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Originally posted by Flying grenade:
2010 p3 qn 1d
Sir ultima can help check my drawingsFor cpd A, the cl is substituted at that position is it because it is most stable there because attached to 3 mildly electron donating C group ah? Markovnikov’s rule ah?
For D, im asking this as an extension
When the Br is being eliminated, are there 2 other by-product as drawn?
Eh don't anyhow misuse his name. Markovnikov (Chinese for "horse cough you also cough") would turn in his grave if he heard you say that. Markovnikov's rule is only for hydrohalogenation via electrophilic addition across a double bond (if you know why it usually works to predict the major product, you'll also know when it'll fail, for challenging tricky qns). This is halogenation via free radical substitution, nothing to do with Markovnikov (he'll say 关我�事!). Because free radical substitution cannot be controlled, you'll get a mixture of many monohalogenated products. To determine which monohalogenated intermediate was used for this qn, observe the structures of the final product upon dehydrohalogenation (ie. this qn isn't asking you to state which is the major product, but rather identify the intermediate that could give you the final products drawn). But assuming you're jus curious if that's the major product of halogenation via free radical substitution, it *is* the major product, based solely upon the factor of stability of alkyl radical intermediate (in actuality, you need to mathematically combine both factors at a specified temperature to determine the actual major product). Then yes, your reasoning (though you wrongly labeled your explanation using Markovnikov's name) is correct : tertiary alkyl radical is more stable than secondary alkyl radical is more stable than primary alkyl radical. This factor is more pronounced for Br, less for Cl, and even less so for F (I is irrelevant here, because iodination via free radical mechanism cannot be carried out; you know why? asking you to explain why, was a Singapore TYS A level exam qn).
No, the dienes you drew cannot be generated. Why? Asking you to explain why, was 2014's Singapore TYS A level exam qn. Hints (any or all of these terms may be used in your explanation) : orbital hybridization, electron geometry, ring strain, angle strain, thermodynamic instability, thermodynamic infeasibility. -
Ohh!!
I only thought of that, due to the factor of ring strain.
Becos sp hybridised should have a angle of 180°, however in that case its bent at an angle.
Hence the diene cannot be formed
So yes, only 1,3diene can be formed right?
Can help improve and add in the rest of the factors into the answer pls
Thank u ultima!
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Why iodine cannot undergo FRS? Cos its nt gaseous / I-I bond too strong?
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Which 2 factors?? Alkyl stability and temperature uh?
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Originally posted by Flying grenade:
Ohh!!
I only thought of that, due to the factor of ring strain.
Becos sp hybridised should have a angle of 180°, however in that case its bent at an angle.
Hence the diene cannot be formed
So yes, only 1,3diene can be formed right?
Can help improve and add in the rest of the factors into the answer pls
Thank u ultima!
You got the idea.
Cumulated dienes can indeed be generated, just not within a small ring like this molecule, because of the resultant ring strain due to angle strain (due to the significant deviation from the ideal bond angles about the sp hybridized C atom as predicted by VSEPR theory, which is to maximize thermodynamic stabilities by minimizing electron pair electrostatic repulsions) making it thermodynamically unstable and thus its formation thermodynamically unfeasible.
Conjugated dienes (such as the final product in this Cambridge qn) are thermodynamically favored due to the resonance stabilization energy (for the same reason why only benzene, but not the non-resonance mythical version, cyclohexa-1,3,5-triene, exists), which results in the resonance hybrid enjoying overall stronger C-C bonds (ie. 3 C-C bonds all with partial double bond characters, is overall still stronger and still has a more exothermic formation enthalpy compared to 2 full double bonds and 1 strictly single bond), a more exothermic formation enthalpy, and thus more thermodynamically stable and formation more thermodynamically favored. -
Originally posted by Flying grenade:
Why iodine cannot undergo FRS? Cos its nt gaseous / I-I bond too strong?
Wrong / Wrong.
Go google up on this yourself. -
Originally posted by UltimaOnline:
You got the idea.
Cumulated dienes can indeed be generated, just not within a small ring like this molecule, because of the resultant ring strain due to angle strain (due to the significant deviation from the ideal bond angles about the sp hybridized C atom as predicted by VSEPR theory, which is to maximize thermodynamic stabilities by minimizing electron pair electrostatic repulsions) making it thermodynamically unstable and thus its formation thermodynamically unfeasible.
Conjugated dienes (such as the final product in this Cambridge qn) are thermodynamically favored due to the resonance stabilization energy (for the same reason why only benzene, but not the non-resonance mythical version, cyclohexa-1,3,5-triene, exists), which results in the resonance hybrid enjoying overall stronger C-C bonds (ie. 3 C-C bonds all with partial double bond characters, is overall still stronger and still has a more exothermic formation enthalpy compared to 2 full double bonds and 1 strictly single bond), a more exothermic formation enthalpy, and thus more thermodynamically stable and formation more thermodynamically favored.Another godly answer.., thank you so much..!!!
Iodine is such a big molecule so it's unreactive?
The equation c2h5• +I• -> c2h5I +HI , is endothermic is it?
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Originally posted by Flying grenade:
Which 2 factors?? Alkyl stability and temperature uh?
1st factor : No. of H atoms substitutable
2nd factor : Stability of alkyl radical intermediate, aka, reactivity of primary vs secondary vs tertiary H atoms.
Your own JC school teacher never teach you meh?
Go google up on how these factors influence product distribution during halogenation via free radical substitution, and the reactivity-selectivity principle (in the context of halogenation via free radical substitution, which requires an understanding of the Hammond postulate). -
Isnt no. Of H atoms available for substitution same as tertiary, secondary or pri alkyl radical intermediate?
Is temperature a factor for FRS?
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Originally posted by Flying grenade:
Another godly answer.., thank you so much..!!!
Iodine is such a big molecule so it's unreactive?
The equation c2h5• +I• -> c2h5I +HI , is endothermic is it?
BedokFunland JC ftw!
Oi, just saying "Iodine is a big molecule so it's unreactive" is a sure way of making the Cambridge examiner shake his/her head, "Singapore students only know how to smoke", don't xia suay Singapore hor.
Yes, correct, you must link to the fact that the bonds formed are weaker than the bonds broken (quote Data Booklet values whenever relevant), and hence the reaction is thermodynamically unfeasible. -
Originally posted by Flying grenade:
Isnt no. Of H atoms available for substitution same as tertiary, secondary or pri alkyl radical intermediate?
Is temperature a factor for FRS?
Not the same. In fact, the two factors are opposing. Take propane for example. Based on each factor, what is the expected major product? 1 factor predicts 1-halopropane as the major product, while the other factor predicts 2-halopropane as the major product.
If the Cambridge A level exam question requires you to calculate out the overall product distribution mathematically (which has appeared in past CIE A levels and Pre-U papers), the question has to provide you with the relative reactivities of a pri vs sec vs tert H atom (corresponding to the relative stabilities of a pri vs sec vs tert alkyl radical) in numerical values, for you to apply mathematically (ie. multiply both factors together). Such values differ across halogens, and across temperatures.
If your school JC teacher didn't teach you about these 2 factors (see the benefits of a Chemistry education @ BedokFunland JC ?), you can google up yourself, and/or read the following webpages :
http://orgchem.chem.uconn.edu/2443s2011/2443-041811.pdf
http://www.masterorganicchemistry.com/2013/10/31/selectivity-in-free-radical-reactions-bromine-vs-chlorine/
http://www.mhhe.com/physsci/chemistry/carey/student/olc/ch04radical.html
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Thank you so much for helping me and the community..
Yes and i hope to have chemistry education at ur place someday
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pH lower than pka, compound protonates
pH greater than pka, compound deprotonates
Which chapter is this??
Acid base equilibria?
Does this have any link to equilibria Le chatelier's principle shift left shift right?
I just memorise ^ only!
Hope i can draw links to reinforce
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2010 p3 qn 2b
Need draw all the protein structures to answer this qn?
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Originally posted by Flying grenade:
pH lower than pka, compound protonates
pH greater than pka, compound deprotonates
Which chapter is this??
Acid base equilibria?
Does this have any link to equilibria Le chatelier's principle shift left shift right?
I just memorise ^ only!
Hope i can draw links to reinforce
Yes of course it's acid-base equilibria, and your school JC teacher *did* teach you this, didn't he/she? It's basic (no pun intended) acid-base equilibria. Of course any equilibrium can be linked to Le Chatelier's principle, and also to Organic Chemistry (eg. acid-base equilibria of amino acids). Everything can be linked, once you understand Chemistry deeply enough.
I've 2 BedokFunland JC questions on acid-base equilibria of amino acids, but if it confuses you (ie. any JC student reading this) then you should skip it (since most Singapore JC students seem to find both acid-base equilibria and amino acids one of the tougher topics in the H2 syllabus, let alone combining both topics). It's more effective to discuss such questions (and students' queries) properly during tuition with my students, compared to online on a forum. Besides, A level exams is the day after tomorrow liao, and it's not a good idea to confuse or worry yourself at the 11th hour.
H2Chem_Acid-Base_Equilibria_of_Arginine
H2Chem_Acid-Base_Equilibria_of_Histidine
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Originally posted by Flying grenade:2010 p3 qn 2b
Need to draw all the protein structures to answer this qn?
Diagrams of structures are optional (as indicated in the question phrasing). Cambridge will award you the marks if the required points are present in your answer, whether in words, in balanced equations, in your diagrams, or all 3. Unless you're supremely confident you can include all the required points are present in words alone, or equations alone, or diagrams alone, it's usually a good idea (assuming your time management is good enough) in such questions to use at least 2 out of the 3 forms in your answers (even if the question doesn't compulsorily require it), to ensure all the required points are present in your answer to secure full marks. So to address your question directly, no need, as long as the required marking points are present (go see a TYS publisher's answer for this). -
From
http://sgforums.com/forums/2297/topics/414979
2010 H2 Chem P3 a killer paper
The problem with FO2, is that F (being in period 2), cannot expand its octet (since it doesn't have vacant, energetically accessible 3d orbitals to use). Hence, it cannot form a similar structure as ClO2, which has expanded its octet (4 bond pairs + 1.5 lone pairs on Cl = 11 electrons in terms of an expanded octet).
Should it be 1.5 lone pairs on Cl?
Btw may i ask,
When will have formal '-' charge and partial δ -ve charge?
Is it one lone electron will result in the molecule/atom having partial -ve charge, while a pair of electrons gained will have full -ve charge?
When will full -ve charge happen? Is it when a lone pair of electrons dative bond is donated to a acceptor atom?
And also when an atom ionises in water, gaining a lp of e-?? Dont really know how ions gain the -ve charge also
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Also, for the case of ClO2,
It can potentially form a structure of 2 dative bonds bonded to 'O'
But it is doubly bonded to both O atoms,
Because double bond more stable so preferentially form right?
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Also, i think one term is confusing. The term of 'number of electron bond pairs'
Say for like, example, NO2.
So we want to consider the bond angle for it
So we say it has one lone pair (one lone e-) , and 2 bond pair, with respecr to N atom
But in the above post, the sentence ' clo2 has 4 bond pairs'
So when do we say a double bond , comprises of 2 bond pairs?
We treat a double bond as 1 bond pair, when we're considering molecular shapes is it?
How about other situations?
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Wtf sia
I hope this website forum wont crash or database lost etc. The wealth of knowledge will be lost just like :
https://en.m.wikipedia.org/wiki/Burning_of_books_and_burying_of_scholars
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Mr heng may i ask u, is the one drawn with pencil correct
https://www.dropbox.com/s/avl731ngdwlyxmb/20151105_134421.jpg?dl=0
Then, if one were to draw the 9 e- ard cl atom, would it be wrong?
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For FO2 to have single bonds between the F and two O atoms, F will gain a 2+ charge and each O atom with gain a -ve charge, and this separation of charge is destabilizing, since F is more electronegative than O.
Wouldn't '+' and '-' attract?
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Originally posted by Flying grenade:
Mr heng may i ask u, is the one drawn with pencil correct
https://www.dropbox.com/s/avl731ngdwlyxmb/20151105_134421.jpg?dl=0
Then, if one were to draw the 9 e- ard cl atom, would it be wrong?
Edited : Yes, my typo, 1.5 lone pairs not 2.5 lone pairs.
I'll only reply to your last qn. For all the other qns, you need to come for my tuition to properly discuss and understand it. Furthermore, there's no time to address it now liao, it's best not to confuse yourself further just a few hours away from the A level exam.
Yes, your doubly bonded ClO2 is acceptable (unless the Cambridge qn specified conditions such as the presence of a dative bond, etc), and is actually 1 of the resonance contributors for ClO2 (the only one without separation of formal charges), no worries. -
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
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Originally posted by UltimaOnline:
Edited : Yes, my typo, 1.5 lone pairs not 2.5 lone pairs.
I'll only reply to your last qn. For all the other qns, you need to come for my tuition to properly discuss and understand it. Furthermore, there's no time to address it now liao, it's best not to confuse yourself further just a few hours away from the A level exam.
Yes, your doubly bonded ClO2 is acceptable (unless the Cambridge qn specified conditions such as the presence of a dative bond, etc), and is actually 1 of the resonance contributors for ClO2 (the only one without separation of formal charges), no worries.Mr heng i wan come ur tuition !!!
Mr heng wont confuse lah i love to seek knowledge
Wont confuse, i understand and absorbed everything you've posted hehe
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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?
