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Chemistry question about Organic Chemistry

Asked Jul 28, 2011, 01:37 AM — 9 Answers
Could you explain to me why I ended up with two marks out of a possible 6 in these questions? I think that I answered them correctly but I am not sure so I would like your help so that then I may tell my instructor.

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9 Answers
Unknown008's Avatar
Unknown008 Posts: 8,147, Reputation: 3745
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#2

Jul 28, 2011, 09:47 AM
1) For the first question, you get 3 marks for those:
1 mark for the reagent A
1 mark for the conditions
1 mark for the name of B.

You got the first only, and that... Wasn't completely right. Your answers are 'somewhat' good, but cannot be given full marks there.
1. HCN
2. In the presence of NaOH as catalyst.
3. The name is 2-hydroxy-2-methyl-propane nitrile. You forgot the hydroxy part.

2) Same mark distribution here.
1. Same thing here, you forgot the hydroxy part. 2-hydroxy-2-methyl-propanoic acid.
2. Reagent: concentrated H2SO4 (which gets your mark, though this might be doubtful given the way (s)he underlined your 'concentrated')
3. Conditions: Boiled at about 180 C under reflux.
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DrBob1's Avatar
DrBob1 Posts: 425, Reputation: 445
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#3

Jul 28, 2011, 02:37 PM
Unknown008 is correct in the errors in your names. Compound B is more commonly referred to as "acetone cyanohydrin". You will find many more references to that name.
To form the cyanohydrin, you really need to use NaCN or KCN ---- you refer to NaOH as a catalyst but, of course, it will react to form the salt. The CN- ion is a much better nucleophile for the addition reaction, and you don't want to use HCN for safety reasons.
The step to dehydrate the hydroxy acid to MAA seems overly robust, but I can't find any information on the proceedure.
You never answered the question about the addition of HBr to MAA to form compouond E.
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Roddilla's Avatar
Roddilla Posts: 140, Reputation: 5
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#4

Jul 28, 2011, 11:27 PM
Comment on Unknown008's post
Quote:
Originally Posted by Unknown008 View Post
1) For the first question, you get 3 marks for those:
1 mark for the reagent A
1 mark for the conditions
1 mark for the name of B.

You got the first only, and that... Wasn't completely right. Your answers are 'somewhat' good, but cannot be given full marks there.
1. HCN
2. In the presence of NaOH as catalyst.
3. The name is 2-hydroxy-2-methyl-propane nitrile. You forgot the hydroxy part.

2) Same mark distribution here.
1. Same thing here, you forgot the hydroxy part. 2-hydroxy-2-methyl-propanoic acid.
2. Reagent: concentrated H2SO4 (which gets your mark, though this might be doubtful given the way (s)he underlined your 'concentrated')
3. Conditions: Boiled at about 180 C under reflux.
I always mix up alcholic with alkaline - that's why I said alcoholic HCN; I wanted to say alkaline HCN i.e. HCN in alkaline solution
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Roddilla's Avatar
Roddilla Posts: 140, Reputation: 5
Junior Member
  
#5

Jul 28, 2011, 11:29 PM
Comment on Unknown008's post
Quote:
Originally Posted by Unknown008 View Post
1) For the first question, you get 3 marks for those:
1 mark for the reagent A
1 mark for the conditions
1 mark for the name of B.

You got the first only, and that... Wasn't completely right. Your answers are 'somewhat' good, but cannot be given full marks there.
1. HCN
2. In the presence of NaOH as catalyst.
3. The name is 2-hydroxy-2-methyl-propane nitrile. You forgot the hydroxy part.

2) Same mark distribution here.
1. Same thing here, you forgot the hydroxy part. 2-hydroxy-2-methyl-propanoic acid.
2. Reagent: concentrated H2SO4 (which gets your mark, though this might be doubtful given the way (s)he underlined your 'concentrated')
3. Conditions: Boiled at about 180 C under reflux.
Thank you very much for now
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Unknown008's Avatar
Unknown008 Posts: 8,147, Reputation: 3745
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#6

Jul 29, 2011, 08:37 AM
DrBob, I think that the last part is on part (c) and I would guess that Roddilla got it right from the sketches

Only thing might be the nomenclature again
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DrBob1's Avatar
DrBob1 Posts: 425, Reputation: 445
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#7

Jul 29, 2011, 03:18 PM
I think you added the Br atom to the wrong carbon, Roddilla. There goes another point!
(HCN in an alkaline solution doesn't exist; it is, and should be called, NaCN or KCN).
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Unknown008's Avatar
Unknown008 Posts: 8,147, Reputation: 3745
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#8

Jul 29, 2011, 11:30 PM
Hmm... I find nothing wrong with the products after addition of HBr...

Or I'm starting to seriously forget my chemistry
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DrBob1's Avatar
DrBob1 Posts: 425, Reputation: 445
Full Member
  
#9

Jul 30, 2011, 11:29 AM
Methacrylic acid + HBr yields 3-bromo-2-methylpropionc acid (the answer provided on the question sheet seems to imply a mixture, so presumably both isomeric acids are probably formed.)
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Unknown008's Avatar
Unknown008 Posts: 8,147, Reputation: 3745
Uber Member
  
#10

Jul 30, 2011, 11:41 AM
Yes, there are two sketches, one above, the 3-bromo-2-methyl-propanoic acid and the second is 2-bromo-2-methyl-propanoic acid. Glad I'm not as rusty as I though
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