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chemistry question!!
Date Posted: 4 years ago
Views: 256
I suspect it’s potassium manganate. Both of the compounds have an alcohol group, so both X and Y allow the decolourisation reactions. You won’t be able to “distinguish” between the two.
Bromine decolourises the C = C double bond in X, but does not decolourise Y (it does not react with the C = O double bond).
Sodium carbonate reacts with Y and gives off effervescence, even though Y is a relatively weak acid, but it has no visible reaction with X.
For the same acid reason, Y turns litmus red, but it will not show a red result with X.
(In actual fact, I think X is also slightly acid due to the alcohol group itself, but this is not taught at your O levels).
Bromine decolourises the C = C double bond in X, but does not decolourise Y (it does not react with the C = O double bond).
Sodium carbonate reacts with Y and gives off effervescence, even though Y is a relatively weak acid, but it has no visible reaction with X.
For the same acid reason, Y turns litmus red, but it will not show a red result with X.
(In actual fact, I think X is also slightly acid due to the alcohol group itself, but this is not taught at your O levels).
Alcohol groups are taken to be neutral since their pKa is usually near water.
For both, there is no group to stabilise the conjugate base upon dissociation (unlike phenol, where the phenolate ion is resonance stabilised), so the conjugate base is very unstable due to the negative charge on the O- being unable to be dispersed. So dissociation is highly disfavoured and both's alcohol group would be considered neutral.
The answer is A as both X and Y 's alcohol groups will be oxidised to ketone for Y and carboxylic acid for X. Decolorisation of purple acidified KMnO4 occurs so we can't distinguish between the two.
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)/13%3A_Structure_and_Synthesis_of_Alcohols/13.05%3A_Acidity_of_Alcohols_and_Phenols
For further reading
For both, there is no group to stabilise the conjugate base upon dissociation (unlike phenol, where the phenolate ion is resonance stabilised), so the conjugate base is very unstable due to the negative charge on the O- being unable to be dispersed. So dissociation is highly disfavoured and both's alcohol group would be considered neutral.
The answer is A as both X and Y 's alcohol groups will be oxidised to ketone for Y and carboxylic acid for X. Decolorisation of purple acidified KMnO4 occurs so we can't distinguish between the two.
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)/13%3A_Structure_and_Synthesis_of_Alcohols/13.05%3A_Acidity_of_Alcohols_and_Phenols
For further reading
Answer is A, for the O-level syllabus, where you learn that only primary alcohols (with the -OH functional group at one of the carbons at the end of a organic molecule) can be oxidised to a carboxylic group (-CO2H). What Nicholas and J comment about is correct.. but leave that knowledge to the appropriate level of education.
From his other comment, the student thought that Y wasn't oxidisable from the so had to explain oxidation to ketone and that decolorisation of purple acidified KMnO4 also occurs.
Otherwise, he would have wrongly concluded that option A could be used to distinguished between the two compounds
Otherwise, he would have wrongly concluded that option A could be used to distinguished between the two compounds
Not wrong for the O level syllabus. Just like students will learn dot and cross diagrams are incorrect at A levels and beyond. Matter of imparting appropriate knowledge at the correct stage.
If that's considered not wrong for O levels, how would the student pick the correct answer for this question? He would think all 4 options work.
As mentioned. The expected answer is A, as they only learn that primary alcohols can be oxidised to carboxylic acids.
But the question is asking which cannot be used to distinguish between the two.
If he only has the knowledge that primary OH can be oxidised to COOH , he would conclude that Y cannot be oxidised as it doesn't have primary OH (not having the knowledge secondary OH is able to be oxidised as well). Then he would conclude there's no decolorisation here.
But X has a primary alcohol so he would think yes, it can be oxidised to COOH and result in decolorisation.
He would then think that A is not the option to choose since he would conclude A can be used to distinguish between the two.
Then since the other 3 options work as well, he would think that none of them are the answers to pick and end up being stuck.
If he only has the knowledge that primary OH can be oxidised to COOH , he would conclude that Y cannot be oxidised as it doesn't have primary OH (not having the knowledge secondary OH is able to be oxidised as well). Then he would conclude there's no decolorisation here.
But X has a primary alcohol so he would think yes, it can be oxidised to COOH and result in decolorisation.
He would then think that A is not the option to choose since he would conclude A can be used to distinguish between the two.
Then since the other 3 options work as well, he would think that none of them are the answers to pick and end up being stuck.
Whoops.. misread the question. B, C and D definitely can differentiate the two, so likely the student may be from a IP or SAP school that teaches about secondary alcohols being oxidised to a ketone. Thanks J.
Welcome Arnold.
I'm just thinking that from his other comment, it seems to imply that he doesn't know about oxidation of secondary OH to ketones (resulting in decolorisation). If he is from a IP or SAP school that teaches this, then he wouldn't have a problem answering it, unless his content mastery isn't there yet.
I'm guessing the student is on the O level track, but the question is a higher order question that expects the student to make an educated guess about the oxidation of secondary OH.
They're not taught about ketones yet, but would know that oxidation of primary OH 'removes' 1 H, 'turns' the other H into OH and the C-O bond becomes C = O, and thus can use this info to deduce that Y's OH is oxidisable as well, just that it doesn't have the extra H to be 'turned' into OH
I'm just thinking that from his other comment, it seems to imply that he doesn't know about oxidation of secondary OH to ketones (resulting in decolorisation). If he is from a IP or SAP school that teaches this, then he wouldn't have a problem answering it, unless his content mastery isn't there yet.
I'm guessing the student is on the O level track, but the question is a higher order question that expects the student to make an educated guess about the oxidation of secondary OH.
They're not taught about ketones yet, but would know that oxidation of primary OH 'removes' 1 H, 'turns' the other H into OH and the C-O bond becomes C = O, and thus can use this info to deduce that Y's OH is oxidisable as well, just that it doesn't have the extra H to be 'turned' into OH
MCQ and paper 2 questions over the last few years need a better understanding of concepts... but student should know that B C and D are definitely possible to use to differentiate between X and Y. So perhaps the student was meant to derive the answer by elimination. No definition questions have appeared only for O level Chemistry, for the last 2 years, but if the student does not understand definitions well, they would have difficulty answering some questions. Personally, I doubt a secondary alcohol would appear, as O level students only learn about the oxidation of alcohols to carboxylic acids by atmospheric oxygen or by acidified KMnO4. This is only applicable to primary alcohols.
I see. Thanks for the sharing
Likewise J. Always good to have a constructive discussion. :)
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Date Posted:
4 years ago
but the position of the OH grp in Y. Y cannot be oxidised to form carboxylic acid.
Actually, the OH group does not have to be at the last carbon. That position is also oxidisable. The OH can be on any of the carbons, not just the cornermost carbons. It is, however, a fact that the carbon in which the OH is attached to is very important, but this is only taught in the A Levels.
When you proceed to the A Levels, you will learn that substances such as X is called a primary alcohol, while Y is called a secondary alcohol. This is because the carbon attached to an OH group in X is attached to two other H as well, while the carbon attaches to an OH group in Y is attached to one other carbon (non-H) and one H.
Potassium manganate will be able to react with both primary alcohols and secondary alcohols, but not with what we call tertiary alcohols, where the carbon attached to the OH group is not directly attached to any H.
Primary alcohols get oxidised to carboxylic acids, while Secondary alcohols gets oxidised to something called ketones. Ketones are not covered in your syllabus.
When you proceed to the A Levels, you will learn that substances such as X is called a primary alcohol, while Y is called a secondary alcohol. This is because the carbon attached to an OH group in X is attached to two other H as well, while the carbon attaches to an OH group in Y is attached to one other carbon (non-H) and one H.
Potassium manganate will be able to react with both primary alcohols and secondary alcohols, but not with what we call tertiary alcohols, where the carbon attached to the OH group is not directly attached to any H.
Primary alcohols get oxidised to carboxylic acids, while Secondary alcohols gets oxidised to something called ketones. Ketones are not covered in your syllabus.
Litmus paper turns green in neutral solution? I think you mean pH paper / pH test strips
Also, it should be purple with X , red with Y instead of the other way round which it was written
Also, it should be purple with X , red with Y instead of the other way round which it was written
Xi, do take note that acidified KMnO4 is a very powerful oxidising agent. At A levels it is taken be unable to oxidise tertiary alcohols, but in actual fact it can.
It does this first via dehydration of the alcohol to the alkene, then subsequent oxidation to the carboxylic acid results. You will learn this at university level chemistry.
It does this first via dehydration of the alcohol to the alkene, then subsequent oxidation to the carboxylic acid results. You will learn this at university level chemistry.
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