Lurkers, register with SgForums and ask ur H2 Chem qns here!

• Light5

  19 Aug 15, 23:18

  General question : What would be the equation for the reaction of ICl (iodo chloride) with KBr ....now Cl is more reactive than Br so it should replace it but Iodine in ICl isnt so would I be converted to I2 or will it remain in ICl.....?

   

• UltimaOnline

  20 Aug 15, 02:04

  Originally posted by Light5:

  General question : What would be the equation for the reaction of ICl (iodo chloride) with KBr ....now Cl is more reactive than Br so it should replace it but Iodine in ICl isnt so would I be converted to I2 or will it remain in ICl.....?

   

  
  I-Cl may be conceptualized as I+ and Cl-, due to the greater electronegativity of Cl over I. Br- from KBr is a reducing agent (since it's electron rich), and can be oxidized to Br2, while reducing I+ to I2.
  
  I+ from I-Cl is an oxidizing agent (since it's electron deficient) and can be reduced to I2, while oxidizing Br- to Br2.
  
  Br- isn't able to further reduce I2 to I-, because I- is itself a stronger reducing agent (due to having a larger ionic radius and hence also having a more positive oxidation potential compared to Br-).
  
  I+ isn't able to further oxidize Br2 to positive OSes of +1, +3, +5 or +7, which requires a stronger oxidizing agent and/or self-disproportionation under alkaline conditions and/or heating.
  
  Hence the overall reaction for this question should be 2 I-Cl + 2 KBr --> I2 + 2 KCl + Br2.
• Light5

  21 Aug 15, 05:44

  Q1.) 

       a.) How the interchain hydrogen bonds in Nomex differ from those in Kevlar?

       b.) How might the flexibility of Nomex compare to that of Kevlar?

  Q2.) I know that cycloalkenes from C3 to C7 dont exhibit cis-trans isomerism due to ring strain but CycloOctene does show cis trans isomerism. However, if , lets say CylcoOctene was attached with other rings in a larger structure, will it still exhibit cis trans isomerism?

  Q3.) From which book or website can i read about how different groups attached to benzene affects its reactivity with certain compounds and how to predict whether these groups will be 2,4 directing or 3-directing?

  Thank You.

• UltimaOnline

  21 Aug 15, 11:04

  Originally posted by Light5:

  Q1.) 

       a.) How the interchain hydrogen bonds in Nomex differ from those in Kevlar?

       b.) How might the flexibility of Nomex compare to that of Kevlar?

  Q2.) I know that cycloalkenes from C3 to C7 dont exhibit cis-trans isomerism due to ring strain but CycloOctene does show cis trans isomerism. However, if , lets say CylcoOctene was attached with other rings in a larger structure, will it still exhibit cis trans isomerism?

  Q3.) From which book or website can i read about how different groups attached to benzene affects its reactivity with certain compounds and how to predict whether these groups will be 2,4 directing or 3-directing?

  Thank You.

  
  Q1. The hydrogen bonds in Nomex are longer, more strained, and hence weaker compared to those in Kevlar ; hence Nomex possesses greater flexibility compared to Kevlar.
  
  Q2. Depends on the larger structures and the other rings to which cyclooctene was attached to. If the cis isomer generates an unacceptable level of ring strain in the other connected rings, then it's a no-go.
  
  Q3. https://www.khanacademy.org/science/organic-chemistry/aromatic-compounds/deffect-2013-07-03T14:44:54.851Z/v/ortho-para-directors-i
• BCME

  24 Aug 15, 12:42

  With reference to JJC 2014 P3 Q2(c)(i),

  Can we say that due to more electron donating akyl groups on TEA, the lone pair of electron on N is more available for hydrogen bonding as compared to DEA? 

• UltimaOnline

  24 Aug 15, 13:24

  Originally posted by BCME:

  With reference to JJC 2014 P3 Q2(c)(i),

  Can we say that due to more electron donating akyl groups on TEA, the lone pair of electron on N is more available for hydrogen bonding as compared to DEA? 

  
  No, as that's an insignificant point relative to the more relevant points required by the mark scheme.
  
  However, the JJC mark scheme can nonetheless be further improved in 2 aspects :
  
  1) Intramolecular hydrogen bonding is possible for both species, hence you should not use it as a reason for why TEA has a higher boiling point than DEA.
  
  2) To explain the more extensive hydrogen bonding between TEA molecules : while both DEA and TEA have 3 protic groups that can function as both hydrogen bond donors and acceptors, but only TEA has an additional 4th group that can function as a hydrogen bond acceptor.
• Mrworry

  24 Aug 15, 21:52

  Regarding Organic chem made easy by George Chong

  Page 228, Example 7.17

  For isomer III why CH3CH2CH(OH)CH3 is not allowed?

  Xie xie ~

• UltimaOnline

  24 Aug 15, 23:05

  Originally posted by Mrworry:

  Regarding Organic chem made easy by George Chong

  Page 228, Example 7.17

  For isomer III why CH3CH2CH(OH)CH3 is not allowed?

  Xie xie ~

  
  Because your described structure fulfills the conditions for isomer I, and the question describes isomers I, II and III as *different* structures, hence the use of the term "isomer".
• Light5

  27 Aug 15, 01:22

  Q1.)

  (a) State the colour you would expect for the interhalogen compound BrCl

  (b.) Predict the equation of the reaction between BrCl (interhalogen) and AgNO3.

  (c.) Describe what you would observe when the products of (a) are added to Dilute NH3 in a test tube.

  Q2.) http://gceguide.com/papers/A%20Levels/Chemistry%20(9701)/9701_s15_qp_11.pdf

     Please explain how to solve Q13 and Q24 from the above paper. 

  Mark Scheme :         http://gceguide.com/papers/A%20Levels/Chemistry%20(9701)/9701_s15_ms_11.pdf

  Thank You !

• UltimaOnline

  27 Aug 15, 02:08

  Originally posted by Light5:

  Q1.)

  (a) State the colour you would expect for the interhalogen compound BrCl

  (b.) Predict the equation of the reaction between BrCl (interhalogen) and AgNO3.

  (c.) Describe what you would observe when the products of (a) are added to Dilute NH3 in a test tube.

  Q2.) http://gceguide.com/papers/A%20Levels/Chemistry%20(9701)/9701_s15_qp_11.pdf

     Please explain how to solve Q13 and Q24 from the above paper. 

  Mark Scheme :         http://gceguide.com/papers/A%20Levels/Chemistry%20(9701)/9701_s15_ms_11.pdf

  Thank You !

  
  Where did you get the interhalogen qn from? Is it a CIE 'A' level Chem qn, or a school's exam qn (if so, which country is the school based in)?
  
  Q1a) Somewhere between the colours of Cl2 (yellow-green) and Br2 (reddish-brown).
  
  Q1b) Although AgBr is less soluble than AgCl, but due to the greater electronegativity of Cl over Br, we have Br+ (not Br-), hence the products would be AgCl(s) and (covalent) BrNO3. BrCl is a strong oxidizing agent, but neither Ag+ nor NO3- (with N already in it's most positive OS) can be readily oxidized (it is difficult to oxidize Ag+ and O atoms having OS of -2).
  
  Q1c) The AgCl ppt would dissolve in NH3(aq), as the soluble ionic diamminesilver(I) coordination complex is generated. The Br atom of BrNO3 would also be nucleophilically attacked by NH3 and H2O, but the reaction products are not readily observable (hence irrelevant to the question).
  
  Q13. This qn tests you on the thermal stabilities of Group 2 carbonates (determined by cationic charge densities). As a fire retardant, you would want two things : greater moles of CO2(g) generated per mole of retardant, and lower thermal decomposition temperature to release the CO2(g) more readily.
  
  Q24. The mechanism involves 2 x nucleophilic substitution (first SN2 on the less sterically hindered terminal C atom, followed by deprotonation, then intramolecular SN2 or SN1 attack on the 5th C atom, followed by deprotonation) of the Br atoms on the 1st and 5th C atoms, by the NH3 nucleophile, to generate the required product Coniine.
• Light5

  27 Aug 15, 03:53

  Originally posted by UltimaOnline:

  
  Where did you get the interhalogen qn from? Is it a CIE 'A' level Chem qn, or a school's exam qn (if so, which country is the school based in)?
  
  
  Q1b) Although AgBr is less soluble than AgCl, but due to the greater electronegativity of Cl over Br, we have Br+ (not Br-), hence the products would be AgCl(s) and (covalent) BrNO3. BrCl is a strong oxidizing agent, but neither Ag+ nor NO3- (with N already in it's most positive OS) can be readily oxidized (it is difficult to oxidize Ag+ and O atoms having OS of -2).
  
  Q1c) The AgCl ppt would dissolve in NH3(aq), as the soluble ionic diamminesilver(I) coordination complex is generated. The Br atom of BrNO3 would also be nucleophilically attacked by NH3 and H2O, but the reaction products are not readily observable (hence irrelevant to the question).
  
  Q13. This qn tests you on the thermal stabilities of Group 2 carbonates (determined by cationic charge densities). As a fire retardant, you would want two things : greater moles of CO2(g) generated per mole of retardant, and lower thermal decomposition temperature to release the CO2(g) more readily.
  

  Sorry if the interhalogen question was vague....i actually constructed the question by myself because cambridge has a knack of asking weird questions from time to time. They have already tested displacement reactions of interhalogen compounds so i wondered if they could test a question like this as well.

  In q1(c) , we would obviously have a colourless test tube as all nitrates are soluble and AgCl is also soluble but what would be the products of the nucleophilic attack of NH3 and H2O on BrNO3?

  In q13(from the CIE paper), how can i predict the temperature required for decomposition of the 3 fire retardants since there are 2 Group 2 elements in one fire retardant.

  Thanks Again.

• UltimaOnline

  27 Aug 15, 04:17

  Originally posted by Light5:

  Sorry if the interhalogen question was vague....i actually constructed the question by myself because cambridge has a knack of asking weird questions from time to time. They have already tested displacement reactions of interhalogen compounds so i wondered if they could test a question like this as well.

  In q1(c) , we would obviously have a colourless test tube as all nitrates are soluble and AgCl is also soluble but what would be the products of the nucleophilic attack of NH3 and H2O on BrNO3?

  In q13(from the CIE paper), how can i predict the temperature required for decomposition of the 3 fire retardants since there are 2 Group 2 elements in one fire retardant.

  Thanks Again.

  
  No prob.
  
  Q1c) NH2Br + HNO3, and BrOH + HNO3, all of which are colourless.
  
  Q13. Each metal carbonate will decompose according to their own temperature, and you can treat each mineral as a mixture of two separate metal carbonates. Eg. For Mg3Ca(CO3)4, magnesium carbonate decomposes first, followed by calcium carbonate. Hence BaCa(CO3)2 is the worst fire retardant because Ba2+ has the lowest cationic charge density and thus requires the highest temperature to decompose, so by the time sufficient CO2(g) is released, your house would have burnt down completely.
• Mrworry

  27 Aug 15, 13:31

  Hi again :D

  For Organic chem made easy page 258,

  Example 8.4, part b)

  Why cannot it be:

  1. Add conc H2SO4(aq) 170C

  2.Cl2 (aq), heat

  3. Add Ethanolic KCN, heat

  4. HCl (aq), reflux

  xie2=3

• UltimaOnline

  27 Aug 15, 14:31

  You really love George Chong's book, doncha? He's busy climbing Mont Blanc right now even as we type, but up there in the lonely peaks of Mont Blanc, he'll surely appreciate the support of a loyal fan! ;Þ
  
  Your proposed synthesis pathway will score some marks, but not full marks.
  
  You'll be penalized *mostly* because your proposed synthesis pathway isn't the shortest possible (ie. George Chong's answer). Always give the shortest possible pathway, because using a longer-than-necessary synthesis pathway will result in a lower atom economy and/or equilibrium yield, and hence will always be penalized by Cambridge. (But hey, if you can't think of any shorter pathway, getting some marks is certainly preferable to getting no marks, so go ahead and just write the shortest pathway you can think of, even if it's 1 or 2 steps longer than specified by the qn).
  
  In addition (this is a minor point), although your step 2 will indeed obtain the required intermediate (used in your step 3) as the major product, but nonetheless there will still be a mixture of products, which should still be avoided in synthesis qns whenever possible (it's perfectly fine to use it when there's no other choice, but in this qn, there *is* another, even shorter pathway, ie. George Chong's answer).
  
  But I emphasize that the main problem with your answer is that it isn't the shortest possible synthesis pathway. If it had the same no. of steps as the shortest possible pathway, then Cambridge (in most cases) wouldn't penalize you for using electrophilic addition of aqueous halogen. No worries.
  
  On this note, using halogenation via free radical substitution will always get you penalized for synthesis qns (as it generates a variety of products, hence low overall atom economy which means wasteful on resources), unless there is absolutely no other choice for that qn.
  
  

  Originally posted by Mrworry:

  Hi again :D

  For Organic chem made easy page 258,

  Example 8.4, part b)

  Why cannot it be:

  1. Add conc H2SO4(aq) 170C

  2.Cl2 (aq), heat

  3. Add Ethanolic KCN, heat

  4. HCl (aq), reflux

  xie2=3

  
  
• Light5

  29 Aug 15, 00:45

  Q.) http://gceguide.com/papers/A%20Levels/Chemistry%20(9701)/9701_s15_qp_11.pdf

  In q 6 of the above paper, if volume is increasing from 1 to 2 , then pressure should decrease therefore equilibrium must shift to the RHS to increase No of Moles of PCl3 and Cl2 and decrease No of Moles of PCl5. This is reflected in both answer choices A and B yet the examiner has given B as the answer in the marking scheme? Can you please clarify this?

   

   

• UltimaOnline

  29 Aug 15, 02:54

  You're right that the position of equilibrium has shifted to the RHS, generating a *slightly* greater number of moles of RHS species at equilibrium.
  
  However, because the volume has doubled (ie. *greatly* increased), hence even with the *slight* increase in moles of RHS species, the new equilibrium molarities of the RHS species still decrease (though the decrease is not as much as the LHS species, whose moles and molarities *both* decrease).
  

  Originally posted by Light5:

  Q.) http://gceguide.com/papers/A%20Levels/Chemistry%20(9701)/9701_s15_qp_11.pdf

  In q 6 of the above paper, if volume is increasing from 1 to 2 , then pressure should decrease therefore equilibrium must shift to the RHS to increase No of Moles of PCl3 and Cl2 and decrease No of Moles of PCl5. This is reflected in both answer choices A and B yet the examiner has given B as the answer in the marking scheme? Can you please clarify this?

   

   

  
  
• Mrworry

  30 Aug 15, 15:18

  Hi

  A compound C3H6O2 which produces CHI3 and reduce Fehling solution. Does CH2(OH)COCH3 fulfill this requirement? Thanks 

• Mrworry

  30 Aug 15, 16:45

  c3h6o2 which does not produce triodomethane but reduce fehling soln

  CH2(OH)CH2OCH 

  Does this fulfill the req, if not why? Thanks 

• UltimaOnline

  30 Aug 15, 16:45

  Originally posted by Mrworry:

  Hi

  A compound C3H6O2 which produces CHI3 and reduce Fehling solution. Does CH2(OH)COCH3 fulfill this requirement? Thanks 

  
  Yes it does. This qn would kill A level H2 Chem students though ;Þ
• UltimaOnline

  30 Aug 15, 16:57

  Originally posted by Mrworry:

  c3h6o2 which does not produce triodomethane but reduce fehling soln

  CH2(OH)CH2OCH 

  Does this fulfill the req, if not why? Thanks 

  
  Your structure is in error. Ensure that you satisfy the valencies of O and C such that all atoms have a stable octet but no formal charge.
  
  Update : MrWorry, you probably got the correct structure, but you expressed the condensed structural formula wrongly. You should write it as follows (ie. the way I wrote it below).
  
  The correct answer should be : CH2(OH)CH2CHO
• Mrworry

  30 Aug 15, 17:04

  Xiexie~

  Btw I am not sure why but the answer given for first part is CH3COCOH and second part is CH3OCH2CHO. @.@ 

  Are my answers accepted? 

• UltimaOnline

  30 Aug 15, 17:07

  Originally posted by Mrworry:

  Xiexie~

  Btw I am not sure why but the answer given for first part is CH3COCOH and second part is CH3OCH2CHO. @.@ 

  Are my answers accepted? 

  
  Yes, your answers are fine. The given answer is fine for the 2nd qn, but wrong for the 1st qn (count the no. of H atoms).
• Light5

  18 Sep 15, 05:58

  http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_w14_qp_43.pdf

  ^Q3(d) .... I know that the observed colour is the complement of the colour absorbed but i cant seem to understand how to interpret this graph....there is high absorption in the blue-violet region which should yield a green-yellow colour according to the colour wheel. Furthermore, the high absorption in the orange-red region should yield a blue-green colour. I cant understand what the final colour will be since there are so many being absorbed and so many combinations possible.

  MS : http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_w14_ms_43.pdf

  http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_w09_qp_41.pdf

  ^Q3(b) from above paper : Same issue as before...a range of wavelengths is being absorbed. For the Copper-Ammonia Complex, max absorption is in green-orange region which yields violet,dark blue and blue according to colour wheel. For the aqua-copper complex there is no absorption at all in both blue AND green regions so shouldnt the colour observed be bluish-green. The MS states it will be pale blue but how can i interpret it by only using the graph.

  MS: http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_w09_qp_41.pdf

   

  Please give a detailed explanation.

  Thank You Very Much !

   

• UltimaOnline

  18 Sep 15, 09:22

  Originally posted by Light5:

  http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_w14_qp_43.pdf

  ^Q3(d) .... I know that the observed colour is the complement of the colour absorbed but i cant seem to understand how to interpret this graph....there is high absorption in the blue-violet region which should yield a green-yellow colour according to the colour wheel. Furthermore, the high absorption in the orange-red region should yield a blue-green colour. I cant understand what the final colour will be since there are so many being absorbed and so many combinations possible.

  MS : http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_w14_ms_43.pdf

  http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_w09_qp_41.pdf

  ^Q3(b) from above paper : Same issue as before...a range of wavelengths is being absorbed. For the Copper-Ammonia Complex, max absorption is in green-orange region which yields violet,dark blue and blue according to colour wheel. For the aqua-copper complex there is no absorption at all in both blue AND green regions so shouldnt the colour observed be bluish-green. The MS states it will be pale blue but how can i interpret it by only using the graph.

  MS: http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_w09_ms_41.pdf

   

  Please give a detailed explanation.

  Thank You Very Much !

   

  
  1st paper : Cambridge is challenging the candidate to think more directly : since almost all colors of the spectrum are absorbed except green / yellow, the color observed is green / yellow. You don't even need to use the color wheel to check the complementary color.
  
  2nd paper : Cambridge is testing whether the candidate notices the exact phrasing of the question : (i) says "State" (ie. from memory, without using spectra graph), while only (ii) then says "Using the spectra above".
• BCME

  20 Sep 15, 02:41

  Hi, the following qns are asked with reference to VJC_P3_2014

  1. 1(a)(iii) not very sure how to calculate the ratio 

  2. 2(d) how do we know how to construct the equation between P and NaOH? 

  3. 3(i) why KMnO4 cannot be used?

  4. 4(b) I know that J contains COOH. However when it is reduced won't it be converted to primary alcohol and alcohol will react with conc sulfuric acid so I am a bit confused. 

  Thanks