AIEEE and Earlier JEE Main Questions: Reactions

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. An unknown compound has a molar mass of one hundred and eighty. We treat it with ethanoyl chloride, and the product has a molar mass of three hundred and ninety. We know nothing about its structure. So can you still tell how many amino groups it has? Sounds impossible, right? By the end of this video, you will do it in one line. Today we solve the past questions on this page, from the AIEEE papers and the early JEE Main papers. They are built on two reactions of amines: the carbylamine test and acylation. Both are worth knowing well, so let's begin.

0:40Idea 1: the carbylamine test

So, idea number one: the carbylamine test. We heat the amine with chloroform, CHC l three, and alcoholic KOH, that is, potassium hydroxide dissolved in alcohol. The small numbers written low are subscripts; they count the atoms. Okay? Here is the equation for a primary amine, RNH two, where R is an alkyl or aryl group. It reacts with chloroform and three KOH, and the arrow means gives. We get RNC, plus three K C l and three water molecules. Right? Now look at the nitrogen. In going to RNC, the amine RNH two loses both hydrogens from its nitrogen. So let me ask you this: does a secondary amine, with only one hydrogen on its nitrogen, give this test? The answer is no. Only primary amines respond, aliphatic or aromatic, and secondary and tertiary amines give nothing. Clear? And this RNC is an isocyanide, with a very offensive smell. So the test is really a smell test, and it is also called the isocyanide test. Got it?

1:52Two quick past-paper checks

Now two quick questions from the past papers on this page. The JEE Main paper of twenty fourteen asked this: what organic compound forms on heating an aliphatic primary amine with chloroform and ethanolic potassium hydroxide, that is, potassium hydroxide dissolved in ethanol? Many students answer an alkyl cyanide. Is that right? That is exactly the trap, so be careful here. The answer is an alkyl isocyanide, RNC, where the nitrogen is joined to R, the alkyl group. Okay? Compare a cyanide, RCN; the word in brackets, nitrile, is its other name. In a cyanide it is the carbon that is joined to the R group. That is a different compound, with a pleasant smell, and you will meet it later in this unit. With me? Next, from the AIEEE paper of two thousand and three: chloroform and alcoholic KOH react with p toluidine; the small p in italics means para. In the brackets, p toluidine is four methylaniline, an aromatic primary amine. So does it give the carbylamine reaction? Yes, it does, because aromatic primary amines respond too. The product is four methylphenyl isocyanide; its formula on screen has the methyl group, C H three, and the isocyanide group, N C, on a benzene ring. Clear?

3:18Example 1: AIEEE 2007, ethanamine with CHCl3 and KOH

Time for Example one, from the AIEEE paper of two thousand and seven. In this equation, C H three C H two N H two reacts with chloroform and three KOH to give two products, A and B, plus three water molecules. Identify both of these products. Pause the video and try it first. Done? Okay, let's check. The amine here is ethanamine, an ethyl group joined to N H two. It is a primary amine, so the carbylamine reaction takes place. Right? Now compare it with our general equation. Here R is the ethyl group, C two H five. So what must the first product be? Yes, exactly, the first product, A, is C two H five N C, ethyl isocyanide. That is the compound with the terrible smell. Okay? Now the second product, called B here. In the general equation, three KOH gives three K C l, so B is three K C l, potassium chloride. Got it? Let us check the balance. On the left, carbon is two plus one, which is three, and hydrogen is seven plus one plus three, which is eleven. On the right, C two H five N C has three carbons and five hydrogens, and three water molecules add six more hydrogens, giving eleven. Nitrogen is one, and chlorine, potassium and oxygen are three each on both sides. So the equation is balanced. With me?

4:53Idea 2: acetylation adds 42 per amino group

Now idea number two: acylation. Primary and secondary amines react with acid chlorides to give amides. Okay? Take ethanoyl chloride, C H three COC l, also called acetyl chloride. RNH two plus ethanoyl chloride gives the amide on screen, plus H C l, hydrogen chloride. Right? Look at what changed: one hydrogen on the nitrogen has been replaced by the acetyl group, COCH three. This is called acetylation. Clear? Now here is the key point. The acetyl group has a mass of twelve plus sixteen plus twelve plus three, which is forty three. So for each amino group, the molar mass rises by forty three, right? That is the trap, because one hydrogen also leaves, with mass one. So forty three comes in, one goes out, and the net rise is forty two for every N H two group that is acetylated. Remember this number, forty two. Got it? And tertiary amines have no hydrogen on the nitrogen, so they are not acylated at all. With me?

6:02Example 2: molar mass 180 to 390, count the amino groups

Now Example two, from the JEE Main paper of twenty thirteen, and look, it is our puzzle from the start! Read it with me: a compound of molar mass one hundred and eighty is acylated with C H three COC l, ethanoyl chloride, to give a compound of molar mass three hundred and ninety. How many amino groups does the first compound contain? Pause the video and try it first. Done? Okay, let's check. Every amino group that is acetylated raises the molar mass, so first find the total rise. Three hundred and ninety minus one hundred and eighty gives two hundred and ten. Okay? Now, by which number should we divide two hundred and ten? We divide by forty two, yes, exactly, the net rise per acetylation from idea two. Two hundred and ten over forty two gives five. Right? And if you divided by forty three, the mass of the acetyl group, two hundred and ten over forty three gives about four point eight eight; the wavy equals sign means approximately. No molecule has a fraction of an amino group, so a number that is not whole tells you that you fell into the trap. Clear? So the first compound contains five amino groups. We solved our puzzle in one line, without knowing a single atom of its structure. Easy, isn't it?

7:30Quick revision: carbylamine test, isocyanide, 42 per NH2

So, quick revision. The carbylamine test works only for primary amines, aliphatic or aromatic. Its product is an isocyanide, RNC, not a cyanide, with a very offensive smell. And in acetylation, the molar mass rises by forty two for every amino group that is acetylated. Now try all four past questions on this page on your own, before you read their solutions. All the best!