Reactions of the C–O Bond

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Three bottles of alcohol have lost their labels. All three have the formula C four H ten O, but one is primary, one secondary and one tertiary, okay? You get one reagent and a clock. Can you tell which bottle is which? By the end of this video, you will. Today we break the C O bond of alcohols, the bond between carbon and oxygen. This comes up often in JEE, so let's begin.

0:32Alcohols to alkyl halides

So, the first reaction. ROH is the alcohol, where R is the alkyl group, and H X is a hydrogen halide, where X is the halogen. The arrow means gives: R X, the alkyl halide, plus water, H two O, with the small two written low. The halogen replaces the O H group, so the C O bond breaks. With me? Now let me ask you this. Tertiary, secondary or primary alcohol: which one reacts fastest? Yes, exactly, the tertiary alcohol reacts fastest! The small circle written high is read as degree, so three degree is tertiary, two degree secondary and one degree primary, and the greater than signs give the order. Among the acids, the order is H I, then H B r, then H C l. Okay? Now, the O H group is protonated first. The hydroxide ion is a strong base and a very poor leaving group, but water is a weak base, and it leaves easily. Clear? Other reagents work too: phosphorus pentachloride, P C l five, phosphorus trichloride, P C l three, phosphorus tribromide, P B r three, red phosphorus, the capital P, with iodine, I two, and thionyl chloride, SOC l two. And notice the last words: no rearrangement. Got it?

1:59The Lucas test: a clock for alcohols

Now, the test for our puzzle. Lucas reagent is concentrated hydrochloric acid, H C l, with anhydrous zinc chloride, Z n C l two. The short form on screen means concentrated, and anhydrous means without water. Right? The alcohol dissolves in the reagent, but the alkyl chloride formed does not, so the clear solution turns cloudy: turbidity. Now tell me, which alcohol turns cloudy first? The tertiary one, because it forms its chloride fastest! Tertiary alcohols turn turbid at once, secondary ones within about five minutes, and primary ones only on heating. So the clock tells you the class. Clear?

2:42Dehydration: alcohol to alkene

Now the second reaction: dehydration, which removes water to give an alkene. We heat the alcohol with concentrated sulphuric acid, H two S O four, or phosphoric acid, H three P O four. Or we pass its vapour over alumina, A l two O three, at six hundred and twenty three kelvin; the capital K is kelvin. Okay? For ethanol, a primary alcohol, we use concentrated sulphuric acid at four hundred and forty three kelvin. Remember this temperature. Got it? Now think about this. Secondary and tertiary alcohols: do they need harsher conditions, or milder ones? They need milder conditions, much milder! Secondary alcohols use eighty five percent phosphoric acid at four hundred and forty kelvin, and tertiary ones just twenty percent, at three hundred and fifty eight kelvin. With me? So the ease of dehydration is tertiary, greater than secondary, greater than primary.

3:46How it happens: the E1 mechanism

Clear? Now, the figure shows the mechanism for ethanol, and the short dashes are single bonds. The two dots on the oxygen are a lone pair, used to pick up a proton, H plus, with the plus written high. This gives the protonated alcohol, C H three, C H two, O H two plus. The pair of half arrows means this fast step is reversible. Okay? Next, water leaves, shown as minus H two O, and a carbocation remains, C H three, C H two plus: a carbon with a positive charge. This step is slow, so it decides the rate. Right? Then a proton is lost quickly, giving ethene, C H two, double bond, C H two; the equals sign is the double bond. The proton comes back, so the acid is regenerated. The slow step involves one species, so this is E one: elimination, unimolecular. With me? And here is the twist. Before losing the proton, the carbocation may rearrange to a more stable one. You will see this in Example two. Got it?

4:50Which alkene? And when does an ether form?

Often, two alkenes are possible. The major one is the more substituted alkene, with more alkyl groups on its double bond carbons. This is Saytzeff's rule. Okay? Now a quick question. Ethanol with concentrated sulphuric acid always gives ethene, right? No, not always, and that is a trap, because the temperature matters! At four hundred and thirteen kelvin, with the alcohol in excess, ethanol gives diethyl ether, C two H five, O, C two H five. At four hundred and forty three kelvin, it gives ethene. Got it? The ether forms when one alcohol molecule attacks a protonated one, and water leaves. That is an S N two attack, with the small N written low: substitution, nucleophilic, bimolecular. Clear?

5:40Example 1: Lucas test to classify three C4H10O alcohols

Now, back to our three bottles. Three alcohols, C four H ten O, are shaken separately with Lucas reagent at room temperature. Which gives turbidity at once, which in about five minutes, and which not at all? Pause the video and try it first. Done? Okay, let's check. C H three in brackets with a small three means three methyl groups on the carbon that carries the O H group. So two methylpropan two ol is tertiary, and it turns cloudy at once. Okay? Next, C H three, C H with O H in brackets, C H two, C H three: butan two ol. Its O H carbon is joined to two carbons, so it is secondary, and it turns cloudy in about five minutes. Right? So the third alcohol is primary, and it never reacts with Lucas reagent at all, right? No, that is the trap, so be careful! It does not react at room temperature, but on heating, it does. So it is butan one ol, or two methylpropan one ol; both are primary. And that solves our puzzle.

6:51Example 2: dehydrating 3-methylbutan-2-ol, hydride shift

Clear? Now a harder one, closer to JEE level. Name the major alkene formed when three methylbutan two ol is heated with concentrated sulphuric acid. Here is a hint: watch the carbocation. Pause the video and try it first. Done? Okay, let's check. The alcohol is C H three, C H with the O H, then C H with two methyl groups, the C H three in brackets with a small two. It is protonated and loses water, so the plus charge sits on carbon two: a secondary carbocation. With me? So this cation loses a proton from carbon one, written C one on screen, giving three methylbut one ene, right? No, no, wait, that is the trap! First, the hydrogen on carbon three moves to carbon two with its electron pair, which is a hydride shift. Now the plus charge is on a carbon joined to three carbons: a more stable, tertiary carbocation. Okay? This cation loses a proton from the C H two next to it, and the equals sign shows the new double bond. It carries three alkyl groups, so this is the more substituted alkene, as Saytzeff's rule says: two methylbut two ene. Right? And notice one more thing. Losing the proton directly from carbon three, C three, gives the same alkene. So the answer is two methylbut two ene either way. Got it?

8:20Quick revision: alkyl halides, Lucas test, E1 dehydration

So, quick revision. Alcohols give alkyl halides, tertiary fastest, and H I reacts fastest among the acids. Lucas test: tertiary at once, secondary in about five minutes, primary only on heating. Dehydration goes through a carbocation that may rearrange, and gives the more substituted alkene. Now go and try the practice problems on this page. All the best!