[kraver00]

Hello.

I have a few questions regarding the problems shown attached.. These are practice problems provided by my professor.

(1) In problems 4 and 5, they asked to "Choose the isomer". How do we know which isomer to choose?
At first, I thought we choose the isomer with the less steric hindrance. However, this is NOT the case. I believe it has something to do with anti-periplanar between the leaving group (Cl) and hydrogen. But, I am still a bit confused...

(2) What is the difference between problems 4 and 7, besides that they use different reagents and solvents?

These are the ONLY questions that I am really having trouble with in Organic Chemistry at the moment. If anyone can help me with these three practice problems, I would not be able to thank you enough.

Thank you so much for your time and consideration.

Thank you. Thank you. Thank you. I really look forward for any feedback or replies. Thank you. 

[kraver00]

Problems 4 and 5

[orgopete]

@kraver

I agree with the anti-periplanar rational. If we start with Q5, the choices fit perfectly. A can give two products and C, none. As indicated, B can give one as it only has a single anti-periplanar proton to eliminate.

Q4 is similar and presumably gives information about Q7. The drawn conformation shows the isomer with an anti-periplanar hydrogen and it can only give a single product.

I am reading Q7 differently. It seems to hint that one forms a single product while the other forms two. This could be correct. However, Q4 teaches us that a conformation change will result in methyl chloride giving a single isomer. Presumably neomenthyl chloride, with its axial chloride and two anti-periplanar hydrogens gives two isomers.

This is how I interpret this result. The neomenthyl chloride, with an axial chloride, can give substitution and elimination. Substitution reactions are faster for axial halides than equatorial. The menthyl chloride could also give substitution and elimination as shown in Q4. However, it must undergo a conformation change to achieve the more rapid axial chloride. I think the conditions for this reaction in DMSO do not achieve this conformation or enable attack in sufficient concentration to enable attack by azide. Therefore, I am surmising that menthol chloride simply reacts too slowly to give any products.

[kraver00]

@orgopete

Hello. Thank you so much for your reply!

For Q5, I guess I didn't read the question right. It asked to "form a single alkene product". Which makes sense with what you said.

For Q4 and Q7, so are you saying that DMSO would not cause a ring flip?
I have also noticed that DMSO is a polar aprotic solvent, while Ethanol is polar protic. Would this have anything to do with conformation changes?

[kraver00]

Also, I've been wondering for a bit on Q4 or Q7.

Why doesn't the Cl atom (halogen) have the higher priority than the isopropyl group? Based on the chair conformations, it indicates that the isopropyl has the highest priority, and the chlorine atom second.

[spirochete]

Also, I've been wondering for a bit on Q4 or Q7.

Why doesn't the Cl atom (halogen) have the higher priority than the isopropyl group? Based on the chair conformations, it indicates that the isopropyl has the highest priority, and the chlorine atom second.

What do you mean by priority?

If I'm interpreting you correctly, you are noting the fact that there is in fact a mistake in the way the answers to one of your worksheets are laid out, in the answer to number four. It suggests that the equilibrium favors the conformation where chlorine and the alkyl substitutents are axial. This is false.

It is true the molecule can only eliminate from this conformation, but the equilibrium for the starting material favors putting all groups equatorial. In this case the higher energy conformation is the reactive one (Google curtin hammet postulate if you care).

[orgopete]

For Q4 and Q7, so are you saying that DMSO would not cause a ring flip?
I have also noticed that DMSO is a polar aprotic solvent, while Ethanol is polar protic. Would this have anything to do with conformation changes?

*oops, sorry about my spelling errors*

Cause? No. I think it is simpler than that. Q4 shows an axial chlorine will react. If you compare menthyl chloride with its isomer, menthyl chloride is all equatorial. It will not adopt the required conformation as easily. It will react slowly.

If you had a 1:1 mixture of cis and trans-1-tBu-4-Cl-cyclohexane and reacted it with 0.5 equivalents of sodium azide, the axial isomer is virtually all that will react. It is simply a faster reaction.

For Q5, I'm assuming similar reaction conditions. For Q4, the formation of a single alkene from menthyl chloride will be a slower reaction than neomenthyl chloride. That is a different question.

I didn't make the problem. I'm simply trying to explain how these results could be achieved. I think this problem emphasizes a principle more than it is technically correct. Equatorial halides will react, just more slowly and syn eliminations also occur, but more slowly. The problem is easier to understand if you assume the reactions are all or nothing.