Naming and Structure

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle to start. Ethanal and benzaldehyde are both aldehydes, ending in the same CHO group. Yet ethanal always exists in equilibrium with a tiny amount of a second form, its enol, while benzaldehyde has no enol at all. So what does ethanal have that benzaldehyde does not? Today we learn to name aldehydes and ketones, see how the carbonyl group is built, and solve this puzzle. These ideas come up often in JEE, so let's begin.

0:34The carbonyl group

So, everything today is built on the carbonyl group: a carbon joined to an oxygen by a double bond, the two short lines. The little arrow head on the left, like a greater than sign, stands for the carbon's two other bonds. Okay? If at least one of those bonds carries a hydrogen, we get an aldehyde, RCHO, where R is the rest of the molecule and the dash is a bond. If both carry carbon groups, we get a ketone, RCOR prime; the small mark, prime, means a second group. Right? The carbon right next to the carbonyl group is the alpha carbon, this Greek letter, and the one after it is the beta carbon, the Greek letter beta. Remember it; it solves our puzzle later. Clear?

1:17Naming: alkanal and alkanone

Now let us learn the naming rules. For an aldehyde, alkane becomes alkanal; the arrow means becomes. In a chain, the CHO carbon is always carbon one, written C one. So acetaldehyde, a common name, is ethanal. With me? For a ketone, alkane becomes alkanone, and the carbonyl carbon, C double bond O, gets the lowest possible number. So acetone is propanone. Okay? Now, suppose the CHO group hangs on a ring, like cyclohexane. Is the CHO carbon one of the ring carbons? No, it sits outside the ring! So we keep the ring name and add the suffix carbaldehyde, giving cyclohexanecarbaldehyde. Got it? And the last rule: when the carbonyl is only a branch, a substituent, the double bonded oxygen, written equals O, is called oxo, and the CHO group is called formyl. Right?

2:11Structure: planar and polar

Now let us look at the structure. The carbonyl carbon is s p two hybridised, with the small two written high. It makes three sigma bonds, this Greek letter, and one pi bond to oxygen, the last Greek letter on the line. Clear? The three sigma bonds lie in one plane, about one hundred and twenty degrees apart; the small circle means degrees. The left of the figure shows this. Okay? Now, oxygen is more electronegative than carbon. So in this double bond, which atom becomes slightly positive? The carbon, yes, exactly! Oxygen pulls the electrons towards itself, so the carbon gets delta plus and the oxygen delta minus; delta, the small Greek letter, means a partial charge. So the carbon is electrophilic, and the oxygen nucleophilic. The rest of the figure, a nucleophile attacking that carbon, is for a later lesson. With me? This polarity gives a large dipole moment, two point three to two point eight debye, with the capital D for debye, larger than for ethers.

3:16Boiling point and solubility

Right? Next, let us look at boiling points. Aldehydes and ketones boil above hydrocarbons and ethers of similar mass, because their polar molecules attract one another; we call it dipole dipole attraction. Got it? But even with such a polar group, they boil below alcohols of similar mass. Why do you think that is? Hydrogen bonding, exactly! Alcohols have an O H group, so their molecules hydrogen-bond to one another. Aldehydes and ketones have no O H group, so no hydrogen bonds form between their own molecules. Right? Here are two facts to remember. Methanal is a gas, and its forty percent solution in water, with the percent sign you see, is called formalin. Ethanal boils at two hundred and ninety four kelvin, the capital K for kelvin. Clear? Now let us look at solubility. Methanal, ethanal and propanone, the lower members, mix with water, because their oxygen accepts hydrogen bonds from water. So, no hydrogen bonds among themselves, but yes with water. As the chain grows, the solubility falls. Okay?

4:26Keto–enol tautomerism

Now, back to our puzzle. Many carbonyl compounds, written as the keto form on the left, exist in equilibrium with an enol form; the two half arrows mean equilibrium. The enol has a C double bond C, with the O H, in brackets, on one of those carbons. Got it? In the enol, the oxygen has gained a hydrogen, and the new double bond sits between the carbonyl carbon and its neighbour. So which hydrogen must have moved? The hydrogen on the alpha carbon, yes! We call it the alpha hydrogen. So this tautomerism needs an alpha hydrogen, and usually the enol is only a tiny proportion. With me? But a beta diketone, with two carbonyl groups one carbon apart, has a large proportion of enol. Its new C double bond C is conjugated with the other carbonyl group, and its O H hydrogen-bonds to that carbonyl oxygen in the same molecule, an intramolecular H-bond. Clear?

5:20Example 1: name 3-methylbutanal and 3-methylbutan-2-one

Now let's try Example one. Give the IUPAC names of these two compounds. The brackets mark a branch, and the small numbers written low count atoms, so the last small two means two methyl groups. Pause the video and try it first. Done? Okay, let's check. In the first compound, the longest chain has four carbons, so it is a butanal. Counting from the left, the branch is on carbon two. So the name is two methylbutanal, right? No, no, wait, that is exactly the trap! The CHO carbon must be carbon one, so we count from the right: CHO is one, C H two is two, and the branched carbon is three. So the name is three methylbutanal. Okay? Now let us name the ketone. Its longest chain also has four carbons. Numbering from the end nearer the carbonyl gives the C double bond O carbon two, not three, and then the methyl branch sits on carbon three. So the name is three methylbutan two one. Got it?

6:26Example 2: which carbonyls show keto-enol tautomerism

Now Example two, our puzzle again. Which of these show keto enol tautomerism: methanal, ethanal, benzaldehyde, propanone, benzophenone, and two comma two dimethylpropanal? Pause the video and try it first. Done? Okay, let's check. Methanal, HCHO, has only one carbon, the carbonyl carbon itself, so there is no alpha carbon and no enol. Right? Ethanal, C H three CHO: its C H three is the alpha carbon, with three hydrogens. Propanone, C H three COCH three, has two such carbons. So both show tautomerism. Clear? Now benzaldehyde and benzophenone. Here the alpha carbon is a ring carbon with no hydrogen, so neither forms an enol. And that answers our puzzle: benzaldehyde has no alpha hydrogen! With me? Last one, two comma two dimethylpropanal. It has nine hydrogens on its three methyl groups. So it must form an enol, right? Oh no, careful, that is the trap! C H three in brackets with a small three means three methyl groups on one carbon, the carbon holding the CHO group. That is the alpha carbon, and it has no hydrogen; the nine sit on beta carbons, okay? So the answer is only ethanal and propanone. Count the hydrogens on the alpha carbon, not in the whole molecule. Got it?

7:56Quick revision: naming, carbonyl structure, keto-enol rule

So, quick revision. Aldehydes are alkanals, with the CHO carbon as carbon one; ketones are alkanones, with the lowest number for the carbonyl carbon; on a ring, use carbaldehyde. The carbonyl group is polar, s p two and planar; that polarity makes them boil above ethers, but with no O H they boil below alcohols. And keto enol tautomerism needs an alpha hydrogen. Now go and try the practice problems on this page. All the best!