j4yman
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a few questions...
Hey i was wondering which reagents would serve as the basis for a simple chemical test to distinguish between benzoic acid and benzamide?
My options are
1. Cold dilute sodium bicarbonate
2. Cold dilute sulfuric acid
3. Cold dilute sodium hydroxide
I was thinking 1 and 3 because an acid would not work....
This other question in the textbook has got me stumped too.
When treated with sodium iodide, a solution of (S)-2-iodooctane in acetone.....(more than 1 correct answer)
1. does not react
2. loses its optical activity
3. gives both (R) - and (S) -2-iodooctane
4. is converted to (R) -2-iodooctane
I'm really struggling with this question and don't know where to start.
Also, treating t-butyl chloride (CH3)3C-Cl with a mixture of H2O CH3OH at room temperature would yield what?
I have a few options, and there mayb more than 1 correct answer.
1. CH2=C(CH3)2
2. (CH3)3COH
3. (CH3)3COCH3
4. CH3Cl
My guess would be 2 and 4.....
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bereal511
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Well, for your first question, I believe that your answer would only be sodium bicarbonate. I'm not too sure if there's any color change with change
in pH for benzoic acid or benzamide, but sodium bicarbonate definitely would react with the benzoic acid to produce visible carbon dioxide bubbling.
As an adolescent I aspired to lasting fame, I craved factual certainty, and I thirsted for a meaningful vision of human life -- so I became a
scientist. This is like becoming an archbishop so you can meet girls.
-- Matt Cartmill
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The_Davster
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I am not going to do them for you but I will give you the necessary info to solve them.
First: Benzoic acid has an acidic proton, what compound in that list would react with an acidic proton to make an obvious result?
Second: Iodine is a pretty good leaving group, so there will be a sort of equilibrium between the iodooctane and the positive carbocation of
octane(carbocation in the spot iodine was removed from) Furthermore, iodines ability as a leaving group is enhanced by the fact that the iodine is on
a secondary carbon. Think SN1. So if another iodide ion were to come in at the carbocation, what would be the result?
Third: Again SN1 type mech this time leaving a tertiary carbocation, much more stable than primary or secondary. Now any lone pair on a molecule will
get sucked over to that carbocation. Which two things in that list are a result of a lone pair going at the carbocation? Product will likely be a mix
of two things there, but you must consider the groups attached to the atom with the lone pair on it, the one which is electron donating will increase
the negative charge on the atom with lone pair and preferentially react with the carbocation.
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turd
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So what? Would it change anything if it was SN2?
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j4yman
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| Quote: | Originally posted by rogue chemist
I am not going to do them for you but I will give you the necessary info to solve them.
First: Benzoic acid has an acidic proton, what compound in that list would react with an acidic proton to make an obvious result?
Second: Iodine is a pretty good leaving group, so there will be a sort of equilibrium between the iodooctane and the positive carbocation of
octane(carbocation in the spot iodine was removed from) Furthermore, iodines ability as a leaving group is enhanced by the fact that the iodine is on
a secondary carbon. Think SN1. So if another iodide ion were to come in at the carbocation, what would be the result?
Third: Again SN1 type mech this time leaving a tertiary carbocation, much more stable than primary or secondary. Now any lone pair on a molecule will
get sucked over to that carbocation. Which two things in that list are a result of a lone pair going at the carbocation? Product will likely be a mix
of two things there, but you must consider the groups attached to the atom with the lone pair on it, the one which is electron donating will increase
the negative charge on the atom with lone pair and preferentially react with the carbocation. |
Thanks for your help.
So the first answer is sodium bicarbonate?
2nd answer is it loses its optical activity and is converted to (R) -2-iodooctane?
3rd answer is 2 and 4?
Please correct me if i'm wrong
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The_Davster
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second: It looses it optical activity because a SN1 type reaction does not have a stereochemical preferance.
third: mostly 3, but sugnificant 2 will also form.
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turd
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| Quote: | Originally posted by rogue chemist
second: It looses it optical activity because a SN1 type reaction does not have a stereochemical preferance. |
It would also lose optical activity with SN2 type reactions.
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The_Davster
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With a true SN2 reaction there would be a single optical isomer produced. With this however, nomatter whether it goes by SN1 or SN2 due to both
'reactant' and 'product' being present, optical activity is lost.
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