Alkylation and Acylation

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Take ethanamine, a primary amine, and N ethylethanamine, a secondary amine. Add ethanoyl chloride, and both of them give amides. Now take N, N diethylethanamine, a tertiary amine. Its nitrogen still has a lone pair, ready to react. But this time, no amide forms at all! So why does the tertiary amine refuse? Reactions like these come up often in JEE, so let's find out.

0:32Alkylation: the lone pair as a nucleophile

So, everything in this chapter starts from the lone pair on nitrogen. Nitrogen can use this electron pair to form a new bond, so the amine acts as a nucleophile. Okay? Now, alkylation. Treat a primary amine, RNH two, with an alkyl halide, R X, written above the arrow; R is an alkyl group, and X is a halogen. An alkyl group replaces one hydrogen on nitrogen, giving the secondary amine, R two N H; the small numbers written low are counts. Right? So, does the reaction stop here? Look at the secondary amine: does its nitrogen still have a lone pair? It does, yes, exactly, so it reacts again with R X and gives R three N, the tertiary amine. Clear? And the tertiary amine still has its lone pair, so it reacts once more. Now nitrogen carries four R groups and a positive charge, the raised plus, with the halide ion, X minus, as its partner. This is a quaternary ammonium salt, and here the chain ends. Got it? So with excess alkyl halide, the reaction runs all the way to the quaternary salt. This is called exhaustive alkylation. With me?

1:46Acylation: an acyl group replaces H on nitrogen

Now, acylation. Here a primary amine, RNH two, reacts with an acid chloride, R prime COC l; the little dash after R is read R prime, and it means another group. We get RNHCOR prime, an amide, and H C l, hydrogen chloride; the arrow means gives. Okay? Acid anhydrides and esters do the same job. In every case, the acyl group, R prime C O, replaces a hydrogen on nitrogen, and we get an amide. Right? Now, the reaction also gives hydrogen chloride, an acid, and the amine is a base. So what will this acid do to the amine that has not reacted yet? That is correct, it turns the amine into its salt, which cannot react. So we add a base, like pyridine or sodium hydroxide, to remove the hydrogen chloride as it forms and drive the reaction to completion. Clear? Now back to our puzzle. The tertiary amine, R three N, has a lone pair, but how many hydrogens does it have on nitrogen? None at all, and that is the answer! The small raised circle is read as degree, so three degree means tertiary, and the dash in N H is the bond. Acylation needs a hydrogen on nitrogen to replace, so a tertiary amine is not acylated, even with its lone pair. So our puzzle is solved. With me?

3:14Acetylation, benzoylation and imines

Now, two acylations you must know by name. First, acetylation, with ethanoyl chloride, the first formula, or with ethanoic anhydride, two C H three C O groups joined through an oxygen; the small two after the bracket means two of them. Okay? Each N H two group becomes NHCOCH three; the dash in front is the bond to the rest of the molecule. And remember this number: for each such group, the molar mass rises by forty two. Right? But wait. The acetyl group, COCH three, has a mass of forty three. So shouldn't the rise be forty three? That is the trap, because we also lost a hydrogen! Forty three comes in, a hydrogen of mass one goes out, and forty three minus one gives forty two. Clear? Second, benzoylation, with benzoyl chloride, C six H five COC l, where C six H five is the phenyl group. Aniline gives benzanilide, whose formula is on screen: a phenyl group on one side of the N H, and a benzoyl group on the other. Got it? And one more reaction, only for primary amines; with the raised circle again, one degree means primary. With aldehydes and ketones, primary amines give imines, also called Schiff's bases. Okay?

4:39Example 1: which amine is not acetylated, and why

Now Example one, an easy one. Which of these is not acetylated by ethanoic anhydride: ethanamine, N methylethanamine, N, N dimethylethanamine, or aniline? Pause the video and try it first. Done? Let's check, and the key is the hydrogen on nitrogen. Ethanamine, C two H five N H two, is primary. In N methylethanamine, the italic N says the methyl group sits on nitrogen, so it is secondary. Both have N H bonds, so both are acetylated. Right? N, N dimethylethanamine has two methyl groups and an ethyl group, all on nitrogen. It is tertiary, with no N H bond, just like the amine in our puzzle. Clear? Now aniline, the aromatic amine. It is a much weaker base than ethanamine, so it will not be acetylated either, right? That is exactly the trap of this question, so be careful! Aniline, C six H five N H two, is a primary amine with two N H bonds, so it is acetylated, and it gives acetanilide. What decides acylation is the hydrogen on nitrogen. Got it? So the only one not acetylated is N, N dimethylethanamine, because its nitrogen carries no hydrogen. With me?

6:03Example 2: count N-H groups from the +42 acetylation rise

And now Example two, a harder one. Read it with me: a polyamine of molar mass one hundred and three gives, on complete acetylation, a product of molar mass two hundred and twenty nine. How many amino groups carrying hydrogen does it have? Suggest a structure too. Pause the video and try it first. Done? Good, let's check. The rise in molar mass is two hundred and twenty nine minus one hundred and three, which gives one hundred and twenty six. Okay? Now, each acetyl group adds forty two. So how many acetyl groups went in? One hundred and twenty six over forty two gives three, yes, exactly! So the polyamine has three amino groups carrying hydrogen. Right? Now one possible structure: an N H two group, two C H two groups, an N H group, two more C H two groups, and another N H two at the end; the short dashes are bonds. Clear? Let us check it. Its formula is C four H thirteen N three. The carbons give forty eight, the hydrogens give thirteen, and the nitrogens give forty two. Forty eight plus thirteen plus forty two gives one hundred and three, as given. Got it? But look, this structure has only two N H two groups, one at each end. So only two groups get acetylated, right? The answer is no, and that is the second trap of the day! The N H in the middle is a secondary amine, and secondary amines are acylated too. So two N H two groups and one N H group take three acetyl groups, and three times forty two is one hundred and twenty six, exactly our rise. With me?

7:54Quick revision: alkylation, acylation and the +42 rule

So, quick revision. Alkylation takes a primary amine to the secondary, then the tertiary amine, and finally the quaternary ammonium salt. In acylation, an acyl group replaces a hydrogen on nitrogen, a base removes the hydrogen chloride, and tertiary amines are not acylated. Acetylation adds forty two per acetyl group, and benzoylation turns aniline into benzanilide. Now try the practice problems on this page, and in every question, first count the hydrogens on nitrogen. All the best!