Formulas, Isomer Counts and Analysis

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. Ethanamine has a molar mass of forty five, aniline has ninety three, and pyridine has seventy nine. All odd numbers, right? But ethane one two diamine has sixty, and urea also has sixty, both even. Is that just luck? And one more puzzle: given only the molar mass of a saturated amine, can you find its full molecular formula? By the end of this video you will answer both. Questions like these come up often in JEE, so let's begin.

0:35The formula of a saturated amine

So, let us build the formula ourselves. Start with a primary amine, RNH two, where N is nitrogen, H is hydrogen, and R is an alkyl group. R equals C n, H two n plus one, with C for carbon. The small letters and numbers written low are subscripts; they count the atoms. Okay? Now count all the hydrogen atoms in RNH two. There are two n plus one in the alkyl group, and some more on the nitrogen. So how many hydrogens in total? It is two n plus three, yes, exactly! Two more hydrogens sit on the nitrogen. So a saturated, open chain amine with one nitrogen atom is C n, H two n plus three, N, whatever its class, as the brackets say. But keep all three conditions in mind: saturated, open chain, one nitrogen. Right? Now the molar mass. Carbon is twelve, hydrogen is one, and nitrogen is fourteen. So we get twelve times n, plus two n plus three in brackets, plus fourteen. That adds up to fourteen times n, plus seventeen. Write this line down; it does most of today's work. Got it?

1:46Degree of unsaturation

Next, a quick check on any formula: the degree of unsaturation. For a compound with c carbons, h hydrogens and n nitrogens, shown as subscripts, it is two times c, plus two, plus n, minus h, all over two. The long line is a fraction bar, so the whole top is divided by two. Okay? Now let me ask you this. What does the formula give for our saturated amine, C n, H two n plus three, N? It comes out as zero, very good! Careful with the letters here: in this line n counts the carbons, and the plus one is our single nitrogen. So on top, two n plus two plus one, minus the bracket, two n plus three, cancels to zero. Zero means no ring and no double bond, which is exactly what saturated and open chain mean. Clear? Now pyridine, which is C five H five N, okay? Two times five is ten, so the top is ten plus two plus one, minus five, which gives eight. And eight over two gives four. That four comes from one ring and three double bonds. With me?

2:51Odd molar mass? Count the nitrogens

Now, back to our first puzzle. Ethanamine, aniline and pyridine each have one nitrogen atom, and their molar masses are forty five, then ninety three, then seventy nine. Okay? Ethane one two diamine and urea each have two nitrogen atoms, and both have a molar mass of sixty. So what decides whether the molar mass is odd or even? It is the number of nitrogen atoms, exactly! For a compound of carbon, hydrogen, oxygen and nitrogen, an odd number of nitrogen atoms gives an odd molar mass, and also an odd number of hydrogen atoms. So it was never luck. Right? And our formula agrees. Fourteen times n is always even, and seventeen is odd. Even plus odd is odd. So a saturated monoamine, an amine with one nitrogen, always has an odd molar mass. Got it?

3:43Counting isomeric amines, class by class

Now the most important part: counting isomers. Take one class at a time. One degree, with the small circle written high for degree, means primary, RNH two. Each different alkyl group gives a different primary amine, so we simply count the alkyl groups, C n H two n plus one. Okay? And here are the counts. When n is one, there is one methyl group, and when n is two, one ethyl group. Three carbons give two propyl groups, four carbons give four butyl groups, and five carbons give eight pentyl groups. Right? Two degree means secondary, and three degree means tertiary, with two or three groups on the nitrogen. So we share the carbons between those groups, and every group needs at least one carbon. Clear? Let's try it. How many amines have the formula C three H nine N? Count all three classes. The answer is four, very good! Primary: there are two propyl groups, so two amines. Secondary: three carbons shared as one and two, a methyl and an ethyl, so one amine. Tertiary: three methyl groups, so one amine. Two plus one plus one gives four. With me?

4:53Example 1: amine of molar mass 87, find its formula

Now, remember our second puzzle? One number, and the whole formula. Read it with me: a saturated, open chain amine with one nitrogen atom has a molar mass of eighty seven. Find its molecular formula. Pause the video and try it first. Done? Okay, let's check. Saturated, open chain, one nitrogen, so the molar mass is fourteen times n, plus seventeen. Set that equal to eighty seven. Then fourteen times n equals seventy, so n equals five. The double arrow means which gives. Okay? So the formula is C five, H thirteen, N, because two times five, plus three, gives thirteen hydrogens. Now check it: five carbons give sixty, thirteen hydrogens give thirteen, and nitrogen gives fourteen, which adds up to eighty seven. So yes, one number was enough! And eighty seven is odd, just as our rule said. Clear?

5:52Example 2: vapour density 36.5, C4H11N has 8 isomers

Now a slightly harder one. The vapour density of a saturated amine is thirty six point five. Find its formula, and the number of isomeric amines with this formula. Here is a hint: capital M, the molar mass, is two times the vapour density. Pause the video and try it first. Done? Okay, let's check. The molar mass is two times thirty six point five, which gives seventy three. Then fourteen times n, plus seventeen, equals seventy three, which gives n equal to four. So the formula is C four, H eleven, N, okay? Now the isomers. There are four butyl groups, so the answer is four amines, right? Oh no, careful, that is a common trap! Four is only the primary amines, one for each butyl group. Secondary and tertiary amines have the same formula too, so we must count them as well. Got it? Secondary first: two groups on the nitrogen, sharing four carbons. One and three means a methyl and a propyl group, and there are two propyl groups, so two amines. Two and two means two ethyl groups, so one amine. That makes three secondary amines. With me? Tertiary next: three groups sharing four carbons, so one carbon, one carbon and two carbons. That is two methyl groups and one ethyl group, and there is only one way to do it. So we get one tertiary amine. Clear? So the total is four plus three plus one, which gives eight. Eight amines, not four. So always count all three classes, and you will not fall into this trap.

7:29Quick revision: CnH2n+3N, M = 14n+17, isomer counts

So, quick revision. Saturated, open chain, one nitrogen: the formula is C n, H two n plus three, N, and capital M, the molar mass, is fourteen times n, plus seventeen. Okay? For compounds of carbon, hydrogen, oxygen and nitrogen, an odd number of nitrogen atoms gives an odd molar mass and an odd number of hydrogen atoms. Right? And count isomers class by class. C two H seven N has two, C three H nine N has four, and C four H eleven N has eight. C five H thirteen N has seventeen: eight primary, six secondary and three tertiary. Now go and try the practice problems on this page. All the best!