Part A: JEE Main Level

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Take one point one six grams of acetone, which is propanone, and add iodine and alkali. Out comes a yellow solid, and it weighs several times more than the acetone you started with. More mass out than you put in? Hmm, so where does all that extra mass come from? This is Part A of the unit problem bank, at JEE Main level. We will sharpen two tools on the iodoform test, then solve two of its questions. So, let's begin.

0:37Tool 1: who gives the iodoform test?

So, tool one. The iodoform test uses I two plus N a O H, iodine and sodium hydroxide; the two in I two is written low, as a subscript. We can also write the reagent as N a O I, sodium hypoiodite. If the test is positive, we get a yellow solid, CHI three, which is iodoform. Okay? Now, which compounds respond? They need the group C H three C O, or the group C H three C H, O H in brackets, where the O H hangs on that second carbon. The dash at the end is a bond, and it must be joined to a carbon or to a hydrogen. Right? So who says yes? Ethanal and ethanol, the methyl ketones, and alcohols of the type C H three C H, O H in brackets, R, where R is a carbon group. Clear? Now, a quick trap. Ethanoic acid also carries a C H three C O group, as the badge says. So does ethanoic acid give the iodoform test? No, it does not, and that is the trap! In the acid, the C H three C O is joined to an O H group, not to a carbon or a hydrogen. The same goes for its esters and its amides. So always check what the group is joined to. Got it?

1:53Tool 2: what the reaction gives

Now, tool two: what does the reaction actually give? First, all three hydrogens of the C H three are replaced by iodine. So C H three C O, dash R, goes to C I three C O, dash R; the arrow means gives, and R is the rest of the molecule. With me? Then hydroxide, O H minus, breaks the carbon carbon bond, the C dash C on screen. The C I three part ends up as iodoform, CHI three, and the rest becomes the acid salt, RCOO minus; the minus written high is a negative charge. This is the haloform reaction. Okay? So let me ask you this: one C H three C O group gives how many molecules of iodoform? Exactly one, that is correct! The methyl carbon itself becomes the carbon of iodoform. So the acid left behind has one carbon fewer. We will need this one to one ratio very soon. Clear? For example, take butan two one, with four carbons. With iodine and sodium hydroxide, then acid, it gives propanoic acid, with three carbons, plus iodoform. Right? Now, one more trap. Suppose we take sodium iodide, N a I, with sodium hydroxide, instead of iodine. Would that also give iodoform? No, it would not, and many students miss this one! The reaction needs an agent that supplies positive iodine, and hypoiodite does that. The iodide ion cannot, so no iodoform forms. Got it?

3:30Example 1: mass of iodoform from 1.16 g of propanone

Now, example one, question eleven of Part A, and it is our puzzle from the start. Find the mass of iodoform obtained from one point one six grams of propanone, with excess iodine and alkali; the small g means grams. In brackets, I equals one hundred and twenty seven, the atomic mass of iodine. Pause the video and try it first. Done? Okay, let's check. Propanone is C H three COCH three: three carbons, six hydrogens and one oxygen. So its molar mass, capital M, is three times twelve, plus six times one, plus sixteen, which gives fifty eight grams per mole; mol to the power minus one means per mole. Okay? The moles of propanone are one point one six over fifty eight, which gives zero point zero two mole. Got it? Now look at propanone again. It has a C H three group on each side of the C O, two of them. So each molecule gives two iodoform, and we get zero point zero four mole, right? Oh no, careful, that is the trap! Once one C H three C O group reacts, the other methyl is left in the ethanoate, the salt of ethanoic acid, and from tool one, ethanoic acid does not respond. So one C H three C O group gives one iodoform, which means zero point zero two mole of CHI three. With me? Now the molar mass of iodoform. Twelve for the carbon, plus one for the hydrogen, plus three times one hundred and twenty seven for the iodines, which gives three hundred and ninety four grams per mole. Clear? So the mass is zero point zero two times three hundred and ninety four, which gives seven point eight eight grams. And there is our puzzle, solved! Most of that mass is the three iodine atoms, carried in from the reagent. Well done if you got it!

5:19Example 2: how many C6H12O ketones give the iodoform test

And now, example two, which is question nine of Part A, from the same bank. How many ketones of formula C six H twelve O give the iodoform test? The six and twelve are subscripts, counting the atoms. Pause the video and try it first. Done? Okay, let's check. To give the test, the ketone must have the group C H three C O, so it is a methyl ketone. That uses two carbons, three hydrogens and the oxygen, leaving C four H nine, a butyl group. So we count the ways to build C four H nine. Okay? First, the straight chain: C H three C O, joined to C H two, C H two, C H two, C H three. That is hexan two one. Right? Now branch it. Join the C O to a C H that carries a methyl, the C H three in brackets, and then an ethyl. That is three methylpentan two one. Clear? Move the branch one carbon further: C H two, then a C H with two methyls, the C H three in brackets with a subscript two outside. That is four methylpentan two one. Got it? And last, the most branched one: a carbon carrying three methyl groups, the tert butyl group. That is three, three dimethylbutan two one. With me? So we have four so far. But hexan three one is also C six H twelve O, with a C H three at each end of its chain. So should we count it as well? No, no, wait, that is exactly the trap! In hexan three one, the C O sits on carbon three, so both its neighbours are C H two groups. The C H three must sit directly on the C double bond O, and the equals sign on screen is that double bond. So the answer is four ketones. If you got four, that is really good work!

7:15Quick revision: iodoform groups, 1:1 ratio, methyl ketones

So, quick revision. First, the iodoform test needs a C H three C O group, or a C H three CHOH group, joined to a carbon or a hydrogen. Second, one C H three C O group gives one iodoform, plus an acid with one carbon fewer. Third, when you count methyl ketones, the C H three must sit right on the C double bond O, the carbonyl group. Now try the other questions of Part A on this page as a test, before you read the solutions. All the best!