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

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    • ‘A’ Level Qn.

      ~~
      Hydroxylamine NH2OH is a toxic substance to many aquatic organisms, and it has chemical properties similar to ammonia. For example, it reacts with hydrochloric acid to form the salt hydroxylammonium chloride NH3OHCl.
      (a) In an experiment, 25.0cm3 sample of the hydroxylammonium chloride solution was titrated against a solution of 0.100M sodium hydroxide, and 14.40 cm3 of sodium hydroxide was needed for complete neutralisation.
      (i) With the aid of an eqn, determine if the nature of the end point of titration is neutral, basic or acidic.
      (ii) Calculate the PH value at the equivalence point of the titration. The base dissociate value, Kb of hydroxylamine is 1.10×10^-8 mol/dm3.
      ~

      Points to note :

      (Don’t worry about the counterbalancing ion Cl-, its just a spectator ion; it doesn’t undergo hydrolysis much because of its low charge density and stability from its strong ion-dipole interactions with water).

      The HONH3+ ion is obviously acidic, as it has an available proton to donate, and no lone pairs available to accept protons. It’s a relatively weak acid though, not a strong one, just like NH4+ losing protons to become NH3 (since HONH2, like NH3, will again have the tendency to accept protons back, and become HOHN3+ or NH4+ again; to properly understand why, we have to look at both conjugate species, charge densities, stabilities, solute-solvent interactions and thermodynamics; not the focus of your question here. Suffice that students familiar with NH3 and NH4+ system are aware this is a weak base and its weak* conjugate acid, and extrapolate its similiarities to the HONH2 and HONH3+ system).

      (*Many students just memorize without understanding “weak base = strong conjugate acid” without understanding, which is unfortunate (that most Sg students just memorize without understanding and appreciating what they study). I make sure my tuition students understand why, and therefore the limitations of applying such ideas. For instance, understanding why Markonikov’s Rule usually works (ie. stability of carbocation intermediate species), will enable you to recognize and understand when it actually fails (ie. when the major product is actually the anti-Markonikov product). Back to the “weak base = strong conjugate acid” idea, it’s definitely true but to a relative degree, based on relative stabilities. But the fact is, while NH3 is a weak base, NH4+ certainly isn’t a strong acid. If you truly understand Chemistry, rather than blindly memorizing, this fact shouldn’t confuse you at all.)

      Titrating HONH3+ (again, you can ignore the Cl- counterbalancing spectator ion) against NaOH, a strong alkali, we end up with HONH2 (ie. deprotonated conjugate base form of HONH3+) at equivalence point (and spectator ions are Na+Cl-, ignore these harmless buggers), and HONH2 is obviously basic (due to an available lone pair on N; rather than the less available lone pair on O), as its 2 remaining protons are not at all acidic at this stage/form (since the conjugate base of HONH2, which is HONH-, is exceedingly unstable), and it is a lone pair available (on the N atom) to accept a proton from water.

      Hence at equivalence point, the species present HONH2, will undergo hydrolysis (ie. reaction with water), in a Bronsted-Lowry acid-base proton transfer reaction, proton transferred from water to HONH2, to generate HONH3+ and OH-.

      Therefore, your Initial Change Equilibrium (ICE) table will focus on to what extent (as specified by the Kb value), hydrolysis will occur to generate how much OH- ions; so you can calcuate molarity of OH- ions, and thus pOH, and therefore pH.

      As a general guideline for acid-base equilibria questions, do an ICF (Initial Change Final) table first, in number of moles; then do an ICE (Initial Change Equilibrium) table next, in molarities. Bear in mind when applying Ka or Kb expression, it is always in molarities of all species, and take care to factor in changing volumes at different stages (eg. 1st equivalence point, 2nd equivalence point, etc) when calculating molarities.

      Also note that the hydroxy group of HONH2 / HONH3+ is neither strongly acidic nor basic. Although there can be reaction pathways to forcefully protonate or deprotonate the hydroxy group (after being done with the amine group), but in water, no significant hydrolysis of the hydroxy group will occur. Treat this hydroxy group as you would an alcohol – neither strongly acidic or basic; inert as far as this titration question is concerned.

      Also, in case the exam question requires you to calculate initial pH, you need to be able to calculate the initial molarity of HONH3+.

      (a) In an experiment, 25.0cm3 sample of the hydroxylammonium chloride solution was titrated against a solution of 0.100M sodium hydroxide, and 14.40 cm3 of sodium hydroxide was needed for complete neutralisation. ~

      Find the number of moles of OH- ions required for complete neutralization.

      This is thus the number of moles of the weak monoprotic HONH3+ acid present.

      Using the formula Molarity = Moles / Volume (dm3), you can find the molarity of the species HONH3+ at initial.

      For equivalence point, the no. of moles of the species HONH2 is exactly the same as the no. of moles of HONH3+ at initial, but the volume will be different (since you added NaOH solution), hence the molarity of HONH2 at equivalence point is certainly different.

      Analogy (for a diprotic acid) to understand why no. of moles of the species is always the same, just not molarity :
      You begun with 10 teddy bears with 2 arms (representing protons) each. During titration neutralization, you’re chopping the arms off the teddy bears. At 1st equivalence point, there are still 10 teddy bears, only now they’re all one-armed. At 2nd equivalence point, there are still 10 teddy bears, only now they all have no arms left. Bear in mind while the number (of moles) of the teddy bears remain the same throughout, the molarity of the teddy bears keep changing, as the volume of the solution keeps increasing (as you keep adding the alkali).

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    • ‘O’ & ‘A’ Level Qn.

      What is the average distance between two air molecules in front of your face? (assume you’re at sea level and at room temperature.)

      Answer :
      3.42 X 10-7 cm
      or
      3.42×10-8 dm
      or
      3.42×10-9 m

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    • ‘O’ & ‘A’ Level Qn.

      a) Is zinc hydroxide soluble in water? How about in excess OH-(aq)?
      b) If we heat NaOH, do we get Na2O and water? How about Zn(OH)2?

      Solution :

      Zn(OH)2 is an insoluble ppt, but can form a soluble complex ion of tetrahydroxozincate(II) ion [Zn(OH-)4]2-(aq), in excess OH-(aq). The complex ion is soluble because it is capable of ion-dipole interactions with water, unlike Zn(OH)2(s).

      NaOH is stable (as Na+ has relatively low charge density), and will not lose water to form Na2O. When excess water is evaporated away, you will get white solid NaOH(s).

      On the other hand, Zn2+ has relatively higher charge density, is more unstable and hence Zn(OH)2 will lose water when heated to form ZnO(s).

      Zinc oxide is yellow when hot, white when cold; the yellow colour arises from interactions in the crystal lattice, and not due to d-d* transition, since Zn2+ has no partially filled d-orbitals and doesn’t qualify as a transition metal.

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    • ‘A’ Level Notes.

      Redox potentials refer to reduction potentials (if a reduction half-equation is written) or oxidation potentials (if an oxidation half-equation is written).

      As long as the equation is given as reversible (ie. equilibrium double half arrow notation used), it means you can reduce one species (eg. Fe3+) to a reduced state (eg. Fe2+), and you can also oxidize the reduced state species (eg Fe2+) to the oxidized state species (eg. Fe3+).

      Hence, for such reversible equations, you can reverse the half equations and the standard potentials.

      When written as a reduction half-equation, the redox potential is called the (standard) reduction potential (if standard conditions). When written as an oxidation half-equation, the redox potential is called the (standard) oxidation potential (if standard conditions).

      The positive/negative sign of the redox potential is reversed when re-writing a reduction half-equation as an oxidation half-equation, or vice-versa.

      Whichever way (reduction or oxidation) you choose to write it, the more positive the standard redox potential, the more the position of equilibrium lies to the right. The more negative the standard redox potential, the more the position lies to the left.

      You can Google/Wikipedia the technical physics definition of redox/reduction/oxidation potentials (which is not required for ‘A’ level Chemistry). Technical physics definition aside, it is helpful for you to think of the meaning of “potential” as “tendency” or “inclination” or “propensity”.

      Hence, the Reduction Potential numerical value (including positive/negative sign) describes the potential or tendency/inclination/propensity for a species to be reduced. The more positive the value, the more the position of equilibrium lies to the right (ie. the reactant species eg. Ag+ prefers to be reduced).

      (Note : due to formatting problems, the word equation is given instead)
      Eg. Silver cation + electron = (ie. left-to-right arrow) = Silver atom ( + 0.80V )
      ( + 0.80V means Silver cation likes to be reduced to Silver atom)
      which can be also be written as :
      Silver atom = (ie. left-to-right arrow) = Silver cation + electron ( – 0.80V )
      (Indeed, – 0.80V means Silver atom doesn’t like to be oxidized to Silver cation)

      Likewise, the Oxidation Potential numerical value (including positive/negative sign) describes the potential or tendency/inclination/propensity for a species to be oxidized. The more positive the value, the more the position of equilibrium lies to the right. (ie. the reactant species prefers to be oxidized).

      (Note : due to formatting problems, the word equation is given instead)
      Eg. Potassium atom = (ie. left-to-right arrow) = Potassium cation + electron ( + 2.92V )
      ( + 2.92V means Potassium atom loves to be oxidized to Potassium cation)
      which can be also be written as :
      Potassium cation + electron = (ie. left-to-right arrow) = Potassium atom ( – 2.92V )
      ( – 2.92V means Potassium cation hates to be reduced to Potassium atom)

      Standard cell potential (which must be positive to be feasible, ie. for the current of a Galvanic/Voltaic cell to flow) is calculated as :

      Oxidation potential at Anode + Reduction potential at Cathode

      or

      Reduction potential at Cathode – Reduction potential at Anode

      As long as standard cell potential is positive, the reaction is feasible, and the current in the Galvanic/Voltaic cell will flow.

      Edited by UltimaOnline 08 Sep `09, 11:49PM
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    • ‘A’ Level Qn.

      lianamaster asked :
      Q1) What relative amounts of 1-bromobutane and 2-bromobutane are actually formed when butane is brominated?
      Q2) If all C-H bonds had the same homolytic dissociation energy, what would be the relative amounts of 1-bromobutane and 2-bromobutane produced in the bromination of butane?

      My reply / the Solution :

      For Q1.

      The reactivity (radical intermediate stability) factor depends on temperature, but bromination is a lot more selective than chlorination.

      At 125 deg C, a tertiary radical is formed 1600x faster, and a secondary radical 82x faster, compared to primary radical.

      You needn’t memorize these values; the exam question will provide the relevant values if they require you to factor this in.

      Relative amount of 1-bromobutane :
      Number of H atoms substitutable x reactivity (radical stability)
      = 6×1 = 6

      Percentage yield = 6/(328+6) = 1.80%

      Relative amount of 2-bromobutane :
      Number of H atoms substitutable x reactivity (radical stability)
      = 4×82 = 328

      Percentage yield = 6/(328+6) = 98.2%

      For Q2.

      Assuming all C-H bonds had the same homolytic dissociation energy, in other words, ignoring the reactivity (radical stability) factor, simply count the number of hydrogens substitutable to obtain the two isomeric monobrominated products :

      Relative amount of 1-bromobutane :
      Number of H atoms substitutable
      = 6

      Percentage yield = 6/(6+4) = 60%

      Relative amount of 2-bromobutane :
      Number of H atoms substitutable
      = 4

      Percentage yield = 4/(6+4) = 40%

      youphoria queried :
      is q1 an a level qn? i have never seen it before. particularly regarding “reactivity (radical stability)” factor. only qn 2 seen before.

      My reply :

      It’s actually University level Organic Chemistry.

      At ‘A’ levels, the exam question will usually allow you to explain and apply only the simpler factor of number of hydrogens substitutable.

      If the exam question at ‘A’ levels requires you to factor in primary, secondary, tertiary radical stability (or equivalently, dissociation energies of primary, secondary, tertiary C-H bonds), the question will clue you in with relevant data (a brief explanation, together with relative rates of free radical substitution involving primary, secondary, tertiary alkyl radical intermediates, eg. 5.0 > 3.8 > 1.0 for chlorination at 298K).

      For the 2007 ‘A’ level H2 Chemistry exam, for instance, the question required the candidate to “describe, explain and calculate based on any ONE factor that determines isomeric product distribution of monochlorination of butane” (where the two factors are of course, number of hydrogens substitutable, and stability of alkyl radical intermediates).

      So don’t worry too much about this.

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    • ‘A’ & ‘O’ Level Qn.

      50cm3 of aqueous sulfuric(VI) acid of unknown molarity was mixed with 75cm3 of 0.45mol/dm3 of aqueous potassium hydroxide. 55cm3 of the resulting alkaline solution A was mixed with 80cm3 of 0.35mol/dm3 of phosphoric(V) acid. 90cm3 of the new resulting acidic solution B required 39.97cm3 of aqueous sodium carbonate of unknown molarity, for complete neutralization. In a separate experiment, 625cm3 of the same aqueous sulfuric(VI) acid solution required 192.3cm3 of the same aqueous sodium carbonate solution for complete neutralization.

      a) Calculate the molarities of the sulfuric(VI) acid solution and the sodium carbonate solution.

      b) Draw the full displayed structural formulae of sulfuric(VI) acid, potassium hydroxide, phosphoric(V) acid, sodium carbonate, and name all salts present in solution A and solution B.

      Answers :
      [H2SO4] = 0.2 mol/dm3
      [Na2CO3] = 0.65 mol/dm3

      Edited by UltimaOnline 24 Sep `09, 10:48PM
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    • ‘A’ & ‘O’ Level Qn.

      On heating, Group I metal nitrate(V) compounds decompose giving the metal nitrate(III) and oxygen gas, while Group II metal nitrate(V) compounds decompose giving the metal oxide, nitrogen dioxide and oxygen gas. 15.35g of a mixture of sodium nitrate(V) and magnesium nitrate(V) was heated in a fume cupboard until no more gases were evolved. The water soluble part of the residue was dissolved in 1dm3 of water to give solution A. 10cm3 of solution A was mixed with 20cm3 of 0.02mol/dm3 acidified potassium manganate(VII) solution, giving solution B. The excess potassium manganate(VII) in 15cm3 of solution B required 6cm3 of 0.05mol/dm3 of sodium ethandioate solution for complete reaction.

      a) Write all half-equations, balanced redox equations, and thermal decomposition equations relevant to the problem above.

      b) Calculate the sample masses of sodium nitrate(V) and magnesium nitrate(V) in the mixture.

      c) Draw the full displayed structural formulae of
      sodium nitrate(III), magnesium nitrate(V), potassium manganate(VII) and sodium ethandioate.

      Answers :
      Sample mass of NaNO3 = 3.4g
      Sample mass of Mg(NO3)2 = 11.95g

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    • ‘A’ Level Qn.

      By reference to the images provided on this Wikipedia URL ( http://en.wikipedia.org/wiki/Phenolthalein ), draw the mechanisms involved for the 4 structures of the phenolphthalein molecule ( IUPAC name 3,3-bis(4-hydroxyphenyl)isobenzofuran-1(3H)-one ) as the aqueous environment progresses with increasing pH from 0 to 12.

      http://en.wikipedia.org/wiki/Phenolthalein

      I’ll give some tips to help you along :
      (but you should attempt to solve it by yourself first).

      1 to 2
      deprotonation of carboxylic; lone pair on carboxylate attacks trigonal planar carbocation.

      2 to 3
      deprotonation of both phenolic groups; lone pair on phenoxide forms pi bond; pi electrons in ring shifts down through ring to form pi bond with sp3 carbon; sigma bond cleaves to become lone pair on carboxylate.

      3 to 4
      hydroxide nucleophile attacks sp2 carbon outside rings; pi bond outside ring shifts into ring to regenerate the phenoxide.

      Edited by UltimaOnline 25 Sep `09, 2:07PM
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    • ‘A’ Level Qn.

      limywv asked :
      “How to draw NO2 correctly? Guys, this question has been a major bugbear for me for quite some time now…I’ve searched the internet but no dot-and-cross diagrams were offered… can anyone help me on this?”

      Solution (plus my comments).

      A couple years ago, my tuition student was just about the only one in her cohort who could correctly draw the structure of NO2 in a major exam. My point is (not about boasting, but about) most JC teachers don’t focus (not blaming them though, they have it tough enough) on teaching the relevant, fundamental and crucial aspects of Chemistry that empower students to understand, appreciate and enjoy the Science as an Art. JC students most often fail to allow themselves to do the above (not blaming them though, they have it tough enough), and instead blindly memorize their way through JC.

      Specifically (for this question), I refer to the concept of formal charges and resonance (that is ill-taught in JCs), without an understand of these, drawing structures become senseless and hopeless.

      To answer your question :

      Central atom is N (with a +ve charge), single (and dative) bonded to an O atom (with a -ve charge), and double bonded to a (neutral) O atom. The N atom thus has 3 bond pairs and an unpaired electron, the (-ve charged) O atom has 3 lone pairs and 1 bond pair, the (neutral) O atom has 2 lone pairs and 2 bond pairs.

      The electron geometry is trigonal planar, the molecular geometry is v-shape or bent or non-linear.

      The resonance hybrid sees each of both N to O bonds as having bond lengths & bond strengths equivalent to 1.5 bonds. The charge on each O atom is an averaged -1/2 charge.

      For ‘A’ levels exams, it will suffice to draw a single resonance contributor.

      limywv replied with :
      “so it’s O=N—>O? with the Natom having 7 electrons while the 2 O atoms have full octet structures? So is this the reason why they can form diamers occassionally?”

      I replied :
      Yes, that’s correct on both counts (N atom has 7 electrons in terms of an octet, and 4 electrons in terms of formal charge). If the question tasks you to draw NO2, it might also require you to draw N2O4 in the same qn. The unpaired electron in one NO2 molecule combines with another unpaired electron in another NO2 molecule to form N2O4.

      Edited by UltimaOnline 16 Nov `09, 7:21PM
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    • ‘O’ & ‘A’ Level Qn.

      “Aqueous iodine” actually refers to the triiodide ion, as molecular iodine isn’t soluble in water.
      10cm3 of a solution of iodate(V) ions of unknown molarity, was mixed with excess iodide ions. Both iodate(V) ions and iodide ions reacted with each other (in a redox reaction) to generate molecular iodine. The iodine generated required 30cm3 of 0.1 mol/dm3 of thiosulfate ions for complete redox reaction.
      a) Write all half-equations and balanced redox equations relevant to the problem above.
      b) Calculate the molarity of the iodate(V) solution used.
      c) Explain why molecular iodine is insoluble, and why the triiodide ion is soluble, in water.
      d) Draw the full displayed structural formulae of the iodate(V) ion, the thiosulfate ion, the tetrathionate ion and the triiodide ion.
      e) State the ionic geometry of the triiodide ion as predicted by VSEPR model; and calculate the oxidation states for the individual sulfur atoms in the thiosulfate and tetrathionate ions.

      Solution :
      0.05 mol/dm3
      Permanent dipole – induced dipole is weaker than hydrogen bonding; ion-dipole interactions stronger than hydrogen bonding.
      Triiodide ion is linear; OS of sulfur atoms are +4 and 0 in thiosulfate, and +5, 0, 0, +5 in tetrathionate.

      Edited by UltimaOnline 04 Oct `09, 1:34PM
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    • ‘A’ Level Qn.

      Given that the Ka of ethanoic acid is 1.8×10^-5, calculate the pH at equivalence point in the titration of 0.1 mol/dm3 of ethanoic acid with 0.1 mol/dm3 of sodium hydroxide.

      Answer :
      pH = 8.72

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    • ‘O’ & ‘A’ Level Qn.

      A 0.1246g sample of a compound of Cr and Cl was dissolved in water. All of the chloride ions were then captured by silver ion in the form of AgCl ppt. A sample mass of 0.3383g of AgCl(s) was obtained. Determine the empirical formula of the compound of Cr and Cl.

      Answer :
      CrCl3

      Hints :
      Since the valency of chlorine when ionically bonded with metals is already known, let the cationic charge of Cr be x, ie. empirical formula be CrClx. Calculate moles of CrClx in terms of x. Calculate moles of Cl- present (precipitated as AgCl). Form an equation in x, solve for x.

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    • ‘A’ Level Qn.

      Suggest why calcium sulfate(VI) decomposes at a higher temperature than calcium carbonate.

      Solution :

      From the lower thermal decomposition temperature, we may deduce that the electron charge clouds on the carbonate anion is more easily polarized and distorted by the dipositive Ca2+ cation, as compared to the sulfate(VI) anion.
      (Further explanation, beyond the requirements of the H2 syllabus)
      The reason for the greater polarizability and distortability of the carbonate anion, in comparison with the sulfate(VI) anion, is due to the greater excess electron density on the oxygen atoms for the carbonate anion, over the sulfate(VI) anion. Notice that the resonance hybrid sees a -1/2 charge on each O atom for the sulfate(VI) anion (ie. dinegative anionic charge dispersed by resonance over 4 oxygen atoms), in contrast to a -2/3 charge on each O atom for the carbonate anion (ie. dinegative anionic charge dispersed by resonance over only 3 oxygen atoms).

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    • ‘A’ Level Qn.

      Explain the differences in the stereostructures of the complex ions formed for the cobalt(II) ion when complexed with water ligands, versus with chloride ligands.

      Solution :
      The hexahydroxocobalt(II) ion has an octahedral stereostructure, while the tetrachlorocobaltate(II) ion has a tetrahedral stereostructure. This is due to the steric considerations : the O atoms (of water ligands) are significantly smaller than the Cl- ion ligands; hence (six) water ligands can pack closer to the Co2+ ion (to give an octahedral stereostructure), while there is not enough space to pack six Cl- ion ligands to give a similar octahedral structure.

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    • ‘A’ Level Qn.

      The solubility products of some Group II fluorides, at rtp, are given below:

      Numerical values of Ksp
      BaF2 1.84×10–7
      CaF2 3.45×10–11

      A student accidentally mixed 25.0 cm3 of 0.100 mol dm–3 CaCl2 solution with 25.0 cm3 of 0.100 mol dm–3 BaCl2 solution in the laboratory. To separate the two metal ions, he added just enough solid KF to precipitate the maximum amount of CaF2 from the mixture, without precipitating BaF2.

      (i) Determine the [F–] required for this separation.
      (ii) Determine the concentration of Ca2+ remaining in the final solution.

      Answers :
      (i) 1.92×10–3 mol dm–3
      (ii) 9.38×10–6 mol dm–3

      Solution (pun intended) Hints :
      - Since equal volumes used, molarities are halved.
      - Based on Ksp BaF2, work out [F-]. The molarity of fluoride ions must be slightly LESS than this calculated molarity, so as to avoid precipitating BaF2.
      - Based on Ksp CaF2, work out [Ca2+]. This works out to be the molarity of aqueous Ca2+ ions, because for any given temperature which remains constant during the reaction, the product of the molarities of the aqueous cations and aqueous anions, each raised to the powers of their stoichiometric coefficients, MUST be equal to (or less than, but never more than) this ionic compound’s Solubility Product (ie. Ksp) at that given temperature. Any additional aqueous ions are precipitated out to allow this value to hold true.

      Edited by UltimaOnline 05 Nov `09, 11:46PM
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    • ‘A’ Level Qn.

      1. Solid NaCl and conc. H2SO4 can be used to generate HCl in situ for the reaction with alkenes to produce choloroalkanes.
      (a) Explain why 2-chloropropane is formed as the major product instead of 1-cholorpropane.
      (b) Will a high yield of 2-iodopropane be obtained by a similar method by replacing NaCl with NaI? explain your answer with appropriate equations.

      Solution :

      (a) As predicted by Markonikov’s Rule, the reaction pathway involving the more stable secondary carbocation intermediate resulting in the generation of 2-chloropropane, will be the major product. The reaction pathway involving the less stable primary carbocation intermediate resulting in the generation of 1-chloropropane, will be the minor product.

      (b) A redox reaction occurs between the iodide ions and sulfuric(VI) acid :

      H2SO4 + NaI —> NaHSO4 + HI
      8HI + H2SO4 -
      > 4I2 + H2S + 4H2O

      Hence, a lower yield of 2-iodopropane will result.

      (Notice that the O.S. of sulfur decreases from +6 to -2, due to the stronger reducing power of iodide ions. If bromide ions were used instead (eg. H2SO4 + NaBr), the O.S. of sulfur decreases only from +6 (H2SO4) to +4 (SO2). The difference in reducing powers of the various halide ions, indicated by their different redox potentials in the Data Booklet (quote these values in the exam!), is explained by the difference of the distance between the positively charged nucleus and the valence shell, for chloride vs bromide vs iodide.)

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    • ‘A’ & ‘O’ Level Qn.

      In the industrial electrolysis of brine (concentrated sodium chloride solution), a diaphragm cell is used. Write equations that occur at the cathode and anode. What are the ratio of useful products generated in this process? What is the purpose of the diaphram? Write equations for the undesirable reactions that would occur if the diaphram was absent.

      Solutions :
      Cl2, H2 and NaOH are produced in the molar ratio of 1:1:2.

      This industrial process is used to manufacture 3 different useful products of chlorine, hydrogen and sodium hydroxide.

      As to the purpose of the diaphragm :
      “In diaphragm cell electrolysis, an asbestos (or polymer-fiber) diaphragm separates a cathode and an anode, preventing the chlorine forming at the anode from re-mixing with the sodium hydroxide and the hydrogen formed at the cathode.” – http://en.wikipedia.org/wiki/Chlorine_production

      Hydrogen reacts with chlorine to form hydrogen chloride gas (which undergoes hydrolysis to form hydrochloric acid). Chlorine reacts with sodium hydroxide to form sodium chlorate(I) (at 15 deg C) and sodium chlorate(V) (at 70 deg C), and additionally reforms sodium chloride and water.

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    • 2009 ‘A’ Level Exam Qn.

      When methylbenzene is reacted with Cl2 and AlCl3, a monochloro compound K is formed. Treatment of K with more Cl2 in the presence of light produces compound L. When L is heated with NaCN in ethanol, compound M, C8H6ClN is formed. M can be converted into an acidic compound N by heating under reflux with dilute H2SO4. Heating L with NaOH(aq) produces compound P, C7H7ClO. When a mixture of N and P is heated with a small amount of concentrated H2SO4, compound Q, C15H12Cl2O2, is produced.

      Elucidate K to Q. State the type of each reaction described above.

      Solution :
      For questions such as these, where it is not clearly specified what they mean by “type of reaction”, students are advised to give both the type of reaction in terms of the mechanism, and in terms of the end product. You will never be penalized for giving both, unless one of them is wrong and/or contradicts the other.

      To produce K, the reaction is : electrophilic aromatic substitution resulting in chlorination of benzene.
      To produce L, the reaction is : free radical substitution resulting in chlorination of alkyl side chain.
      To produce M, the reaction is : SN2 nucleophilic substitution (cyanide ion as nucleophile) resulting in the formation of a cyano compound.
      To produce N, the reaction is : repeated nucleophilic substitions during hydrolysis (water as nucleophile) to generate a carboxylic acid.
      To produce P, the reaction is : SN2 nucleophilic substitution (hydroxide ion as nucleophile) resulting in the formation of a primary alcohol.
      To produce Q, the reaction is : addition-elimination (not nucleophilic substitution) resulting in the removal of water (ie. condensation reaction) and the formation of an ester (ie. esterification reaction).

      Edited by UltimaOnline 15 Nov `09, 1:08PM
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    • 2009 ‘A’ Level Exam Qn.

      Methylbenzene requires concentrated sulfuric(VI) and nitric(V) acids and 30 degrees celsius to undergo nitration whereas phenol only requires dilute nitric acid. Give a reason for the difference in conditions and reagents.

      Solution :

      Phenolic hydroxy group is electron donating by resonance and is hence a strong activator, enriching the pi electron density of the benzene ring, making it a much stronger nucleophile (than ubsubstituted benzene). Hence only dilute nitric(V) acid is required for electrophilic aromatic substitution and nitration to occur.

      Methyl group is electron donating by induction (and hyperconjugation) and is hence a weak activator, only slightly enriching the pi electron density of the benzene ring, making it only a slightly stronger nucleophile (than ubsubstituted benzene). Hence a nitrating mixture of concentrated nitric(V) and sulfuric(VI) acids are required to generate the nitronium cation (NO2+) for electrophilic aromatic substitution and nitration to occur.

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    • 'A' Level Qn.

       

      M has formula C13H16O. it decolourises aq bromine and produce white fumes with PCl5. it reacts with hot KMnO4 to produce N (C8H8 O), P (C3H2O5) and Q (C2H4O2). N does not react with fehling but react with aq alkaline iodine to give yellow ppt + salt R (C7H5O2Na). P gives effervescence with solid sodium carbonate and gives orange ppt with dinitrophenylhydrazine. Q gives effervescence with solid sodium carbonate but no ppt with dinitrophenylhydrazine. identify M N P Q R.

       

       

      Answer :

      N - CH3 CO C6H5

      P - HOOC CO COOH

      Q - CH3 COOH

      M - C6H5 C (CH3) = C (CH2OH) CH = CH CH3

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    • 'A' Level Qns.

       

       

      Qn1.

      When a mixture of the sodium salts of two mono-carboxylic acids is electrolyzed, a mixture of alkanes is produced.

       

      RCOO-    +    R’COO-    à    R-R    +    R-R’    +    R’-R’                 (note : the equation is not balanced)

       

      Electrolyzing a mixture of the sodium salts of two mono-carboxylic acids M and N produced three alkanes O, P and Q, which could be separated by fractional distillation.

       

      On titration, a solution containing 0.10g of the acid N required 16.7cm3 of 0.10mol/dm3 NaOH(aq) to reach equivalence point.

       

      A gaseous sample of 0.10g of O took up a volume of 54cm3 at a temperature of 380K and a pressure of 1 atm.

       

      Complete combustion of a sample of P, yields 10.12g of CO2 and 6.21g of H2O.

       

      When Q was reacted with bromine under UV light, only two isomeric monobromo compounds R and S were produced. Compound R rotates plane polarized light.

       

      Elucidate the structures of  M, N, O, P, Q and S.  Explain your reasoning. Draw a “mirror image” diagram to show the two stereoisomeric forms of R.   

       

       

       

      Qn2.

      Compounds T and U are isomers. Both are gaseous hydrocarbons with slightly different boiling points.

      In an experiment, 15cm3 of T was combusted with 120cm3 of oxygen gas at a temperature of 150°C. The residual gases that remained at the end of the combustion process, when cooled to room temperature, experienced a reduction of 60cm3 in gas volume. When bubbled through aqueous sodium hydroxide, only 30cm3 of residual gas remained.

       

      When U was reacted with chlorine under UV light, exactly three structurally isomeric monochloro compounds V, W and X were produced. Compound V exists in two enantiomeric stereoisomeric forms, V1 and V2.

       

      Based on number of H atoms substitutable, X would be generated in greater quantities compared to W. Based on stability of alkyl radical intermediates, W would be generated in greater quantities compared to X.

       

      Upon undergoing the same reaction, T produced only one monochloro product Y1. Upon further exposure to chlorine and UV light, T can be dichlorinated to produce Y2, which exists as 3 possible structural isomeric forms.

       

      Elucidate T, U, V1, V2, W, X, Y1, and the 3 isomeric forms of Y2. Explain your reasoning.

       

      Name the two enantiomeric forms of V using IUPAC nomenclature.

       

       

       

       

      .

      .

      .

      .

      .

      .

      Answers :

       

      Qn1

      M is CH3CH2CH2COOH

      N is CH3COOH

      O is CH3CH2CH2CH3

      P is CH3CH3

      Q is 2,3-dimethylbutane

      S is 2-bromo-2,3-dimethylbutane

      R is 1-bromo-2,3-dimethylbutane, which exists in two enantiomeric isomeric forms :

      Based on dextrorotatory & levorotatory classification;

      (+)-1-bromo-2,3-dimethylbutane (or d-1-bromo-2,3-dimethylbutane)

      and (-)-1-bromo-2,3-dimethylbutane (or l-1-bromo-2,3-dimethylbutane)

      Based on rectus & sinister classification; R-1-bromo-2,3-dimethylbutane and S-1-bromo-2,3-dimethylbutane.

       

      Qn2

      T is cyclobutane

      U is 1-methylcyclopropane.

      W is 1-chloro-1-methylcyclopropane

      X is 1-chloromethylcyclopropane

      Y1 is 1-chlorocyclobutane

      V is 1-chloro-2-methylcyclopropane, which exists in two enantiomeric isomeric forms :

      Based on dextrorotatory & levorotatory classification;

      (+)-1-chloro-2-methylcyclopropane (or d-1-chloro-2-methylcyclopropane)

      and (-)-1-chloro-2-methylcyclopropane (or l-1-chloro-2-methylcyclopropane)

      Based on rectus & sinister classification; R-1-chloro-2-methylcyclopropane and S1-chloro-2-methylcyclopropane.

      The 3 isomeric forms of Y2 are 1,1-dichlorocyclobutane, 1,2-dichlorocyclobutane and 1,3-dichlorocyclobutane.

      Note that overall, W would far outweigh X in terms of quantity generated, as a tertiary alkyl radical intermediate is very much more stable than a primary alkyl radical intermediate. (Furthermore, the ratio of H atoms substitutable for X and W is a relatively low ratio of 3 to 1, as opposed to, for instance, 9 to 1.)

       

      Edited by UltimaOnline 09 Jan `10, 12:23AM
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    • 'O' level / 'A' level Qn

       

       

      Someone posted :

      "NH3 + H2SO4 --> ??
      NH3 is an alkali right? Alkali + Acid --> Salt + Water. Online sources show the ammonia being absorbed by the acid to form [NH4]2 SO4, but I don't really understand why.
      "

       

      I replied :

       

      Don't blindly memorize without understanding why acid + base = salt + water. (Well, if you're 'O' levels, most of it is blind memorization without understanding, but if you're 'O' levels and you've exceptional interest and ability, or if you're 'A' levels H2/H1/H3 / IB Diploma / Poly Chemical Sciences Diploma, then you should do your best to understand rather than blindly memorize).

      Wikipedia out the 3 definitions of an acid and a base : Arrhenius, Bronsted-Lowry, Lewis.

      For most purposes in acid-base chemistry, we use the Bronsted-Lowry definition. Ammonia is a proton acceptor. Sulfuric(VI) acid is a proton donor. (Where protons = H+ ions)

      The reason why acids + alkalis generate a salt and water, is because the protons (Lewis acid) from the acid combine with the hydroxide ions (Lewis base) from the alkali, to form water. What's left over (the spectator ions from acid and alkali), is called the salt. In terms of the mechanism, we say the hydroxide ions donate a lone pair to form a dative sigma bond with the proton.

      Ammonia is a weak base, and therefore it is a weak alkali. A base (if you use the Bronsted definition of a base) is a proton acceptor. When it accepts protons from water, the conjugate acid of the protonated base is the NH4+ ammonium cation. The conjugate base of water is the hydroxide ion.

      NH3 + H2O <---> NH4+ + OH-

      However, because the hydroxide ion is a significantly stronger base than the ammonia molecule, the position of equilibrium lies strongly to the left. Which means to say that in a bottle labelled as "ammonium hydroxide NH4OH(aq)", the majority of the species present is really NH3(aq) rather than NH4OH(aq), and hence the bottle is more properly labelled "aqueous ammonia NH3(aq)".

      If you wish, you can still write the standard acid + base = salt + water equation as follows :
      2NH4OH + H2SO4 --> 2H2O + (NH4)2 SO4.

      However, as explained in my preceding paragraphs, technically it is more correct (ie. relevant) to write the equation as :
      2NH3 + H2SO4 --> (NH4)2 SO4
      (assuming experimental molar ratios allow for this.)

      There are many alternative equations that are also relevant and/or acceptable, such as those involving hydrolysis of the base to generate hydroxide ions (ie. NH3 + H2O --> NH4+ + OH-), hydrolysis or proton dissociation of the acid to generate hydroxonium ions H3O+ (ie. H2SO4 + H2O --> H3O+ + HSO4-), and the 'main' equation itself could be H3O+ reacting with the OH-, or H3O+ reacting with the NH3, or H2SO4 reacting with the OH-, or H2SO4 reacting with NH3, etc etc.

      Acid-base chemistry can certainly be fascinating, interesting and fun to work with. But depending on your current level of study, eg. if you're 'O' levels now, it may be more practical to mentally accept a fair bit of blind memorization first, and look forward to the deeper and more correct understanding at higher levels, at 'A' levels H2/H1/H3, IB Diploma, Chemical Sciences Poly Diploma, University levels (you'll study varying aspects/levels/areas of Chemistry in almost all Science-based degree courses, eg. Medicine, Engineering, Biology, etc) at various levels, Bachelors Honors Masters Doctorate, etc.


      Enjoy your academic life ahead!

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    • 'O' & 'A' Level Qn

       

      It has been found that the chemical BPA (bisphenol A) commonly found in many plastics (eg. drinking bottles, food packaging) causes heart disease, liver disease and diabetes.

       

      Google these keywords "chemical in plastic BPA heart disease diabetes" for more information. 

       

      For 'A' level students, given that the IUPAC name is 4,4'-dihydroxy-2,2-diphenylpropane (it might help to consider an alternative name 2,2-para,para'-diphenolpropane), draw the compound, then check your answer here :

      http://en.wikipedia.org/wiki/File:Bisphenol_A_skeletal.png

       

      For 'O' & 'A' level students, by referring to the structure given above, count :

      a) number of valence electrons in one molecule of BPA

      b) number of (valence) non-bonding electrons in one molecule of BPA

      c) number of (valence) bonding electrons in one molecule of BPA

      d) number of non-valence electrons in one molecule of BPA

      e) total number of electrons in one molecule of BPA

       

       

      Answers :

      Number of electrons (highlight with mouse)

      a) 88

      b) 8

      c) 80

      d) 34

      e) 122

       

      Edited by UltimaOnline 16 Jan `10, 12:48PM
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    • 'A' Level Qn.


      Explain the difference in rates of (ie. susceptibility towards) hydrolysis of silicon tetrachloride, SiCl4, versus carbon tetrachloride (aka tetrachloromethane), CCl4.

      -----------------------------

      "longchamp3" wrote :
      hey, i dont understand the hydrolysis of silicon tetrachloride and its acidity. Is hydrolysis the breaking up of hte H-OH bond in water? and it states that hydrolysis is bossible because of the energetically accessible vacant 3d orbitals available for dative bonding with water molecules. i know that the O atom will have a dative bond with Si, but isnt the O atom supposed to be with the H atom in the OH- anion? and does that mean that the Cl atoms leave when there is dative bonding with the O atom? gosh im confused =(

      "kineticenergy" wrote :
      In the case for SiCl4, Si is able to accommodate lone pairs of e-, due to energetically accessible vacant 3d orbitals. This means that in water, lone e- pair in O atom of H2O will fit into the empty 3d orbitals of Si, due to attraction between the nucleus of Si and e- pair of O atom. The attraction is so great that one of O-H bond is weakened till it breaks, forming H+ and OH-, with the OH- bonded to Si causing one Cl- to leave. This process is repeated, eventually forming SiO2. The H+ that has left will form H3O+ with a H2O molecule, which results in the solution being acidic.

      "longchamp3" wrote :
      When u say that the H+ will form H30+ with other H2O, resulting in an acidic solution, u mean to say that H3O+ is acidic? And i thought the equation for hydrolysis for SiCl4 is SiCl4 + 2H2O -> SiO2 + 4HCl? so what is the H attached to the O in the OH- still doing down there?

      I wrote :
      HCl(aq) can equivalently be written as H+(aq) and Cl-(aq) which can equivalently be written as H3O+(aq) + Cl-(aq).

      So do you see that it's ok to write either HCl(aq) on the RHS, or H3O+(aq) and Cl-(aq) on the RHS? Both are correct and acceptable.

      Just to add, there is another (secondary) reason that CCl4 is resistant to hydrolysis while SiCl4 is susceptible to hydrolysis, other than the existence of energetically accessible vacant 3d orbitals available in Si (in contrast to C) TO ACCEPT dative covalent bonding FROM nucleophilic water molecules during hydrolysis (HYDROLYSIS is definted as the REACTION of any particular species WITH WATER), thereby facilitating or increasing the rate of hydrolysis reaction by allowing for a more stable intermediate.

      This (secondary) reason has to do with the relatively better shielding effect (ie. shielding against the water nucleophile) of the partially negatively charged Cl atoms (due to electronegativity difference) of CCl4, compared with SiCl4, due to the shorter bond lengths of C-Cl versus Si-Cl.

      Note that for 'A' levels, just the (primary) reason of having energetically accessible vacant 3d orbitals available in Si (in contrast to C) to accept dative covalent bonding from nucleophilic water molecules during hydrolysis, will suffice for the H2 syllabus.

      Edited by UltimaOnline 16 Jan `10, 1:15PM
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    • 'A' Level Qn (with some aspects of the discussion going beyond the requirements of the H2 syllabus)

       

      Originally posted by TenSaru:

      If have a question too! (sorry for invading your tread)

      If protic solvents form hydrogen bonds with nucleophiles and slow down Sn2 reaction tremendously then why is it that most of the nucleophilic substitution (of halogenoalkanes in h2 chem) reagents and conditions involve ethanol as solvent?

       

      First, some additional info for H2 Chem students, to serve as a background for this discussion.


      Solvents may be classified as protic polar, aprotic polar and aprotic non-polar (for obvious reasons, 'protic non-polar' solvents do not exist).

       

      If a reactant in the rate-determining step is charged (eg. OH- ion in SN2), increasing the polarity of the solvent will decrease the rate of the reaction. If none of the reactants in the rate-determining step are charged (eg. alkyl halide in SN1), increasing the polarity of the solvent will increase the rate of reaction.

       

      This occurs because :

       

      For charged reactants (in the rate determining step, eg. SN2), a polar solvent will stabilize the (charged) reactants more extensively than it will stabilize the transition state (which is relatively less charged; as charge is spread out over more atoms), increasing the activation energy and decreasing the rate of reaction.

       

      For non-charged reactants (in the rate determining step, eg. SN1), a polar solvent will stabilize the (charged) transition state more extensively than it will stabilize the (uncharged) reactants, decreasing the activation energy and increasing the rate of reaction.

       

      For SN2 reactions, the nucleophile is presumably charged (otherwise, SN1 would more likely predominate). Consequently, a protic polar solvent stabilizes the nucleophile more extensively (by way of strong hydrogen bonds) and hence decreases the rate of reaction for reasons discussed above.

       

      Since using protic polar solvents slow down SN2 reactions, shall we use (aprotic) non-polar solvents (eg. CCl4) instead? The problem with using non-polar solvents, is that the negatively charged nucleophile (eg. OH- ion) will not dissolve in non-polar solvents, nor be miscible with the (relatively) non-polar alkyl halide substrate, as they would rather form ionic (electrostatic) interactions with each other instead (eg. Na+ and OH-), hence the reaction is unable to proceed.

       

      So do we have no better solution (pun intended) to this dilemma? Actually, we do. The solvent to dissolve (pun intended again) this difficulty, lies in a middle-man so to speak. Something polar enough to dissolve the charged nucleophile, but not solvate it so extensively (via hydrogen bonds) that it slows down the rate of the SN2 reaction.

       

      In short, we need an aprotic polar solvent, such as dimethyl sulfoxide (DMSO). (See Wikipedia for a list of aprotic polar solvents). Thus the rate of an SN2 reaction with a negatively charged nucleophile will be greater in an aprotic polar solvent than in a protic polar solvent.

       

      Consequently, an aprotic polar solvent is the solvent of choice for an SN2 reaction in which the nucleophile is negatively charged, while a protic polar solvent is used if the nucleophile is a neutral molecule. For SN1 reactions in which the uncharged alkyl halide is the only reactant in the rate determining step, a protic polar solvent is most effective in increasing the rate of the SN1 reaction.

       

      Now, to address TenSaru's question on why despite slowing down the SN2 reaction, ethanol is often used as a solvent in the H2 Chem syllabus.

       

      First of all, in regards to the H2 Chem syllabus, I'll like to point out that for the hydrolysis (ie. reaction with water) of alkyl halides (a.k.a. halogenoalkanes) to form alcohols, which is a nucleophilic substitution reaction (it could be SN2 or SN1, depending on whether the akyl halide is primary, secondary or tertiary; ie. tertiary alkyl halide presents greatest steric hinderance for nucleophile in SN2 *and* greatest stability of carbocation intermediate of SN1; conversely primary alkyl halide presents least steric hinderance for nucleophile in SN2 *and* least stability of carbocation intermediate of SN1), the H2 Chem student is expected to use aqueous solvent.

       

      (The reason why it is called 'hydrolysis' is because water may function as the nucleophile here; after the nucleophilic attack, the loss of a proton to rid the positive formal charge on the O atom consequently yields an alcohol. However, to speed up the rate of reaction, hydroxide ions are introduced in the form of an alkali. Notice that water is the protonated conjugate acid of the hydroxide ion, and hence the initial hydrolysis product is the protonated conjugate acid of the alcohol generated; subsequent loss of a proton generates the same alcohol product that the hydroxide ion nucleophile would directly generate.)

       

      In contrast, for elimination or dehydrohalogenation of alkyl halides to form alkenes, the H2 Chem student is expected to use alcoholic solvent (eg. ethanol).

       

      The reason for this has to do with the behaviour of the hydroxide ion. Although both water and alcohols are protic polar solvents, but water affords more extensive ion-dipole and hydrogen bond interactions for the hydroxide ion, compared to an alcohol solvent (due to the one less partially positively charged H atom available, and the presence of the non-polar alkyl group of the alkanol/alcohol).

       

      Consequently, the hydroxide ion is (relatively) more unstable in alcoholic solvent than in aqueous solvent. The more unstable a negatively charged species is, the greater its propensity to behave as a base. (Since a base accepts protons in a bid to stabilize itself).

       

      In summary, this is the reason why alcoholic solvent is used for elimination reactions of alkyl halides, while aqueous solvent is used for hydrolysis reaction of alkyl halides (as far as the H2 Chem syllabus is concerned).

       

      (Unfortunately, JCs do not teach H2 Chem students what I've just explained, which means JC students resort to blind memorization without understanding. I make it a point to share with my tuition students these and other relevant explanations, so that they can understand more clearly and consequently can appreciate Chemistry more deeply, than they would in their JC environment where blind memorization is the unfortunate norm.)

       

      But back to TenSaru's question, why use aqueous or alcoholic solvents for SN2 reactions, since water and alcohols are protic polar solvents and will slow down SN2 reactions? Wouldn't an aprotic polar solvent work better?

       

      Several points.

       

      1. The H2 Chem syllabus does not explore the role of protic polar, aprotic polar, and aprotic non-polar solvents, and how they affect SN2 and SN1 reactions. Hence, the H2 Chem student has only two choices - aqueous or alcoholic.

       

      2. Hydrolysis (nucleophilic substitution reactions with water or hydroxide ions as the nucleophile) of alkyl halides, is not differentiated into SN2 vs SN1 in the old (ie. before 2010) H2 Chem syllabus, and therefore generally employ water (a protic polar solvent) as the solvent of choice, ie. aqueous conditions. Even though this works better for SN1 than for SN2.

       

      Even though the new H2 Chem syllabus starting 2010 does require the H2 Chem student to draw both the SN2 and SN1 nucleophilic substitution mechanisms, but the only factor that influences SN2 vs SN1 that they need to know would be tertiary vs secondary vs primary alkyl halide (ie. tertiary alkyl halide presents greatest steric hinderance for nucleophile in SN2 *and* greatest stability of carbocation intermediate of SN1; conversely primary alkyl halide presents least steric hinderance for nucleophile in SN2 *and* least stability of carbocation intermediate of SN1), and not the role of the protic polar, aprotic polar and aprotic non-polar solvent. Hence water is still employed as the solvent of choice for hydrolysis reactions, while alcohols are employed as the solvent of choice for elimination reactions and nucleophilic substitutions involving non-polar reactants (my next point).

       

      3. In the H2 Chem syllabus, for the formation of nitriles, the H2 Chem student is required to heat or reflux alcoholic KCN with alkyl halides, to generate nitriles in a nucleophilic substitution reaction (whether by SN2 or by SN1). There is yet another reason for alcohol as the choice of solvent, rather than water (even though both are protic polar solvents; as stated earlier, the H2 Chem syllabus is not concerned over this aspect).

       

      The reason being :

      The alkyl halide electrophile (eg. bromoethane) is largely non-polar. The nucleophile is ionic (eg. K+ and CN-). If water is used, yes the nucleophile reactant is miscible with water, as ion-dipole interactions are established. But the electrophile (the alkyl halide) reactant, being (largely) non-polar (and hence incapable of either ion-dipole, hydrogen bonding or even the weaker permanent dipole-dipole interactions to any significant extent), will not dissolve well in water. Hence, the reaction is unable to proceed.

       

      But if alcohol solvent is utilized (in spite of it being a protic polar solvent; as discussed earlier, whether by SN2 (unfavourable) or SN1 (favourable), the role of protic polar vs aprotic polar vs aprotic non-polar solvents are ignored by the H2 Chem syllabus), in place of aqueous solvent, then because the alcohol molecule consists of both a polar (hydroxy) group and a non-polar (alkyl) group, it is able to effective function as a middle-man, someone that is able to form favourable interactions with both reactants.

       

      As an analogy, the ionic nucleophile (eg. CN- and its counter ion K+) are the ruling elite snobbish aristocratic class, while the non-polar alkyl halide (eg. bromoethane) are the lower working class peasants or serfdom. The elite snobs woud rather interact with each other than with the lower class (ionic bonding is incomparably stronger than ion - induced dipole interactions). So the reaction (which requires both reactants to meet and react with each other) is unable to proceed to any appreciable extent. However, if you've a bourgeoisie middle class that could get along with both the elite snobs and the lower class, you might actually have a chance to get them to meet and the reaction to proceed.

       

      Enter the bourgeoisie alcohol. Part elite polar (hydroxy group), part working class non-polar (alkyl chain). The hydroxy group is capable of ion-dipole and/or hydrogen bonding with the nucleophile (eg. CN- ion), and the alkyl chain is capable of instantaneous dipole - induced dipole van der Waals interaction with the electrophile (eg. bromoethane).

       

      Hence, utilizing an alcoholic solvent is a cunning manipulative trick, to get the two otherwise immiscible reactants (ionic CN- ion, and non-polar alkyl halide) to interact, and the reaction to proceed.

       

      (Again, tragically, JCs do not explain these underlying rationales to the H2 Chem students. Which only serves to perpetuate the myth that Organic Chem is all tough memory work. Nonsense. My tuition students have the privilege of understanding these underlying rationales, and hence have the opportunity to enjoy Organic Chem without having to suffer torturous blind memorizations.)

       

      So finally again, to summarily address TenSaru's question :

      >>> If protic solvents form hydrogen bonds with nucleophiles and slow down Sn2 reaction tremendously then why is it that most of the nucleophilic substitution (of halogenoalkanes in h2 chem) reagents and conditions involve ethanol as solvent? <<<

       

      Because :

       

      1) Ethanol is useful in allowing both reactants (usually one non-polar and one ionic) to be more miscible, and hence for the reaction to proceed. Between the two protic polar solvents of water and alcohol, the alcohol is the superior choice in this regard; even though (see next point)

       

      2) Indeed, an aprotic polar solvent (eg. DMSO) would be better for SN2 reactions, but the H2 Chem syllabus does not concern itself with the role of protic vs aprotic polar solvents. Nor as much with SN1 vs SN2 pathways. So ethanol is fine for the H2 syllabus (even though DMSO would be better for SN2).

       

      3) Just a reminder that in the H2 syllabus, water or aqueous solvent is used for hydrolysis (to generate alcohols), while alcoholic solvents are used for other nucleophilic substitution reactions (eg. to generate nitriles), and for elimination reactions (to generate alkenes).

      Edited by UltimaOnline 01 Feb `10, 2:09AM
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