IIT-JEE Questions (1981–1995): Reactions

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Ethanal and benzaldehyde are both aldehydes, okay? Put ethanal in dilute alkali, and two molecules join into one bigger molecule. Heat benzaldehyde with concentrated alkali, and two molecules give two different products, an alcohol and the salt of an acid. Same family, completely different results. Why does that happen? Hmm, hold that thought. Today we solve two questions from old IITJEE papers, from nineteen eighty four and nineteen eighty seven. These two ideas are worth knowing well. So, let's begin.

0:39Idea 1: alpha hydrogen decides aldol or Cannizzaro

So, idea number one. Look at the carbonyl group, C double bond O; the two lines on screen are the double bond. The carbon right next to it is the alpha carbon, and alpha is this Greek letter. Any hydrogen on that carbon is an alpha hydrogen. Okay? Now take ethanal, C H three, CHO; the small numbers written low count the atoms, and the dash is a bond. How many alpha hydrogens does it have? It has three of them, yes, exactly! The C H three carbon sits right next to the carbonyl carbon, and it carries three hydrogens. With me? Now the top row of the figure: two molecules of ethanal, that is the two in front, in dilute sodium hydroxide, and dil on the arrow means dilute; the pair of half arrows shows that this step can go both ways. The alkali removes an alpha hydrogen, giving the enolate ion, which adds to the carbonyl group of a second molecule. The product is three hydroxybutanal, the aldol. This aldol reaction needs at least one alpha hydrogen. Clear? And on heating, the aldol loses water, H two O, and gives but two enal, with a carbon to carbon double bond. That is aldol condensation. Got it? Now benzaldehyde, C six H five, CHO, where C six H five is the phenyl group, a benzene ring. Does benzaldehyde have an alpha hydrogen? No, it has none! Its alpha carbon is a ring carbon, with no hydrogen on it. So there is no aldol. Instead, with concentrated alkali, and conc means concentrated, it gives the Cannizzaro reaction in the bottom row. One molecule is reduced to benzyl alcohol, and another is oxidised to potassium benzoate, COO minus, K plus. Right?

2:34Idea 2: reading the clues

Now idea number two. Old questions often describe a compound only through its tests. First, Tollens' reagent, ammoniacal silver nitrate. Every aldehyde, aliphatic or aromatic, gives a silver mirror, but ketones do not. Okay? Next, Fehling's solution. Benzaldehyde gives the silver mirror, so it reduces Fehling's solution too, right? No, no, careful, that is a trap! Fehling's solution gives a red-brown precipitate of C u two O only with aliphatic aldehydes. Aromatic aldehydes like benzaldehyde do not respond to it at all. With me? Third, sodium hydrogensulphite, N a HSO three. It forms a crystalline bisulphite addition compound with aldehydes, methyl ketones and unhindered cyclic ketones. So a bisulphite compound alone does not prove an aldehyde. Clear? And fourth, the iodoform test, with iodine and alkali. Methyl ketones, with the C H three C O group, and also ethanal, give a yellow precipitate of iodoform, CHI three. The dash joins the group to the rest of the molecule, so the C H three must sit on the carbonyl carbon. Alcohols that iodine can oxidise to such a group, like ethanol and propan two ol, give iodoform too. Got it?

3:57Example 1: which aldehydes give aldol (IIT-JEE 1984)

Time for Example one, from the IITJEE paper of nineteen eighty four. Which of these undergo the base catalysed aldol reaction? Propanal, benzaldehyde, two methylpropanal, and two two dimethylpropanal. Here is a hint: hunt for the alpha hydrogens. Pause the video and try it first. Done? Okay, let's check. Propanal is C H three, C H two, CHO, and its alpha carbon is the C H two, with two hydrogens. So propanal gives the aldol reaction. Right? Benzaldehyde, we just saw, has no alpha hydrogen, so there is no aldol for it. Okay? Now two two dimethylpropanal. It clearly has an alpha carbon, right next to the CHO group. So it gives the aldol reaction, right? No, no, wait, that is the trap! In the formula, C H three is in brackets, and the small three after the bracket means three methyl groups, all on one carbon, which is joined to the CHO group. So that alpha carbon carries not a single hydrogen. An alpha carbon is not enough; we need an alpha hydrogen. With me? And two methylpropanal has two methyl groups on a carbon that still keeps one hydrogen, joined to the CHO group. That one alpha hydrogen is enough, so it gives the aldol reaction. Clear? So the answer is propanal and two methylpropanal, these two. Got it? And now our puzzle is solved. Benzaldehyde and two two dimethylpropanal have no alpha hydrogen, so with concentrated alkali they take the Cannizzaro path instead, while ethanal, with its alpha hydrogens, goes the aldol way. One small hydrogen decides it all. Okay?

5:46Example 2: identify C5H10O from its tests (IIT-JEE 1987)

Now the harder one, Example two, from the nineteen eighty seven paper. An organic compound contains sixty nine point seven seven percent carbon, eleven point six three percent hydrogen, and the rest oxygen. Its molar mass is eighty six. It does not reduce Fehling's solution, but it forms a bisulphite addition compound and gives a positive iodoform test. What are its possible structures? Pause the video and try it first. Done? Okay, let's check. First, the oxygen: one hundred, minus sixty nine point seven seven, minus eleven point six three, equals eighteen point six zero percent. Okay? Now divide each percentage by its atomic mass: twelve for carbon, one for hydrogen, and sixteen for oxygen; each fraction is one such division, and the colons show a ratio. We get five point eight one, then eleven point six three, then one point one six. With me? Now divide all three numbers by the smallest one, which is one point one six. What ratio do you get? You get five, ten and one, very good! So the formula is C five H ten O, and here is the check: sixty for five carbons, plus ten, plus sixteen, makes eighty six. That is the molar mass exactly. Clear? Now the clues. It forms a bisulphite addition compound, so it must be an aldehyde, right? Careful now, because that is exactly the trap! Methyl ketones form bisulphite compounds too. And it does not reduce Fehling's solution. With only five carbons there is no room for a benzene ring, so any aldehyde here would be aliphatic and would reduce it. Then the positive iodoform test settles it: a methyl ketone, with the C H three C O group. Got it? Now, pentan three one is also a ketone with the formula C five H ten O, okay? Does it fit all the clues? It does not fit, and here is why. Pentan three one is C H three C H two, then C O, then C H two C H three, an ethyl group on each side of the carbonyl. There is no C H three C O group, so no iodoform. Right? So our compound is C H three C O, joined to three more carbons. In a straight chain, C H three C O, C H two, C H two, C H three, that is pentan two one. Branched, C H three C O, then a C H carrying two C H three groups, shown in brackets, that is three methylbutan two one. Both are methyl ketones, both fit every clue, and these are the two possible structures. Okay?

8:42Quick revision: aldol, Cannizzaro, Fehling's, iodoform

So, quick revision. An alpha hydrogen and dilute alkali give aldol; no alpha hydrogen and concentrated alkali give Cannizzaro. Fehling's solution answers only to aliphatic aldehydes, and Tollens' reagent to all aldehydes. Bisulphite adds to aldehydes and methyl ketones, and the iodoform test picks out the C H three C O group and ethanal. There are twenty four more solved questions on this page, so try each one yourself before reading its answer. All the best!