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Current time:0:00Total duration:7:07

Elimination vs substitution: primary substrate

Video transcript

this video we're going to look at primary substrates and figure out if the reaction is a substitution or an elimination so for this primary alkyl halide we know an sn1 reaction is out because that would require a stable carbo cation and we can't make one from this primary alkyl halide an sn2 reaction is possible because of the decreased steric hindrance of our primary alkyl halides an e1 reaction is out again for the same reason as sn1 we can't form a stable carbo cation and an e2 mechanism is possible so now the next step is to look at our reagent and figure out what the reagent is going to do so for this reaction we have a sulfur nucleophile which we know is going to act only as a nucleophile and not as a base and since it's going to act as a nucleophile that means the e2 is out right we need a strong base for an e2 reaction so this must be an sn2 mechanism which we know is a concerted mechanism our nucleophile attacks our electrophile at the same time that we get loss of a leaving group so we're going to form a bond between the sulfur and this carbon here in red at the same time these electrons come off to form the iodide leaving group so let's draw in our final product so we have four carbons and then we have a bond to our sulfur which is bonded to a hydrogen so the carbon in red is this one and so this is our product and we don't have to worry about any stereochemistry here since we don't have any chiral centers so again we have a primary alkyl halide and we've seen that sn1 and e1 are out so we only have to decide between sn2 and e2 so we look at our reagent and we know that dbn is a strong base so therefore this is going to be an e2 reaction and not an sn2 reaction so dbn 1/2 the function as a strong base and we know that the halogen is directly connected to this carbon so that must be our alpha carbon and an e2 mechanism we're going to take a proton from a beta carbon so this is a data carbon right here and there's one there's one hydrogen on this beta carbon the dvn we know is a neutral base so I'll draw in here just a generic base with a lone pair of electrons this base is going to take this proton and these electrons going to move into here at the same time these electrons come off to form the iodide anion so let's draw the final product you have this long carbon chain here so double bonds in this position and then we're also going to form a double bond in this position so let's look at those electrons so these electrons here in magenta would move in here to form our double bonds for this primary alkyl halides again we've seen sn1 and e1 or out so we're deciding between sn2 and e2 so next we look at our reagent and we have potassium hydroxide and we know that the hydroxide ion is a strong nucleophile and a strong base so that's the category and we saw in an earlier video and for a primary alkyl halide that is unhindered the sn2 reaction is going to win out so a small nucleophile can attack this carbon at the same time these electrons come off onto the bromine to form our bromide anion so let's see our carbon in red is this one let's draw out our carbon chain we have 1 2 3 4 5 carbons and this carbon is the one in red and attached to the carbon in red is going to be our Oh H so we're going to form an alcohol from this sn2 reaction so this alkyl halide again we're deciding between sn2 and e2 now we have potassium at tert-butoxide as our reagent and this looks similar to the previous problem we have potassium hydroxide right of negative charge on this oxygen so you would think this could act as a strong nucleophile or a strong base but potassium tert-butoxide is sterically hindered because of this large group over here and because it's sterically hindered it can't get close enough to act as a nucleophile so an sn2 reaction is out which means this must be an e2 reaction so let me redraw our alkyl halides here just so we can see things a little better but the bromine going down in this direction and we know that the carbon directly connected to the bromine here is our alpha carbons all mark that and the carbon that bonded it to the alpha carbon is our beta carbons that's our beta carbon and we have two beta hydrogen's I'll just draw one in here and we know in our eetu mechanism our strong base is going to take that beta proton so our base takes this proton at the same time these electrons move into here and these electrons come off to form our bromide ion so our final products let's draw in our carbons here so we should have five carbons and then a double bond between these two carbons so our electrons in magenta moved in here to form our double bond for last example let's look at FL bromide reacting with water we know it's not sn1 we know it's not a one so we're deciding between sn2 and e2 we know that water is a weak nucleophile and it's also a weak base and since water is a weak base and e2 reaction is out so this must be an sn2 reaction so our nucleophile attacks our alkyl halides and the bond forms between the oxygen and this carbon here which I'll mark in red so let's draw the results of that this happens at the same time the electrons come off onto the bromine to form the bromide anion so let's draw what we have in here so this oxygen forms a bond with the carbon in red and a lone pair of electrons on the oxygen which I'll make magenta would form this bond the oxygen is still bonded to two hydrogen's so I'll draw in these two hydrogen's here and it's still the lone pair of electrons on that oxygen so that gives the oxygen a plus 1 formal charge to get to a neutral product we need to deprotonate this so another molecule of water comes along and acts as a base to take let's say this proton here leaving these electrons behind on the oxygen to give us our our final product so I'll just draw an H and here is weed form ethanol now this reaction would need a lot of heat it would need a lot of times it's not exactly the most practical reaction because an sn2 reaction needs a strong nucleophile and water is not that great of a nucleophile because this reaction is so slow you might see some textbooks say it won't happen because it isn't very practical