Orgo questions

[Bigbirdo]

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1. For question 27, why is the answer C not E? The hydrogen next to the triple bond is also an acidic hydrogen?

2. From Chad's video, I know that for allene, if both sides have two different groups, it will be chiral. What about a alkene with two different groupss on both sides? Is it also chiral and optically active?

3. For question 119, how do you turn a Newman projection into Fischer projection? Is there any way you can tell the chirality of a molecule in its Newman projection?

 

[Illfavor]

27- yes, a terminal alkyne has a pKa around 25 and is relatively acidic.
119- Just assign R/S configurations.

 

[Bigbirdo]

27- yes, a terminal alkyne has a pKa around 25 and is relatively acidic.
119- Just assign R/S configurations.

What are some Pka values you would recommend to memorize?
For 119, which carbon in the Fisher projection does the front carbon correspond to?

 

[Illfavor]

The most useful ones (I suppose) would be: CO2H, Phenol, Generic alcohol, Water, Ammonia, Alkyne/Alkene/Alkane. They are mostly a convenience that improves speed, and my OC teacher made us learn like 30 😴

The bottom carbon of the Fischer proj. = the front carbon of Newman proj.

 

[denttal]

1) Can you all explain why you are looking at acidity for this question? .. These are my thoughts on the question... I am not sure what you are implying by the hydrogen next to the triple bond is acidic? The making of a Grignard reaction is one that is not usually covered much, so I wouldn't worry much about it. However, to make a Grignard, the Carbon that is attached to the halide is the electrophilic carbon. In the question, choice C is the best answer b/c Cl is more EN than Iodide. Therefore, the carbon attached to the Cl is more electrophilic. The other possible answers are all eliminated b/c they have hydrogen bonding in them and a Grignard reagent will never exist with Hydrogen bonding on it, b/c the -MgX is highly basic!
Can you let me know where I am going wrong?

2) An allene (w/ 2 different R groups on each end) is the only molecule that can be chiral DUE TO THE DOUBLE BONDS. An alkene will never be chiral due to the double bond. That being said, if there is a chiral center somewhere else in the alkene, the alkene can be chiral.

3) I believe the answer is identical (each molecule has an R and an S chiral center, and the chiral centers have the same subs, so that would make them identical). Can you let me know if I am correct? I can then explain how to figure it out.

 

[Bigbirdo]

The most useful ones (I suppose) would be: CO2H, Phenol, Generic alcohol, Water, Ammonia, Alkyne/Alkene/Alkane. They are mostly a convenience that improves speed, and my OC teacher made us learn like 30 😴

The bottom carbon of the Fischer proj. = the front carbon of Newman proj.

What about the positions of the Cls? According to the answer, when the Newman projection given in the question is converted Fischer, the two Cls are opposite to each other. Does that mean when I convert Newman to Fischer, everything on the same side turn to opposite, and everything opposite should be on the same side?

 

[Illfavor]

What about the positions of the Cls? According to the answer, when the Newman projection given in the question is converted Fischer, the two Cls are opposite to each other. Does that mean when I convert Newman to Fischer, everything on the same side turn to opposite, and everything opposite should be on the same side?

Your questions is a bit unclear to me. All you need to know when converting between the two is the R/S configuration (if there are two chiral centers) and make sure they stay the same.

 

[Bigbirdo]

Never mind, I get it. Thank you!

 

[Psa]

Never mind, I get it. Thank you!

Haha can you please try to explain it? I'm confused about this same problem. Here's the solution to that problem. I don't get how you go from Newman to Fischer

 

[Bigbirdo]

I am just going to say in the given question, the molecule on the left is A, the molecule on the right is B. For this problem, all you need to know is that "The bottom carbon of the Fischer proj. = the front carbon of Newman proj". You then just assign R and S to the newman projection of B. Because you know how the carbons correspond to each other, you can just compare the chirality of B to the chirality of A (the fischer projection.)

 

[Psa]

Then what's with the bolded bonds in the solution? Cuz accoring to that, "The top carbon on fischer = front carbon on Newman"

 

[Bigbirdo]

Then what's with the bolded bonds in the solution? Cuz accoring to that, "The top carbon on fischer = front carbon on Newman"

Interesting......I am confused again...

 

[Psa]

Same here. Hopefully someone will come to the rescue.

 

[ChrisM07]

I think the bolded bonds are just to keep track of what carbon is which since they both contain a methyl group and a Cl. The bolded guy (the front one) keeps the conformation while the one in the back is flipped and becomes the bottom one. Why it gets turned, I don't know, I wondered the same thing myself.

 

[JDHK]

The bolded parts are really misleading. The front carbon in the newman is the bottom carbon in the fischer.

 

[Psa]

The bolded parts are really misleading. The front carbon in the newman is the bottom carbon in the fischer.

Even if the front carbon in newman is the bottom carbon in fischer, the Cl and the H is still swapped.

Btw, I've never actually read anywhere that the front one in newman must be the "lower" one in fischer. Can you please explain why? I assume the book made the bolded parts one purpose.

 

[JDHK]

Even if the front carbon in newman is the bottom carbon in fischer, the Cl and the H is still swapped.

Btw, I've never actually read anywhere that the front one in newman must be the "lower" one in fischer. Can you please explain why? I assume the book made the bolded parts one purpose.

The Cl and H are not swapped. If you rotate the single bond (because you can) when you mentally convert the bottom carbon of the fischer model, it rotates to exactly the same as the one depicted by the newman projection. Try drawing out the aftermath of the rotation.

 

[Bigbirdo]

1) Can you all explain why you are looking at acidity for this question? .. These are my thoughts on the question... I am not sure what you are implying by the hydrogen next to the triple bond is acidic? The making of a Grignard reaction is one that is not usually covered much, so I wouldn't worry much about it. However, to make a Grignard, the Carbon that is attached to the halide is the electrophilic carbon. In the question, choice C is the best answer b/c Cl is more EN than Iodide. Therefore, the carbon attached to the Cl is more electrophilic. The other possible answers are all eliminated b/c they have hydrogen bonding in them and a Grignard reagent will never exist with Hydrogen bonding on it, b/c the -MgX is highly basic!
Can you let me know where I am going wrong?

2) An allene (w/ 2 different R groups on each end) is the only molecule that can be chiral DUE TO THE DOUBLE BONDS. An alkene will never be chiral due to the double bond. That being said, if there is a chiral center somewhere else in the alkene, the alkene can be chiral.

3) I believe the answer is identical (each molecule has an R and an S chiral center, and the chiral centers have the same subs, so that would make them identical). Can you let me know if I am correct? I can then explain how to figure it out.

Are you sure the carbon is electrophilic in the Grignard reaction? if that is true, I think you explanation also works. The reason i mentioned the acidity of the terminal alkyne is because the more acidic the alkyne, the more likely it is going to give out its hydrogen, which makes itself unsuitable for a Grignard reaction. Here is the website you can refer to http://www.mhhe.com/physsci/chemistry/carey5e/Ch09/ch9-2.html