Overview

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Methanoic acid is a carboxylic acid. Yet, with Tollens' reagent, it gives a shining silver mirror, exactly like an aldehyde. So how can an acid behave like an aldehyde? Hold that thought. This chapter gives you three sets of mixed problems from the whole unit, and questions of this kind are typical of JEE Main. So, let us see how to use these sets, and sharpen the tests they need. Let's begin.

0:32How this chapter works

So, what is inside this chapter? Three sets, each with fifteen original problems, in the style of recent JEE Main papers. And each set mixes all the chapters of the unit: alcohols, phenols, ethers, aldehydes and ketones, and carboxylic acids. Okay? Now, the timing. Give yourself twenty five minutes for each set, and only then check the solutions. So, twenty five minutes for fifteen problems: how much time do you get for each one? Yes, one hundred seconds, exactly! Twenty five minutes is fifteen hundred seconds, with s for seconds, and fifteen hundred over fifteen gives one hundred; the division sign means over. So the tests must be ready in your head. Right?

1:18Tool 1: alcohols, phenols and acids

Now, tool one, the tests for alcohols, phenols and acids. Lucas reagent is concentrated H C l, plus anhydrous zinc chloride, Z n C l two; conc. is short for concentrated, and the small two written low is a subscript. Okay? The alcohol dissolves in it, but the alkyl chloride it forms does not, so the solution turns cloudy. We call that turbid. With me? Now tell me: primary, secondary or tertiary, which alcohol turns turbid first? It is the tertiary alcohol, yes, exactly! Tertiary turns turbid at once, secondary within about five minutes, and primary only on heating. The small circle means degree, so three degree is tertiary. Clear? Phenols have their own test. With neutral ferric chloride, F e C l three, a phenol gives a violet colour. Got it? Now, phenol is acidic. So it gives carbon dioxide, C O two, with sodium bicarbonate, N a HCO three, right? No, no, careful, that is a trap! Sodium bicarbonate gives carbon dioxide with carboxylic acids, but not with simple phenols like phenol itself. Only phenols with several nitro groups, like picric acid, are exceptions. So this one test separates an acid from a phenol. Okay?

2:41Tool 2: aldehydes and ketones

Now, tool two, for aldehydes and ketones. Two four DNP, short for two four dinitrophenylhydrazine, gives a yellow, orange or red precipitate with any aldehyde or ketone. So it finds the carbonyl group, but it cannot tell an aldehyde from a ketone. Right? Tollens' reagent gives a silver mirror with all aldehydes, both aliphatic and aromatic. Okay? Now, Fehling's solution also detects aldehydes. So, does benzaldehyde, an aromatic aldehyde, give Fehling's test too? No, it does not, and many students lose marks here! Fehling's solution gives a red-brown precipitate of C u two O only with aliphatic aldehydes. Aromatic aldehydes, like benzaldehyde, do not respond. Got it? Next, the iodoform test, with I two plus N a O H, iodine and sodium hydroxide, in brackets. It gives yellow CHI three, which is iodoform, from the group C H three C O, or C H three C H, O H in brackets. The dash at the end is the bond to the rest of the molecule. Clear? Now, ethanoic acid also has the C H three C O group. So it gives iodoform, right? Oh no, careful, it does not! The group must be joined to carbon or hydrogen only. In ethanoic acid it is joined to an O H group, so there is no yellow precipitate. Ethanol and ethanal do respond. With me?

4:16Example 1: which compounds give a Tollens silver mirror

Now, Example one, a medium problem from Set one. How many of these give a silver mirror with Tollens' reagent: methanal, benzaldehyde, propanone, methanoic acid, ethanoic acid, and two methylpropanal? Pause the video and try it first. Done? Okay, let's check. Methanal, benzaldehyde and two methylpropanal are all aldehydes. Tollens' reagent works with all aldehydes, aromatic too, so benzaldehyde counts. That makes three. Okay? Propanone is a ketone, and ethanoic acid is an ordinary acid, so neither gives a mirror. With me? And methanoic acid? It is an acid, just like ethanoic acid, so no mirror, right? No, no, wait, this is our opening puzzle! Methanoic acid is H, joined to COOH; the short dash is a single bond. So it contains an aldehydic unit, H, C, double bond, O, and the equals sign is that double bond. That makes methanoic acid a reducing agent, and it reduces Tollens' reagent. So our puzzle is solved. Clear? So the answer is four: the three aldehydes, plus methanoic acid. If you wrote three, you fell for the trap. Got it?

5:37Example 2: identify C9H10O using DNP, Tollens, iodoform

Now, Example two, a hard one, also from Set one. We have a compound, C nine H ten O, which gives a two four DNP derivative, but no silver mirror and no iodoform. Clemmensen reduction gives propylbenzene. Identify the compound. Pause the video and try it first. Done? Okay, let's check, one clue at a time. The two four DNP derivative tells us there is a carbonyl group, C double bond O, so it is an aldehyde or a ketone. Okay? Clemmensen reduction uses zinc amalgam, written Z n, dash, H g, with concentrated hydrochloric acid. It turns the C double bond O into C H two, and leaves the carbon skeleton as it is; the arrow means gives, and the angle sign in front of each carbon stands for its two other bonds. Right? So our compound has the skeleton of propylbenzene: C six H five, the benzene ring, joined to C H two, C H two, C H three, a chain of three carbons. Clear? Now think about this: the carbonyl carbon must be one of those three chain carbons. Which position fits all the clues? At the end of the chain, it would be an aldehyde, CHO, and that gives a silver mirror. The question says no mirror, so this one is ruled out. Okay? On the middle carbon, it is a methyl ketone, with the C H three C O group, and that gives iodoform. So that one is ruled out too, right? So the carbonyl group sits right next to the ring: C six H five, C O, C H two, C H three. That is one phenyl propan one one, also called propiophenone. Count it: nine carbons, ten hydrogens and one oxygen, exactly C nine H ten O, so everything fits. With me?

7:40Quick revision: Lucas, FeCl3, NaHCO3, DNP, Tollens, iodoform

So, quick revision. Lucas reagent: tertiary at once, secondary in about five minutes. Phenol gives a violet colour with ferric chloride, but no carbon dioxide with sodium bicarbonate. Two four DNP finds any carbonyl group; Tollens' reagent, all aldehydes and methanoic acid; Fehling's solution, aliphatic aldehydes only. And iodoform needs C H three C O, or C H three CHOH, joined to carbon or hydrogen. Now take Set one, start a twenty five minute timer, and check the solutions only at the end. All the best!