Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: Twickel on May 18, 2012, 04:44:27 AM
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Hi
I am having trouble with the following, there is one more which I will attach soon.
Let me say what I think the answers are and please correct me if I am wrong.
1) Is the second option?
2) I have no idea, im guessing the first step will involve NaOEt though
3) I will post an image
4) nuc armoatic sub, aromatics cant form carbocations to unstable?
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that should be a pentagon and the double bond that got cut should have an Oxygen at the end.
For number2, you need to form an enolate ion, I think
so I believe this is the correct set of steps NaOEt in dry EtOh ii PhCH2Br ii Oh-/h2o/heat iv H+/H2o/heat?
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For number 2 have you not included an extra C atom in the starting material?
If so then it a hydrogenation of the aromatic ring, if not then I have no idea.
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Twickel, can you please spare 30 seconds to proof read your posts before pressing the "post" button? I mean, this is nonsense isn't it?
that should be a pentagon and the double bond that got cut should have an
For number2, you need to form an enolate ion?
I
or NaOEt in dry EtOh ii PhCH2Br ii Oh-/h2o/heat iv H+/H2o/heat?
If you are not willing to put in the effort to ask a coherent question, you should not be too optimistic about coherent responses.
Anyway,
1. As discodermolide asked, check structures for Q1, I think you copied them down incorrectly.
2. Your sequence looks quite good: (i) NaOEt, EtOH; (ii) PhCH2Br; (iii) NaOH(aq), heat; (iv) H3O+, heat.
You probably won't need step (iii), as the esters will hydrolyse under acidic conditions anyway.
3. Can you redraw/rewrite this question? You ask what the reaction will be when "cyclopentane" is subjected to the conditions - do you really mean cyclopentane? Alkanes will not react with any of the reagents in that sequence. Is the structure you drew in your second post an attempt at the answer or is it the starting material?
4. I agree with your answer.
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I am sorry, I was in a rush, I will be sure to double check my questions. It was meant to be cyclopentanone.
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OK, so what is your answer for Q1? Which of those 4 starting materials would you expect to give the product.
For Q3, I don't agree with your answer. Can you draw a mechanism for its formation?
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For Q1 I said it was number 2, because you want the ketone to form the enolate not the aldehyde ( to stop self condensation) option one doesnt form a five membered ring I think ( ihave trouble forming rings).
In regards to Q3, I am unsure as to what role the EtOCO2Et does, is this a claisen condensation that does not go to completion with a Sn2 reaction adding to the enolate formed?
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For Q3 is it one of these? If not I do not know
I think its
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From your abysmal picture it looks like number 2!
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I am really sorry, my scanner keeps chopping bits of for some reason, I re scanned it three times. Was it number 2 for the cyclopentanone?
Btw, how about the intramolecular aldol condensation, was it the first or second one?
Thanks
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I am really sorry, my scanner keeps chopping bits of for some reason, I re scanned it three times. Was it number 2 for the cyclopentanone?
Btw, how about the intramolecular aldol condensation, was it the first or second one?
Thanks
Sorry I can't define which reaction you are referring to.
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Sorry for not being specific, in my first image with the first molecule, where I asked what was the starting product of this reaction ( intra molecular aldol condensation)
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For Q1 I said it was number 2, because you want the ketone to form the enolate not the aldehyde ( to stop self condensation) option one doesnt form a five membered ring I think ( ihave trouble forming rings).
It is not a choice of 1 or 2 - you posted 4 candidate starting materials in your second posting of this question. Which of those 4 do you think it is - can you draw the mechanism?
For Q3:
You have deprotonated cyclopentanone and jumped straight in with methyl iodide. The first step utilises diethyl carbonate - what do you think that's doing?
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Actually, I can not. How pathetic, I tried but its gone way wrong some where.
I would say its the solvent ( in regards to diethyl carbonate) but going from your tone Im guessing its a reagent.... not too sure sorry
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I think then that its the second option for the intra molecular reaction....
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I think then that its the second option for the intra molecular reaction....
Your product in the latest picture has a pentavalent carbon atom!
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:@:@. damn it. I am not thinking,to be honest I do not know what the answer is, can someone explain it to me?
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:@:@. damn it. I am not thinking,to be honest I do not know what the answer is, can someone explain it to me?
Sorry, I'm going to bow out here. All these pictures have totally confused me so that I do not know what you require to know.
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Can I just re draw the ONE question?
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I think I see some correct thinking occurring. That should render Q1 to a correct solution. However, this step makes no sense.
I think if you were to simply attempt an aldol cyclization from each of the possible enolates should answer the question.
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I didnt know the correct positions of the two carbonyls relative to each other, when I first drew it out, they were not alpha to one another but Beta, so I just made that up to get the hydrogens to be alpha to the carbonylss
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I think I see some correct thinking occurring. That should render Q1 to a correct solution. However, this step makes no sense.
I think if you were to simply attempt an aldol cyclization from each of the possible enolates should answer the question.
The problem is my text book, only give one example.... I do not know how to do this reaction. Can anyoneplease just show me the correct mechanism I will give it one more shot, realised another mistake ive been making. It has to be B? is it possible to have mixed intra molecular aldol condensations?
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Is it B, I worked with retro arrows, so the new pi bond alpha and beta to the carbony carbon is broken, I stick an Oh group on the Beta carbon, break that bond and I get something that looks like B,
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Go one step at a time. For Q1, take each of the 4 substrates in turn, draw an aldol condensation mechanism and see what you get as the product. You cannot simply pick one because it "looks right", go through the intramolecular condensations for each starting material - the appropriate starting material is the one that gives you the product by a reasonable mechanism. Also - check the structures again! You posted the 4 options twice now, but compound C has changed from a dialdehyde to a ketoaldehyde, and D has changed from a ketoaldehyde to a diketone (see below) - which one is it?
Take your time, and check your work before posting it up. Slow down and don't cut corners - it is clear you have the concepts required to solve these problems, I just get the impression you are just rushing too much.
For Q3, what kind of reagent is diethyl carbonate - acid, base, nuclophile, electrophile?
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I think it is a weak acid, because it can lose a proton and the conjugate base can be stabilised through resonance.
In regards to Q1, I can not do it.... can someone please walk me through it. I will post the mechanism for A and from that it is wrong....
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Look I can not do it I have tried so many times, it is not working, I need help, someone please show me the correct mechanism, for some reason I get 6 membered rings.
But I think its the aldehyde that has to be enolised, because the carbonyl thats first enolised is not the one that gets protonated to form OH
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Look I can not do it I have tried so many times, it is not working, I need help, someone please show me the correct mechanism, for some reason I get 6 membered rings.
But I think its the aldehyde that has to be enolised, because the carbonyl thats first enolised is not the one that gets protonated to form OH
If I understand correctly your product has an aldehyde attached to a 5 membered ring with a double bond in it, correct? As in my picture?
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Yes that is the product, I need to know where that came from
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Yes that is the product, I need to know where that came from
OK then your starting material must have two CH2CHO groups in it. Like the picture.
Perhaps you can work it out from there?
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Thank you. The problem I have is forming the actual ring, I form the enolate ion on the bottom aldehyde. I then count the carbons number 1 being the carbon bonded to the negatively charged oxygem, number 6 being the carbon of the other aldehyde, that gives me 6. How do I correctly number the carbons when forming rings?
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Thank you. The problem I have is forming the actual ring, I form the enolate ion on the bottom aldehyde. I then count the carbons number 1 being the carbon bonded to the negatively charged oxygem, number 6 being the carbon of the other aldehyde, that gives me 6. How do I correctly number the carbons when forming rings?
It does not matter here where the enolate is formed as it is symmetrical. It is the enolate carbon which reacts not the carbonyl carbon.
Try again before I post the answer.
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I am obviously doing something wrong when closing the ring, as I can not get the carbonyls to be alpha to one another, so I can not form the final conjugated product.
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I am obviously doing something wrong when closing the ring, as I can not get the carbonyls to be alpha to one another, so I can not form the final conjugated product.
You are closing the ring with the wrong carbon! In your drawing it is carbon number 1 which attacks the other aldehyde.
Try again before I post the answer
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but is carbon 1 not the negatively charged carbon?
If I draw a resonance diagram, I thought it was the alpha carbon which is negatively charged, now I got that carbon bonded to five things again.
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but is carbon 1 not the negatively charged carbon?
If I draw a resonance diagram, I thought it was the alpha carbon which is negatively charged
It is the carbon at the end of the enolate C=C-O- which attacks.
Try again before I post the answer.
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I hope this is it. i had to scribble around so the scanner would scan everything.
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Bingo it doesn't look bad.
Here is my scheme, without the curly arrows which you can put in.
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Thank you, I now know it is much easier to draw the enolate, with the negative charge on the carbon, rather then on the oxygen, but only for intra molecular reactions....
Mind helping me with one more questions? Its in the thread but I need to re write its been lost.
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Thank you, I now know it is much easier to draw the enolate, with the negative charge on the carbon, rather then on the oxygen, but only for intra molecular reactions....
Mind helping me with one more questions? Its in the thread but I need to re write its been lost.
Re-write it , but I have to go out soon so don't expect a rapid answer.
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Cyclopentanone reacts with i) NaOEt/ EtoCO2Et/ dry EtOH
ii) CH3I
iii) H+/H20/Heat
What is the product?
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Cyclopentanone reacts with i) NaOEt/ EtoCO2Et/ dry EtOH
ii) CH3I
iii) H+/H20/Heat
What is the product?
What is the diethyl carbonate EtOCO2Et doing here? Make the enolate from cyclopentanone and take it from there!
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Sorry for the late reply, thats the thing I dont know what the ester does.. is it an acid?
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Sorry for the late reply, thats the thing I dont know what the ester does.. is it an acid?
No it is not an acid it's an electrophile. React the cyclopentanone enolate with the carbonyl of the diethyl carbonate and see what you get.
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Ok, if that is right, where do I attack the CH3 from
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Ok, if that is right, where do I attack the CH3 from
In your product draw it as an ester then see where you go next. Look at my picture and compare yours!
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(Quote is link to reply #43)
@Twickel, As frustrating as you are finding the chemistry, we are finding a similar frustration. I used to advise students to number the atoms to avoid making a wrong connection. In one of your posts, you have indeed numbered your starting material and yet failed to draw the correct product. For me, that is frustrating.
For your last problem, I could say to eliminate the weakest base, but with the structure so sloppily drawn, I am not sure what might be the result. If it is the result of a correctly drawn intermediate, it would be the correct product (to that point).
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(Quote is link to reply #43)
@Twickel, As frustrating as you are finding the chemistry, we are finding a similar frustration. I used to advise students to number the atoms to avoid making a wrong connection. In one of your posts, you have indeed numbered your starting material and yet failed to draw the correct product. For me, that is frustrating.
For your last problem, I could say to eliminate the weakest base, but with the structure so sloppily drawn, I am not sure what might be the result. If it is the result of a correctly drawn intermediate, it would be the correct product (to that point).
I'm not sure Mr Twickel knows what the weakest base is. He is panicking and making errors. He needs to slow down and think.
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Ok, yes I have been drawing my products poorly I guess I am rushing too much and this is effecting my thinking, getting anxious because we have only had lectures on carbonyl chemistry, we have no tutorials ( do not know why) the tuts jumped straight from aromatics to quantam chemistry. This is also my first ever organic chem course I have taken.
So I formed the tetrehedral product, I then lose OEt-, not sure if I am correct but I thought, the Oet- that has left the tetrehedral carbon is a strong enough base to remove a hydrogen alpha to the di carbonyls, this is the driving force of the reaction, but there is a step somewhere in which you can add alkyl halides via Sn2, let me try I will draw as neat as possible.
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Looks good, but you ought to draw out the mechanism for the hydrolysis/deprotection step for your class.
I will add a little to orgopete's comment: The most frustrating thing for me here was the fact that you were posting several different versions of the same question. For example, you posted 3 versions of Q1, and each version had different multiple choice answers. Only one of the versions you posted actually had the correct answer as an option, because you copied the question down incorrectly. Take your time, read the question twice, and be careful - it will save you time in the long run.
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Yes, I did not realise until you mentioned it today, will try to organise myself better in the future.
Thank you to everyone for all the help.
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Yes, I did not realise until you mentioned it today, will try to organise myself better in the future.
Thank you to everyone for all the help.
Here is my answer to this question.Once again without curly arrows, which you can supply.
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Thanks for those diagrams, I print them fill them in and stick them to my notes.