Practice-Problem: Three-Reaction Pathway Professor Dave Explains English English

Practice-Problem: Three-Reaction PathwayYouTube↗ IA↗(long shot)
Education 1.6K views 47 likes 0 dislikes 06 Nov 2018

Professor Dave Explains 4.3M subscribers United States YouTube↗

okay guys this practice problem is very straightforward we have a starting

material and we have a sequence of three reaction conditions we are going to

subject it to and we just have to look at what these are recognize what they're

going to do and try to draw the product of this sequence of reactions so give it a shot.

so let's go through these reactions one at a time what is the

first thing we're doing here so we've got bromine and hv what is that? that's

that's light so that's indicating that there's a certain frequency of light

that is going to be relevant here and so while at first we would see Br2 we

would think okay that's gonna react with a pi bond maybe do a dihalogenation

but in fact that's that's not what's gonna happen because although we do have

pi bonds here that's benzene and an aromatic ring like benzene is not going

to do addition with with bromine so instead we recall that these are the

conditions for radical halogenation so this is a way that we can this is the

way that we can halogenate at an aliphatic position we can do this on an

alkane and if we remember something about the regioselectivity of radical

halogenation in terms of bromine radical bromination is highly regioselective as

opposed to radical chlorination where the chlorine will just end up in a lot

of different places bromine when we're broman ating as a radical the bromine is

gonna go to the most substituted carbon so we know that we're gonna put a

bromine on this molecule and it's gonna go there okay we know that that is

what's going to happen so so for the first reaction let's just call this

reaction one right we know that there's we don't need to show the whole involved

mechanism we remember that radical halogenation mechanism from the Organic

Chemistry series and so we can draw this and we have this here this is not this

is not going to be stereoselective but we don't care so it wouldn't also be the

enantiomer but we'll just draw one of these for now

so we know that we're getting a bromine at the most substituted position on the

molecule so bromine is definitely going to go there exclusively. now for number

two all right we're looking at reaction to what's going to happen here well

we've got we've gotten methoxide in methanol okay so that's one thing and we

know it's pretty hot so we've got this alkyl bromide we've got a strong base

we've got methoxide so we know that probably SN2 or E2 are going to be

likely candidates but a couple of things number one it is hot and we know that

hot temperatures favored elimination over substitution

second of all we can't do substitution there we cannot do SN2 there so we know

that we're gonna do elimination but we know we're gonna do E2 so we know

that that's what's gonna happen here but there are different protons that are

available we could be getting some of the protons here right we know that

there are some protons here but there are also protons here that are beta that

are in the correct position there but what do we know about methoxide

methoxide is we need to understand that it's not very sterically hindered right

methoxide is not sterically hindered it has equal access to any of these protons

so it's going to go the thermodynamic route it's going to want to produce the

most thermodynamically favorable product so it's gonna go and get one of these

guys we're gonna kick that off that way so that will give us

this here okay so that's going to be our alkene product of this E2 reaction.

we're gonna get that double bond there and not there, if this had been tert

butoxide then we would have ended up with the double bond here because we

would have have to get these more sterically available protons but

methoxide no problem and then lastly for reaction 3 we know that this is going to

be a dibromination right so whereas before we didn't have any pi bond I mean

apart from the pi bonds in the benzene ring which are not going to be reactive

towards addition reactions we didn't have any other pi bonds available so we

didn't do dihalogenation, and hv, that tipped us off that we were doing radical

bromination. over here we see the standard conditions for dihalogenation

right Br2, CCl4 and we do have a pi bond available for an addition reaction

so we can just do regular dihalogenation here and so that is going

to give us a mixture of products because we can have right we can have we can

have the bromonium ion intermediate if we had the bromonium ion intermediate

form on the wedge then the bromide would come in on the dash

or we could do it the other way and we could have right we could put this

methyl on the wedge and this would be as though the bromonium ion intermediate

was on the dash therefore the bromide came from the front and opened that up

that way so that is an anti addition reaction those two bromines have to be

trans to one another and so those are the stereochemical possibilities there

so once again radical bromination sites if E2 elimination and then di

halogenation is going to give us the product of this sequence of reactions.

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