JEE Advanced and IIT-JEE Questions: Preparation

Transcript

0:00Introduction

Hello everyone! Let me start with a question from JEE Advanced twenty sixteen. One benzene compound with two side chains is taken down two different paths, and you must name the final product of each path. And here is the surprise: one of the two answers does not even have a benzene ring. So how does that happen? By the end of this video, you will solve it yourself. Past questions on preparing amines test only a few ideas. So, let's begin.

0:31Idea 1: count the carbons first

So, idea number one: before you think of any reagent, count the carbons. R X is an alkyl halide, where R is the alkyl group and X is a halogen. If the amine keeps the same number of carbons as the halide, use the Gabriel synthesis, or ammonolysis, which means heating the halide with ammonia. Right? One carbon more than the halide? Use KCN, potassium cyanide, to make the nitrile, then reduce it. The cyanide brings its own carbon. With me? The same number of carbons as an amide? Then reduce the amide with lithium aluminium hydride, L i A l H four; the small four written low is a subscript that counts hydrogen atoms. Clear? Now suppose the amine must have one carbon fewer than the amide. Which reaction will you choose? It is the Hoffmann bromamide degradation, yes, exactly! That means bromine, B r two, together with N a O H, sodium hydroxide. Keep these four lines in mind, because many of these questions start right here. Got it?

1:38Idea 2: what Hoffmann questions test

Now idea number two, the Hoffmann reaction itself. RCONH two is the amide, with C O the carbonyl group. With bromine and four N a O H, the arrow gives RNH two, plus N a two C O three, sodium carbonate, plus two N a B r, sodium bromide, plus two water molecules. Okay? Now count the carbons: the amine has one carbon fewer than the amide. So where did that carbon go? Look at the products. It went into sodium carbonate, very good! The carbon of the C O group, C double bond O on screen, leaves as carbonate. That is why the chain steps down by one. Right? Questions also ask for the intermediates. First, RCONHB r, the N bromoamide; the italic N says the bromine is on nitrogen. Then its anion, and then the isocyanate, RNCO, where each equals sign is a double bond and the dash a single bond. With me? On the way, the group R moves from the carbon to the nitrogen, within the same molecule; it never breaks away. We call this intramolecular. And notice the two hydrogens on the nitrogen of RCONH two. The reaction needs both of them, so only RCONH two reacts; N substituted amides, with a carbon group on the nitrogen, do not. Clear?

3:06Idea 3: the Gabriel phthalimide synthesis

Now idea number three, the Gabriel phthalimide synthesis. In phthalimide, the N H group sits between two C O groups, and that makes its hydrogen acidic. So KOH, written over the arrow, removes it, giving potassium phthalimide, N minus K plus; the raised signs are the charges. Okay? This anion attacks the alkyl halide, R X, written over the next arrow, and pushes out the halide, giving N alkyl phthalimide. Then N a O H, with a q in brackets, meaning aqueous sodium hydroxide, hydrolyses it and releases the amine, RNH two. Right? The big advantage is that the amine comes out pure, a primary amine, free of secondary and tertiary amines. So let me ask you this: can the Gabriel synthesis make aniline? No, it cannot, and this is a common trap, so be careful. Aryl halides, with the halogen on a benzene ring, do not react with the phthalimide anion. So the Gabriel synthesis gives no aromatic primary amines. With me? And tertiary halides give elimination instead, so they are out too. Clear?

4:19Example 1: acetamide to methylamine, IIT-JEE 1983

Time for Example one, an easy one from the IITJEE paper of nineteen eighty three. Acetamide is treated separately with four reagents: P C l five, phosphorus pentachloride; sodium hydroxide with bromine; soda lime; and hot concentrated sulphuric acid, H two S O four. Which one gives methylamine? Pause the video and try it first. Done? Okay, let's check, and count first. Acetamide is C H three CONH two: a methyl group plus the C O carbon, so two carbons. Okay? Methylamine is C H three N H two, with only one carbon. So we need an amine with one carbon fewer than the amide. Which of the four reagents does that? Yes, it is sodium hydroxide with bromine, correct! One carbon fewer than the amide means the Hoffmann bromamide degradation, so the answer is N a O H with bromine. Clear?

5:23Example 2: diamide to diamine and alanine, JEE Advanced 2016

Now the harder one, Example two, from JEE Advanced twenty sixteen. An ortho dialkylbenzene, labelled O, has two alkyl groups on neighbouring carbons of the ring; that is what ortho means. K M n O four, potassium permanganate, with H plus after the slash, which means in acid, oxidises it to P, okay? Heating P with ammonia gives Q, and Q with bromine and sodium hydroxide gives the compound R, right? Strong heating of Q gives S instead, and S with ethyl two bromopropanoate and KOH, then acid, gives T, and we must find R and T, both of them. Pause the video and try it first. Done? Okay, let's check. Acidified permanganate oxidises each alkyl group on the ring to a COOH group, a carboxylic acid. Two acid groups on neighbouring carbons make phthalic acid: C six H four, with COOH in brackets, and the two outside the bracket means two such groups. Right? Heating with ammonia turns each acid group into an amide group, CONH two, so Q is the diamide, with two amide groups on the ring. With me? Now Q meets bromine and sodium hydroxide, the Hoffmann reagents. Does the reaction act on one amide group, or on both of them? It acts on both of them, exactly! Each CONH two loses its carbon and becomes N H two, so eight carbons become six. So R is benzene one two diamine, C six H four with two N H two groups on neighbouring carbons. That is the first answer. Clear? Now the other path: strong heating turns the diamide into S, phthalimide, the ring with an N H between two C O groups. And phthalimide means idea three, the Gabriel synthesis. Got it? Ethyl two bromopropanoate, on screen, is a chain of three carbons with the bromine on the middle carbon and an ethyl ester group at the end. That carbon is not tertiary, so substitution works: KOH makes the phthalimide anion, and its nitrogen replaces the bromine on that carbon. Okay? Then the hydrolysis frees the amino group. So T still carries the ethyl ester, right? Oh no, careful, because that is exactly the trap! The alkali hydrolyses the ester too, and acidification then gives the free COOH group. So T is two aminopropanoic acid, with N H two on the middle carbon: alanine. And look, alanine has no benzene ring at all. One ring, two paths, a diamine and an amino acid: our puzzle is solved! Very good.

8:13Quick revision: carbon count, Hoffmann, Gabriel limits

Clear? So, quick revision. Count the carbons first: the same as the halide means Gabriel, one more means potassium cyanide, and one fewer than the amide means Hoffmann. Hoffmann goes through the N bromoamide and the isocyanate, and the carbonyl carbon leaves as carbonate. Gabriel gives pure primary amines, but not from aryl or tertiary halides. Now go and try the practice problems on this page. All the best!