JEE Advanced Questions (2013–2015): Alcohols

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle: the name one one dimethylethan one ol. Draw it, and you get a perfectly real alcohol with four carbons. So, is the name correct? Most students say yes, because the molecule exists. But JEE Advanced twenty fourteen said no, and today you will see why. So let's begin.

0:24Naming an alcohol: the rules

So, first the naming rules from the course. The parent is the longest carbon chain. Drop the final e of that alkane and add o l: ethane becomes ethanol, and the arrow on screen means becomes. Okay? Now this alcohol: C H three, C H, C H two, C H three, with O H in round brackets on the second carbon. The dashes are single bonds, and the small numbers written low count the hydrogens. Which end is carbon one? The end nearer the O H group, yes, exactly! So the O H is on carbon two: butan two ol, never butan three ol. Clear?

1:05Making alcohols: the reagent decides

Now our toolkit, beside the course map of alcohols: making alcohols on its left, reactions on its right. First, an alkene with water and H plus; the raised plus is the charge, and the slash means together. It is Markovnikov, through a carbocation that may rearrange. With me? Second, mercuric acetate, H g, O A c in brackets with a small two, and water. The semicolon means then, so next comes sodium borohydride, N a B H four. Markovnikov again, with no rearrangement. Okay? Now let me ask you this. Which route puts the O H group on the end carbon, against Markovnikov's rule? Hydroboration oxidation, very good! Diborane, B two H six, then hydrogen peroxide, H two O two, with hydroxide, O H minus; the small raised minus is its charge. It gives the anti-Markovnikov alcohol. Right? And the Grignard reagent, R M g X, where R is an alkyl group, M g is magnesium and X is a halogen, adds to a ketone; then water gives a tertiary alcohol. The small raised circle is read as degree. Got it?

2:24Alcohol to alkene: acid and heat

Now the reverse: alcohol to alkene, with acid and heat. In the figure, the two dots on the oxygen of ethanol are a lone pair, which takes up H plus; the two half arrows mean both ways. In the slow step, water leaves, and a carbocation is left; the small plus is its charge. Okay? In the fast last step, the carbocation loses a proton, H plus, giving the alkene, ethene. As the note under the figure says, the carbocation may rearrange before that. Clear? So, tell me this: primary, secondary or tertiary, which alcohol loses water most easily? Tertiary, yes, exactly! The order is tertiary, greater than secondary, greater than primary; those pointed signs mean greater than. Right? When the proton can leave from more than one carbon, the major alkene is the more substituted one, with more groups on its double bond: Saytzeff's rule. With me?

3:24Example 1: correct IUPAC names of C4H10O alcohols

Alright, example one, from JEE Advanced twenty fourteen. Which pairs of names correctly describe isomeric alcohols of formula C four H ten O? Part a, tert-butanol and two methylpropan two ol. Part b, tert-butanol and one one dimethylethan one ol. Part c, n-butanol and butan one ol. Part d, isobutyl alcohol and two methylpropan one ol. Pause the video and try it first. Done? Okay, let's check. Tert-butanol is C H three in brackets with a small three, then C, O H: three methyls on the O H carbon. Its longest chain has three carbons, with the O H and a methyl on carbon two: two methylpropan two ol. Part a is correct. Okay? Next, part b: this name puts the O H and two methyl groups on carbon one of ethane; C dash one means carbon one. Draw it, and you get the very same molecule as tert-butanol! So part b is correct too, right? Oh no, careful, that is the trap! One of those methyls simply continues the chain, so the longest chain has three carbons, not two, and it was not chosen. So the name is wrong, though the molecule is real, and our opening puzzle is solved! Clear? In part c, n-butanol is the straight chain, C H three, three C H two groups, then O H, so it is butan one ol; this pair is correct too. With me? In part d, isobutyl alcohol is C H three in brackets with a small two, then C H, then C H two O H: a three carbon chain with a methyl on carbon two, so two methylpropan one ol. Right? So the answer is a, c and d. One rule did it all: find the longest chain first. Got it?

5:22Example 2: phenylethyne road map, hydroboration and Grignard

Now the harder one, from JEE Advanced twenty fifteen. C eight H six with H two over P d, B a S O four, gives C eight H eight; then B two H six, and H two O two with N a O H, give X; C eight H six with water, H g S O four and H two S O four gives C eight H eight O; then ethylmagnesium bromide, C two H five M g B r, with water, and then acid and heat, give Y, the major compound. Identify both, and pause the video to try it first. Done? Okay, let's check. Take a phenyl ring, C six H five, out of C eight H six, and C two H is left: a triple bond at the end. So it is phenylethyne: C six H five, C, triple bond, C H; the three short lines are the triple bond. Okay? Over the arrow, hydrogen with palladium on barium sulphate: a partial hydrogenation, which turns the alkyne into an alkene and stops there. We get C six H five, C H, double bond, C H two, styrene, C eight H eight, as the question says. Right? Now the diborane route on styrene. Where does the O H group go: on the carbon next to the ring, or on the end carbon? On the end carbon, exactly! Hydroboration oxidation is anti-Markovnikov. So X is C six H five, C H two, C H two, O H, which is two phenylethanol. Put the O H next to the ring, and you fell for Markovnikov's rule. Clear? Now the second road: water with mercuric sulphate and sulphuric acid hydrates the alkyne the Markovnikov way, putting the oxygen on the carbon next to the ring. That gives the ketone acetophenone, C six H five, C O, C H three, which is C eight H eight O, as given. With me? Then the ethylmagnesium bromide adds an ethyl group to the C O carbon, and water gives the alcohol. That carbon holds the ring, a methyl, an ethyl and the O H group: two phenylbutan two ol, a tertiary alcohol. Got it? Now acid and heat, and tertiary alcohols lose water easily. The proton can leave from the methyl, or from the C H two of the ethyl group. So which alkene is the major one? The more substituted one, by Saytzeff's rule! Losing the proton from the C H two gives C six H five, C, C H three in brackets, double bond, C H, C H three: three groups on the double bond, conjugated with the ring. So Y is two phenylbut two ene, and X was two phenylethanol!

8:22Quick revision: naming, Markovnikov routes, dehydration

So, quick revision. To name an alcohol, find the longest chain first, then number from the end nearer the O H group. From alkenes, the acid and mercury routes are Markovnikov, and the diborane route is anti-Markovnikov. In dehydration, tertiary alcohols lose water most easily, and the more substituted alkene wins. Now try the twenty thirteen question on this page, on choosing the reagent for each conversion. You did really well today, all the best!