Set 3

Transcript

0:00Introduction

Hello everyone! Let us start with a puzzle from Set three. Take a carboxylic acid with a molar mass of eighty eight. Dissolve it in benzene, measure the molar mass again, and the experiment says one hundred and sixty. So did the molecules suddenly get heavier? Strange, right? Hold that thought. Today we warm up with quick problems from Set three, then solve two JEE Main pattern problems in full, including this one. So, let's begin.

0:30Quick fire: making alcohols, carbonyl reactivity

So, first a quick fire round. With N a B H four, sodium borohydride, an aldehyde gives a primary alcohol, and a ketone gives a secondary alcohol. The small numbers written low count the atoms, the arrow means gives, and the little circle after the number means primary or secondary. Okay? Now, four alcohols share the formula C four H ten O, shown in the badge. How many of them can be made by reducing an aldehyde or a ketone this way? You might say all four of them, right? No, no, careful, only three of them can be made! The two primary alcohols come from aldehydes, and butan two ol, which is secondary, comes from a ketone. The fourth one is tertiary, three with the little circle on screen, and reducing an aldehyde or a ketone never gives a tertiary alcohol. Got it? Next, arrange ethanal, propanone, trichloroethanal and benzaldehyde in decreasing reactivity towards nucleophilic addition. The order is trichloroethanal, ethanal, benzaldehyde, then propanone; the greater than sign means more reactive than. Three chlorine atoms pull electrons away, making that carbonyl carbon the most positive. In benzaldehyde the ring reduces the charge by resonance, and propanone has two alkyl groups, which hinder the approach of the nucleophile and reduce the charge further. Right?

1:59Quick fire: tests and the Cannizzaro reaction

Now, but one yne is warmed with dilute sulphuric acid and mercuric sulphate, H g S O four. Does the product give the iodoform test? And Tollens' test? First the product: water adds across the triple bond the Markovnikov way, through an enol, so every alkyne except ethyne gives a ketone. Here it is butanone, C H three COCH two C H three; the reagents over the arrow are the conditions, and dil. means dilute. Okay? Butanone is a methyl ketone, so the iodoform test is positive. But Tollens' reagent is reduced by aldehydes, and simple ketones like this one do not reduce it. So, yes for iodoform, and no for Tollens'. Clear? And the last one in this round: two moles of benzaldehyde, written two m o l in the badge, undergo the Cannizzaro reaction completely. How many moles of sodium hydroxide are consumed? Is it two moles, one for each mole of aldehyde? Oh no, careful, check the equation first! Benzaldehyde has no alpha hydrogen, so with concentrated alkali one molecule becomes the alcohol, and another becomes the salt of the acid. Two C six H five CHO plus N a O H gives benzyl alcohol, C six H five C H two O H, plus sodium benzoate, C six H five COON a, okay? Only the salt takes up sodium, so two moles of benzaldehyde use just one mole of N a O H, not two. With me?

3:44Two acids, one hydrocarbon?

Now let us turn to acids. Heat the sodium salt of an acid with soda lime, that is sodium hydroxide with calcium oxide, and the COON a group is replaced by hydrogen. With R for the alkyl group, RCOON a gives R H, okay? Here is a hard one from Set three. The sodium salts of three methylbutanoic acid and two two dimethylpropanoic acid are heated separately with soda lime. Two different acids, so two different hydrocarbons, surely? What do you think? Let us check. The first salt is C H three bracket two, C H, C H two, COON a; the small two after the bracket means two methyl groups. Replace COON a by H, and we get C H three bracket two, C H, C H three. Right? The second salt is C H three bracket three, C, COON a. The same swap gives C H three bracket three, C H, a carbon with three methyl groups and one hydrogen. With me? Now look closely: the first product is also a carbon with three methyl groups and one hydrogen. So both give the same molecule, two methylpropane. Different acids, same hydrocarbon. Clear?

5:06Example 1: mass of ether from 9.2 g ethanol at 50% yield

Now, Example one, a medium problem from Set three. Nine point two grams of ethanol is heated with concentrated sulphuric acid at four hundred and thirteen kelvin; g means grams, and capital K means kelvin. If the yield is fifty percent, find the mass of ethoxyethane formed. Pause the video and try it first. Done? Okay, let's check. At four hundred and thirteen kelvin, with the alcohol in excess, ethanol gives the ether, not ethene. Two C two H five O H gives C two H five O C two H five, plus water, H two O, okay? The molar mass, capital M, of ethanol is forty six, so nine point two over forty six gives zero point two mole; the slash means divided by. Right? So zero point two mole of ethanol gives zero point two mole of ether. Is that it? Oh no, careful, look at the two in front of the ethanol! Two ethanol molecules make one ether molecule. So zero point two over two gives zero point one mole of ether at full yield. Clear? But the yield is only fifty percent, so we get zero point zero five mole. And the molar mass of ethoxyethane is seventy four. Got it? So the mass is zero point zero five times seventy four, which gives three point seven grams; the cross means times. Forget that two, and you get seven point four, double the right answer. With me?

6:41Example 2: fraction of an acid present as dimers in benzene

Now, Example two, the hard one, and it is our puzzle from the start. Take a monocarboxylic acid, with one COOH group, of molar mass eighty eight. In benzene it shows an apparent molar mass of one hundred and sixty. What fraction of it is present as dimers? Pause the video and try it first. Done? Okay, let's check. The molecules did not get heavier. In benzene, carboxylic acids pair up as cyclic dimers, held by two hydrogen bonds per pair. Fewer particles, so the experiment sees a bigger molar mass. Okay? Now the van 't Hoff factor, small i, is the normal molar mass over the observed molar mass. Eighty eight over one hundred and sixty gives zero point five five. Right? When n molecules associate into one, with degree alpha, the Greek letter alpha, i equals one minus, in brackets, one minus one over n, times alpha. For dimers n is two, so i equals one minus alpha over two. Clear? So the fraction present as dimers is one minus zero point five five, which is zero point four five, right? No, no, wait, that is the trap! Zero point four five is alpha over two, not alpha. One minus alpha over two equals zero point five five, so alpha over two is zero point four five, and alpha is zero point nine; the double arrow means which gives. Got it? So the fraction present as dimers is zero point nine. Puzzle solved: the molecules never got heavier, they simply paired up. With me?

8:24Quick revision: ether yield, dimers, Tollens, Cannizzaro

So, quick revision. Read the balanced equation: two ethanol molecules make one ether molecule. The van 't Hoff factor is the normal over the observed molar mass, and for dimers, i equals one minus alpha over two. Reducing a carbonyl group never gives a tertiary alcohol, and simple ketones fail Tollens' test. Soda lime swaps COON a for H, and in the Cannizzaro reaction two moles of aldehyde use one mole of sodium hydroxide. Now try the rest of Set three on this page. Well done today, and all the best!