AIEEE and JEE Main Questions

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Benzonitrile is a nitrogen compound with no carbonyl group at all. Add a Grignard reagent, then a little acid, and it gives a liquid with a carbonyl group. So which test will that liquid answer: a test for aldehydes, a test for methyl ketones, or a test for amino acids? We will settle it in Example one. Today we solve past JEE Main and AIEEE questions from this chapter. Each one checks a rule from this chapter, so let's begin.

0:34Nitriles and isocyanides: what they turn into

So, first our tools, from the course's figure. In a nitrile, R, an alkyl or aryl group, sits on carbon; the dash is a single bond, and the three lines are a triple bond to nitrogen. In an isocyanide, R sits on the nitrogen instead. Okay? Now, R prime M g X, in the question on screen, is a Grignard reagent: M g is magnesium, X is a halogen, and the prime marks a second group. So tell me, what does a nitrile give with a Grignard reagent and then dilute acid? It gives a ketone, yes, exactly! The acid is H three O plus, with a subscript three and a raised plus for the charge, and the ketone is RCOR prime. So R and R prime sit together on one carbonyl carbon. Right? Now compare: stannous chloride, S n C l two, with H C l, where the slash means together with, then water. This time the nitrile gives an aldehyde, RCHO, in the Stephen reaction. Mixing up these two rules is a common trap. Clear? Now the isocyanide, at the bottom of the figure. In dilute acid, the nitrogen keeps R and becomes an amine, RNH two, and the end carbon leaves as methanoic acid, HCOOH; the plus sign means and. Got it? Now a quick past question, from AIEEE two thousand and three: ethyl isocyanide, C two H five N C, is hydrolysed in acid. What does it give? It gives an ethylamine salt and methanoic acid, very good! The ethyl group stays on the nitrogen, and in acid the amine forms its salt. The trap is to treat it like a nitrile, but in an isocyanide that end carbon leaves as methanoic acid. With me?

2:30Urea, biuret and the Bhopal gas

Now our second tool, urea. N H two in brackets with a small two outside means two N H two groups, both on one carbonyl group, so urea is the diamide of carbonic acid. It has forty six point seven percent nitrogen; remember that number for Example two. Okay? Now we heat urea gently. Two molecules join together, and they lose one small molecule. Which one do you think? They lose ammonia, N H three, perfect! The two in front means two molecules of urea, the arrow means gives, and the long product is biuret, two amide units joined through one N H group. Right? And biuret gives a violet colour with alkaline copper sulphate: the biuret test, which peptides and proteins also give. This violet colour is a clue in Example two. Clear? Now a quick past question on a serious subject, from JEE Main twenty thirteen. Which gas leaked from the storage tank of the Union Carbide plant in the Bhopal tragedy? The gas was methyl isocyanate. The double lines are double bonds: nitrogen to carbon, and carbon to oxygen. Careful with the name: isocyanate has that oxygen, so do not confuse it with methyl isocyanide. Got it?

3:50Example 1: benzonitrile with CH3MgBr, which test P answers

Now Example one, from JEE Main twenty twenty one. Benzonitrile reacts with one equivalent of methylmagnesium bromide, C H three M g B r, and the product is hydrolysed to give a liquid, P, okay? Which test does P answer: the iodoform test, Schiff's test, the ninhydrin test, or Tollens' test? Pause the video and try it first. Done? Okay, let's check. Benzonitrile is C six H five C N, a phenyl group on the nitrile carbon. The reagents over the arrow follow our Grignard rule: R is phenyl and R prime is methyl, so we get the ketone C six H five COCH three, with both groups on one carbonyl carbon. With me? So P is acetophenone, a methyl ketone, because a methyl group sits right on the carbonyl, and that methyl group came from the Grignard reagent. Okay? Now, many students think like this: benzonitrile is a nitrogen compound, so the answer must be the ninhydrin test, right? That would be a mistake, and it is the trap here. The ninhydrin test is for amino acids, and look at P on screen: there is no nitrogen left in it at all. Clear? What about Tollens' test and Schiff's test? Both are tests for aldehydes, and simple ketones like P give neither. You would pick them only if you mixed this up with the Stephen reaction. Okay? Now the iodoform test: methyl ketones, C H three COR, give yellow iodoform, CHI three. The dash before R shows where the rest of the molecule is attached. Got it? P has exactly that group, with phenyl as R, so the answer is the iodoform test, and our opening puzzle is solved. Right?

5:41Example 2: identify urea from percent data and biuret test

And now Example two, a harder one, from AIEEE two thousand and five. An organic compound of molar mass sixty contains twenty percent carbon, six point six seven percent hydrogen, and forty six point six seven percent nitrogen, the rest being oxygen. On heating, it gives ammonia and a solid residue that gives a violet colour with alkaline copper sulphate. Identify the compound. Pause the video and try it first. Done? Okay, let's check. The question gives percentages for carbon, hydrogen and nitrogen only, so we work with these three, right? No, and that is the trap: the rest is oxygen, and we must count it. Oxygen is one hundred minus the total in the brackets, which gives twenty six point six six percent. Okay? Now divide each percentage by its atomic mass: twelve for carbon, one for hydrogen, fourteen for nitrogen and sixteen for oxygen. The colons on screen mean a ratio. Clear? Twenty over twelve gives one point six seven, hydrogen stays six point six seven, forty six point six seven over fourteen gives three point three three, and oxygen gives one point six seven. Divide by the smallest, and the ratio is one to four to two to one. Got it? So the empirical formula is C H four N two O, where each small number counts atoms. Now, is this the molecular formula itself, or do we double it? Add the masses: twelve plus four plus twenty eight plus sixteen gives sixty, equal to the molar mass. So there is no doubling, and the formula stays as it is. Right? Now the chemistry clue: ammonia on heating, and a residue that turns alkaline copper sulphate violet, which is biuret. That is exactly what urea gives on gentle heating. With me? So the compound is urea. And here is a topper's trick: forty six point six seven percent nitrogen matches the forty six point seven percent we remembered for urea. So you can spot it at once; the calculation just confirms it. Okay?

7:54Quick revision: nitrile to ketone, iodoform, urea, biuret

So, quick revision. Nitriles with a Grignard reagent and acid give ketones, while isocyanides in acid give an amine and methanoic acid. The iodoform test is for methyl ketones, and Tollens' and Schiff's tests are for aldehydes, not simple ketones. And urea has forty six point seven percent nitrogen, and gentle heating gives biuret, which turns alkaline copper sulphate, C u S O four, violet. Now try the other past JEE questions in this chapter; they will feel easy now. All the best!