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Do we have to specify/indicate the polarity of an electrolytic cell /electrochemical cell?
E.g. 2010 p3 qn 5
When do we have to?
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Originally posted by Flying grenade:
SOLUBILITY OF SULFATES, CARBONATES, NITRATES DECREASES DOWN THE GROUP
Whereas SOLUBILITY OF HYDROXIDES INCREASES DOWN THE GROUP
How to account for this? I only know this
Less exothermic DelHsoln signifies a less soluble salt
Less exo DelHhyd decreases solubility
Less exo L.E. enhances solubility
No time liao to bother with this (and you need to come for my tuition to discuss this properly, it's too inefficient back and forth on an online forum). But for those who feel they're prepared for tmrw's paper liao, and are keen on understanding this (it's *possible* Cambridge can ask this), can read : http://www.chemguide.co.uk/inorganic/group2/problems.html -
Originally posted by Flying grenade:
Do we have to specify/indicate the polarity of an electrolytic cell /electrochemical cell?
E.g. 2010 p3 qn 5
When do we have to?
Wah lau eh, if in doubt just specify lor! If the qn need u to specify, it'll say so mah! -
Originally posted by Flying grenade:
2010 p3 qn 4 gii
Photo is here
https://www.dropbox.com/s/morxcgb8519dx60/20151105_145329.jpg?dl=0
How can secondary alcohol oxidise to Carboxylic acid??
Thought secondary alcohol can only oxidise to ketone at maximum?
There are only 3 ways for a ketone to be further oxidized :
If it's part of a benzene ring side chain (with a benzylic H atom).
If you use a chemical oxidizing agent stronger than KMnO4, eg. concentrated HNO3.
If you burn it. -
You rock my world,
Thank you. Angel sent from heaven
Sludge man!
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I'm logging off this forum at 9pm sharp until tmrw afternoon. Any serious and urgen qns, you guys (ie. JC students taking A levels tmrw) better post now. I won't reply any posts after 9pm. Tell u first hor.
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Logging off for tonight liao. Cya guys tomorrow here in this forum after P2 to discuss the P2 questions! Have fun! ;D
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Since amide and ester can react with both naoh and hcl, can it be said that they are amphoteric /amphiprotic?
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No.
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When/what can say amphoteric/amphoprotic species
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Originally posted by Flying grenade:
When/what can say amphoteric/amphoprotic species
When the species (eg. HCO3-) can both donate and accept a proton in an acid-base reaction. An ester or amide can only *kena hydrolyzed* by acid or alkali in a *hydrolysis* reaction, not acid-base reaction. -
Originally posted by Flying grenade:
When/what can say amphoteric/amphoprotic species
And it's "amphiprotic" lah, still write what "amphoprotic" leh, make Cambridge laugh at Singapore students.
Edited : ok I noticed you spelled it correctly in an earlier post, so it's just your typo. Notwithstanding, typos can still make Cambridge laugh at Singapore students. -
Oh, in case you're wondering (or if Cambridge decides to play punk and asks) "What's the difference between "amphoteric" and "amphiprotic"?", go read Jim Clark's http://www.chemguide.co.uk/physical/acidbaseeqia/theories.html
Bonus : If Cambridge asks "What's the difference between "amphoteric" and "amphiprotic"? Illustrate your explanation with an example of a species that is amphoteric but not amphiprotic."
Well, Al2O3 is an example of a species that is amphoteric but not amphiprotic. Can you figure out why? ;) -
Hi for MJC Prelim 2014 P2/Q3b(iii)
the answer written for the oxidation state of cobalt in salt A is +4, but I cannot see why? From drawing the structure shouldn't it be 0?
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Originally posted by Vitalitylx:
Hi for MJC Prelim 2014 P2/Q3b(iii)
the answer written for the oxidation state of cobalt in salt A is +4, but I cannot see why? From drawing the structure shouldn't it be 0?
The question didn't even ask for the OS of Co in salt A, only in salt B. So if the question didn't ask, how can there be an answer (for a question that wasn't asked)? But yes you're right, the OS of cobalt in salt A is 0 (formal charge 4-) and in salt B is +3 (formal charge 3-). -
Originally posted by UltimaOnline:
The question didn't even ask for the OS of Co in salt A, only in salt B. So if the question didn't ask, how can there be an answer (for a question that wasn't asked)? But yes you're right, the OS of cobalt in salt A is 0 (formal charge 4-) and in salt B is +3 (formal charge 3-).On my paper it says state the oxidation state of both salts though, and the answers gave +4 in A and +3 in B. But okay thank u!
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Originally posted by Vitalitylx:
On my paper it says state the oxidation state of both salts though, and the answers gave +4 in A and +3 in B. But okay thank u!
Ah ok, the formatting on the version I have is screwed up, with overlaps, so I couldn't read the question properly. Yes, the OS of Co in that salt A is definitely zero. What probably happened : the MJC teacher wrote the wrong answer in an earlier version of the Prelim paper with the wrong OS of +4 (that you have), subsequently when other MJC teachers pointed out the error, the MJC teacher released a later version of the Prelim paper with the correct OS of 0. Good for you that you realized the given answer couldn't have been correct.
Another frequently encountered neutral coordination complex (you cannot say "complex molecule", there is no such thing, unlike "complex ion" which is an acceptable shortcut for "ionic coordination complex") is the iron(0) pentacarbonyl coordination complex, in which Fe has an OS of 0 and formal charge of 5-, C has an OS of +2 and formal charge of 0, and O has an OS of O is -2 and a formal charge of 1+. Overall ionic charge (which is sum of formal charges and also the sum of oxidation states) is therefore zero, and that's why it's a neutral coordination complex.
https://en.wikipedia.org/wiki/Iron_pentacarbonyl -
Al2o3 is a amphoteric species cos it can react with both acids and alkalis?
It is not amphiprotic because it does not accept nor donate protons?
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Originally posted by Flying grenade:
Al2o3 is a amphoteric species cos it can react with both acids and alkalis?
It is not amphiprotic because it does not accept nor donate protons?
Uhh yeah, but to get full marks, you gotta quote the relevant acid-base equations for Al2O3 that the H2 syllabus requires you to memorize. -
For reduction of nitrile to form amine,
One of the ways is to use lialh4 in dry ether, r.t.
Do we need to have a 2nd step 'followed by h2o'????
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Hi Ultima, was just hoping you could provide some advice on these questions:
1. This is pertaining to the 2 factors that affect the basicity of amines:
1i. Factor 1 would be the availability of the lone pair of electrons on N. Generally, we learn that it can be affected by the presence of electron donating alkyl groups. In applying what we've learnt about steric hindrance, I was wondering if this factor can also be affected by bulky alkyl groups not because of their electron donating capabilities but also if they can "obstruct the protons from reaching the lone pair of electrons for coordinate bonding" due to their size / steric hindrance. Is that a flawed reasoning or will it be accepted?
1ii. Factor 2 would be on amine/ammonium salt solubility. On doing papers I realised there are 2 different explanations for this factor, one would be that the presence of bulky non-polar groups makes the amine insoluble and thus less basic (regardless of whether it has been protonated or not) (case in point used by the solutions being why benzylamine is less basic than ethylamine). Another explanation would be based on the ability to form hydrogen bonds, where if an amine has a lack of H atoms attached to the N atom (case in point being N,N,N-trimethylamine), it would be less basic . Both still address the main factor of solubility. However, which explanation of this Factor 2 would be more acceptable in your opinion?
2. This is pertaining to the acidic hydrolysis of acid anhydrides. Following the "pattern" of the hydrolysis of esters, why does the carbonyl segment of the acid anhydride not become a H2CO3 but a CO2 instead? Under what circumstances would H2CO3 exist more preferentially then?
3. This is pertaining to the oxidation of alcohols. Generally, we learn that tertiary alcohols do not tend to undergo oxidation, but why does a tertiary benzyl alcohol (pardon for the loose nomenclature) undergo oxidation?
4. This is pertaining to the aqueous bromination of phenols. When this common question is asked in papers, the generally accepted answer / product for such a reaction is tribromophenol. However, must we take into consideration the ionisation / deprotonation of the tribromophenol in the given aqueous condition to give tribromophenoxide instead? Is that overthinking and how should we reconcile such "environmental" factors in determining the end product of reactions?
5. This is related to Question 4 on tribromophenols. The general baseline is that tribromophenols are highly insoluble and give a white ppt. But with more and more permutations given in exams (with other groups present on the benzene such that we only get dibromo or even monobromo), do nontrihalogented phenols still give a white ppt? Can we say that as long as the benzene seems "highly substituted" with non-hydrogen groups (besides Br and halogens), it will be a white ppt? In what cases will it then be soluble? Also do we consider a dibromophenol soluble?
Thanks so much!
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Originally posted by Flying grenade:
For reduction of nitrile to form amine,
One of the ways is to use lialh4 in dry ether, r.t.
Do we need to have a 2nd step 'followed by h2o'????
For A level purposes, no need. -
Originally posted by Sugarfortress:
Hi Ultima, was just hoping you could provide some advice on these questions:
1. This is pertaining to the 2 factors that affect the basicity of amines:
1i. Factor 1 would be the availability of the lone pair of electrons on N. Generally, we learn that it can be affected by the presence of electron donating alkyl groups. In applying what we've learnt about steric hindrance, I was wondering if this factor can also be affected by bulky alkyl groups not because of their electron donating capabilities but also if they can "obstruct the protons from reaching the lone pair of electrons for coordinate bonding" due to their size / steric hindrance. Is that a flawed reasoning or will it be accepted?
1ii. Factor 2 would be on amine/ammonium salt solubility. On doing papers I realised there are 2 different explanations for this factor, one would be that the presence of bulky non-polar groups makes the amine insoluble and thus less basic (regardless of whether it has been protonated or not) (case in point used by the solutions being why benzylamine is less basic than ethylamine). Another explanation would be based on the ability to form hydrogen bonds, where if an amine has a lack of H atoms attached to the N atom (case in point being N,N,N-trimethylamine), it would be less basic . Both still address the main factor of solubility. However, which explanation of this Factor 2 would be more acceptable in your opinion?
2. This is pertaining to the acidic hydrolysis of acid anhydrides. Following the "pattern" of the hydrolysis of esters, why does the carbonyl segment of the acid anhydride not become a H2CO3 but a CO2 instead? Under what circumstances would H2CO3 exist more preferentially then?
3. This is pertaining to the oxidation of alcohols. Generally, we learn that tertiary alcohols do not tend to undergo oxidation, but why does a tertiary benzyl alcohol (pardon for the loose nomenclature) undergo oxidation?
4. This is pertaining to the aqueous bromination of phenols. When this common question is asked in papers, the generally accepted answer / product for such a reaction is tribromophenol. However, must we take into consideration the ionisation / deprotonation of the tribromophenol to give tribromophenoxide instead? Is that overthinking and how should we reconcile such "environmental" factors in determining the end product of reactions?
5. This is related to Question 4 on tribromophenols. The general baseline is that tribromophenols are highly insoluble and give a white ppt. But with more and more permutations given in exams (with other groups present on the benzene such that we only get dibromo or even monobromo), do nontrisubstituted phenols still give a white ppt? Can we say that as long as the benzene seems "highly substituted" with non-hydrogen groups (besides Br and halogens), it will be a white ppt? In what cases will it then be soluble?
Thanks so much!
Q1i. Steric hindrance affects nucleophilicity much more than it affects basicity (as H+ or H3O+ ions are small and thus relatively unaffected by steric hindrance*). So use steric hindrance to explain reduced nucleophilicity rather than basicity. Note that any Bronsted-Lowry base can potentially also function as a nucleophile, and vice-versa, as they're all Lewis bases.
* Evidence : The deliberately built-in steric hindrance of lithium diisopropyl amide forces it to function only as a Bronsted-Lowry base, but not as a nucleophile. It's function as a Bronsted-Lowry base is uncompromised, but it's function as a nucleophile is completely nullified.
Q1ii. Solubility only indirectly affects basicity, but hydrogen bonding directly affects basicity. Because the more stable the conjugate acid, the stronger the base. Magnitude of hydrogen bonding stabilization energy for primary ammonium > secondary ammonium > tertiary ammonium, conjugate acids.
Q2. When 1 mole of acid anhydride is hydrolyzed (no need for acidic or alkaline conditions), the only products are 2 moles of carboxylic acids. No carbonic(IV) acid or CO2 is generated. On a separate note, carbonic(IV) acid exists in equilibrium with, and hence can decompose into, gaseous carbon dioxide and liquid water, with thermodynamically favorable positive entropy change. Moreover, gaseous carbon dioxide leaves the reaction mixture, pulling the position of equilibrium over to the RHS. Which is why (unless you're actively pumping in carbon dioxide gas to shift the position of equilibrium over to the LHS*), the position of equilibrium always lies on the RHS, and any carbonic(IV) acid will instantaneously decompose (mechanism : after intramolecular Bronsted-Lowry acid-base proton transfer reaction, elimination of H2O yields CO2) to carbon dioxide gas.
* Even then, H2CO3 will undergo Ka1 proton dissociation (or hydrolysis) to yield HCO3- and H+ (or H3O+). Molecular H2CO3 can only be isolated under exceptional conditions, not under standard aqueous conditions.
Q3. As long as a benzene ring has a benzylic H atom, the functional group (as far as oxidation by KMnO4 is concerned) is no longer considered a typical alcohol or ketone or alkyl group, but falls under "oxidation of benzene ring side chain (provided a benzylic H atom is present) to benzoic acid".
Q4. It is precisely the Ka partial proton dissociation of the phenolic OH group, that makes the phenolic OH group strongly electron donating by resonance, due to the negative formal charge present on the O atom of the phenoxide C6H5O- ion. With a negative formal charge on the O atom, the electron-rich O- atom is no longer electron withdrawing by induction (unlike the OH phenolic group), allowing for the full effect of its electron donating by resonance capabilities. When you state the product as "tribromophenol", you're already implicitly stating "tribromophenol (and it's conjugate base in an acid-base equilibrium)" because anyone with any chemistry sense knows that all acids exist in equilibrium with their conjugate base (provided, of course, that the solvent employed allows for this).
Q5. As you're aware, aqueous solubility of halogenated phenols (and phenylamines) decreases significantly with the multiplicity of halogenation. Hence aqueous solubility of mono-halogenated phenol > di-halogenated phenol > tri-halogenated phenol. Similarly, for other substituted groups other than halogens (which of course, you must take into consideration the polarity, proticity (can accept and/or donate H bonds?) and presence/absence of formal/ionic charge, of these other substituents, as such would contribute to the overall solubility/insolubility of the entire molecule). So whether a ppt is observed or not, depends on its molarity, its molar solubility and the temperature. In conclusion there isn't a simplified rule for this (ie. case-by-case basis, but for A level purposes it's quite safe to state that tri-halogenated phenols usually forms a ppt under aqueous conditions, while mono-halogenated phenols usually do not), and because Cambridge knows this, they'll set reasonable MCQ options (or for P2 & P3, accept reasonable answers in this context), eg. by eliminating the other more obviously wrong options. -
Originally posted by UltimaOnline:
Q1i. Steric hindrance affects nucleophilicity much more than it affects basicity (as H+ or H3O+ ions are small and thus relatively unaffected by steric hindrance). So use steric hindrance to explain reduced nucleophilicity rather than basicity. Note that any Bronsted-Lowry base can also function as a nucleophile, and vice-versa, as they're all Lewis bases.
Q1ii. Solubility only indirectly affects basicity, but hydrogen bonding directly affects basicity. Because the more stable the conjugate acid, the stronger the base. Magnitude of hydrogen bonding stabilization energy for primary ammonium > secondary ammonium > tertiary ammonium, conjugate acids.
Q2. When 1 mole of acid anhydride is hydrolyzed (no need for acidic or alkaline conditions), the only products are 2 moles of carboxylic acids. No carbonic(IV) acid or CO2 is generated. On a separate note, carbonic(IV) acid exists in equilibrium with, and hence can decompose into, gaseous carbon dioxide and liquid water, with thermodynamically favorable positive entropy change. Moreover, gaseous carbon dioxide leaves the reaction mixture, pulling the position of equilibrium over to the RHS. Which is why (unless you're actively pumping in carbon dioxide gas to shift the position of equilibrium over to the LHS*), the position of equilibrium always lies on the RHS, and any carbonic(IV) acid will instantaneously decompose (mechanism : after intramolecular Bronsted-Lowry acid-base proton transfer reaction, elimination of H2O yields CO2) to carbon dioxide gas.
*Even then, H2CO3 will undergo Ka1 proton dissociation (or hydrolysis) to yield HCO3- and H+ (or H3O+). Molecular H2CO3 can only be isolated under exceptional conditions, not under standard aqueous conditions.
Q3. As long as a benzene ring has a benzylic H atom, the functional group (as far as oxidation by KMnO4 is concerned) is no longer considered a typical alcohol or ketone or alkyl group, but falls under "oxidation of benzene ring side chain (provided a benzylic H atom is present) to benzoic acid".
Q4. It is precisely the Ka partial proton dissociation of the phenolic OH group, that makes the phenolic OH group strongly electron donating by resonance, due to the negative formal charge present on the O atom of the phenoxide C6H5O- ion. With a negative formal charge on the O atom, the electron-rich O- atom is no longer electron withdrawing by induction (unlike the OH phenolic group), allowing for the full effect of its electron donating by resonance capabilities. When you state the product as "tribromophenol", you're already implicitly stating "tribromophenol (and it's conjugate base in an acid-base equilibrium)" because anyone with any chemistry sense knows that all acids exist in equilibrium with their conjugate base (provided, of course, that the solvent employed allows for this).
Q5. As you're aware, aqueous solubility of halogenated phenols (and phenylamines) decreases significantly with the multiplicity of halogenation. Hence aqueous solubility of mono-halogenated phenol > di-halogenated phenol > tri-halogenated phenol. Similarly, for other substituted groups other than halogens (which of course, you must take into consideration the polarity, proticity (can accept and/or donate H bonds?) and presence/absence of formal/ionic charge, of these other substituents, as such would contribute to the overall solubility/insolubility of the entire molecule). So whether a ppt is observed or not, depends on its molarity, its molar solubility and the temperature. In conclusion there isn't a simplified rule for this (ie. case-by-case basis, but for A level purposes it's quite safe to state that tri-halogenated phenols usually forms a ppt under aqueous conditions, while mono-halogenated phenols usually do not), and because Cambridge knows this, they'll set reasonable MCQ options (or for P2 & P3, accept reasonable answers in this context), eg. by eliminating the other more obviously wrong options.Thanks for the prompt reply!
So sorry I got confused by the acid anhydride, I actually wanted to ask about the hydrolysis of - this "functional group"(not too sure what it is called) -o(c=o)o- which would yield co2 when hydrolysed.
But thanks for the clear explanation for h2co3 and co2! :)
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Originally posted by Sugarfortress:
Thanks for the prompt reply!
So sorry I got confused by the acid anhydride, I actually wanted to ask about the hydrolysis of - this "functional group"(not too sure what it is called) -o(c=o)o- which would yield co2 when hydrolysed.
But thanks for the clear explanation for h2co3 and co2! :)
Totally welcome, Sugarfortress :)
Here ya go : https://en.wikipedia.org/wiki/Carbonate_ester