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Current time:0:00Total duration:6:59

Video transcript

let's look at how to prepare sulfides from tiles so over here on the Left I have my thigh all and to that thigh I'm going to add sodium hydroxide and the sodium hydroxide is going to deprotonate the Thyle which is then going to react with this alkyl halide in the second step of the reaction to produce my sulfide as my product so here is my sulfide right here this reaction is the analog of the Williamson ether synthesis which we've seen in an earlier video so in in that video we started off with an alcohol and we reacted our alcohol with a strong base in the first step in an alkyl halide in the second step and we formed an ether as our product so we can go ahead and draw our ether in here like that so the Thyle is the sulphur analogue to an alcohol and a sulfide is the sulphur analogue to an ether let's look at the mechanism to make sulfides so if I start with my file right here so I have carbon bonded to sulfur bonded to a hydrogen and then two lone pairs of electrons on that sulfur if I think about the difference in electronegativity between carbon and sulfur there's actually not much of a difference in terms of in terms of numbers so this is not a very polar bond actually that's different from what we see with an alcohol so up here if we look at the alcohol I know that oxygen is much more electronegative than carbon so this oxygen here we get a partial negative and this carbon on the Left it's mana two would get a partial positive so there's much more of an electronegativity difference in alcohols and dials not there's not really that much of a difference but thiols can still function as nucleophiles because these lone pairs of electrons are located further away from the nucleus in the lone pair of electrons and oxygen because sulfur is a larger atom so those electrons are more are more polarizable and so thiols are actually excellent nucleophiles so if we go back here to our mechanism alright we're now going to add sodium hydroxide which is a base so we can go ahead and put a china sover here so the hydroxide anion is going to function as a base and a lone pair of electrons are going to take this proton and leave these electrons behind and on the sulfur so let's go ahead and draw the conjugate base to the Thyle so we now have carbon bonded to sulfur and this sulfur now has three lone pairs of electrons giving it a negative 1 formal charge so this is called a thiolate anion so let me just go ahead and write that a thigh Oh late anion in here and thiolate anions are very stable that negative charge on the sulfur since sulfur as is it as a large as a large atom you can you can spread out that negative charge over a very large area so the thiolate anion is relatively stable and that makes thiols more acidic than alcohols okay so alcohols don't have the same type of stabilization since alcohols are smaller so thiols are actually very acidic and that's why we can use sodium hydroxide here to deprotonate our Thyle to form the thiolate anion in the second step right we add our alkyl halide and so here's my alkyl halide and the alkyl halide does have a polarized bond right the difference in electronegativity between a halogen and the carbon atom is is fairly large so this is halogen here is going to get a partial negative charge and this carbon is going to get a partial positive charge so thiols are good nucleophiles thiolate anions are even better nucleophiles and so the thiolate anion is going to function as a nucleophile the partially positive carbon is going to function as the electrophile and we're going to get an sn2 type mechanism where our strong nucleophile attacks our electrophile and kicks these electrons off here onto the halogen and we can go ahead and form our product right so this is an sn2 type mechanism and we end up with the sulfur right now bonded to two R groups they they could obviously be the same R groups or they could be different R groups and so we formed our sulfide like that let's do let's do an example of the preparation of sulfide so we're going to start with let's see let's start with this one right here alright so we start with this molecule and to that thye all we're going to sodium hydroxide in the first step so let's go ahead and write sodium hydroxide here now na plus and then Oh H minus like that and then in the second step we're going to add an alkyl halide so let's add this as our alkyl halide so ethyl bromide so when I think about the mechanism right I know the first step is an acid-base reaction the the electrons on the hydroxide anion so one of these electron pairs here I'm going to take this proton that's the acidic proton on my Thyle leaving these electrons behind on the sulfur so I can go ahead and draw the resulting thiolate anion so I can go ahead and draw that let me see what we have here we have our ring and then we have our sulfur and our sulfur now has three lone pairs of electrons around like that phyllite anions are excellent nucleophiles and when I look at my alkyl halide once again I know the electronegativity difference between bromine and this carbon here are going to give bromine a partial negative charge and this carbon is going to be partially positive so the thiolate anion is going to act as a nucleophile in an sn2 type mechanism and a lone pair of electrons here are going to attack my electrophile and we can go ahead and form our product right so now we have we have our ring here which is connected to our sulfur and our sulfur now just picked up two more carbons right because these electrons in here are going to kick off onto the bromine and to end up putting an ethyl group on to that sulfur so there are two carbons now on that sulfur like that and the sulfur has two lone pairs of electrons and so we formed our sulfide now if I were to name this sulfide it's it's a lot like naming ethers so I could use the common way of naming this and and treat those as alkyl groups and if I look on the right this would be this would be an ethyl group so I could go ahead and start naming it I could say it's ethyl and if I look at the alkyl group on the left side alright this is a phenyl group right here so it's ethyl and then phenol since I'm following the alphabet rule here EB 4 EB 4 P and then finally I know it's a sulphide so I can just go ahead and finish the nomenclature by saying sulphide here so ethyl phenyl sulphide is the sulphide produced in this analogue of the Williamson ether synthesis