How to prepare sulfides from thiols. Sulfides are like ethers, but with a sulfur instead of an oxygen atom. Created by Jay.
Want to join the conversation?
- At about0:45, Jay compared thiols and sulfides to alcohols and esters. The difference is that Sulfur is directly below Oxygen on the Periodic Table, therefore they have similar properties and react in a related manner.
If we look at the Periodic Table, we see that Silicon is directly below Carbon. Is there a silicon-based chemistry analogous to the carbon-based chemistry that is organic chemistry? I know that some science fiction uses silicon-based life forms. I'm not expecting this to evolve into a branch of biochemistry, but is there a branch of chemistry based in Silicon that mirrors organic chemistry?(12 votes)
- The chemistry of silicon is similar but not identical to that of carbon.
Like carbon atoms, silicon atoms can join together to form chains. As in hydrocarbons, these chains (called silanes) progressively grow in length as additional silicon atoms are added. But there is a very quick end to this trend. The largest silane, called hexasilane, has six silicon atoms (Si6H14 compared to C6H14).
Hexasilane is the largest possible silane because Si-Si bonds are not particularly strong. In fact, silanes are rather prone to decomposition by reaction with oxygen.Think of the stability of SiO2 (sand).
Silanes also have a tendency to swap out their hydrogens for other elements and alkyl groups and become organosilanes. An example is dichlorodimethylsilane, (CH3)2SiCl2.
There are chemists who spend their lives studying the chemistry of silicon, but silicon chemistry does not mirror organic chemistry, mainly because the Si-Si bonds are so weak.(18 votes)
- When naming the last product, are the names "benzyl" and "phenyl" synonomous?(4 votes)
- No, its kind of odd, actually. This picture comparison explains it better than I can: http://bbruner.org/obc/phen_fig/phenyl2.gif(6 votes)
- The thiol is tertiary carbon but how can it be SN2 mechanism?(2 votes)
- For this reaction, the thiolate ion is acting as the reagent (or the nucleophile) and the alkyl halide is the substrate (or the molecule to accept the attack of the nucleophile). It is the substrate in a SN2 reaction which should not be tertiary to promote this type of reaction since bulkier groups will block the nucleophile's attack. Here the alkyl halide substrate is primary so it'll be a good substrate for SN2.
Side note: not that it matters, but the thiol is secondary since the carbon bearing the sulfur is bonded to two other carbons.
Hope that helps.(2 votes)
- Can these reactions also occur through the SN1 mechanism if we have a tertiary alkyl group attached to a good LG?(2 votes)
- A base like S- would give an E2 reaction as the major product on a tertiary alkyl group with a good leaving group. Under solvolysis conditions(solvents that are not bases), SN1 and E1 reactions could occur, but these are more of a theoretical reaction since performing them in the lab gives a mixture of products.(2 votes)
- Are there any other reactions like this that can be replaced with a thiol instead of an alcohol, as in stated at0:48which forms a similar product to the original one in organic chemistry?(2 votes)
- What is ideal reaction medium or solvent for the above reaction ? Guessing should it be something that is both polar but not water, like butanol ?(2 votes)
- There was a question which i couldn't solve...
A black mineral A on heating in air gives a gas B.The mineral A on reaction with H2SO4 gives a gas C and a compound D.Bubbling C into an aqueous solution of B gives white turbidity.The aqueous solution of compound D, on exposure to air with NH4SCN gives a red compound E.The compounds A and E are?
a) PbS and Pb(SCN)2
b)NiS and Ni(SCN)2
c)FeS and Fe(SCN)2
d)CoS and Co(SCN)2
After solving a bit I think B is SO2 and C is H2S...and aqueous solution of B would mean H2SO3 or H2SO4.And all the four sulphides in the answer options are black in colour.(2 votes)
- Is there any particular way to remember that sulfide is a functional group and not exactly a Sulfer anion? I understand what sulfide is and all that but I keep getting confused when I hear sulfide because I automatically think of S- instead of R-S-R'. Just wondering if there is a simple way to think about it?(2 votes)
- At5:25does the phenyl thiolate have resonance? If so, what are the consequences of that for the reaction?
(The same question could be asked for phenyl with an O-, instead of S-)(2 votes)
- To my knowledge it would have resonance. I know phenoxide has resonance because the O⁻ is a strongly electron donating substituent. To see a similar example in action, see https://www.khanacademy.org/science/organic-chemistry/aromatic-compounds/directing-effects/v/ortho-para-directors-i and skip to about5:07, when Jay is showing the how the methoxy group can help to stabilize the molecule when a nitro group has been added ortho to it. I think the phenyl thiolate would also have resonance, though my gut is telling me that the resonance wouldn't be as impactful compared to an oxygen atom due to poorer overlap of sulfur's orbitals with carbon due to the size mismatch in atoms.
Nevertheless, due to the resonance that sulfur would have with the aromatic ring, sulfur would activate the ring (making it more nucleophilic and cause electrophilic aromatic substitution to occur faster), cause substituents to add ortho- and para- to the thiolate in electrophilic aromatic substitution, and finally shield the carbons and hydrogens in the ring during NMR, decreasing their chemical shifts.(1 vote)
- What would be the IUPAC name for ethyl-phenyl-sulfide?(1 vote)
- It is a "Thiophenetole". This site might good for you to find IUPAC name with a structure.
Let's look at how to prepare sulfides from thiols. So over here on the left, I have my thiol. And to that thiol, I'm going to add sodium hydroxide. And the sodium hydroxide is going to deprotonated the thiol, 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's my sulfide right here. This reaction is the analog of the Williamson ether synthesis, which we've seen in earlier videos. So in that video, we started off with an alcohol, and we reacted our alcohol with a strong base in the first step and 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 thiol is the sulfur analog into an alcohol, and a sulfide is the sulfur analog to an ether. Let's look at the mechanism to make sulfide. So if I start with my thiol right here-- so I have carbon bonded to sulfur bonded to a hydrogen. And then two lone pairs of electrons on that sulfur. And if I think about the difference in electronegativity between carbon and sulfur, there's actually not much of a difference 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 would get a partial negative. And this carbon on the left that it's bonded to would get a partial positive. So there's much more of an electronegativity difference in alcohols. In thiols, 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 than the lone pair of electrons in oxygen, because sulfur is a larger atom. So those electrons are more polarizable, and so thiols are actually excellent nucleophiles. So if we go back here to our mechanism, we're now going to add sodium hydroxide, which is a base. So we can go ahead and put OH minus over 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 on the sulfur. So let's go ahead and draw the conjugate base to the thiol. So we now have carbon bonded to sulfur, and this sulfur now has three lone pairs of electrons, giving it a negative one formal charge. So this is called a thiolate anion. So let me just go ahead and write that. And thiolate anions are very stable. That negative charge on the sulfur-- since sulfur is a large atom, 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. OK? 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 deprotonated our thiol to form the thiolate anion. In the second step, 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 a carbon atom is fairly large. So this 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 the nucleophile. The partially positive carbon is going to function is 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. So this is an SN2-type mechanism. And we end up with the sulfur now bonded to two R groups. And they could, obviously, be the same R groups. Or they could be different R groups. And so we've formed our sulfide like that. 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. All right. So we start with this molecule. And to that thiol, we're going to add sodium hydroxide in the first step. So let's go ahead and write sodium hydroxide here. A Na plus and then OH 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 bromine. So when I think about the mechanism, I know the first step is an acid base reaction. The electrons on the hydroxide anion-- so one of these electron pairs here-- are going to take this proton-- that's the acidic proton on my thiol-- 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 it like that. Thiolate anions are excellence nucleophiles. And when I look at my alkyl halide, once again, I know the electronegativity difference between bromine and this carbon here. I'm going to give bromine a partial negative charge. This carbon is going to be partially positive. So the thiolate anion's going to act as a nucleophile. And 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. So now we have our ring here, which is connected to our sulfur. And our sulfur, now, just picked up 2 more carbons, right? Because these electrons in here are going to kick off onto the bromine, and we end up putting an ethyl group onto that sulfur. So there are 2 carbons now on that sulfur like that. And the sulfur has 2 lone pairs of electrons. And so we formed our sulfide. Now, if I were to name this sulfide, it's a lot like naming ethers. So I could use the common way of naming this and treat those as alkyl groups. And if I look on the right, 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-- this is a phenyl group, right here. So it's ethyl. And then phenyl since I'm calling the alphabet rule here-- e before p. And then finally, I know it's a sulfide. So I can just go ahead and finish the nomenclature by saying sulfide here. So ethyl phenyl sulfide is the sulfide produced in this analog of the Williamson ether synthesis.