Part C: Concept Checks

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Treat ethanamine with an acid chloride, and you get an amide. Treat it with benzenesulphonyl chloride, and you get a sulphonamide. Both products still have one hydrogen on the nitrogen. And yet only the sulphonamide dissolves in alkali. So what makes the difference? Example two will answer it. This is Part C of the unit problem bank: short concept checks, where every answer needs a reason. Questions like these come up often in JEE, so let's begin.

0:33Check 1: basic strength and boiling points

Check number one has two parts. First, why are amines stronger bases than alcohols and ethers? The amine offers a lone pair on nitrogen, the others a lone pair on oxygen. So which atom holds its lone pair less tightly, nitrogen or oxygen? Yes, exactly, it is nitrogen. On screen, N is nitrogen and O is oxygen, and nitrogen is the less electronegative of the two. So it holds its lone pair less tightly. Okay? And a base gives its lone pair to a proton, H plus, with the small plus written high. Nitrogen does this more readily, so amines are the stronger bases. Right? Now part two: primary amines boil higher than isomeric tertiary amines, with the same molecular formula. So what can the primary amine do that the tertiary one cannot? Look at the hydrogens on nitrogen. The small raised circle is read as degree: one degree means primary, and three degree means tertiary. The primary amine has two N H bonds, the dash showing the bond, and the tertiary amine has none. With me? Yes, exactly, it can form hydrogen bonds. So primary amine molecules associate, holding on to one another through hydrogen bonds, and they boil higher. With no hydrogen on nitrogen, tertiary amines cannot do this. Clear?

1:59Check 2: limits of the Gabriel and Hoffmann routes

Now check number two, on making amines. The Gabriel phthalimide synthesis gives pure primary amines. Aniline is a primary amine too. So can we make aniline this way, starting from an aryl halide? Let us see how the Gabriel synthesis works. In it, the phthalimide anion displaces the halide from R X, an alkyl halide, where R is the carbon group and X the halogen, okay? But aryl halides, with the halogen on a benzene ring, do not undergo this nucleophilic substitution with the phthalimide anion. So the answer is no, the Gabriel synthesis cannot give aniline, or any aromatic primary amine. Be careful here, because this is a common trap. Right? Now part two, the Hoffmann bromamide reaction. The amide, RCONH two, with bromine and four N a O H, gives the amine, RNH two, plus sodium carbonate, two sodium bromide and two water. The small numbers written low count the atoms, and the arrow means gives. Now count the carbons on both sides. What do you notice? The amine has one carbon fewer, that is correct. The carbon of the C double bond O group, the equals sign being the double bond, is lost as carbonate. Losing a carbon is why we call it a degradation. Clear?

3:23Check 3: diazotisation and coupling

Now check number three, on diazonium salts. First, diazotisation: a primary aromatic amine, Ar N H two, where Ar is an aryl group, with sodium nitrite, N a N O two, and H C l, at two seventy three to two seventy eight kelvin. The figure shows the salt, N two plus C l minus, coupling with phenol in weakly alkaline solution, O H minus, and with aniline in weakly acidic solution, H plus. The products are orange and yellow azo dyes, with an N double bond N link; the italic p in their names means para. Okay? Now the first question: why do we use an excess of mineral acid when we diazotise aniline? Here is a hint: look at the lower row of the figure. That lower row shows aniline coupling with the diazonium salt. So the excess acid keeps any unreacted aniline protonated. Right? And protonated aniline cannot couple with the diazonium salt as it forms, so this unwanted coupling is stopped. With me? Second question: in the figure, phenol and aniline both couple, but plain benzene does not. So what do these two rings have that benzene lacks? The note under the figure gives you a clue. They are strongly activated, by the O H and the N H two groups, that is correct. As the note under the figure says, the diazonium ion is only a weak electrophile. So it attacks only strongly activated rings, not plain benzene. Got it?

5:00Example 1: why secondary amines fail the carbylamine test

Now the two examples. Example one: why do secondary amines not give the carbylamine reaction? Here is a hint: count the hydrogens on nitrogen. Pause the video and try it first. Done? Okay, let's check. The primary amine, RNH two, heated with chloroform, CHC l three, and three KOH in alcohol, gives RNC, an isocyanide with a very offensive smell, along with three K C l and three water molecules. Okay? The reactive species is dichlorocarbene, C C l two, and the two dots in front are a lone pair on its carbon. Now compare RNH two with RNC, the isocyanide. How many hydrogens has the nitrogen lost? It has lost both of them, exactly. The reaction needs two hydrogen atoms on nitrogen, and they are lost together with the two chlorine atoms of dichlorocarbene. Right? Now the secondary amine, R two N H, has only one hydrogen on nitrogen. So it cannot form the isocyanide, and only primary amines, aliphatic or aromatic, respond. Got it?

6:14Example 2: why Hinsberg uses a sulphonyl, not acid, chloride

Now Example two, a step harder: our opening puzzle. Why is a sulphonyl chloride, and not an acid chloride, used to tell the three classes of amines apart? The figure shows Hinsberg's test: shake the amine with benzenesulphonyl chloride, C six H five S O two C l, and aqueous KOH, an alkali. Pause the video and try it first. Done? Okay, let's check, class by class. The primary amine gives C six H five S O two NHR, with one hydrogen left on nitrogen, and the strongly electron withdrawing sulphonyl group makes that hydrogen acidic. So this product is soluble in alkali. Okay? The secondary amine gives C six H five S O two N R two, with no hydrogen left on nitrogen, so it is insoluble in alkali. Right? And the tertiary amine has no hydrogen on nitrogen to lose, so it does not react at all. The notes on the right of the figure say what you actually see in the test tube. With me? Now here is the trap. An acid chloride does not react with a tertiary amine either, for the same reason: no hydrogen on nitrogen to lose after the acyl group attaches. So why not use an acid chloride? Look at the other two classes. With an acid chloride, primary and secondary amines both react, and both give amides. Okay? And the N H of an ordinary amide is not acidic enough to dissolve in alkali. So the amides from the primary and the secondary amine both fail to dissolve in alkali, and you cannot tell them apart. Clear? Only the sulphonyl chloride gives three different results. And that solves our puzzle: the sulphonyl group makes the N H acidic enough to dissolve in alkali, while an amide's N H is not. Got it?

8:06Quick revision: the seven amine reasons in one place

So, quick revision. Nitrogen holds its lone pair less tightly than oxygen, so amines are stronger bases. Primary amines hydrogen bond, so they boil higher than their tertiary isomers. Gabriel gives no aromatic amines, and Hoffmann gives one carbon fewer. In diazotisation, excess acid stops coupling with unreacted aniline. The carbylamine test is for primary amines only, and a sulphonyl chloride tells all three classes apart. Now go and try the practice problems on this page. All the best!