Organic Chem!

[zany]

Okay, we are looking at ionic reactions, and I am having trouble with a few.
I am supposed to complete and balance the equations. (All the #'s are subscripts and I have charges directly after anything ionized, with no spaces)

1) (CH3)3N + CH3(CH2)11Br ---->

2) CH3CH2CH2CH2I + _______ ----> CH3CH2CH2CH2SH + Na+I-

------Cl
------|
3) CH3CHCH3 + (CH3)3CO-K+ ----> (sorry for the dashes, the Cl is branched off the central C)


Now, here are my answers

1) ----> CH3(CH2)11N(CH3)3 + Br- (On this one I think the lone pair of N would go to the positive end after Br- comes off but I don't know if it would keep all 3 CH3 groups, or lose one to make CH3Br, and double bond to the closest C)

2) NaSH (I think this would work.. but is that even an available compound?)

-------------OC(CH3)3
-------------|
3) ----> CH3CHCH3 + K+Cl- (I have no idea if a substitution with something branched like this is even possible)


Anyhow, any help whatsoever would be much appreciated.

[FireWall]

Finally, a good homework thread.

Okay, we are looking at ionic reactions, and I am having trouble with a few.
I am supposed to complete and balance the equations. (All the #'s are subscripts and I have charges directly after anything ionized, with no spaces)

1) (CH3)3N + CH3(CH2)11Br ---->

The substrate (i.e., 1-bromododecane) is a primary alkyl bromide. That eliminates the likelihood of an SN1 or E1 reaction because the carbocation intermediate would be too unstable. Furthermore, the nucleophile is a weak base. That rules out the likelihood of an E2 reaction. Consequently, the reaction occurs via an SN2 mechanism. If you disregard the transition state, the reaction occurs in a single step. There is no inversion of stereochemistry because the substrate is achiral.

1) (CH3)3N + CH3(CH2)11Br ----> CH3(CH2)11N(CH3)3 + Br-

2) CH3CH2CH2CH2I + _______ ----> CH3CH2CH2CH2SH + Na+I-

The correct answer is NaSH (sodium thiol). Yes, it is a real compound. It is analoguous to NaOH because sulfur is directly beneath oxygen on the periodic table.


------Cl
------|

3) CH3CHCH3 + (CH3)3CO-K+ ----> (sorry for the dashes, the Cl is branched off the central C)

The substrate is a secondary alkyl chloride. That eliminates the likelihood of an SN2 reaction (especially with a strong hindered base). An SN1 reaction is unfavorable too because the base is simply too large to effectively negotiate the steric hindrance of the methyl groups, and the base is too strong. It is much easier for the base to simply deprotonate the substrate. A biomolecular elimination mechanism is favored because of the secondary structure of the substrate and the strength of the base. Consequently, the reaction occurs via an E2 pathway, and the resultant alkene has a trigonal planar molecular geometry at each of the sp2 hybridized carbons.

CH3CHClCH3 + (CH3)3CO-K+ ----> CH2=CHCH3 + (CH3)3COH + K+

The key to correctly determining the reaction mechanism above is recognizing that the nucleophile is a very strong hindered base. You do not need to see a pKa value to determine that, either. All that you need to realize is that tert-butoxide is the conjugate base of an extremely weak acid (i.e., tert-butyl alcohol). The alcohol hydroxyl group does not enjoy deprotonation.

Now, here are my answers

1) ----> CH3(CH2)11N(CH3)3 + Br- (On this one I think the lone pair of N would go to the positive end after Br- comes off but I don't know if it would keep all 3 CH3 groups, or lose one to make CH3Br, and double bond to the closest C)

2) NaSH (I think this would work.. but is that even an available compound?)

-------------OC(CH3)3
-------------|
3) ----> CH3CHCH3 + K+Cl- (I have no idea if a substitution with something branched like this is even possible)

You are on the right track. Just make sure that you study the various substitution and elimination mechanisms thoroughly. If you have any additional questions, please do not hesitate to ask me.

[zany]

Awesome post, thanks a lot. I understand definitely understand those now... I had completely thrown eliminations out of my head when I was working on this, so the third one makes much more sense now. I am still a little confused on the stereochemistry of #1 though. I don't really see how 1-bromododecane can be superimposed on its mirror image... unless you can just flip it vertically? Actually, most of what my professor covered regarding stereochemistry was pretty unclear, so I am still in the dark.

[FireWall]

Awesome post, thanks a lot. I understand definitely understand those now... I had completely thrown eliminations out of my head when I was working on this, so the third one makes much more sense now. I am still a little confused on the stereochemistry of #1 though. I don't really see how 1-bromododecane can be superimposed on its mirror image... unless you can just flip it vertically? Actually, most of what my professor covered regarding stereochemistry was pretty unclear, so I am still in the dark.


No problem. With regard to chirality, you accidently misinterpreted what I wrote. 1-bromododecane is achiral, which means that it does not have any enantiomers. You have the right idea. Just make sure that you do not confuse superimposibility with superposibility. Superimposition does not require that every part match up completely. Superposition does. I was completely unaware of the difference until I noticed a tip in my textbook.

[zany]

Ah okay, I think I see what is going on now. 1-bromododecane is achiral because it has a plane of symmetry correct? And chiral molecules (enantiomers) simply cannot be superimposed on their mirror image, while achiral molecules can? But when I put it like that, I get confused as to what diastereomers are... hum

[FireWall]

Ah okay, I think I see what is going on now. 1-bromododecane is achiral because it has a plane of symmetry correct? And chiral molecules (enantiomers) simply cannot be superimposed on their mirror image, while achiral molecules can? But when I put it like that, I get confused as to what diastereomers are... hum

Diastereomers are stereoisomers that are not mirror images of each other. The best examples are cis and trans isomers. Cis and trans isomers are certainly stereomisers. Their molecular formulas are the same, and their condensed structural formulas are the same. However, they are not the same compound because the spacial arrangement of their atoms is different. In other words, they are not superposable.

That leaves two possibilities: they are either enantiomers or diastereomers. Both enantiomers and diastereomers are nonsuperposable. However, enantiomers are nonsuperposable mirror images. Diastereomers are not mirror images. And since cis and trans isomers are not mirror images, they have to be diastereomers.

Keep in mind that you can have diastereomers that are not cis and trans isomers too.

[zany]

Awesome, gotta love a good example. I do believe this finally clicked for me, thank you much.

[jurassic]

Wow, I'm also taking Ochem this quarter, but we just started on Alkenes and such.

If I ever have any questions I'm coming to you firewall :D

[zany]

I just decided to bump this thread since it is the same topic, but I have a few questions on a couple problems I am looking at.

One question deals with the rate of elimination reactions for two stereoisomers--
trans-4-chloro-tert-butylcyclohexane
cis-4-chloro-tert-butylcyclohexane
Now, the question asks which compound will eliminate HCl faster when treated with potassium tert-butoxide, in tert-butyl alcohol. I am pretty confused as to how the rate is actually going to be affected by the different stereoisomeric forms.

Another question involves two different reactions with HC=-CCH2CH2CH3 (1-pentyne, I don't know how else to represent a triple bond)

The first reaction is 1-pentyne treated with NaNH2. I am pretty sure that the NH2- ion will just jack the lone H, and form the deprotonated (is that the right word?) anion of 1-pentyne.

The second reaction is 1-pentyne treated with CH3Br. Here I am not sure what would happen because I didn't the the Br would leave the methyl group without being attacked by a nucleophile, and I don't see the H leaving 1-pentyne without something else... so I am guessing I don't know exactly what is going on here, or perhaps these are successive reactions, and the solution is treated with CH3Br after the anion has already been created.

[Streetwolf]

Eh I'm busy with my math homework so I'll just say that for the first one I'd guess the cis isomer because it's the more hindered one.

[Forgotten]

I'm not clearly visualizing those cis- trans- steroisomers, but like streetwolf said, the reason is sterics. That giant t-butyl substituent with the giant t-butoxide base will govern the speed.

For the second one, I'd probably say the same as you. Have the NH2 snatch the H to form NH3 and the anion, but draw the Na+ next to it. Even though it's not really a bond, it's important to note the interaction.

And for the third one, I'd probably just do it like electrophilic addition. I'd add the CH3 to the first carbon and the Br to the second, going from a triple to a double bond. If they are successive reactions, then I'd still add the CH3 to the first carbon to replace the Na+, and then show Na+Br- next to eachother.

My orgo is really shaky. I'm sure Firewall will chime in with more help.

[zany]

Thanks for the help guys, I see now why the cis- stereoisomer will react faster. I had to draw it in a chair conformation to see that the trans- formation results in an axial Cl, which in turn makes the axial Hydrogens less reactive. However I am still somewhat confused on the second question. I think I see generally what would happen with the addition of CH3Br, but I am not sure which H would be more acidic in the 1-pentyne chain.

[Forgotten]

The hydrogen on the first carbon is definitely the most acidic. The triple bond draws electrons away from that side and gives it some acidic character. All the other hydrogens really don't have any. If you look at a pKa table, a hydrogen adjacent to triple bonded carbons has a value of about 24, while the other ones are probably in the 40-50 range.

[Streetwolf]

It was said earlier. The NaNH2 will react with the H on the triple bond to produce a salt, and then the CH3Br will be involved in an SN2 reaction to form CH3C(triple bond)C-CH2-etc. Seems like the questions were connected.

[zany]

I believe I've got it now. Thank you sirs, I appreciate it.