Preparation of Alcohols

Transcript

0:00Introduction

Hello everyone! Let me start with a small puzzle. Take propene, and add water across its double bond. One student says the O H group goes to carbon two. Another says it goes to carbon one. So, who is right? By the end of this video, you will make both right, just by choosing the reagent. This idea comes up often in JEE, so let's begin.

0:26From alkenes: water adds the Markovnikov way

We start with alkenes. In acid-catalysed hydration, water adds with an acid catalyst, in the Markovnikov sense, through a carbocation. So, back to our propene. Which carbon gets the O H group? Carbon two, yes, exactly! On screen, the dashes are single bonds, the equals sign is the double bond, and the small numbers written low count the atoms. H plus over the arrow, with its plus sign written high, is the catalyst. The proton goes to the end carbon, which has more hydrogens, so the carbocation, and the O H, end up on the middle carbon. The O H in brackets hangs on that carbon: propan two ol. Clear? But careful, here is the trap. The carbocation may rearrange, by shifting a group from the next carbon, to become more stable. Then the O H lands where you did not expect. Got it? To avoid that, we use oxymercuration demercuration. First, mercuric acetate, written H g, then O A c in brackets with a small two, together with water; the slash means together. Then sodium borohydride, N a B H four. You get the Markovnikov alcohol, with no rearrangement. With me?

1:44From alkenes: the anti-Markovnikov route

Now the third method, hydroboration oxidation. First diborane, B two H six, or B H three in THF, the solvent tetrahydrofuran. Then hydrogen peroxide, H two O two, with hydroxide, O H minus; the small minus sign written high is its charge. Okay? So, propene again. This time, which carbon gets the O H group? Carbon one, the end carbon, exactly! We get C H three, C H two, C H two, O H, which is propan one ol. So there is our puzzle, solved: both students were right, each with a different reagent. Clear? This is anti-Markovnikov and syn addition: the H and the O H add to the same face of the double bond, and there is no rearrangement. Got it? Also from alkenes, cold, dilute, alkaline potassium permanganate, K M n O four, called Baeyer's reagent, gives a diol, two O H groups, again by syn addition. Okay?

2:51From carbonyl compounds: reduction

Now the second family: carbonyl compounds. Reducing an aldehyde gives a primary alcohol, and a ketone gives a secondary one. The small circle written high is read as degree: one degree means primary, and two degree means secondary. Okay? The reagents are hydrogen with nickel, platinum or palladium, sodium borohydride, or lithium aluminium hydride, L i A l H four. Suppose a molecule has a carbon carbon double bond and a CHO group. Hydrogen with nickel, or sodium borohydride: which one reduces only the CHO group? Yes, sodium borohydride is the one! Hydrogen with nickel would attack the double bond too, but sodium borohydride leaves it untouched. It also does not reduce acids or esters. Right? So acids and esters need lithium aluminium hydride, written over the arrow, to give primary alcohols. Clear?

3:50Grignard reagents: the carbonyl decides the alcohol

Now a very useful method, the Grignard reagent, R M g X, where R is an alkyl group, M g is magnesium, and X is a halogen, okay? It adds to the carbonyl group, C double bond O, and hydrolysis then gives the alcohol. Clear? With methanal, HCHO, you get a primary alcohol, RCH two O H, okay? With any other aldehyde, R prime CHO, you get a secondary alcohol; R prime, with the small dash written high, is a second alkyl group. Right? Now you can see the pattern. So what does a ketone give? Yes, exactly, a tertiary alcohol! The ketone is R prime COR double prime, where the two small dashes mark a third group. It has no hydrogen on its carbonyl carbon, so the alcohol carbon gets three groups. Right? Careful, this is where marks are lost. An ester or an acid chloride takes two molecules of R M g X, and gives a tertiary alcohol with two identical R groups. We will see why in example two. With me? Two more rows: methanoate esters give a secondary alcohol, R two CHOH, while oxirane, ethylene oxide, gives a primary alcohol, RCH two C H two O H, two carbons longer. Got it?

5:14Other methods; methanol and ethanol

Here are two more methods. An alkyl halide, R X, with aqueous KOH or moist silver oxide, A g two O, gives the alcohol, and so does a primary aliphatic amine with nitrous acid, HNO two. An aromatic amine like aniline gives a diazonium salt instead. Okay? Methanol, C H three O H, is made from carbon monoxide and hydrogen, C O plus two H two, over zinc oxide with chromium oxide, Z n O, C r two O three, at five hundred and seventy three to six hundred and seventy three kelvin, and two hundred to three hundred atmospheres. And remember, it is highly poisonous. Right? Ethanol comes from the fermentation of sugars: invertase converts sucrose into glucose and fructose, and zymase converts these into ethanol. Clear?

6:10Example 1: but-1-ene by acid hydration and hydroboration

Alright, Example one, your turn. Name the alcohol formed from but one ene by, part a, acid-catalysed hydration, and part b, hydroboration oxidation. Pause the video and try it first. Done? Okay, let's check. But one ene is C H two, double bond, C H, C H two, C H three. Right? Let us do part a first. The proton goes to carbon one, so a secondary carbocation forms on carbon two; C dash two means carbon number two. Now remember the trap. Will this carbocation rearrange? No, it does not rearrange here. Any shift from carbon three only gives another secondary carbocation, nothing more stable. So the O H goes to carbon two: butan two ol. Okay? Part b is anti-Markovnikov, so the O H goes to carbon one, giving butan one ol. Same alkene, two alcohols, just like our propene puzzle! If you got both, very good. Clear?

7:17Example 2: ethyl ethanoate with two equivalents of EtMgBr

Now a harder one, Example two. Ethyl ethanoate is treated with two equivalents of ethylmagnesium bromide, and then with dilute acid. Name the product, and explain why two equivalents are used. Pause the video and try it first. Done? Okay, let's check. Ethyl ethanoate is C H three COOC two H five, and the Grignard reagent, C two H five M g B r, is written over the arrow. The first equivalent gives a ketone, C H three COC two H five, which is butan two one. Right? Now, here is the key question. Does the reaction stop at this ketone? No, it does not stop there! The ketone is more reactive than the ester, so it takes up the second equivalent. That is why two equivalents are used. Clear? After dilute acid we get C H three, C, then O H in brackets, then C two H five in brackets with a small two, meaning two ethyl groups on that carbon. The longest chain, through the central carbon, has five carbons. The O H and the methyl group sit on carbon three, so it is three methylpentan three ol, a tertiary alcohol with two identical ethyl groups, just as our table said. If you explained the two equivalents, well done!

8:42Quick revision: alkene hydration, reduction, Grignard

So, quick revision. From alkenes, the acid and mercury routes give the Markovnikov alcohol, and the diborane route the anti-Markovnikov one. Reduction turns aldehydes into primary and ketones into secondary alcohols, and sodium borohydride spares the double bond. With Grignard reagents, methanal gives primary, other aldehydes secondary, and ketones or esters tertiary alcohols, except methanoate esters, which give secondary. Now you know which reagent gives which alcohol. Go and try the practice problems on this page. All the best!