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Originally posted by Flying grenade:
repost on top, just before u posted, i posted
when decrease temp for endo rxn, position of eqm shift left (exothermic side). k-forward ( of endo rxn) decreases less extent than k-backward (of exo rxn) ?
when decrease temp for exo rxn, position of eqm shift right (exothermic side). k-forward (of exo rxn) decreases to a larger extent than k-backward (of endo rxn)?
Partly correct, partly wrong. Figure out yourself, which part you got right, which part you got wrong. -
Hi UltimaOnline, Sorry to trouble you again, but here are some questions which I have
NJC/2012/P1/Q22
Why can’t B be an answer, given that acyl chlorides can be made by reacting thionyl chlorides with carboxylic acid?
NJC/2012/P1/Q25
I understand that C is the answer (http://www.chemguide.co.uk/organicprops/haloalkanes/making.html), but why can’t the other options be prepared with distillation?
NJC/2012/P1/Q27
I think there are 3 ways which the elimination can take place given that there are 3 B hydrogens. Applying Saytzeff’s Rule, I can only eliminate the option where the hydrogen is abstracted from the methyl substituent. The other options (where the hydrogen is abstracted from the CH2, and where the hydrogen is abstracted from the cyclohexane substituent) seem equally substituted to me. So what would be the major product of the reaction? Further, wouldn’t either of the products contain 2 double bonds, thereby making the total number of stereoisomers 4?
Once again, thank you for your help. :)
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Originally posted by gohby:
Hi UltimaOnline, Sorry to trouble you again, but here are some questions which I have
NJC/2012/P1/Q22
Why can’t B be an answer, given that acyl chlorides can be made by reacting thionyl chlorides with carboxylic acid?
NJC/2012/P1/Q25
I understand that C is the answer (http://www.chemguide.co.uk/organicprops/haloalkanes/making.html), but why can’t the other options be prepared with distillation?
NJC/2012/P1/Q27
I think there are 3 ways which the elimination can take place given that there are 3 B hydrogens. Applying Saytzeff’s Rule, I can only eliminate the option where the hydrogen is abstracted from the methyl substituent. The other options (where the hydrogen is abstracted from the CH2, and where the hydrogen is abstracted from the cyclohexane substituent) seem equally substituted to me. So what would be the major product of the reaction? Further, wouldn’t either of the products contain 2 double bonds, thereby making the total number of stereoisomers 4?
Once again, thank you for your help. :)
Hi Gohby, no prob :)NJC/2012/P1/Q22 : B is incorrect because the amine group present is even more nucleophilic than the OH of COOH, and hence would react even more readily with the thionyl halide electrophile, in this case the primary amine would undergo 2 consecutive nucleophilic substitutions with the thionyl bromide to generate the N=S=O group, eliminating 2 molecules of H-Br. This is beyond the A level H2 syllabus, and so isn't completely fair, but nonetheless Cambridge *could* ask this question as a challenging A grade question, to determine if the candidate understands the concepts of relative nucleophilicities and electrophilicities, and choose the best possible option among the 4 presented.
Cambridge has done indeed so many times in the past, especially in recent years, deliberately presenting 2 likely options : 1 option going beyond the A level H2 syllabus so candidates cannot be sure if it's correct, the other option within the A level H2 syllabus so competent candidates should be sure if it's correct or wrong, and therefore choose accordingly.
NJC/2012/P1/Q25 : Options A & D can be carried out at room temperature, and hence doesn't require this setup (and therefore should not be used). B requires heating under reflux. The setup shown is heating with distillation to separate out the desired product which has a lower boiling point compared to the reactants.
NJC/2012/P1/Q27 : This reaction involves competition between the Zaitsev aka Saytzeff elimination, and the Hofmann elimination. While the Bronsted-Lowry base employed is the sterically unhindered OH- ion, nonetheless the cyclohexane ring does pose some degree of steric hindrance, but in this case the Zaitsev product should ultimately still dominate (and conveniently also because the Hofmann elimination isn't taught in the H2 syllabus).
The major product which is the Zaitsev product, is therefore the product that has the double bond with the cyclohexane ring, as this product would be a more substituted internal alkene (with no H atoms bonded to its sp2 C atoms), compared to the alternative product that has the double bond with the alkyl side chain (with 1 H atom bonded to its sp2 C atoms).
The number of stereoisomers of the major product is therefore 2 (ie. as a result of the pre-existing alkene double bond in the longer side chain), instead of 4 (cis-cis, cis-trans, trans-cis, trans-trans).
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tryptophan's N atom cannot accept a proton because it's lone pair residing in sp2 unhybridised orbital is delocalised by resonance?
why histidine's N atom can ? isn't the lone pair on histidine's N atom residing in the sp2 unhybridised orbital, and is also delocalised by resonance?
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ok i know why liao, can help check correct or not
for tryptophan, the secondary amine's lp of e- is delocalised by resonance.
for histidine,
only the R group 2° amine's lone pair is slightly delocalized by resonance before protonation, but the R group imine's lone pair isn't, and it is the R group imine which is protonated (not the amine). Thereupon, the R group amine's lone pair then becomes extensively delocalized, in order to allow the positive formal charge on the conjugate acid to be effectively delocalized, across both N atoms.
but why the imine's lone pair didnt, or cannot delocalise by resonance??
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for Histidine,
pKa of α-COOH is 1.8
pKa of α-NH3+ is 9.3
pKa of R group is 6.0
for Rgroup imine to be protonated, pH <6.0
for α-NH2 to be unprotonated, pH >9.3
for α-COOH to be deprotonated, pH <1.8
how to solve for the isoelectric point for histidine??
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Originally posted by Flying grenade:
ok i know why liao, can help check correct or not
for tryptophan, the amine's lp of e- is delocalised by resonance.
for histidine,
only the R group amine's lone pair is slightly delocalized by resonance before protonation, but the R group imine's lone pair isn't, and it is the R group imine which is protonated (not the amine). Thereupon, the R group amine's lone pair then becomes extensively delocalized, in order to allow the positive formal charge on the conjugate acid to be effectively delocalized, across both N atoms.
but why the imine's lone pair didnt, or cannot delocalise by resonance??
Identify the hybridization of the imine N atom, and the orbital that the lone pair resides in, and you'll have your answer. If you still don't understand, go ask your school teacher or private tutor. -
Originally posted by Flying grenade:
for Histidine,
pKa of α-COOH is 1.8
pKa of α-NH3+ is 9.3
pKa of R group is 6.0
for Rgroup imine to be protonated, pH <6.0
for α-NH2 to be unprotonated, pH >9.3
for α-COOH to be deprotonated, pH <1.8
how to solve for the isoelectric point for histidine??
What do you think? Google up how to calculate the isoelectric points of amino acids in general, then apply to Histidine. See if you are competent enough to settle this on your own.Cambridge can challenge you to calculate the isoelectric points of difficult amino acids (ie. with acidic or basic R groups) as an A grade question in the A levels.
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Originally posted by UltimaOnline:
Identify the hybridization of the imine N atom, and the orbital that the lone pair resides in, and you'll have your answer. If you still don't understand, go ask your school teacher or private tutor.sp2, resides in hybridised p orbital ah?
CORRECTION
the hybridisation of imine N atom is sp2, the lone pair of imine N atom resides in sp2 hybridised orbital, and therefore cannot delocalise by resonance!
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Originally posted by Flying grenade:
sp2, resides in hybridised p orbital ah?
Wrong. -
Originally posted by UltimaOnline:
Identify the hybridization of the imine N atom, and the orbital that the lone pair resides in, and you'll have your answer. If you still don't understand, go ask your school teacher or private tutor.lone pair on N atom of tryptophan resides in a unhybridised p orbital. the lone pair of e- is extensively delocalised by resonance to the benzene ring to the right, as well as to the alkene to the left. it is also not nucleophilic due to the steric hindrance brought upon by benzene ring.
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Hi Ultima, hope u can help out here,
my answers for TPJC 2015 P1 is
1. B
2. C
3. C
4. C
5. C
6. A
7. C
8. D
9. B
10. C
11. C
12. D
13. C
14. C
15. B
16. C
17. A
18. C
19. D
20. D
21. C
22. A
23. B
24. C
25. A
26. C
27. A
28. B
29. C
30. C
31. D
32. D
33. A
34. A
35. C
36. C
37. C
38. D
39. D
40. D
hope u could help check and post the solns. Thank you !
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Originally posted by Flying grenade:
Hi Ultima, hope u can help out here,
my answers for TPJC 2015 P1 is
1. B
2. C
3. C
4. C
5. C
6. A
7. C
8. D
9. B
10. C
11. C
12. D
13. C
14. C
15. B
16. C
17. A
18. C
19. D
20. D
21. C
22. A
23. B
24. C
25. A
26. C
27. A
28. B
29. C
30. C
31. D
32. D
33. A
34. A
35. C
36. C
37. C
38. D
39. D
40. D
hope u could help check and post the solns. Thank you !
1 B --- 11 D --- 21 A --- 31 D
2 C --- 12 D --- 22 A --- 32 B
3 C --- 13 B --- 23 C --- 33 D
4 C --- 14 C --- 24 D --- 34 B
5 C --- 15 D --- 25 D --- 35 B
6 A --- 16 C --- 26 D --- 36 D
7 C --- 17 A --- 27 A --- 37 B
8 D --- 18 A --- 28 B --- 38 A
9 B --- 19 B --- 29 C --- 39 D
10 B --- 20 C --- 30 A --- 40 D -
did and constructed the equation for disproportionation of Cl2 in hot aq.NaOH in a test, instead of memorising
i wrote Cl2 + 6OH- -> Cl- + ClO3- + 3H2O (completely balanced)
Cstoh's book (and other sources)
3Cl2 + 6OH- -> 5Cl- + ClO3- + H2O
is my eqn wrong? ? it's balanced correctly !!
notice the possibility of varying the stoichiometric coefficients of Cl2 and Cl- ?
e.g.
1 , 1
2, 3
3, 5
4, 7
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Originally posted by Flying grenade:
did and constructed the equation for disproportionation of Cl2 in hot aq.NaOH in a test, instead of memorising
i wrote Cl2 + 6OH- -> Cl- + ClO3- + 3H2O (completely balanced)
Cstoh's book (and other sources)
3Cl2 + 6OH- -> 5Cl- + ClO3- + H2O
is my eqn wrong? ? it's balanced correctly !!
notice the possibility of varying the stoichiometric coefficients of Cl2 and Cl- ?
e.g.
1 , 1
2, 3
3, 5
4, 7
You really can't figure out why you're wrong, and need me to point it out to you??? Seriously??? -
3Cl2 + 6OH- -> 5Cl- + ClO3- + H2O
is the correct equation because the charges on both side of equation is the same, and hence same number of Counter ions (Na+) is the same for both side of eqn.
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Originally posted by Flying grenade:
3Cl2 + 6OH- -> 5Cl- + ClO3- + H2O
is the correct equation because the charges on both side of equation is the same, and hence same number of Counter ions (Na+) is the same for both side of eqn.
Ok, here's a BedokFunland JC challenge for you, solve it now on the forum to see if you've learnt something from what you've been discussing :Write the ionic equation for the reaction between Cu2+ ions and potassium iodide solution, to generate copper(I) iodide precipitate and molecular iodine.
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2Cu2+(aq) + 4I-(aq) -> 2CuI(s) + I2(aq)
I2 (aq) state, not solid state.
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Originally posted by Flying grenade:
is it
2Cu2+(aq) + 4I-(aq) -> 2CuI(s) + I2(s) ??
Almost entirely correct, except I2 should be aqueous, not solid. Write the equation (ideally draw out the mechanism) to explain why I2(s) is soluble in KI(aq).Separate Qn : Write the balanced equation for the complete combustion of CxHyOH.
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oh yes !!!
iodine dissolves in aqueous KI solution , due to the formation of I3-(aq) ions. for full detailed explanation, shld draw mechanism.
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is it
CxHyOH + [x + (y+1)/4 -1] O2 -> xCO2 + (y+1)/2 H2O ?
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Originally posted by Flying grenade:
is it
CxHyOH + [x + (y+1)/4 -1] O2 -> xCO2 + (y+1)/2 H2O ?
Almost correct, but your coefficient of O2 is slightly off. Here's the correct working to get the correct coefficient of O2.Qn : Write the balanced equation for the complete combustion of CxHyOH.
2k + 1 = 2x + (y+1)/2
2k = 2x + (y+1)/2 - 1
k = (x + (y+1)/4 - 1/2) -
ahh okay i see, Thanks Ultima !!!!
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page 301 cs toh advanced guide, btm of page
should the arrows reach the H atom?
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Originally posted by Flying grenade:
page 301 cs toh advanced guide, btm of page
should the arrows reach the H atom?
Yes.