user55119

@orthocresol: In one of your posts you mentioned that "we" have an app to format references. Is the app available to all users?

@orthocresol: Take a look at chemistry.stackexchange.com/questions/133838/…. The structures were reduced by adding a "period" to the lower right corner of the CD drawing. The structures are now smaller but the total space used is the same. Not a good solution to save space.

@Hot nitric acid: It's all relative.

@orthocresol: It isn't clear to me how to access the src file in html.

I draw structures with ChemDraw. When I post the jpeg, they are larger than I would like them. I have reduced the size of the ChemDraw structure and the result is the same. Any suggestions?

@Zhe: I decided to show the identity of the 2 epoxides the way I did. To put bromine on the bottom of the bromonium ion might have been confusing because it looks different from the top epoxide.

e:

@Zh

@Zhe: Thanks for the kind comment.

@Gaurang Tandon: Anything I can do to make the answer more comprehensible to you.

@TanYongBoon: Second link is this one: pubs.acs.org/doi/abs/10.1021/ja00060a006

@Tan Yong Boon: Regarding enolization of ketones with organometallic reagents, here are two links to the problem in total syntheses. pubs.acs.org/doi/pdf/10.1021/ja00992a032; pubs.acs.org/doi/pdf/10

I have not used mhchem in "comments" thus far. Is it possible to be sure the code is correct prior to posting.

Yes. They are two different compounds. Imagine a hypothetical DA reaction between 5-methylcyclopenta-1,3-diene and ethylene. You can conceptually cycloadd syn or anti to the methyl group.

@Safdar: look here: sigmaaldrich.com/catalog/product/aldrich/…

What manipulation in your answer? The unsaturated cmpds can't be interchanged unless bonds are broken. You can't do it via conformational changes.

@Safdar: The anti-isomer is R-s-S while the syn-isomer is R-r-S. No different from the diastereomers ribitol and xylitol. ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/isomers/…

You could use R/S/r/s. The question gave 3 choices. Clearly, they are not identical nor enantiomers. Thus, diastereomers. Alternatively, the methyl is syn to the double bond; in the other anti. Thus, diastereomers. An analogy: Your two hands are mirror images (the methyl groups; RS). Place the same chiral screw in each hand (the double bond; say RR). The two hands now have a diastereomeric relationship, RR vs. RS. The term diastereomers existed before CIP rules.

@Safdar: Could you rephrase your question?

Put a mirror between the first two structures. The methyl groups reflect; the double bond doesn't. They are diastereomers. In the second set of structures, rotate the first one 180 degrees about an axis that passes through the center of the 6-membered ring and the one carbon bridge. They overlap. They are identical. PS: In the second case, ignore the "pitch" of the bridge carbon. Best to build a model.

Some more. In your case all 4 isomers have a gauche t-butyl/methyl interaction, so that is common to all 4 and it is a wash. Two equatorial methyls is the most stable and lowest heat of combustion. Two axial (syn-pentane interaction) least stable highest combustion heat. The two isomers with methyls eq. and axial are about the same stability and are in between the other two for heat of combustion.

Here is my advice. Read about cyclohexane compounds here. ursula.chem.yale.edu/~chem220/chem220js/…

Get back to basics. Do you understand the effect of an axial methyl group in methylcyclohexane? Do you understand methyl/methyl gauche interaction? (PS: How did you open this conversation?)

Is there a penalty for an axial methyl group? Are there any gauche t-butyl/methyl interactions? Make an effort to solve this qualitatively and amend your post before I would provide an answer.

Do you realize that all 4 isomers give the same amount of CO2 and H2O on combustion? The more stable the isomer, the less the heat of combustion. Your job, not WE, is to draw all four and make a conclusion.

No. For a single group on cyclohexane it is an equilibrium as a function of its effective size. The larger the group, the more equatorial. t-Butyl is essentially always equatorial. This might help: ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS i/Avalues.htm

The t-butyl group will be equatorial then draw the possible diastereomers. Read this: chemistry.stackexchange.com/questions/118093/…

screenshot Apparently CD can center a circle but not arcs. CD does not know what you want to do. You don't have to place the 3 objects inside one another and then center. Put them anywhere on the page and use center commands.

@andselisk: I got [screenshot](URL) to work. Thanks.

As to the screenshot issue, I am still unsure as to how the URL is added in the Comment. I know how to get the URL, but how does one link to it?

Manual adjustment looks better. As to the form/function issue, recall that Picasso got the eyes in the wrong place but, nonetheless, got the message across.

So it is! As to your "perfectionist nuts", it is function that should precede form. Nice chat.

@andselisk: I was able to make a centered structure that now reads cyclohexane. How do you put a screen shot in comments?

@andselisk: About centering.. Draw ring, arc and + inside each other. Select them. Go to Object>align. Select L/R centered then T/B centered.

CD13 or 17 cannot handle "Structure to Name". But that wasn't the question. On an Apple laptop I had no problem (CD13) with orienting the arc or reducing its size. I hope I understood you correctly.

@tatan: Draw each Fischer projection with solid wedges and dotted bonds. C1 has the COOH group attached.

Unconjugated is the kinetic product product in either case. Stronger acid allows carbonyl oxygen protonation, dienol formation and gamma protonation.

My recollection is that the anisole case gives a mixture. I have done 3,4-dimethylanisole, which gives all conjugated. In this case the conjugated isomer is preferred in the presence of aq. mineral acid. Mild acid hydrolysis (I believe aq. oxalic acid) of the enol ether will give the unconjugated isomer.

Your structure is the Birch reduction product of anisole. Hydrolysis of the enol ether gives methanol and a mixture of the conjugated and deconjugated cyclohexenone. as @Zhe has said, you should post this kind of ? on the main site.

I spent some time on Chem. Abst. Apparently, the reacting species is unknown to CA.

@YUSUFHASAN, @Zhe: Who is "they" who say SN2.? I am unsure how the salt is made. Benzoyl chloride and Trimethylamine? Methylation of the amide occurs on O, not N. My instincts and experience agree with Zhe.

I was speaking only to the aromatic example.

Sounds reasonable. Fluoride is strictly induction. The closer to the OH, the better. The hydrogen bonding in the o-nitrophenol is "tying up" the proton relative to the p-isomer. No resonance stabilization for meta-nitro. At least that is the way I look at it. You should look at the pKa's and see how significant the differences are.

I described why benzoate does the SN2 and water goes after the oxonium ion. The oxonium ion is reversible with benzoate while benzoate is a better nucleophile for the SN2. Don't think that benzoate can only do an SN2 reaction but not get involved with the oxonium ion. I think that you are overthinking many of your questions through out ChemSE.

Do you believe that benzoate only does an SN2 in the Prevost rxn because doesn't add to the oxonium ion. If it did it could be reversible on its own or with the aid of Ag+. As to the Woodward variant, water is a poor SN2 nucleophile but once it adds to the oxonium ion, the intermediate probably collapses faster than it reverses. No need for HSAB (not a big fan).;)

@andselisk,@mithoron: The pre-IUPAC name was Claus benzene (1867). Today, we could say "I win at Trivial Pursuit". ;)

@YUSUFHASAN: Continuation of 11:34 post: Since the TS barrier for axial elimination is lower than for "equatorial elimination", then the rate constant k(ax) >>>>k(eq) and the concentration of the axial conformer is less than for the equatorial one. [ax]<[eq]. Therefore, rate(ax)=k(ax)[ax]>>>rate(eq)=k(eq)[eq]. There are no exceptions to the CHP, only irrelevant applications of it.

@YUSUFHASAN: As to the Curtin-Hammett Principle, bromocyclohexane is in rapid equilibrium between the axial and equatorial chair conformations with the axial conformation in lower concentration. Yet, it is the axial conformation that undergoes E2 base elimination faster than the equatorial conformation because of the anti-periplanar overlap of the vicinal C-H and C-Br bonds in the axial conformation. The equatorial conformation does not have that luxury.

The question asks about the reaction sequence. If you stop at the aldol reaction then you willfully ignore the second step, acetal formation.

I've already answered the Cannizzaro question previously: chemistry.stackexchange.com/questions/86573/…. The acetal formation with CH2O/H+ is fine. To obtain hydrolysis requires water in addition to H+. The answer is A.