Dichloromethane can undergo SN2 displacement, so...

[FROGGBUSTER]

---

Members don't see this ad.

I'm wondering if Carbon 4 of my attached pic could also act as a electrophilic center for SN2 displacement, where Br- leaves?

 

[MShopes]

I'm wondering if Carbon 4 of my attached pic could also act as a electrophilic center for SN2 displacement, where Br- leaves?

What are you trying to add? And I don't think the Br- would be the one to leave. The OH would be the one to leave because upon formation of H2O, it is considered a good leaving group and therefore it would rather leave and Br- stays. First protonation would occur to OH forming H2O which is a very good leaving group and then the nucleophile attack from the backside (backside attack) at the same time the H2O (the good leaving group) leaves. There is obviously no carbocation involved and it is a concerted 2 step reaction because it is SN2.

Maybe if you tell me what are you trying to synthesize, I would be able to help better!

 

[FROGGBUSTER]

What are you trying to add? And I don't think the Br- would be the one to leave. The OH would be the one to leave because upon formation of H2O, it is considered a good leaving group and therefore it would rather leave and Br- stays. First protonation would occur to OH forming H2O which is a very good leaving group and then the nucleophile attack from the backside (backside attack) at the same time the H2O (the good leaving group) leaves. There is obviously no carbocation involved and it is a concerted 2 step reaction because it is SN2.

Maybe if you tell me what are you trying to synthesize, I would be able to help better!

Thanks for replying.

I don't know what I'm trying to synthesize lol, it's just a hypothetical question because I'm very unsure if the regular substitution reactions work with molecules other than regular alkanes. I had read that substitution does occur with dichloromethane, which surprised me, so I wanted to know if it was possible for substitution to occur at C4 as well.

So say you threw in a good nucleophile like I- in there. Would it displace Br- in an SN2 rxn?

 

[MShopes]

Thanks for replying.

I don't know what I'm trying to synthesize lol, it's just a hypothetical question because I'm very unsure if the regular substitution reactions work with molecules other than regular alkanes. I had read that substitution does occur with dichloromethane, which surprised me, so I wanted to know if it was possible for substitution to occur at C4 as well.

So say you threw in a good nucleophile like I- in there. Would it displace Br- in an SN2 rxn?

I'm very sorry about my last message I have mixed things up and gave you a wrong answer more or less. In this case, the OH would be a very poor leaving group when you are trying to attack with a nucleophile such as Br-...I just thought about it like if you are adding HI which in this case OH can be protonated to give a good leaving group. But since it is not HI and it is more like Br- (Salt-Na+ I- for example), then OH would be very poor leaving group and wouldn't be displaced. Instead, the I- (excellent nucleophile) would attack the carbon number 4 and displace the Br- which is a good electrophile as well. It is a simple SN1 or SN2 mechanism (depending on the solvent if it is protic or aprotic). But overall, yes to your question. The Br- would be displaced by the I.

 

[salim271]

What are you trying to add? And I don't think the Br- would be the one to leave. The OH would be the one to leave because upon formation of H2O, it is considered a good leaving group and therefore it would rather leave and Br- stays. First protonation would occur to OH forming H2O which is a very good leaving group and then the nucleophile attack from the backside (backside attack) at the same time the H2O (the good leaving group) leaves. There is obviously no carbocation involved and it is a concerted 2 step reaction because it is SN2.

Maybe if you tell me what are you trying to synthesize, I would be able to help better!

The protonation of OH wouldnt cause SN2, it would cause H20 to leave in a slow step and formation of a carbocation which is then attacked by the nucleophile, thats SN1.

Its possible for SN2 to occur with the starting material, the reason it works with chloromethane and here is because there are two electron withdrawing groups, although i believe the resonance possiblities that OH presents would lend a little stabilization, but in the end C4 is slightly positively charged and therefore a good target for nucleophiles. I would say this reaction works, but yield might not be too good because of the presence of three groups attached to the carbon, causing steric hinderance.

 

[Pre-Med Village]

I think what will happen first in neutral conditions will be elimination of bromine to form an α,β unsaturated ketone followed by conjugate nucleophilic addition. Because the carbon to which the nucleophile eventually will bind in the α,β unsaturated ketone is trigonal planar there shouldn't be stereoselectivity, so the product will be racemic.