17 Apr, 05:34PM in sunny Singapore!

BedokFunland JC's A Level H2 Chemistry Qns (Part 1)

Subscribe to BedokFunland JC's A Level H2 Chemistry Qns (Part 1) 202 posts

Please Login or Signup to reply.
  • Moderator
    A Levels H1 & H2 Chemistry Tuition @ BedokFunland JC
    UltimaOnline's Avatar
    4,109 posts since May '05
    • 'A' Level Qn.




      Solution :



      If these units (Pressure in Pa; Volume in m3) are used, the value of the gas constant R is 8.314.

      If these units (Pressure in atm; Volume in dm3) are used, the value of the gas constant R is 0.08206.


      pV = nRT

      pV = (m/Mr)(RT)

      Mr = (m)(R)(T) / (pV)

      Mr = (m)(0.08206)(T) / (pV/1000)   

      Mr = (m)(82.06)(T) / (pV)

      Mr = (m)(22400/273)(T) / (pV)

      Mr = (m)(22400)(T) / (pV)(273)

      Edited by UltimaOnline 07 Jan `09, 10:35AM
    • 'O' Levels & 'A' Levels

      Qualitative Analysis Question


      Originally posted by anpanman:

      Firstly, I want to discuss with everyone some procedures carried out in this experiment involving Qualitative Analysis. A few questions is in store too.


      Objective: Identify element(s) present in solid R and solution S

                     - the Anion present in S


      Solid R contains 2 elements. The metal found in R is also present in solution S


      1a Put about half of Sample R into test tube and heat strongly until no further change is observed. Next, wait for residue to cool off.

      (my observation:Solid changes from brown to yellowish orange. Gas produced relights glowing splint. oxygen is released.)


      2 To new sample R, add small volume of concentrated HCl and warm mixture gently and leave to stand for a few minutes.

      (my observation: bubbled produced. Gas produced turns moist blue litmus paper red and is then bleached. Cl2 gas is evolved.)


      3a To a portion of solution S , add NaOH until change is seen

      (observation: white precipitate formed. )


      3b Add NaOH in excess to mixture in (a)

      observation: white ppt dissolves in excess NaOH to give colourless solution.


      3c To a portion of mixture from (b), add a small piece of aluminium foil and warm gently

      observation: bubbles prodiced. Red litmus paper turns blue. Ammonia gas produced and NO3- ion is present


      4a To a portion of solution S, add aqueous potassium chromate(VI)

      obsevation: solution S turns yellow. Yellow precipitate is produced. Hence S is reducing agent.


      4b Pour small portion of mixture from (a) into boiling tube and add dilute HCl until boiling tube is almost 1/3 filled. Leave mixture to stand for a few minutes

      observation : yellow precipitate turns white. Redox action has occurred.


      What I find strange is that a delivery tube was given to us and not all made use of it. So I assume CO2 gas must have evolved at some point during experiment. Having said that, Qualitative Analysis experiments do not tell us which gas to test for. In this case, the gas to test for for question one is rather unclear. So how then do we know which method for testing should be applied?


      And my friend said the elements present in solid R are lead and oxygen, whereas anion in solution S in NO3-. I think the answer regarding solid R's components are wrong. Can someone try to guess what it might be fromthe observations described above? They may not be entirely correct but I hope just reading through the procedure would give you a brief idea.


      Please help me solve these queries of mine as I am rather eager to learn about the kind of methods for gas test I should use. I can't perform all the tests because this would take up time and maybe, the gas would have been diffused after some time.


      Not forgetting, a big thanks!

      Warning : This experiment may be a health hazard, and should not have been given to students.


      S appears to be lead(II) nitrate, which when mixed with potassium chromate(VI), produces a yellow precipitate of lead(II) chromate (which is fatally toxic and carcinogenic).


      R appears to be lead(IV) oxide (stock name) or lead dioxide (prefix name), PbO2(s), a brown solid, which decomposes to oxygen gas and a yellowish-orange solid Pb3O4(s) which is actually (PbO)2.PbO2(s), a complex of lead(II)oxide and lead(IV)oxide.


      Because lead(IV) oxide is a strong oxidizing agent, chloride ions are oxidized to chlorine gas.


      Again, this experiment involves fatally toxic and carcinogenic materials, and should not be conducted by students.


      In regard to your question on which gas to test for, you have to test for any gas you suspect is present. You'll have to pick up clues from the other tests and their results, and make educated guesses and prioritize your tests accordingly, for maximum efficiency.


      Thermal decomposition of the following solids may produce the corresponding gases (note : there are exceptions) :

      oxides - oxygen (rarely)

      hydroxides - water vapour

      carbonates - carbon dioxide

      sulfates - sulfur dioxide and oxygen

      nitrates - nitrogen dioxide and oxygen


      Efferverscence which occurs when you add a solution to a solid, without heating, may indicate hydrogen gas (eg. metal and acid) or carbon dioxide gas (eg. carbonate and acid).


      When metal halids are heated with concentrated sulfuric acid (usually too hazardous for students to handle), steamy fumes of hydrogen halides and halogen gases may be produced.



      Originally posted by anpanman:

      Your explanation helps a lot. May I know in which part of the experiment indicate that Pb2+ ion was present in solution S and solid R?

      The cation in question gave a white ppt with NaOH(aq), which dissolved in excess NaOH(aq). Hence, the cation could be either Al3+, Pb2+, or Zn2+.


      Of these 3 cations, only Pb2+ qualifies as a cation of a transition metal* (that has partially filled d-orbitals), and hence only Pb2+ forms coloured compounds (eg. the yellow lead(II) chromate ppt).


      (*Note : although zinc is a d-block metal, it is not a transition metal. It does not form coloured compounds.)




      angel.png We Love Qualitative Analysis! angel.png

      Edited by UltimaOnline 08 Jan `09, 11:31PM
    • 'A' Level Qn.


      Originally posted by Bigcable22:

      is BeCl2 covalent or ionic ?   if it is ionic compound with covalent characteristics do i treat it as covalent or ionic for doing mcq?


      >>> is BeCl2 covalent or ionic ? <<<

      It is covalent, due to the high electronegativity and high charge density (both these properties, relative to the rest of Group II metals). Its high electronegativity prevents complete loss of electrons to form cations which are stable on their own (required for formation of ionic compounds). Its high charge density attracts and causes lone pairs from chlorine to form bond pairs with itself.


      As a gas, it is simple covalent. As a solid, it is a polymer macromolecule (with dative covalent bonds between the chlorine and beryllium atoms of adjacent molecules).


      One interesting point to note, about beryllium complex ions, is that being in period 2, it does not have empty 3d orbitals with which to expand its octet. Consequently, beryllium complex ions are 4-coordinated, meaning they can at most have 4 bond pairs, eg. [Be(H2O)4]2+.


      This is in contrast with magnesium (also in Group II), which are 6-coordinated, eg. [Mg(H2O)6]2+.

      Edited by UltimaOnline 08 Jan `09, 11:18PM
    • 'O' & 'A' Level Qn.




      Solution :


      Ar = 39.9g > N2 = 28g > He = 4g, Ne = 20.2g, CH4 = 16g


      Only argon Ar(g), is heavier (hence denser) than nitrogen gas N2(g).

      Edited by UltimaOnline 08 Jan `09, 11:23PM
    • 'A' Level Qn.



      Originally posted by Bigcable22:

      wanna use this thread to ask another question..


      what is the shape of CH3CH=CH2?

      and double check is the shape of HNO3 trigonal planar? as the H ion is dative bonded to 1 of the O atom hence i should consider only the 3bond pairs of electrons around N atom ?


      >>> what is the shape of CH3CH=CH2? <<<

      You have to specify, "about which atom?". Look at how many lone pairs and bond pairs each atom has, to determine the geometries within the molecule.


      Also, you should be able to distinguish between electron geometry (which tells you the hybridization of the atom) versus molecular geometry (which tells you the shape of the molecule).


      For instance, what is the hybridization of N in the ammonia molecule? Since it has 4 electron pairs, the electron geometry is tetrahedral and the N is sp3 hybridized. But since one of the 4 electron pairs is a lone pair, the molecular geometry is trigonal pyramidal.



      For CH3CH=CH2

      Eg. the 1st C (sp3 hybridized) is tetrahedral, the 2nd C (sp2 hybridized) is trigonal planar, the 3rd C (sp3 hybridized) is trigonal planar.



      >>> and double check is the shape of HNO3 trigonal planar? <<<


      HNO3 or nitric(V) acid :





           + N = O




      (+ve formal charge on N atom, since it is in Group V but only has 4 bond pairs (maximum octet as it has no empty d-orbitals to expand its octet); -ve formal charge on 1 O atom, since it is in Group VI but it has 3 lone pairs and 1 bond pair)




      Conjugate base - NO3- or nitrate(V) ion :





       + N = O





      (+ve formal charge on N atom, since it is in Group V but only has 4 bond pairs (maximum octet as it has no empty d-orbitals to expand its octet); -ve formal charge on 2 O atoms, since they are in Group VI but they have 3 lone pairs and 1 bond pair)



      The fact is, to draw Kekule structures of a molecule or polyatomic ion correctly, you need to be familiar with the concept of formal charges, which is often neglected by many JC teachers.


      Similarly, to have a proper understanding of Organic Chemistry, you need to be familiar with the concept of resonance, which again is often neglected by many JC teachers.


      But if you come for my tuition, I will definitely teach it to you (all my students), as these concepts (among many others) are fundamental and integral to a proper understanding and appreciation of Chemistry.  

      Edited by UltimaOnline 09 Jan `09, 1:17PM
    • 'A' Level Qn.


      Which is the better nucleophile? OH- or SH-?

      Which is the better base? OH- or SH-?



      The OH- ion can act as either a base or a nucleophile, depending on the solvent. Explain.



      Solution :


      Often, a strong nucleophile is also a strong base, and vice-versa. However, occasionally you may encounter a species that is a strong nucleophile but is a weak/moderate base, or a strong base, but is a weak/moderate nucleophile.


      Basicity measures stability of a species (a thermodynamic issue), while nucleophilicity measures the rate at which a nucleophile attacks an electrophile (a kinetic issue).


      A base is a species that abstracts a proton (ie. H+ ion).

      A nucleophile is an electron-rich species that has a lone pair available to donate to form a bond pair with an electron-poor species or electrophile.


      Although oxygen is more electronegative compared to sulfur, however because the atomic radius of sulfur is greater (one additional electron shell) than that of oxygen, the negative charge is better stabilized in the SH- ion. Therefore, OH- is a stronger base compared to SH-.


      And also because the atomic radius of sulfur is greater (one additional electron shell) than that of oxygen, the electron density in sulfur is more polarizable (ie. easier to shift electron density around) as compared to oxygen; consequently the lone pairs in SH- is more available for donation as compared to that in OH-. Therefore SH- is a stronger nucleophile compared to OH-.


      As far as the H2 Chemistry syllabus is concerned, the OH- ion acts as a nucleophile under aqueous conditions, but acts as a base under alcoholic conditions. The reasons for this are as follows :


      Water is a protic (ie. protons available for donation), polar solvent. Alcohol is an aprotic (ie. no protons available for donation), (relatively) non-polar solvent. 


      When a water molecule donates a proton to the OH- ion, the water molecule becomes a OH- ion and the OH- ion becomes a water molecule. So status quo is maintained. Hence it is primarily the fact that water is a strongly polar solvent, one which forms stabilizing ion-dipole interactions with the OH- anion, that contributes to the tendency of the OH- to function more as a nucleophile than as a base.


      Under alcoholic solvent conditions, the OH- ion is not stabilized by the solvent; because the alcohol solvent can neither donate protons, nor form stabilizing ion-dipole interactions with the OH- ion (alcohol can only form weaker permanent dipole-induced dipole interactons with the OH- ion), hence the highly unstable OH- anion ion here, has a high tendency to function as a base, to abstract any available proton as soon as possible to stabilize it's negative charge.







    • 'A' level Qn. (Supplementary thermodynamics question; not strictly required in H2 syllabus).


      For the reaction involving a Group I metal and water, briefly describe (qualitatively) the expected difference between energy change and enthalpy change.


      Solution :


      The reaction is exogernic and exothermic, and produces hydrogen gas. Under standard conditions, there is an increase in no. of moles of gas from LHS (none) to RHS (hydrogen gas).


      Therefore, we expect the energy change to be slightly more exogernic as compared to the enthalpy change; this is due to some of the energy (produced from the reaction) being used up (by the hydrogen gas produced) to do work to push back atmospheric air molecules.


      Energy change is measured under constant volume (not constant pressure).

      Enthalpy change is measured under constant pressure (not constant volume).

      For the reaction :


      2Na(s) + 2H2O(l) --> 2NaOH(aq) + H2(g)


      Enthalpy change has been measured to be approximately -367.5kJ/mol (of H2 gas produced).


      Energy change can been calculated to be approximately

      Energy change(kJ/mol) = Ethalpy change(kJ/mol) - (Pressure(atm) x Volume(dm3) change)

      = -367.5kJ - (1.0atm x 24.5dm3)

      = -367.5kJ - (2.45kJ)

      = -369.5kJ/mol (of H2 gas produced).


      These values support our expectation that the energy change of the reaction is slightly more exogernic (ie. stronger) than the enthalpy change measured, as some energy is used up to do work (to push surrounding air molecules).




    • 'A' Level Qns [Biology+Chemistry]

      (Supplementary; beyond the H2 & H3 syllabuses)



      Q1) "Briefly describe the importance of gaseous oxygen to life on Earth."


      Solution :

      Oxygen is vital for eukaryotic life forms, due to its role in cellular aerobic respiration, namely to allow the oxidation of pyruvate via the (Krebs/Citric Acid/TriCarboxylic Acid) cycle, and subsequently to enable oxidative phosphorylation, which uses energy released by the electron transport chain to power ATP synthesis on the mitochondrial membrane (in a process known as chemiosmosis using proton gradient generated), with oxygen functioning as an electron and proton acceptor at the end of the electron transport chain, forming the waste product of water.



      Q2) "Ironically, oxygen is also highly toxic, and constitutes one of the contributive factors responsible for aging, cell damage and tissue death. Explain", OR "Oxygen molecules can be attracted by a magnet. Explain."


      Solution :


      The common Kekule/Lewis/Dot-&-Cross structure of the O2 molecule commonly drawn at 'O' levels and even 'A' levels (each oxygen atom having 2 lone pairs and 2 bond pairs), is actually misleading and inaccurate, due to the limitations of valence bond theory.

       ..  ..



      If the above structure was true, then all electrons would be paired up (ie. having opposite spins), and the O2 molecule would be diamagnetic (ie. non-magnetic).


      Experimental evidence has demonstrated that the O2 molecule is in fact paramagnetic (ie. it is attracted by the magnetic field but does not remain magnetic once it leaves the field).


      Note that Omolecules are most observably attracted to a magnetic in the liquid phase; because in the gaseous phase, O2 molecules have too much kinetic energy and are moving too rapidly to be noticeably attracted to a magnet. You would need a very powerful magnet to attract gaseous Omolecules.


      Watch this video titled "Paramagnetism of Liquid Oxygen" http://www.youtube.com/watch?v=Isd9IEnR4bw


      To account for O2's paramagnetism, we would have to draw its structure as

       ..  ..


        .  .


      However, this is unsatisfactory for 2 reasons. Experimental evidence suggests a double bond, and in this model, both O atoms would lack a stable octet.


      Based on the above discussion, chemists realized that the valence bond theory was inadequate, and the Molecular Orbital theory was required to explain the paramagnetism of the O2 molecule.


      The ground state electron configuration of oxygen is 1s2 2s2 2p4, hence an O2 molecule has 16 elecrons. Using Molecular Orbital theory involving bonding and anti-bonding* orbitals, we see O2 as having (filled in order of ascending energy levels)


      (σ1s)2, (σ*1s)2, (σ2s)2, (σ*2s)2, (σ2px)2, (π2py)2, (π2pz)2, (π*2py)1, (π*2pz)1, (σ*2px)0


      Following the Pauli Exclusion Principle, and filling electrons in order of ascending energy levels (in the O2 molecule), 4 electrons will fill the σ1s and σ*1s orbitals, 4 electrons will fill the σ2s and σ*2s orbitals, 2 electrons will fill the σ2px orbitals, 4 electrons will fill the π2py and π2pz orbitals,and that leaves 2 electrons to fill the π*2py and π*2pz orbitals, and 0 electrons to fill the σ*2px orbitals (which are hence empty).


      According the Hund's rule, the last 2 electrons enter the π*2py and π*2pz orbitals (of the O2 molecule), with parallel spins. Since the rest of the electrons in the molecule are all paired, the remaining two electrons in the π*2py and π*2pz orbitals, give the diatomic molecule a net total spin;

      consequently, the O2 molecule is paramagnetic.


      (Diagram Acknowledgement : Michigan State University's Department of Chemistry. For the molecular orbital diagrams of fluorine and nitrogen, visit http://www.cem.msu.edu/~reusch/VirtualText/illust2.htm )



      Having determined the (free) radical nature of the O2 molecule due to unpaired electrons, the next step in understanding it's destructive effects on living cells and subcellular systems, involves the consequent susceptible formation of even more reactive oxygen free radical species, including the superoxide radical, the peroxide radical, the hydroxyl radical, and the singlet oxygen radical.


      For further information on the harmful biological effects of these reactive oxygen radicals, visit :




      Edited by UltimaOnline 17 Jan `09, 4:14AM
    • 'A' Level Qn.


      Explain why point of maximum buffer capacity exists when pH = pKa.



      Solution :


      When the pH is lower (ie. more acidic) than the pKa value, the majority of the species exists in protonated form; and has the capacity only to donate protons to buffer incoming bases and/or OH- ions.


      When the pH is higher (ie. more alkaline) than the pKa value, the majority of the species exists in deprotonated form; and has the capacity only to accept protons to buffer incoming acids and/or H+ ions.


      When the pH is at pKa value, half of the species exists in protonated form, and half of the species exists in deprotonated form; and hence has the maximum capacity to buffer both incoming acids as well as bases.

    • 'A' Level Qn.

      When polyunsaturated cis-fatty acids are heated during the frying process, a conversion to the trans isomeric form occurs.



      a) why the conversion of a cis-isomer to a trans-isomer is exothermic;

      b) the link between kinetics and equilibrium;

      c) whether frying foods causes a shift in equilibrium to produce more cis-isomers or more trans-isomers;

      d) in terms of thermodynamics and Gibbs free energy, the interconversion of the geometric isomers.


      Solution :


      a) The cis-isomer experiences greater electron repulsion between the (electron clouds of the) bulky groups (in close proximity) on the same side of the double bond. This kind of strain is known as steric strain. Therefore, the conversion of the more unstable (possessing higher potential energy from the electron repulsion of the steric strain) cis-isomer to a more stable (and hence possessing less energy) trans-isomer is exothermic, as the additional energy of the cis-isomer is lost to the environment (as heat), as it converts into the trans-isomer.


      b) The kinetics of the forward vs backward reaction, is determined by the relative magnitudes of the energy barriers (in which activation energy is defined as the energy required to overcome the energy barrier) of the forward vs backward reactions. An exothermic forward reaction has a relatively lower energy barrier and is hence faster than the backward endothermic reaction (for the same reaction), at any given temperature. However, as you increase temperature, although the rates of the exothermic (or forward reaction, here) and the endothermic (or backward reaction, here) reactions both increase (due to increased % of all particles having energy equals or exceeding activation energy required for both forward and backward reactions); however as temperature increases, the rate of increase, of the rate of the endothermic (or backward) reaction, is faster compared to that of the exothermic (or forward) reaction. (This is mathematically expressed by the Arrhenius equation). As such, Le Chatelier's principle predicts an equilibrium shift that results in more cis-isomers being formed. (Note : a shortcut way of remembering the end result for any position of equilibrium shift due to temperature, is to regard heat as either a reactant (endothermic reaction) or a product (exothermic reaction) and apply Le Chatelier's principle accordingly.)


      c) In terms of enthalpy and Le Chatelier's principle, assuming we started off with an equilibrium mixture of cis and trans isomers, we would expect increasing the temperature (when frying foods) to shift the position of equilibrium to form more cis-isomers (applying Le Chatelier's principle, for exothermic reactions, heat may be regarded as a product, see discussion above on link between kinetics and equilibrium). However, because initially unsaturated fatty acids (in raw oils) exist only in the healthy cis-isomeric form and not the unhealthy trans-isomeric form, hence the reaction quotion, Qc, at initial conditions, is 0. And regardless of the endothermic or exothermic enthalpic nature of the backward or forward reactions, when activation energy is provided as heat (in the frying process) to overcome the energy barrier, rotation across the double bonds will begin, and a mixture of cis and trans isomeric forms will consequently result (when initially, in natural raw organic oils, only healthy cis isomers are present).


      Therefore, as we fry foods with unsaturated oils, the problem is not the shifting of position of equilibrium (which as predicted by Le Chatelier's principle, for an exothermic reaction of cis-to-trans isomers in which heat may be regarded as a product, we would see more trans-isomers convert into cis-isomers, which would be wonderful), but rather the fact that our fatty acids, all of which began initially as healthy cis-fatty acids, would now experience double bond rotation and conversion into unhealthy trans-fatty acids.


      d) In terms of thermodynamics and Gibbs free energy, when the (more unstable, hence more repulsion potential energy) cis-isomer is converted to the (more stable, less energetic) trans-isomer, because there is a favourable increase in entropy as well as a favourable exothermic enthalpy change, hence Gibbs free energy predicts that the conversion of cis to trans isomers would be feasible and spontaneous, which is unfortunate in terms of a healthy diet.


      (i) The difference in entropy (at any given temperature) between the cis and trans isomers, relates to the steric strain resulting from the close proximity of the (electron clouds of the) bulky groups on the same side of the double bond. Due to translational, vibrational and rotational restrictions imposed by the electron repulsion from the steric strain, there is consequently an increase in total number of microstates (and hence a corresponding increase in entropy) of the trans-isomer, relative to the cis-isomer.


      (ii) We observe that in the Gibbs free energy formula, the temperature variable is attached to the entropy variable. This is a mathematical way of expressing the concept that an increase in temperature directly, positively and powerfully supports entropy (as heat energy is invariably converted to kinetic energy, resulting in a greater number of microstates or degree of disorderliness). Accordingly or hence, when frying foods at high temperatures, the favourable entropy factor (when cis-isomers are converted to trans-isomers) becomes increasingly important.


      (iii) Furthermore, the enthalpy factor in the Gibbs free energy formula also supports the conversion of the cis-isomer to the trans-isomer, it being an exothermic process which is thermodynamically favourable.


      Conclusion : Avoid fried foods and consume more raw foods. In addition to an entire host of carcinogens (which cause DNA mutations that activate proto-oncogenes and deactivate tumour suppressor genes, thereby causing cancer) that are produced in the deep frying process; polyunsaturated cis-fatty acids are converted to trans-fatty acids (that behave identical to saturated fatty acids as far as biological systems are concerned) in the frying process, which lead to a dramatic decline in integrity of all lipid based biological membranes, including notably the nervous system and the brain (and hence, quality of I.Q., E.Q., and 'A' level grades as well), and consequently a decline in health and quality of life.

      For further information on diet and health, google "Aajonus Vonderplanitz".


      Edited by UltimaOnline 20 Jan `09, 11:16PM
    • 'A' Level Qn




      Solution :


      A is para-(1-methylpropan-1-ol)phenol

      B is the deprotonated sodium phenoxide salt of A

      C is ortho,ortho-dibromo-para-(1-methylpropan-1-ol)phenol

      D is para-(1-methylprop-1-ene)phenol



      Notes on compound A (for which B, C, D is derived), based on information given in the question.


      - Chiral compound, with a chiral (stereogenic) center (usually a carbon atom), since it is optically active.

      - No carboxylic acid group present, since no reaction with sodium carbonate.

      - Phenolic group present (since it reacts with sodium hydroxide in a proton transfer reaction***, to produce a deprotonated sodium phenoxide salt which is soluble due to ion-dipole interactions with polar water molecules)

      - A hydroxy group present that can undergo a SN1**** substitution by HBr*

      - Cannot be oxidized by dichromate(VI)**, hence the alcohol (see preceeding point) must be tertiary.

      - Can undergo bromination (at ortho and para positions, except that the para position is already taken up by the side chain), to liberate fumes of HBr.

      - Can undergo dehydration (OH + H removed from adjacent C atoms) with concentrated sulfuric acid to form an alkene, which possesses geometric cis/trans isomerism.



      (* the hydroxy group is protonated, making it an excellent leaving group of H2O, which leaves as Br- nucleophile attacks the partial-positive alpha carbon, to avoid violating the octet). 

      (** if manganate(VII) was used instead, the entire side chain would be oxidized to give para-hydroxy-benzoic acid)

      (*** the reason why phenol is more acidic than water (ie. why a proton is transferred from phenol to OH- ion), is due to the greater polarity of the O-H bond (because of the electron withdrawing benzene ring) in phenol compared to the O-H bond in water, hence allowing for easier dissociation of the acidic proton in phenol; in addition (or alternatively), the relatively unstable OH- ion will naturally seek to stabilize itself by abstracting (ie. "grabbing") a proton from phenol, which is willing to release the proton, because the negative charge on the conjugate base phenoxide ion, is relatively stabilized by resonance due to delocalization into the benzene ring, in contrast to the more unstable OH- ion which is not stabilized by resonance.)

      (**** we know it is SN1 rather than SN2 because the alcohol is tertiary, and thus presents steric hinderance to the incoming nucleophile.)

      Edited by UltimaOnline 26 Jun `09, 1:09PM
    • 'A' Level Qn.


      (Note : I posted this question several months ago, in this thread ("Q9"), but I've just today (23 Jan 09) updated it with the mechanism diagram I drew, scanned and uploaded.)



      Question :

      When trichloroethanal is heated with aqueous sodium hydroxide, two products are formed in a single reaction pathway. One of the products is trichloromethane. Draw the mechanism for this pathway and hence identify the other product.




      Solution :



      1) Hydroxide ion nucleophile attacks delta +ve carbonyl carbon.

      2) Electron density shifts up; pi-bond pair becomes a lone pair on oxygen.

      3) When electron density shifts back down (lone pair becomes pi-bond pair, reforming the carbonyl group), the trichloromethyl group leaves as an anion.

      4) The trichloromethyl anion is capable of being a leaving group (ie. it is stable enough to exist on its own if only briefly), only because of the 3 highly electronegative and hence strongly electron-withdrawing Cl groups.

      5) However, carbanions are still more unstable compared to alkanoate ions, hence the next step is

      6) proton transfer (ie. acid-base reaction) from methanoic acid to the carbanion (ie. lone pair on carbanion becomes bond pair with proton (H+); bond pair between proton (H+) and oxygen becomes lone pair on oxygen), forming

      7) trichloromethane and sodium methanoate (negative charge of HCOO- counterbalanced by Na+; note that these ions are aqueous meaning dissolved in water, having a hydration shell, ion-dipole interactions with polar water molecules).

      Edited by UltimaOnline 04 May `09, 8:58PM
    • 'A' Level Qns.



      1) Compare the structures of carbon dioxide and silicon dioxide.


      2) Compare the solubilities of calcium fluoride vs calcium chloride (and contrast these, in turn, to the solubilities of silver fluoride vs silver chloride)


      3) Compare the structures of beryllium fluoride vs calcium fluoride.


      4) Compare the structures of beryllium fluoride vs beryllium chloride.



      Solutions :



      1) Compare the structures of carbon dioxide and silicon dioxide.


      At room temperature and pressure, carbon dioxide is a simple (linear) covalent molecule while silicon dioxide has a giant tetrahedral 3D structure.


      The preference for silicon to form 4 single bonds (hence tetrahedral macromolecule), rather than 2 double bonds (as in carbon dioxide) can be traced to the decreased ease of overlap of its 3p orbitals, compared to the 2p orbitals in carbon, to form pi bonds. In other words, due to the greater atomic radius of silicon (compared to carbon) and hence greater inter-atomic distance between silicon and oxygen (compared to carbon and oxygen), it is easier and more stable to form single bonds between silicon and oxygen, while in contrast, it is easier and more stable to form double bonds between carbon and oxygen.



      2) Compare the solubilities of calcium fluoride vs calcium chloride.


      Calcium fluoride is insoluble while calcium chloride is soluble. Let's consider qualitatively why calcium fluoride is insoluble (as compared to calcium chloride).


      In terms of enthalpy, the (endothermic) lattice dissociation enthalpy outweighs the (exothermic) hydration enthalpy, for calcium fluoride; while the (exothermic) hydration enthalpy outweighs the (endothermic) lattice dissociation enthalpy, for calcium chloride.


      This is due to the much higher charge density on the fluoride ion, resulting in a much more strongly endothermic lattice dissociation enthalpy for calcium fluoride (compared to its chloride). The exothermic hydration enthalpy also falls from fluoride (higher charge density) to chloride (lower charge density), but the decrease is not as sharp as the fall in lattice dissociation enthalpy.


      (Note that the difference in hydration enthalpies is apparent when you consider that silver fluoride is soluble while silver chloride, silver bromide and silver iodide are insoluble. In this case, because the silver cation has low charge density, hydration enthalpy changes more significantly than lattice dissociation enthalpy, between fluoride and chloride.)


      Overall, lattice dissociation enthalpy falls more sharply than hydration enthalpy, when we go from calcium fluoride to calcium chloride. Enthalpy argument hence favours the solution/dissolving of the chloride (ie. it's worth the cheap endothermic lattice dissociation investment, we get back even more heat energy from the ion-dipole interactions with water) over the solution of the fluoride (ie. it's not worth it, we can't get back our expensive endothermic lattice dissociation investment).


      In terms of entropy, the fluoride anion has a much higher charge density (compared to chloride) and hence far stronger ordering power over surrounding polar water molecules. Entropy (ie. disorderliness) hence favours the solution of the chloride over the fluoride.


      Interestingly (and fortunately for stressed students), both enthalpy and entropy arguments favour the solution of the chloride over the fluoride.


      Acknowledgement & Reference : 

      Jim Clark's ChemGuide.co.uk : http://www.chemguide.co.uk/inorganic/group2/problems.html



      3) Compare the structures of beryllium fluoride vs calcium fluoride.


      Calcium fluoride has an ionic lattice structure (wherein calcium is coordinated to eight fluoride anions and each F ion is surrounded by four Ca2+ ions). Beryllium fluoride has a covalent structure (even though it's more ionic compared to say, beryllium chloride; due to the fact that fluorine, being the #1 king of electronegativity in the entire periodic table, is even more electronegative than both chlorine and the Be2+ cation), and exists as a simple covalent (linear molecular geometry) molecule in the gas phase, while in the solid crystalline phase as a silicon dioxide-like tetrahedral 3D structure with four coordinate beryllium (ie. one fluoride atom each from 2 neighbouring molecules give dative bonds to the beryllium atom of a central molecule, allowing beryllium to have a stable octet; consequently and structurally, every Be atom is bonded to 4 F atoms, and every F atom is bonded to 2 Be atom; formal charges of -2 and +1 exist on the beryllium and fluorine atoms respectively).



      4) Compare the structures of beryllium fluoride vs beryllium chloride.


      In the gas phase, both beryllium fluoride and beryllium chloride exist as a simple linear covalent molecule (see answer to Q3 above).

      In the solid phase, while beryllium fluoride has a silicon dioxde-like tetrahedral 3D structure (see answer to Q3 above), beryllium chloride polymerises to make long (virtually) 1-dimensional polymer chains. One chlorine atom each from 2 neighbouring molecules will give a dative bond with a beryllium atom from a central molecule, allowing beryllium atoms to enjoy a stable octet. So a similar process for a similar objective occurs between the beryllium atoms and the halogen atoms, in both beryllium fluoride and beryllium chloride, but the end structure is different (akin to the alpha helix vs beta pleated secondary structure of polypeptides due to hydrogen bonding).


      In this case, the end-result difference in beryllium halide structure is naturally related to the difference in properties between fluorine vs chlorine, resulting in an alternative most-stable beryllium halide solid structure (the specifics are beyond the requirements of the 'A' level or H2/H3 syllabuses.)


      Wikipedia image of beryllium chloride long chain polymer





      Edited by UltimaOnline 05 Mar `09, 1:51AM
    • 'A' Level Qn.


      You have in your possession a Singapore 50 dollars note (bill). There are several ink or dirt smudges on it (of more than 1 colour or appearance). Describe and explain, using easily available (ie. non-controlled) chemical substances and/or tools, to determine if the smudges is covalent or ionic in nature.



      Solution :


      Using a tissue paper moistened with water, wipe the smudges. If some of the smudges come off, the nature of the composite pigment could either be soluble ionic, or polar covalent (possibly with the capacity to form hydrogen bonds with water). If the smudges do not come off, the pigment could either be insoluble ionic, or non-polar covalent.


      To determine which, wipe the remaining smudges with an alcohol (eg. isopropyl alcohol a.k.a. propan-2-ol) swab that can be purchased from any pharmacy (eg. NTUC, Guardian, Watsons, etc). If the smudges come off, the pigment must have been non-polar covalent. If the smudges remain, the pigment must be insoluble ionic.


      An ionic compound is soluble in water, if the Gibbs Free Energy for the solution process (ie. lattice dissociation + hydration enthalpies) is feasible. Otherwise, it is insoluble ionic.


      A covalent compound that is soluble in water, is either strongly polar, and/or has the capacity to form hydrogen bonds with water. Water would rather hydrogen bond within itself (ie. water molecules with other water molecules), than interact favourable with non-polar covalent compounds.


      (Indeed, this is the basis for the so-called hydrophobic interactions of non-polar 'R' groups of amino acids in proteins. They get pushed away by water molecules (which prefer to bond within themselves, macham elitist), until they (the non-polar groups) end up in close proximity, afterwich induced dipole-dipole van der Waals interactions take over to hold them together (hence, "hydrophobic interactions").)


      Alcohol is relatively non-polar (to be precise, the longer the non-polar or "fatty" hydrocarbon chain, the more non-polar the alcohol is; this explains why fats are "fatty" - they have long non-polar hydrocarbon chains). Hence, relatively non-polar and/or hydrogen bonding incapable covalent compounds, that are insoluble to water, may be soluble in alcohols (as well as oil/fats), due to favourable induced dipole-dipole interactions between solute and solvent. Which explains why the alcohol swab is able to wipe away non-polar and/or hydrogen bonding incapable covalent pigments that water cannot.


      Tip : for you students who write messily and end up with ink-stained hands after taking an exam paper, wipe your hands with alcohol swabs (which can be bought cheaply from any pharmacy, eg. NTUC, Guardian, Watsons, etc). Often, the ink pigments are relatively non-polar and/or hydrogen bonding incapable covalent compounds, that can be wiped away easily with alcohol swabs.



      Edited by UltimaOnline 01 Feb `09, 11:47PM
    • 'A' Level Qn.


      Assuming we're talking about atoms of the same element (eg. carbon), why is it that a double bond is less than twice the strength of a single bond?



      Solution :


      A single bond consists of 1 sigma bond.

      A double bond consists of 1 sigma bond and 1 pi bond.


      A sigma bond is formed when orbitals from 2 atoms overlap head-to-head (the sigma bond electron density is concentrated between the 2 nuclei and is responsible for holding the atoms together against the mutual repulsion of the positively charged nuclei).

      A pi bond is formed when the p orbitals of the 2 atoms overlap sideways, with the pi bond electron density above and below the sigma bond.


      For C-C, the sigma bond consists of sp3-sp3 orbitals overlapping end-on-end.

      For C=C, the sigma bond consists of sp2-sp2 orbitals overlapping head-to-head; and the pi bond consists of the unhybridized pz-pz orbital overlapping sideways.


      The pi bond is weaker than the sigma bond, due to the following reasons :


      (i) for the pi bond, there is significantly less electron density overlap between the component p-orbitals due to their parallel directional orientation in 3D space (in contrast to the head-on-head overlap of the sigma bond).


      (ii) the pi bond electron density is further away from the positive charge of the nucleus and hence uses more energy to sustain and is therefore less stable (compared to the more stable sigma bond electron density located directly (and "resting lazily") between the two positively charged nuclei).


      (iii) the pi bond orbitals are unhybridized while the sigma bond orbitals are hybridized (eg. sp3 or sp2), which means the energy level of the hybridized sigma orbitals (ie. the weighted average of the different orbital types; recall that the s orbital always has a slightly lower energy than the p orbitals at the same energy level or electron shell) is lower (ie. more stable) compared to the higher energy (ie. more unstable) unhybridized p orbital.


      Although the pi bond by itself is weaker than a sigma bond, pi bonds are often components of multiple bonds, together with sigma bonds. The combination of pi and sigma bond is stronger than either bond by itself.


      For instance, as the following excerpt from the Data Booklet states, the average bond energy (or enthalpy) of C-C single bond is 350kJ/mol, while that of C=C double bond (consisting of 1 sigma bond + 1 pi bond) is 610kJ/mol, which is less than twice the strength of the single bond.


      Edited by UltimaOnline 01 Feb `09, 11:48PM
    • 'A' Level Qn.

      by Dr Rod Beavon (http://home.clara.net/rod.beavon)


      Inorganic Chemistry Problem1



      A is a white compound of lead; on heating it gives yellow B and a colourles gas that turns limewater milky. If B is heated in air for several hours at 470oC it is converted into a scarlet powder C. C contains 90.66% lead and 9.34% oxygen by mass.

      C on heating with dilute aqueous nitric acid gives a colourless solution D and a brown solid E. If sodium hydroxide solution is added to solution D, a white gelatinous precipitate is formed, which with excess sodium hydroxide solution gives a colourless solution G.

      Compound E reacts with concentrated hydrochloric acid to give a white solid I and a green gas J. I is soluble in hot water but insoluble in cold, and forms soluble complex ions such as K with excess chloride ions. With potassium iodide solution J gives a brown solution of L, which on shaking with hexane gives a deep purple organic layer.

      With potassium iodide solution D gives a bright yellow solid M; this is insoluble in cold water, but in hot water gives a colourless solution.

      Identify all the substances A - M, and write equations for all the reactions involved.


      Solution :




      Edited by UltimaOnline 02 Feb `09, 12:45PM
    • 'A' Level Qn.

      by Dr Rod Beavon (http://home.clara.net/rod.beavon)


      Inorganic Chemistry Problem2




      Copper reacts with 50% nitric acid to give a blue solution A and a brown gas B. If the solution A is diluted and sodium hydroxide solution added cautiously, a gelatinous blue precipitate C is obtained, which if warmed forms a black solid G.

      Addition of concentrated ammonia solution to C gives a deep blue solution that contains the ion D. Addition of concentrated hydrochloric acid to C gives a green solution of the ion E, which on dilution with water gives a solution of the ion F.

      If the brown gas B is passed into water a mixture of acids H and I is formed in a disproportionation reaction.

      Identify A - I, giving equations for and explaining each reaction as far as possible.



      Solution :






      Edited by UltimaOnline 02 Feb `09, 12:48PM
    • 'A' Level Qn.

      by Dr Rod Beavon (http://home.clara.net/rod.beavon)


      Inorganic Chemistry Problem3



      A white powder A on heating turns yellow and evolves a gas C that turns limewater milky, as well as water vapour. The yellow residue B turns white on cooling, but will turn yellow again when heated.

      B reacts with dilute sulphuric acid to give a colourless solution D. If dilute sodium hydroxide solution is added to D, a white precipitate E is produced which with excess alkali gives a colourless solution F.

      With dilute ammonia solution D gives a white precipitate G; adding excess ammonia causes the mixture to become clear, giving solution H.

      Identify all the substances A – H, and account as thoroughly as you can for the observations.



      Solution :


      Edited by UltimaOnline 02 Feb `09, 12:50PM
    • 'A' Level Qn.

      by Dr Rod Beavon (http://home.clara.net/rod.beavon)


      Inorganic Chemistry Problem4



      A is a metal that reacts moderately quickly with dilute sulphuric acid to give a pale green solution B and a gas C. If solution B is allowed to crystallise, the pale green solid obtained gives on strong heating a solid H, and two gases I and J. These gases are compounds of non-metallic elements, one of which is in a different oxidation state in each compound. If J is passed into a solution of barium chloride there is a vigorous reaction and a white precipitate K is formed.

      A reacts with steam in an equilibrium reaction to give a solid M, which contains A in two different oxidation states, and the gas C; this reaction was at one time used industrially to make C. A reacts with gaseous HCl on heating to give K, a white solid, and C. Hydrated K is pale green. With chlorine, A gives a brown covalent solid L which sublimes on heating; aqueous solutions of the hydrate of L react with copper metal, and for that reason are used to etch printed-circuit boards in electronics.

      Solution B gives a dirty-green precipitate D if sodium hydroxide is added; this precipitate remains in excess sodium hydroxide. The same reaction is seen with the addition of ammonia. If D is allowed to stand in air it forms a foxy-red compound E, which is also obtained if sodium hydroxide solution is added to solutions of L.

      Acidified solutions of B decolourise potassium manganate(VII) solution giving the manganese-containing ion N as one of the products.

      Identify all of the substances A – N, and deduce as much as you can about the processes occurring.


      Solution :




      Edited by UltimaOnline 02 Feb `09, 1:07PM
    • 'A' Level Qn.

      by Dr Rod Beavon (http://home.clara.net/rod.beavon)


      Organic Chemistry Problem1



      Compound W, C3H8O, reacts with hot concentrated phosphoric acid to give X, C3H6, which rapidly decolourises a solution of bromine in tetrachloromethane. On treatment with ethanoyl chloride, W gives Y, C5H10O2. Oxidation of W with potassium dichromate in dilute sulphuric acid gives Z, C3H6O, which gives an orange precipitate with 2,4-dinitrophenylhydrazine but does not react with an ammoniacal solution of silver nitrate.

      Deduce the structures of W, X, Y and Z, explaining carefully what each piece of evidence tells you.


      Solution :


      Edited by UltimaOnline 02 Feb `09, 12:56PM
    • 'A' Level Qn.

      by Dr Rod Beavon (http://home.clara.net/rod.beavon)


      Organic Chemistry Problem2



      An oily liquid A is insoluble in water, but on heating with aqueous sodium hydroxide solution for half an hour it dissolves. From the reaction mixture a liquid B can be distilled, which gives a yellow precipitate with iodine and sodium hydroxide solution. On careful oxidation B gives an aldehyde, C, which also gives a yellow precipitate with iodine and sodium hydroxide solution. If sulphuric acid is added to the solution obtained from heating A with sodium hydroxide solution, a white precipitate D is obtained. D liberates carbon dioxide from sodium hydrogen carbonate solution, one mole of D reacting with sodium hydrogen carbonate to give one mole of carbon dioxide. Heating D with soda-lime* converts it to benzene.

      Find the structures of A, B, C and D, giving your reasoning.

      * Soda-lime is calcium oxide that has been slaked with sodium hydroxide solution. Carboxylic acids, on heating with soda-lime, are decarboxylated, that is the -COOH group is replaced by -H.


      Solution :


      Edited by UltimaOnline 02 Feb `09, 12:58PM
    • 'A' Level Qn.

      by Dr Rod Beavon (http://home.clara.net/rod.beavon)



      Organic Chemistry Problem3



      Compound A, C4H8O2, gives an orange precipitate with 2,4-dinitrophenylhydrazine reagent, and a silver mirror with ammoniacal silver nitrate. A is converted to B, C4H6O, on treatment with acid. B may exist as a cis-or trans- isomer; it reacts with hydrogen chloride to give C, C4H7OCl, and with chlorine to give D, C4H6OC12.

      Reduction of C gives E, C4H9OCl, which is hydrolysed by dilute alkali to F, C4H10O2. Both A and F give a positive iodoform (CHI3) test, whereas E does not.

      Identify compounds A to F.


      Solution :


      Edited by UltimaOnline 02 Feb `09, 12:59PM
    • 'A' Level Qn.

      by Dr Rod Beavon (http://home.clara.net/rod.beavon)


      Organic Chemistry Problem4



      An alcohol has the molecular formula C4H7OH. Oxidation of this alcohol gives a ketone. Ozonolysis of this alcohol gives methanal as one of the products.

      Giving your reasoning in full, deduce the structural formula for this alcohol and predict whether it will be optically active.

      Ozonolysis is a technique for finding the position of C=C double bonds in a molecule. The compound is reacted with ozone-enriched oxygen in a suitable inert solvent, and the product is hydrolysed with ethanoic acid in the presence of zinc. This produces a mixture of carbonyl compounds, which may be separated and identified, for example by means of their 2,4-DNPH derivatives.


      Solution :


      Edited by UltimaOnline 02 Feb `09, 1:00PM
    • 'A' Level Qn.

      by Dr Rod Beavon (http://home.clara.net/rod.beavon)



      Organic Chemistry Problem5



      Iodoethane CH3CH2I reacts with aqueous sodium hydroxide solution to give A, with ammonia to give B, and with potassium cyanide in ethanolic solution to give C, C3H5N. C reacts with lithium aluminium hydride in dry ether solution to give D. C with aqueous sodium hydroxide solution followed by dilute hydrochloric acid gives E, C3H6O2, which with phosphorus(V) chloride gives F.

      Identify compounds A to F, giving your reasons and equations where appropriate.


      Solution :


      Edited by UltimaOnline 02 Feb `09, 1:00PM
    • 'A' Level Qn.

      by Dr Rod Beavon (http://home.clara.net/rod.beavon)


      Organic Chemistry Problem6



      A, C3H7Cl, reacts with aqueous sodium hydroxide solution to give C, C3H8O. C reacts with phosphorus pentachloride to give steamy fumes, and can be oxidised with acidified potassium dichromate solution to give D, C3H6O, and then E, C3H6O2. D reacts with both 2,4-dinitrophenylhydrazine and ammoniacal silver nitrate. E reacts with phosphorus pentachloride and with sodium hydrogen carbonate solution, in the latter case giving carbon dioxide.

      A will react with benzene in the presence of anhydrous aluminium chloride to give B, C9H12. The structure of this compound is not what might have been expected from A. B can be made from benzene and F as the expected product, however, F being an isomer of A. Sodium hydroxide solution converts F to G, an isomer of C. G can be oxidised with acidified potassium dichromate solution to H, an isomer of D, but no further. H gives a positive result with the 2,4-DNP test but not the ammoniacal silver nitrate test; it also gives a positive result in the iodoform reaction.

      On heating with alkaline potassium manganate(VII) solution B gives an aqueous solution of I, which on treatment with dilute sulphuric acid gives a white precipitate J.

      (a) Identify all of the substances AJ, and explain each of the observations as fully as you can with supporting equations.

      (b) Suggest why A and F give the same product with benzene.


      Solution :


      Edited by UltimaOnline 02 Feb `09, 1:06PM
Please Login or Signup to reply.