Reduction and Oxidation

Transcript

0:00Introduction

Hello everyone! Let us start with some lab magic. Take ammoniacal silver nitrate in a test tube, add an aldehyde, and warm it gently. The inside of the tube turns into a shining silver mirror! Now do the same with a ketone. Nothing happens. Both have a carbonyl group, so why does only one react? Interesting, right? Today we answer that, and we also learn to reduce the carbonyl group, to an alcohol or right down to a hydrocarbon. These reagents and tests come up often in JEE, so let's begin.

0:37Reduction to alcohols

Let us start with reduction to alcohols, using any of three reagents. First, N a B H four, sodium borohydride; the small four written low counts the hydrogen atoms. Second, L i A l H four, lithium aluminium hydride. Third, hydrogen gas, H two, with nickel, platinum or palladium, written N i, P t and P d. Okay? Now tell me. An aldehyde and a ketone: which kind of alcohol does each one give, primary, secondary or tertiary? Yes, exactly, aldehydes give primary alcohols, and ketones give secondary alcohols. We will use this pair in Example one. Clear?

1:23Down to a hydrocarbon: Clemmensen and Wolff–Kishner

Now, sometimes we want the carbonyl group to become a plain C H two group, giving a hydrocarbon. 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. With me? Method one is the Clemmensen reduction. Zinc amalgam, written Z n, dash, H g, for zinc and mercury, with concentrated hydrochloric acid, H C l. It works in strong acid, so it is not for acid-sensitive groups. Okay? Method two is the Wolff-Kishner reduction. First hydrazine, then KOH, potassium hydroxide, in ethylene glycol, with heating. So this one works in hot base. Right? Now suppose your molecule also carries a chlorine atom, a halogen. Which method could damage it? That is correct if you said Wolff-Kishner. Hot base can eliminate or replace a halogen, so Wolff-Kishner is not for base-sensitive groups. So the rule is: acid-sensitive groups, choose Wolff-Kishner; base-sensitive groups, choose Clemmensen. Got it?

2:34Oxidation: aldehydes easily, ketones only by force

Now the other direction, oxidation. An aldehyde, RCHO, where R is the rest of the molecule, becomes RCOOH, a carboxylic acid, with the same number of carbons. And even mild oxidising agents can do this. Okay? So, what about a ketone? Will a mild oxidising agent work on a ketone as well? No, it will not, because ketones resist oxidation. Only under vigorous conditions does a carbon carbon bond break; that is the dash between C and C on screen. Right? And when that bond does break, we get acids with fewer carbons than the ketone. The C double bond O group stays mainly with the smaller alkyl group. Clear?

3:20Tollens' test: ammoniacal silver nitrate

Now the famous tests. Tollens' reagent is ammoniacal silver nitrate, meaning silver nitrate in ammonia. Read the equation with me. RCHO, plus two silver complex ions: in the square brackets, silver, A g, holds two ammonia molecules, N H three in round brackets with a small two outside, and the plus written high is the charge. Plus three O H minus, hydroxide ions. They give RCOO minus, the carboxylate ion, two silver atoms, four ammonia and two water molecules. Okay? Those silver atoms coat the glass as the silver mirror. Tollens' test is given by all aldehydes, aliphatic and aromatic, so benzaldehyde gives it too. With me? Two more give it too: alpha hydroxy ketones, where alpha, the Greek letter on screen, means the O H group sits on the carbon next to the carbonyl group, and methanoic acid. But ordinary ketones do not, and that solves our opening puzzle: the aldehyde is oxidised, the ketone resists. Okay?

4:29Fehling's test: a red-brown precipitate

Next, Fehling's solution, which comes in two parts. Fehling's A is aqueous copper sulphate, C u S O four. Fehling's B is alkaline sodium potassium tartrate, called Rochelle salt. Right? Here is the equation. RCHO, plus two copper ions, C u two plus, with the two plus written high, plus five hydroxide ions, give RCOO minus, plus C u two O, cuprous oxide, plus three water molecules. Clear? Now here comes a trap. Benzaldehyde gave the silver mirror with Tollens' reagent. So with Fehling's solution, it will surely give the precipitate too, right? No, no, careful, aromatic aldehydes do not respond to Fehling's test. The red-brown precipitate of C u two O comes from aliphatic aldehydes. So among aldehydes, Tollens' catches them all, while Fehling's catches only the aliphatic ones. Got it? Benedict's solution uses citrate in place of tartrate, and behaves in a similar way to Fehling's. Okay?

5:35Schiff's reagent, haloform and the ring

Three quick points to finish. With Schiff's reagent, aldehydes restore its pink colour. Right? Next, the haloform reaction. Methyl ketones, and also ethanal, are oxidised by N a O X, sodium hypohalite, where X stands for a halogen. They give the acid with one carbon fewer. Clear? And last, the benzene ring of benzaldehyde. The CHO group deactivates the ring and directs meta, so nitration gives three nitrobenzaldehyde. With me?

6:07Example 1: acetophenone by Clemmensen and by NaBH4

Time for Example one, an easy one. Name the product when acetophenone is treated with, part a, zinc amalgam and concentrated H C l, and part b, sodium borohydride. Pause the video and try it first. Done? Okay, let's check. Acetophenone is C six H five COCH three: a phenyl group, C six H five, and a methyl group, joined to a carbonyl group. So it is a ketone. Right? Part a uses zinc amalgam and concentrated H C l. Does it stop at an alcohol, or go all the way down? All the way down, yes, exactly! This is the Clemmensen reduction, so the carbonyl group becomes C H two. We get C six H five C H two C H three, which is ethylbenzene. Okay? Part b, sodium borohydride, stops at the alcohol. Acetophenone is a ketone, so we get a secondary alcohol: C six H five C H, then O H in brackets, then C H three. Its name is one-phenylethanol. Clear?

7:22Example 2: silver from 0.44 g ethanal with Tollens' reagent

Now Example two, a numerical one. Find the mass of silver deposited when zero point four four grams of ethanal is oxidised completely by Tollens' reagent. Silver's atomic mass is one hundred and eight. Pause the video and try it first. Here is a hint: use the Tollens' equation. Done? Okay, let's check. Ethanal, C H three CHO, has a molar mass, capital M, of forty four: two carbons at twelve, four hydrogens at one, and an oxygen at sixteen. The unit, g mol to the power minus one, means grams per mole. So the moles, small n, are zero point four four over forty four, which gives zero point zero one mole. Right? Now the key step. In the Tollens' equation, one mole of aldehyde gives how many moles of silver? Two moles, yes, exactly, because the equation had two silver atoms on the right. So zero point zero one mole of ethanal gives zero point zero two mole of silver. Clear? So the mass of silver is zero point zero two times one hundred and eight, which gives two point one six grams. Forget that factor of two, and you get exactly half the answer, which is a classic trap. Got it?

8:44Quick revision: reduction, oxidation, Tollens vs Fehling

So, quick revision. Reduction: sodium borohydride or lithium aluminium hydride give alcohols, and Clemmensen in acid, or Wolff-Kishner in base, take the carbonyl group down to C H two. Oxidation: aldehydes go easily and keep their carbons; ketones need vigorous conditions. Tests: Tollens' for all aldehydes, Fehling's for aliphatic aldehydes but not aromatic ones. Now go and try the practice problems on this page. All the best!