Help with Organic Chemistry

[zerg]

This thing is in the presence of sodium methoxide.

I am under the impression that a strong nucleophile favours SN2 reactions. But, since the bromide is bonded to a tertiary carbon, wouldn't that favour SN1? How do you decide which?

In that reaction, which intermediate is formed? If it is SN2, then there is no intermediate, right? And in SN1, the intermediate is a carbocation with (CH3)2C bonded to cyclohexane?

I'm having some trouble with reaction mechanisms, does anyone have any websites or anything to recommend?

[Mr Pete]

It is tertiary so the mechanism is Sn1. This is mainly due to stabilisation of the carbocation intermediate by inductance and hyperconjugation, but a bit of steric hinderance is involved too.

The rate limiting step for Sn1 is loss of the bromide to form the carbocation. The strength of the nucleophile is therefore irrelevant.

You are correct in that a strong nucleophile will accelerate an Sn2 process, but I wouldn't have thought it has much effect on which mechanism is being used (except for cases like secondary halides where Sn1 and Sn2 compete).

When they ask for an intermediate for the Sn2 what they are probably after is the transition state.

What's troubling you regarding the mechanisms?

[zerg]

Thank you sir!

I keep having trouble determining whether a reaction will be SN1 or SN2. So, the strength of the nucleophile has less impact in determining the mechanism than whether the carbon is primary, secondary or tertiary? I must have missed that. I think that makes a lot of my confusion go away. Also, determining the strength of nucleophiles is required for secondary carbons, but I don't quite have that down yet either, so do you mind helping with that too?

One of the questions is:

"Identify the stronger nucleophile in each pair given below.
a) NaSH vs H2S

b) NaOH vs H2O

c) MeO-/MeOH vs MeO-/DMSO"

For (a), I put NaSH, because H2S is more electronegative. Also, NaSH is a stronger base, which tends to mean it is a stronger nucleophile, right? For (b), I put NaOH, with the same reasoning. I'm not entirely sure what is meant by (c). Thank you once again.

[TychoMaudd]

c) MeO-/MeOH vs MeO-/DMSO"

For (a), I put NaSH, because H2S is more electronegative. Also, NaSH is a stronger base, which tends to mean it is a stronger nucleophile, right? For (b), I put NaOH, with the same reasoning. I'm not entirely sure what is meant by (c). Thank you once again.

For c) it's asking about MeO- in methanol vs MeO- in DMSO (dimethylsulfoxide). The strength of the nucleophile doesn't just depend on the nucleophile itself, but also what type of solution it is in. Your answers for a) and b) are correct, generally a stronger base is a stronger nucleophile, but I'm feeling pretty dumb right now because I don't know the answer to c) off the top of my head. I would normally say that a protic solvent would weaken MeO- as a nucleophile because it could pick up a proton from the solvent before reacting, but in this case even if it picks up a proton it would just form MeO-.

[Mr Pete]

Thank you sir!

I keep having trouble determining whether a reaction will be SN1 or SN2. So, the strength of the nucleophile has less impact in determining the mechanism than whether the carbon is primary, secondary or tertiary? I must have missed that. I think that makes a lot of my confusion go away. Also, determining the strength of nucleophiles is required for secondary carbons, but I don't quite have that down yet either, so do you mind helping with that too?

Its a bit handwavy but generally:
Primary halides are Sn2
Tertiary halides are Sn1
And secondary halides are a mixture of Sn1 and Sn2.

One of the questions is:

"Identify the stronger nucleophile in each pair given below.
a) NaSH vs H2S

b) NaOH vs H2O

c) MeO-/MeOH vs MeO-/DMSO"

For (a), I put NaSH, because H2S is more electronegative. Also, NaSH is a stronger base, which tends to mean it is a stronger nucleophile, right? For (b), I put NaOH, with the same reasoning. I'm not entirely sure what is meant by (c). Thank you once again.

a) NaSH would be stronger as it has negative charge, and so wants the electrophile more than the neutral species
b) NaOH, same as above
c) Solvent effects, proton exchange does happen, but I wouldn't have thought it matters that much.
For both the active species (MeO-) would be surrounded by a cage of solvent, part of the activation barrier is removing this cage. So a strong binding solvent will make a nucleophile weaker, while a weakly binding one will make it stronger.

MeOH can hydrogen bond to MeO-, DMSO cannot. MeOH binds stronger and makes a weaker nucleophile. (I think this is correct anyway).

(If you're interested look up phase transfer catalysis with crown ethers as an application of this effect.)

[Scow]

In regards to the question posed by the O.P., "by what mechanism does 'this thing' react with sodium methoxide?" Was this a problem out of a text book where they only give you SN1 and SN2 as possible answers? If so, the answer is SN2 for the same reason the other posts discussed. That aside, this reaction will almost certainly give you an E2 product. SN2 is unlikely due to the methoxide being a strong base and the substrate being tertiary (e.g. hindered substrate prevents the backside attack required for the SN2 mechanism to proceed).

c) Solvent effects, proton exchange does happen, but I wouldn't have thought it matters that much.For both the active species (MeO-) would be surrounded by a cage of solvent, part of the activation barrier is removing this cage. So a strong binding solvent will make a nucleophile weaker, while a weakly binding one will make it stronger.MeOH can hydrogen bond to MeO-, DMSO cannot. MeOH binds stronger and makes a weaker nucleophile. (I think this is correct anyway).

This is correct. However, methoxide is such a powerful base the solvent is unlikely to prevent it from abstracting a proton from neigboring tertiary carbon or the methyl groups, thus following the E2 reaction pathway. Due to the tertiary substrate, the rate of at which the methoxide can negotiate past the methyl groups is simply too slow to yield any major SN2 product.

[zerg]

Yes, SN1 and SN2 were the only choices.

I recently took an exam, and I missed a few questions, so I was wondering if you guys could tell me where I went wrong?

First, how would you change cyclohexane to cyclohexene?

Secondly, the IUPAC name for this:

With these options:

A. (2S, 3S)-2,3-dibromopentane B. (2S, 3R)-2,3-dibromopentane

C. (2R, 3S)-2,3-dibromopentane D. (2R, 3R)-2,3-dibromopentane

I thought D, but apparently not, so is it A?

Three, which of these carbocations are the most stable:

I wasn't really sure, and I put D, but I guess that's wrong.

Four, what step is likely to happen here:

I thought A, but wrong again. Thoughts?

And lastly, which of these is the strongest base:

Here I had no idea, so I put C.

Thanks for all of the help. The xkcd forum is an awesome place.

[Tass]

3A because it is allylic. You can move the electrons of the double bond to move the charge. In reality the charge is on both carbon 2 and carbon 4.

4C. You are almost right, but forgot the strong base added. Removing the proton makes oxygen a much better nucleophile.

5B. Primary amines are much stronger bases than aniline or pyridine derivates.

[Scow]

First, how would you change cyclohexane to cyclohexene?

Radical bromination followed by treatment with a bulky alkoxide to facilitate E2 to get the alkene product. It isn't the most selective reaction in the world but it works.

Secondly, the IUPAC name for this?

I must be missing something because I keep getting D as well. Perhaps I've forgotten how to apply the right hand rule. There could be a mistake in the exam. The most diplomatic way to find out is to ask your instructor to explain the answer to you.

[Aodhan]

The answer for the dibromopentane question is definitely D.

[zerg]

Thank you all. I've emailed the professor, though I don't know that he will respond over the weekend.

Also, since this is now less about the original assignment and more about trying to survive this class in one piece, I thought I would change the title. Might I ask how one goes about doing this? (EDIT: Never mind, it seems I figured it out.)

And, of course, I have some more questions.
"1. Compound X reacts with one equivalent of H2 in the presence of a catalyst to give
methylcyclohexane. Compound X also reacts with dilute H2SO4 to give 1-methylcyclohexanol.
Compound X can be prepared upon treatment of 1-bromo-1-methylcyclohexane with sodium
ethoxide. Identify compound X by providing its IUPAC name."

I'm fairly sure this is just 1-methylcyclohexene, right?

"2. Name the major product of the following reaction:
"

1-bromo-1-ethylcyclohexane? Hydrogen is added, secondary carbocation formed, hydride shift to tertiary carbocation, bromide added.

"3. Provide ALL the reagents (chemical formula) necessary for each of the following transformations.


"


My answers:
(a) HBr, H2O2
(b) H2SO4, H2O
(c) B2H6, then H2O2 in KOH
(d) ICH2ZnI
(e) Br2, then H2O
(f) PhCO3H, then H2SO4
I don't know the answer to (g), so any help would be appreciated.

"4. Identify all the reagents (using chemical formula or chemical names) needed to accomplish each of
the following transformations. List the reagents in the correct sequence that they appear in the
synthesis process.
"

To which I have responded:
(a) H2SO4 to form 3-methylenepentane, then B2H6 followed by H2O2 in KOH to form 2-ethylbutan-1-ol.
(b) Br2

My answer to (b) seems to simple, so I'm guessing I've done something wrong.

I haven't done quite as well as I would have hoped in the class, so am trying to work hard and get a better grade, and you guys are a big help. Thank you.

EDIT: Okay, I think the following will work for (g): Br2 in light to form 1-bromo-1-methylcyclohexane, followed by KOH in acetone to form 1-methylcyclohexene, followed by O3, followed by H2 and Zn.

I'm still somewhat nervous about my answer to 4b, though.

EDIT: Well, it may help if I had the right pictures in the right places. Also, for 4a, I changed my answer to "HCl to form 3-methyl-3-chloropentane, followed by (CH3)3COH, a large base, to form 3-methylenepentane, and then B2H6 followed by H2O2 in KOH to form 2-ethylbutan-1-ol," which I believe is right, because H2SO4 would have formed 3-methylpent-2-ene, right? Which is not correct, and obviously I need to read more carefully. Also, I am quite sure that my answer to 4b is wrong, because just adding bromine would add it across the double bond, which would form 1-bromomethyl-1-bromo-cyclohexane (I doubt that is the right name), which is again not what is supposed to form.

Also, I got a response, and the answer to that exam question was, in fact, D. So, I managed to get some points back. Yay!

EDIT: Yet another edit; for 4b b) H2 with a platinum catalyst to form methylcyclohexane, then Br2 in light to form 1-bromo-1-methylcyclohexane, followed by KOH to form 1-methylcyclohexene, followed by Br2 to form 1,2-dibromo-1-methylcyclohexane should work, right?