Part A: JEE Main Level

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Take benzamide, which has seven carbon atoms, and treat it with bromine and sodium hydroxide. Out comes aniline, and aniline has only six carbon atoms. So where did the seventh carbon go? This is Part A of the unit problem bank, at JEE Main level. We will sharpen two tools, the carbylamine reaction and the Hoffmann bromamide reaction, and then solve two of its questions with moles. So, let's begin.

0:33Tool 1: the carbylamine reaction

So, tool one is the carbylamine reaction. Take a primary amine, and heat it with chloroform, CHC l three, and alcoholic KOH, potassium hydroxide in alcohol. The small three in CHC l three is a subscript, so there are three chlorine atoms. Okay? Here is the full equation. RNH two, plus chloroform, plus three KOH, gives RNC, plus three K C l, plus three water. R stands for the alkyl or aryl group, and the arrow means gives. Right? This product, RNC, is an isocyanide, with R joined to the nitrogen. And it has a very offensive smell, and that smell is the test itself. Clear? Now let me ask you this: which amines give this smell? Primary, secondary, tertiary, or all three? Yes, only primary amines respond. Aliphatic or aromatic, both work, so aniline gives the test too. But secondary and tertiary amines do not respond at all. With me? Now look at the equation once more. One mole of RNH two gives how many moles of RNC? Yes, exactly one mole, so the ratio is one to one. And see, R keeps all its carbons, and the product gets one more carbon, joined to the nitrogen, which comes from chloroform. So the carbylamine reaction adds a carbon, but it does not lengthen the chain of R itself. Got it?

2:05Tool 2: the Hoffmann bromamide reaction

Now, tool two is the Hoffmann bromamide reaction. We take an amide, RCONH two, with bromine, B r two, and four N a O H, which is sodium hydroxide. The short forms in brackets mean aqueous or alcoholic. And the figure on the right shows the steps of this reaction; we will come back to it in a moment. Okay? And it gives RNH two, a primary amine, plus sodium carbonate, N a two C O three, plus two sodium bromide, plus two water. Right? Now compare the two. RCONH two went in, and RNH two came out. So tell me, what is missing? Yes, the C O group is missing, so the amine has one carbon fewer. And where does that carbon go? Look at the figure on the right. The amide passes through an isocyanate, RNCO, and in the last box the carbon leaves as carbonate, C O three, with the two minus written high as its charge. So that is where the seventh carbon of benzamide went! With me? Now, here is a quick trap. N methylethanamide is also an amide. So does it give the Hoffmann reaction as well? No, it does not, and that is the trap! Only RCONH two reacts, with two hydrogens on the nitrogen. The N in italics means a group sits on the nitrogen, and such N substituted amides do not react. Clear? So, for the numbers, benzamide, with seven carbons, gives aniline, with six. And one mole of amide gives one mole of amine, again one to one. Got it?

3:45Example 1: mass of ethyl isocyanide from 0.90 g ethanamine

Now, example one, which is question nine from Part A of the bank. Find the mass of ethyl isocyanide formed when zero point nine zero grams of ethanamine is converted completely by the carbylamine reaction; the small g means grams. Pause the video and try it first. Done? Okay, let's check. Ethanamine is C two H five N H two, with two carbons, seven hydrogens in all, and one nitrogen. So its molar mass, capital M, is two times twelve, plus seven times one, plus fourteen, which gives forty five grams per mole; mol to the power minus one means per mole. Okay? So the moles of ethanamine are zero point nine zero over forty five, which gives zero point zero two mole; the slash means divided by. Right? Now, before we go on, a quick question. Will the isocyanide weigh more or less than the zero point nine zero grams of amine? Yes, it weighs more, and here is why. From tool one, the ratio is one to one, so we get zero point zero two mole of C two H five N C, ethyl isocyanide. Same number of moles, but each molecule has picked up a carbon. With me? Its molar mass is three times twelve, plus five times one, plus fourteen, which gives fifty five grams per mole. Or the quick way: forty five, minus two for the two hydrogens lost, plus twelve for the new carbon, again fifty five. Clear? So the mass is zero point zero two times fifty five, which gives one point one zero grams. That is more than the zero point nine zero grams of amine, just as we said. Well done if you got it!

5:31Example 2: mass of aniline from benzamide at 90% yield

And now, example two, which is question six of Part A, and it is our puzzle from the start. Find the mass of aniline obtained from six point zero five grams of benzamide by the Hoffmann reaction, in ninety percent yield. Pause the video and try it first. Done? Okay, let's check. Ninety percent yield, so some students simply take ninety percent of six point zero five grams, which gives five point four four five grams. Is that the answer? No, that is exactly the trap we must avoid. Six point zero five grams is the mass of benzamide, and aniline is a different molecule, so we must go through moles. Benzamide is C six H five CONH two, with seven carbons, seven hydrogens, one oxygen and one nitrogen. So M is seven times twelve, plus seven times one, plus sixteen, plus fourteen, which gives one hundred and twenty one grams per mole. Okay? So the moles of benzamide are six point zero five over one hundred and twenty one, which gives zero point zero five mole. Right? From tool two, one mole of amide gives one mole of amine. So we get zero point zero five mole of aniline, C six H five N H two. Its molar mass is six times twelve, plus seven times one, plus fourteen, which gives ninety three grams per mole. Clear? If the yield were a hundred percent, the mass would be zero point zero five times ninety three, which gives four point six five grams. See, that is already less than six point zero five grams, because the C O part has left as carbonate. Got it? But the yield is only ninety percent. So four point six five times zero point nine gives four point one eight five, which is about four point one nine grams; the wavy equals sign means approximately. And there is our puzzle, solved: seven carbons went in, six came out, and the lost carbon left as carbonate. Very good if you got it!

7:39Quick revision: carbylamine, Hoffmann and mass steps

So, quick revision. First, in the carbylamine reaction only primary amines respond, and RNH two goes to RNC, one to one, with one carbon more. Second, in the Hoffmann reaction only RCONH two reacts, and it gives RNH two, one to one, with one carbon fewer. Third, for mass problems, change grams to moles, use the ratio, go back to grams, and apply the yield at the end. Now try the other questions of Part A on this page as a test, before you read the solutions. All the best!