[lissydoll206]

Hi.

  I'm working on a final assignment for my Organic Chemistry class and I just had a few questions on some of the problems that I was working on.  If anyone could let me know if I did them right, or give me insight if I did them wrong, it would be greatly appreciated!

  Alright, here goes.  See corresponding pictures for each number.

1.  I have an Isomer of C7H12.  It is reactive with 1 mol equivelent of H2, and yields cyclohexenone after treatment with acidic KMNO4.  
  Is the compound that I have drawn the correct isomer that yields this product?  Also, another product is formed... I said that it is carbon dioxide because the double bond was on the end of the isomer and so it would form carbon dioxide.  Am I right in the way I am thinking this problem through?

2.  I just want to doublecheck to make sure that what I have drawn is a meso compound.  Thanks.

3.  When naming these compounds based on their sterochemistry (S and R), does the O-CH3 group have higher priority than the carbon that is connected to a double bond?   I haven't found anything that can tell me for sure.  However, I thought that you went by the point of first difference....   I think that the OCH3 has higher priority over the carbon connected to the double bond... is my thinking right?

4.  Finally, I have a reaction that takes place, and I have the product drawn in the chair conformer plus the NaCl that forms.  The way I picture the product is that the methyl group and the SCH3 group are in equitorial postion.  However, a classmate that I am working on the assignment with disagrees, and says that the methyl group is equitorial, but the SCH3 group is axial.  Which product will form with the NaCl? Can anyone explain one way or the other?

Thanks for any insight you can give.  I feel like I have a pretty good understanding of these questions, but I am always looking to learn from any mistakes I have made, and to reinforce the what I do know.
 
Thanks again!  Hope everyone is having a great week!

-Melissa

[lissydoll206]

Oops, I don't think the picture attached... if it had, sorry that this posted twice!

[lemonoman]

I suppose I'll do what I can right off the bat here...and let other do the rest  

For the meso-ness of the compound, it depends on the stereochemistry of each carbon that you have a COOH coming off.  If one goes forward and one goes backward, then you won't have a meso compoun.  If they both go forward or both go back ward though, then you do have a meso compound.  Remember that meso compounds have that line-of-symmetry.

As for the R/S configurations, let's do a comparison for OCH3 and C=.

O is heavier than C.  So O is higher priority.  The double bond would come into play IF you had to go one further in the chain to establish priority.  Like, if you can CH2CH3 as one substituent and -CH=CH2 as the other substituent...both have C as the first atom closest to the stereocenter.  Then the other carbon has 2 Hs and 1 C. and the other has 1 H and 2 Cs (THATS what the double bond does - its like being connected to two carbons  )

Long answer for not much reward.  But it's worth it to learn  Hope it helped!

[sdekivit]

I suppose I'll do what I can right off the bat here...and let other do the rest  

For the meso-ness of the compound, it depends on the stereochemistry of each carbon that you have a COOH coming off.  If one goes forward and one goes backward, then you won't have a meso compoun.  If they both go forward or both go back ward though, then you do have a meso compound.  Remember that meso compounds have that line-of-symmetry.

As for the R/S configurations, let's do a comparison for OCH3 and C=.

O is heavier than C.  So O is higher priority.  The double bond would come into play IF you had to go one further in the chain to establish priority.  Like, if you can CH2CH3 as one substituent and -CH=CH2 as the other substituent...both have C as the first atom closest to the stereocenter.  Then the other carbon has 2 Hs and 1 C. and the other has 1 H and 2 Cs (THATS what the double bond does - its like being connected to two carbons  )

Long answer for not much reward.  But it's worth it to learn  Hope it helped!

priority for Cahn-Ingold-Prelog nomenclature is not based on mass, but on atomnumber. O has numer 8, C has 6 and H is 1. So O has the highest priority, because it has the highest atom number.

Double bonds count double. thus -C = CH2 you must count for the priority CCH and not CHH ( you have to make 2 single bonds)

[lemonoman]

Touche.

Hopefully THAT helps as well

[macdonda]

My two cents: I haven't done anything organic for a while so I hope this isn't way off:

Thermodynamics aside, as I understand it, a more stable product will dominate.  I look at the H3C and the SCH3 groups and I picture them as large electron clouds; there's undoubtedly going to be repulsive forces; as I see it, the equatorial position of the SCH3 group is a more stable product (the 'clouds' are further apart).  

[alphahydroxy]

for question 1 I'd reckon your answer is right - the formula is Cn H2n-2, suggesting 2 degrees of unsaturation and a ring with a double bond fits the bill nicely - the reaction with 1 mole of H2 is further evidence for this

reaction with permanganate should form the diol then oxidise to the ketone and carbon dioxide

for question 4, it looks like your nucleophile is   -SMe (i.e. sulfur is nucleophilic atom) and your leaving group is also SMe

now, assuming this is an SN2 reaction, your initial product will have the attacking species in an axial position. However, as this would then release -SMe, which must be equally able to attack at the same position, the equlibrium should favour the isomer where both substituents are equatorial - i.e. the thermnodynamic product would probably be the same as that which you started with.


Having another look at your equation, it seems that maybe the substituent on the ring should actually be chlorine?  in this case, it's unlikely that chloride ion would be a good enough nucleophile to attack at the hetero-substituted carbon to reform the starting material.

If this is indeed the case, the product would then have the sulfur substituent in an axial position (or, at least, it would be opposite to the methyl), as is necessary in irreversible SN2 reactions (inversion of stereochemistry).

Essentially you should be considering the products obtained via kinetic versus thermodynamic control.