From Alcohols

Transcript

0:00Introduction

Hello everyone! Let me start with a small puzzle. Take an alcohol with its O H group on carbon two, and treat it with hydrogen bromide. The bromine simply takes the place of the O H group, right? But for one such alcohol, the main product has its bromine on the next carbon! So how did the bromine jump? We will find out today. This comes up often in JEE, so let's begin.

0:26Hydrogen halides: the direct method

So, on screen, ROH is an alcohol, R being the alkyl group, and H X is a hydrogen halide, X being the halogen. The arrow means gives, and the small two written low in H two O is a subscript. So ROH plus H X gives R X, the alkyl halide, plus water. Now tell me, which reacts fastest: H I, H B r or H C l? Yes, exactly, H I reacts fastest! The sign between them, wide open on the left, is read as greater than. So H I is greater than H B r, greater than H C l. Okay? Now the alcohols. The small circle written high is read as degree, so three degree means tertiary, two degree secondary, and one degree primary. Tertiary reacts fastest because the reaction goes through a carbocation, a positively charged carbon, or something close to one, and tertiary carbocations are the most stable. Right? Primary and secondary alcohols need help. Conc. on screen is short for concentrated. We use concentrated H C l with anhydrous, that is completely dry, zinc chloride, Z n C l two. This mixture is the Lucas reagent. With me? Tertiary alcohols need no help. Just shake them with concentrated H C l at room temperature. Clear?

1:54Where do HBr and HI come from?

Now, H B r is generated from sodium bromide and sulphuric acid, H two S O four. So for H I, we take potassium iodide, K I, with the same sulphuric acid, right? No, no, wait, that is the trap! For H I, we take potassium iodide with phosphoric acid, H three P O four, and not with sulphuric acid. Got it? So why not sulphuric acid? Because concentrated sulphuric acid oxidises H I to iodine: two H I plus H two S O four gives I two, plus S O two, plus two H two O, as shown. Phosphoric acid is a non-oxidising acid, so it is used instead. Clear?

2:40Phosphorus halides

Next, phosphorus pentachloride, P C l five. ROH plus P C l five gives R C l, plus POC l three, plus H C l. Okay? With a phosphorus trihalide, P X three, three ROH give three R X, plus H three P O three. So, with excess alcohol, which gives more alkyl chloride per mole, P C l three or P C l five? Very good if you said P C l three! With excess alcohol, one mole of P C l three gives three moles of R C l, but P C l five gives only one, because POC l three keeps three of its chlorines. Clear? And P B r three and P I three are made in the flask, from red phosphorus, red P on screen, with bromine or iodine. Right?

3:29Thionyl chloride, and the rearrangement problem

Now the best method, thionyl chloride, SOC l two. ROH plus SOC l two gives R C l, plus S O two, plus H C l, and the small arrows pointing up mean these two escape as gases. So why is it the best? Yes, exactly, both by-products are gases, so they escape, and the alkyl chloride is obtained pure! Now, the catch with hydrogen halides. With a secondary or tertiary alcohol, H X reacts through a carbocation, and it may rearrange. Remember our puzzle? This is the clue. But P X three and thionyl chloride do not give rearranged products. Got it? Here is a topper's trick. If a question wants no rearranged isomers, reach for P X three or thionyl chloride, not hydrogen halides. With me?

4:22Why phenols do not react this way

One last question. Ethanol and H C l give chloroethane. So can phenol and H C l give chlorobenzene? No, it cannot, and here is why. C dash O on screen is the carbon oxygen bond. In phenol, it has partial double bond character because of resonance, and the carbon is s p two, with the two written high. So the bond is too strong to break, and phenols give no aryl halide this way.

4:51Example 1: SOCl2 mass and gas volume from 0.5 mol butanol

Clear? Alright, Example one, your turn. Zero point five moles of butan one ol is converted into one chlorobutane with thionyl chloride. Find the mass of thionyl chloride needed, and the total volume of the gases evolved at STP, standard temperature and pressure. Sulphur is thirty two, chlorine is thirty five point five. Pause the video and try it first. Done? Okay, let's check. Butan one ol is C four H nine O H, and it gives C four H nine C l, S O two and H C l. Everything is one to one, as the colon on screen shows. Okay? Now M, the molar mass of thionyl chloride. Thirty two for sulphur, sixteen for oxygen, and two times thirty five point five for chlorine, which gives one hundred and nineteen grams per mole. The minus one written high on mol means per mole. Right? So the mass is zero point five times one hundred and nineteen, which gives fifty nine point five grams. Now the gases. Zero point five moles, so eleven point two litres, right? Oh no, careful, there are two gases, not one! Zero point five moles of S O two plus zero point five moles of H C l makes one mole of gas. Clear? At STP, one mole of gas occupies twenty two point four litres, and that is our total volume. That's it!

6:22Example 2: 3-methylbutan-2-ol with HBr, hydride shift

Now the harder one, and it brings back our puzzle. Example two. Three methylbutan two ol is treated with H B r. Write the structure of the major product, and explain. Pause the video and try it first. Here is a hint: what kind of alcohol is it? Done? Okay, let's check. The bracket with a small two means two methyl groups, C H three, on one carbon, and the dashes are bonds. The O H group is on carbon two, written C dash two, which is joined to two carbons, so the alcohol is secondary. So bromine simply replaces the O H group, right? No, no, wait, remember the catch! Water leaves, and carbon two is left with a positive charge, the small plus written high. This is a secondary carbocation. Okay? Now look at the next carbon, which carries a hydrogen. That hydrogen moves over to carbon two, and this is called a one two hydride shift. The positive charge now sits on a carbon joined to three carbons, a tertiary carbocation, which is more stable. Right? Only now does the bromide ion join, B r with a small minus written high. It bonds to the positive carbon, next to where the O H group was. With me? So the major product is two bromo two methylbutane. And there is our puzzle, solved! The bromine never jumped. The positive charge moved first, and the bromine followed it. Got it?

7:53Quick revision: HX, PCl5, PX3, SOCl2 and rearrangement

So, quick revision. With hydrogen halides, H I reacts fastest and H C l slowest, tertiary alcohols fastest and primary slowest, with the Lucas reagent for primary and secondary ones. Thionyl chloride gives gaseous by-products and a pure alkyl chloride, while P C l five gives one R C l, and P X three gives three. Hydrogen halides may rearrange, P X three and thionyl chloride do not, and phenols do not react this way. Now go and try the practice problems on this page. All the best!