Set 1

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Take zero point nine zero grams of dimethylamine, and let it react completely with nitrous acid. The product weighs more than the zero point nine zero grams of amine you put in! So where does the extra mass come from, and how much product do we get? Keep that question in mind, because we settle it in Example two. This is Set one: fifteen mixed questions from the whole unit, in the JEE Main pattern, and the course suggests twenty five minutes for the set. We build three quick tools from its questions, then solve two together. So, let's begin.

0:40Tool 1: count the carbons

So, tool one: count the carbons. Here is the nitrile route. R X is an alkyl halide: R is the carbon group, and the letter X stands for the halogen. The arrow with KCN over it means potassium cyanide, which gives RCN, the nitrile. Reduction, the second arrow, then gives RCH two N H two, where the small numbers written low count atoms. The cyanide brings its own carbon, so the amine has one C, one carbon, more than the halide. Okay? Now, question four of this set. Which bromoalkane gives pentan one amine through a nitrile? Pentan one amine has five carbons, and the nitrile route adds one, so the halide has four. The double arrow means so: the answer is one bromobutane. Writing one bromopentane is the trap, because it forgets the cyanide's carbon. Clear? Now the opposite move, the Hoffmann bromamide reaction. An unsubstituted amide, RCONH two, with bromine, B r two, and sodium hydroxide, N a O H, over the arrow, gives RNH two. The carbon of the C O group leaves, so the amine has one carbon fewer than the amide. Question five: what does three methylbutanamide give? Three methylbutanamide has five carbons, four in the chain, counting the C O carbon, plus a methyl branch on carbon three. Hoffmann removes the C O carbon, so four carbons are left: a chain of three with N H two at the end, and the methyl on carbon two. That is two methylpropan one amine. Got it?

2:20Tool 2: what can the Gabriel synthesis make?

Now tool two, with question six in the box: how many of these six amines can the Gabriel phthalimide synthesis make? In this method, the phthalimide anion replaces the halogen of an alkyl halide, and hydrolysis releases the amine. So it gives pure primary amines; the small raised circle is read as degree, and one degree means primary. The nitrogen sits on an s p three carbon, with a small raised three, the carbon that carried the halogen. Okay? Now, aniline is a primary amine. So the Gabriel synthesis can make it, right? It cannot, so be careful here. To make aniline, the phthalimide anion would have to replace a halogen on a benzene ring, and aryl halides do not undergo this substitution. So aniline is out, and so is four methylaniline. Right? N methylaniline is out for another reason. The italic capital N in its name says a methyl group sits on the nitrogen, so it is secondary, written two degree. And the Gabriel synthesis gives only primary amines. Clear? That leaves methanamine, propan one amine and benzylamine. Benzylamine has a ring, but its nitrogen is on the C H two group, an s p three carbon, so it behaves as an aliphatic amine. So the answer is three. With me?

3:41Tool 3: who gives the carbylamine test?

Tool three is the carbylamine test, from question ten in the box. The primary amine, RNH two, is heated with chloroform, CHC l three, and alcoholic KOH, potassium hydroxide. It gives RNC, an isocyanide with a very offensive smell, along with potassium chloride and water. Okay? Now look at the box. Four methylaniline was out in the Gabriel question. So it fails this test too, right? No, it passes this one, and that is exactly the trap. The carbylamine test works for every primary amine, aliphatic or aromatic. Secondary and tertiary amines, two degree and three degree, do not respond. Clear? Now check each nitrogen. Butan one amine, four methylaniline and cyclohexylamine each carry N H two, so all three are primary; the ring in cyclohexylamine does not change that. N methylaniline and diphenylamine are secondary, and triethylamine is tertiary. So the answer is three again, but a different three!

4:45Example 1: pKa of ethylammonium ion from pKb 3.29

Got it? Now Example one, an easy one, question eight of this set. The p K b of ethanamine is three point two nine, where the small b written low stands for base. Find the p K a of its conjugate acid. Pause the video and try it first. Done? Okay, let's check. The conjugate acid is the amine with one extra proton: C two H five N H three plus, the ethylammonium ion, and the raised plus is its charge. Okay? At two hundred and ninety eight kelvin, p K a plus p K b equals fourteen, where p means minus log, and the small a stands for acid. Right? Now, ethanamine is a stronger base than ammonia. So its conjugate acid must be strong too, with a small p K a, right? Careful, it is the other way round. Fourteen minus three point two nine gives ten point seven one. That is a large p K a, so this is a weak acid. The weaker the base, the stronger its conjugate acid, so a fairly strong base like ethanamine has a weak conjugate acid.

5:54Example 2: mass of N-nitrosodimethylamine from 0.90 g amine

Clear? And now Example two, question twelve, which is our puzzle from the start. Find the mass of N nitrosodimethylamine formed when zero point nine zero grams of dimethylamine reacts completely with nitrous acid. Pause the video and try it first. Done? Okay, let's check. Dimethylamine has two methyl groups on nitrogen, shown by the bracket with a small two outside, so it is a secondary amine. Its molar mass, capital M, is twenty four for two carbons, plus seven for the hydrogens, plus fourteen for nitrogen, which makes forty five. Okay? Now tell me, what does a secondary amine give with nitrous acid? Yes, exactly, a secondary amine with nitrous acid, HNO two, gives a yellow oily N nitrosoamine. The italic N says the N O group sits on the nitrogen, in place of its hydrogen; the dash is a single bond, and the equals sign a double bond. One mole of amine gives one mole of product. Right? So the moles of amine are zero point nine zero over forty five, which gives zero point zero two moles. One to one, so zero point zero two moles of product too. With me? The product's molar mass is twenty four for two carbons, plus six for the hydrogens, plus twenty eight for two nitrogens, plus sixteen for oxygen, seventy four in all. Zero point zero two times seventy four gives one point four eight grams. Got it? And here is our puzzle solved. The N O group, of mass thirty, took the place of one hydrogen, of mass one. Forty five minus one is forty four, and adding thirty gives seventy four. The extra mass came from the nitrous acid, so nothing was created! Clear?

7:39Quick revision: count carbons, Gabriel, carbylamine, pKa

So, quick revision. First, count the carbons: the nitrile route adds one, and Hoffmann removes one. Second, Gabriel makes only primary aliphatic amines, while the carbylamine test catches every primary amine. Third, p K a plus p K b is fourteen at two hundred and ninety eight kelvin, and a secondary amine with nitrous acid gives an N nitrosoamine. Now try all fifteen problems of Set one on this page in twenty five minutes, then check the solutions. All the best!