoic acid gives four. Propanone, a ketone, and ethanoic acid give nothing. Got it?

2:41Tool 2: count the silver, and the Clemmensen reduction

Now tool two, for the numericals. Here is the Tollens' equation. RCHO, the aldehyde, where R is the rest of the molecule, plus two silver complex ions: in the square brackets, silver, A g, holds two ammonia molecules, N H three, and the plus written high is the charge. Plus three hydroxide ions, O H minus. The arrow gives RCOO minus, the carboxylate ion, then two silver atoms, four ammonia and two water molecules. With me? So, look at the right side. One mole of aldehyde gives how many moles of silver? That is correct if you said two moles, because the equation has two silver atoms on the right. So one mole of aldehyde gives two moles of silver, and m o l is short for mole. Remember this factor of two, okay? Example one depends on it. And one reduction we need for Example two, the Clemmensen reduction. Zinc amalgam, written Z n, dash, H g, for zinc and mercury, with concentrated hydrochloric acid, H C l. On screen, the equals sign between C and O is the double bond, the little angle in front of each carbon stands for its two other bonds, and the arrow means becomes. So the carbonyl group becomes a C H two group, and the carbon skeleton stays the same. Clear?

4:08Example 1: identify an aldehyde from 4.32 g of silver

Time for Example one, which is question fourteen of this set. One point one six grams of a saturated aliphatic aldehyde deposits four point three two grams of silver with excess Tollens' reagent. Identify the aldehyde. The atomic mass of silver is one hundred and eight. Pause the video and try it first. Done? Okay, let's check. Moles of silver first; small n of A g means the number of moles of silver. Four point three two over one hundred and eight gives zero point zero four mole. Right? Now, many students write zero point zero four moles of aldehyde too. Is that correct? No, no, wait, remember the factor of two! Each mole of aldehyde gives two moles of silver. So the moles of aldehyde are zero point zero four over two, which gives zero point zero two mole. Okay? So the molar mass, capital M, is one point one six over zero point zero two, which gives fifty eight. The unit, g mol to the power minus one, means grams per mole. Clear? Now, every saturated aliphatic aldehyde fits the formula C n H two n O, with n carbons and two n hydrogens. Its molar mass is twelve times n, plus two times n, plus sixteen, which is fourteen n plus sixteen. Set it equal to fifty eight, and the double arrow on screen means so: n comes out as three. With me? Three carbons, with the CHO group at the end, is propanal: C H three C H two CHO, okay? So the aldehyde is propanal. Got it?

5:52Example 2: identify C9H10O from three tests and Clemmensen

Now Example two, question ten of this set, and it is our detective story from the start. We have a compound, C nine H ten O, okay? It gives a two four DNP derivative, but no silver mirror, and no iodoform. Clemmensen reduction gives propylbenzene. Identify it. Pause the video and try it first. Here is a hint: use one clue at a time. Done? Okay, let's check, one clue at a time. The two four DNP derivative tells us the compound is an aldehyde or a ketone. Right? Next, there is no silver mirror, and Tollens' reagent catches every aldehyde. So it is not an aldehyde; it must be a ketone. Okay? Now, Clemmensen reduction gives propylbenzene, C six H five C H two C H two C H three: a benzene ring, C six H five, carrying a chain of three carbons. Clemmensen only turns the carbonyl group into C H two, so our ketone has this same skeleton. With me? So now tell me. Where in that chain of three carbons can the C double bond O group sit, and still give a ketone? At the far end of the chain it would be an aldehyde, and that is already ruled out. So it sits in the middle, or next to the ring. Take the middle first: C six H five C H two COCH three. That has a C H three C O group, so it is a methyl ketone, and it would give iodoform. But our compound gives none, so it is ruled out. Clear? So the C O group sits next to the ring: C six H five COCH two C H three. It is a ketone, so two four DNP is positive and there is no silver mirror. Its C O group carries an ethyl group, not a methyl group, so there is no iodoform. And Clemmensen turns it into propylbenzene. Its name is one-phenylpropan-one-one, also called propiophenone, and our detective story is solved. Got it?

8:05Quick revision: 2,4-DNP, Tollens, iodoform, 2 mol Ag

So, quick revision. Two four DNP says aldehyde or ketone, and Tollens' says aldehyde, or methanoic acid. Iodoform needs C H three C O, or C H three CHOH, on a carbon or a hydrogen. And in Tollens' numericals, one mole of aldehyde gives two moles of silver. You solved a full detective problem today, so well done! Now go through the other questions of this set on this page, and try each one before you read the solution. All the best!