The Carbon–Halogen Bond and Physical Properties

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Here are three liquids, all with the same formula, C four H nine B r. Yet one boils at three hundred and seventy five kelvin, the second at three hundred and sixty four kelvin, and the third at three hundred and forty six kelvin. Same formula, but three different boiling points? Strange, isn't it? By the end of this video, you will match each one to its name. Questions on such orders do come up in JEE, so let's begin.

0:33The C–X bond is polar

So, first, the bond itself. In the heading, C is carbon, and X stands for any halogen. Now let me ask you this. In this bond, which atom pulls the shared electrons towards itself, carbon or the halogen? It is the halogen, yes, exactly! It is more electronegative than carbon, so the bond is polar. On screen, the small Greek letter delta with a plus sign, delta plus, on carbon, means a partial positive charge, and delta minus on the halogen means a partial negative charge. The dash is the bond. Okay? So the delta plus carbon is open to attack by nucleophiles. We will use this idea again in the reactions of haloalkanes. Right?

1:17Down the group: longer, weaker C–X bonds

Now, let us go down the group: fluorine, chlorine, bromine and iodine, the symbols F, C l, B r and I on screen. The halogen atom gets larger. So tell me, does the carbon halogen bond get longer, or shorter? It gets longer, very good! C H three F is methyl fluoride; the small three written low means three hydrogen atoms. And p m is picometres. The length rises, as the less than signs show, from one hundred and thirty nine in methyl fluoride, to one hundred and seventy eight in methyl chloride, one hundred and ninety three in methyl bromide, and two hundred and fourteen in methyl iodide. With me? And a longer bond is a weaker bond. Bond enthalpy is the energy needed to break it, in kilojoules per mole; the minus one written high means per mole. It falls, as the greater than signs show, from four hundred and fifty two in methyl fluoride, to three hundred and fifty one in methyl chloride, two hundred and ninety three in methyl bromide, and two hundred and thirty four in methyl iodide. Clear? We write an alkyl halide as R X, where R is the alkyl group. The weakest bond breaks most easily. So, for the same R, which halide reacts fastest? It is the iodide, yes! Reactivity goes R I, greater than R B r, greater than R C l, greater than R F, for the same R, okay? The weakest bond gives the fastest reaction. Got it?

2:42Dipole moment: the fluoride surprise

Now, dipole moment, which tells us how polar the molecule is. Fluorine is more electronegative than chlorine, so methyl fluoride has the largest dipole moment, right? No, no, wait, look at the data first. Capital D is the debye, the unit of dipole moment. Methyl fluoride, one point eight five; methyl chloride, one point eight six; methyl bromide, one point eight three; and methyl iodide, one point six four. So methyl chloride wins, just slightly. Surprised? So the order is methyl chloride, greater than methyl fluoride, greater than methyl bromide, greater than methyl iodide. Putting the fluoride first is the trap here. Okay? Why does this happen? Mu, the Greek letter we use for dipole moment, equals q times d, the charge times the distance between the charges. The charge separation is largest in methyl fluoride, but its bond is so short that q times d comes out slightly smaller than for the carbon chlorine bond. Clear?

3:45Boiling point: what pushes it up or down

Now, boiling points. Take a haloalkane, and a hydrocarbon of similar mass. Which one boils at the higher temperature? It is the haloalkane, that is correct! There are dipole dipole forces, because the bond is polar, and stronger van der Waals forces too. More attraction means more heat to boil it. With me? Next, the boiling point rises with chain length, and with the number of halogen atoms. Now take isomers, with the same formula. Does branching raise the boiling point, or lower it? It lowers it, exactly! On screen, one, two and three with a little circle written high are read primary, secondary and tertiary. So among isomers, primary, greater than secondary, greater than tertiary. Hold on to this; our three mystery liquids need it. Got it?

4:40Melting point, density and solubility

Now, melting points. Take the three dihalobenzenes, a benzene ring with two halogen atoms, as the ortho, meta or para isomer. The para isomer melts much higher, because its symmetrical molecules pack better in the crystal. Right? Next, density. It rises with the number of halogen atoms, and with the atomic mass of the halogen. But careful: for the same halogen, a longer carbon chain makes it slightly less dense, because the extra C H two groups are light. Okay? Now compare with water. Bromo, iodo and polychloro compounds are denser than water, while alkyl fluorides and alkyl monochlorides are lighter. Clear? And solubility. Haloalkanes are only slightly soluble in water, but they dissolve freely in organic solvents. But wait, these molecules are polar. So why do they hardly dissolve in water? Here is the reason. They cannot form hydrogen bonds with water, the H bonds on screen. To dissolve, they must break the hydrogen bonds between water molecules, and they cannot repay that energy. Got it?

5:48Example 1: boiling point order of CH3F, CH3Cl, CH3Br, CH3I

Now, Example one, an easy one. Arrange in order of increasing boiling point: methyl chloride, methyl bromide, methyl iodide and methyl fluoride. Pause the video and try it first. Done? Okay, let's check. All four have the same C H three group, the methyl group, and only the halogen changes. Now, from our data, methyl chloride has the largest dipole moment. So it boils the highest, right? Oh no, careful, that is the trap! From fluorine to iodine, the size and mass of the molecule rise, and so do the van der Waals forces. That decides the order here. Okay? So the increasing order is methyl fluoride, less than methyl chloride, less than methyl bromide, less than methyl iodide. Methyl iodide, the heaviest, boils the highest. Right?

6:40Example 2: match C4H9Br isomers to 375, 364 and 346 K

Now, Example two: our three mystery liquids from the start. The boiling points of three isomeric bromides, C four H nine B r, are three hundred and seventy five kelvin, three hundred and sixty four kelvin, and three hundred and forty six kelvin; capital K is kelvin. Match them with one bromobutane, two bromobutane, and two bromo two methylpropane. Pause the video and try it first. Done? Okay, let's check. They are isomers, so only the shape matters. More branching means a smaller surface area, so weaker van der Waals forces, and a lower boiling point. So which of the three boils the highest? It is one bromobutane, yes, exactly! Its chain is unbranched, with the bromine on the end carbon, a primary bromide with the largest surface area. It gets three hundred and seventy five kelvin. With me? Next, two bromobutane is a secondary bromide, so it comes in the middle, at three hundred and sixty four kelvin. Clear? And two bromo two methylpropane, the tertiary bromide and the most branched, boils the lowest, at three hundred and forty six kelvin. See, our puzzle is solved: the same formula, but different shapes, so different boiling points. Okay?

8:01Quick revision: C-X bond, reactivity, dipole, boiling point

So, quick revision. The C X bond gets longer and weaker down the group, so reactivity goes R I, then R B r, then R C l, then R F, right? For dipole moment, methyl chloride is the highest, just above methyl fluoride. And boiling point goes up with mass and with the size of the halogen, and down with branching. Now go and try the practice problems on this page. You will find them easy. All the best!