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

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    • 'A' Level Qn.

       

      Competition between Substitution and Elimination (or SN1 vs E1 vs SN2 vs E2)

       

      (Note : some of the following discussion is additional to, and not strictly required for, the H2 syllabus; read for your own understanding and not blind memorization.)


      There are two approaches you may use to determine the likely products of a reaction between an alkyl halide and a nucleophile/base (ie. SN1 vs E1 vs SN2 vs E2).
      The first approach initially looks at the strength of a nucleophile/base, while the second approach looks initially at the primary/secondary/tertiary nature of an alkyl halide. Before we look at both approaches, revise your understanding of nucleophilicity versus basicity.

      Note that while many strong nucleophiles are strong bases (and vice-versa), and many weak nucleophiles are weak bases (and vice-versa); there also exist strong nucleophiles which are weak bases (and vice-versa), and strong bases which are weak nucleophiles (and vice-versa).

       

      Read David Klein's "Organic Chemistry" page 234 for nucleophilicity versus basicity.

       

       

      Approach #1  (preferred by some chemists eg. Paula Yurkanis Bruice)


      Factors which determine SN2 / E2 versus SN1 / E1

      A (high molarity of a) strong nucleophile/base promotes SN2 / E2.
      A weak nucleophile/base promotes SN1 / E1.


      (Assuming strong nucleophile/base) Factors which determine SN2 vs E2

      A primary alkyl halide will see both SN2 and E2 ocurring.

      SN1 and E1 does not occur, since a primary carbocation is too unstable to form.
      With nucleophlies/bases that are more strongly nucleophilic (highly polarizable electron density) than basic, SN2 predominates.
      With nucleophiles/bases that are more strongly basic than nucleophilic, E2 predominates. The more unstable the species, the stronger it is as a base (since bases accept protons in a bid to stabilize themselves). Also, bulky or branched nucleophiles/bases are sterically hindered in their approach to the alkyl halide electrophile, and the only thing left they can do is to act as a base and abstract ('pluck off') protons.
      Similarly, if the primary alkyl halide itself is also sterically hindered, E2 predominates. (eg. if the alkyl halide, even though primary, has bulky branching chains a few atoms away that hinder the approach of the nucleophile.)

      A secondary alkyl halide will experience both SN2 and E2, but the main factor that influences which way it goes, is the base/nucleophile. The more unstable (hence a stronger base) and/or bulkier the base/nucleophile used, the greater the propensity towards E2 (ie. acts as a base rather than nucleophile).

      A tertiary alkyl halide will experience only E2, since steric hinderance prevents the approach of the nucleophile via SN2.


      (Assuming weak nucleophile/base) Factors which determine SN1 vs E1

      A primary alkyl halide cannot undergo SN1 or E1, since a primary carbocation is too unstable to form.

      For both secondary alkyl halides and tertiary alkyl halides, a mixture of SN1 and E1 will both occur.
      This is because the rate determining step for both for SN1 and E1 is the same - the formation of the carbocation.

      In addition, based on Gibbs Free Energy, higher temperatures always favour elimination over substitution, since in the Gibbs Free Energy formula, temperature 'T' is married to entropy 'S', and elimination (which has more products) increases entropy more than substitution.

       

      Approach #2 (preferred by some chemists eg. Janice Gorzynski Smith)


      For tertiary alkyl halides.

      With strong bases/nucleophiles, E2 predominates (since a tertiary alkyl halide presents steric hinderance for the approaching nucleophile in SN2) .

      With weak bases/nucleophiles, a mixture of SN1 and E1 will both occur. This is because the rate determining step for both for SN1 and E1 is the same - the formation of the carbocation.


      For secondary alkyl halides.

      With strong bases/nucleophiles, both SN2 and E2 will occur, but the main factor that influences which way it goes, is the base/nucleophile. The more unstable (hence a stronger base) and/or bulkier the base/nucleophile used, the greater the propensity towards E2 (ie. acts as a base rather than nucleophile).

      With weak bases/nucleophiles, a mixture of SN1 and E1 will both occur. This is because the rate determining step for both for SN1 and E1 is the same - the formation of the carbocation.


      For primary alkyl halides.

      SN1 and E1 does not occur, since a primary carbocation is too unstable to form.
      With nucleophlies/bases that are more strongly nucleophilic (highly polarizable electron density) than basic, SN2 predominates.
      With nucleophiles/bases that are more strongly basic than nucleophilic, E2 predominates. The more unstable the species, the stronger it is as a base (since bases accept protons in a bid to stabilize themselves). Also, bulky or branched nucleophiles/bases are sterically hindered in their approach to the alkyl halide electrophile, and the only thing left they can do is to act as a base and abstract ('pluck off') protons.
      Similarly, if the primary alkyl halide itself is also sterically hindered, E2 predominates. (eg. if the alkyl halide, even though primary, has bulky branching chains a few atoms away that hinder the approach of the nucleophile.)


      In addition, based on Gibbs Free Energy, higher temperatures always favour elimination over substitution, since in the Gibbs Free Energy formula, temperature 'T' is married to entropy 'S', and elimination (which has more products) increases entropy more than substitution.

       

      [Note that for the H2 Chem syllabus, which greatly simplifies all the matters discussion above; the H2 Chem student is expected to control the nucleophilic/basic behaviour of the hydroxide ion OH-, by choice of solvent (ie. aqueous vs alcoholic). See previous discussion on effects of solvent.]

       


      References :
      David Klein's Organic Chemistry
      Paula Yurkanis Bruice's Organic Chemistry
      Janice Gorzynski Smith's Organic Chemistry

    • 'O' & 'A' Level Qns.

       

      Qn 1.
      When 1.01g sample of potassium nitrate(V) is heated above its melting point, oxygen gas is evolved. The mass of the sample decreases by 0.16g. Construct a balanced equation for this decomposition reaction, name the residue and (for 'A' level students) draw its structural formula.

       

      Q2.

      Sulfur and chlorine can react together to form disulfur dichloride. When 1.00g of disulfur dichloride was reacted with water, 0.36g of a yellow precipitate of sulfur was formed together with a solution containing a mixture of sulfuric(IV) acid (a.k.a. sulfurous acid), and hydrochloric acid. Given that the most common isotope of sulfur is S8, use the above data to deduce the balanced equation for the reaction between disulfur dichloride and water. (For 'A' level students) draw the structural formulae of disulfur dichloride and sulfuric(IV) acid.

       

       

       

      Solutions (partial).

       

      Qn1. Since 1 mole of potassium nitrate(V) decomposes to give 0.5 mole of oxygen gas, the residue is potassium nitrate(III) {stock name} or potassium nitrite {latin name}.

      For the nitrate(III) ion (aka nitrite ion), uninegative formal charge on a singly bonded O atom; central N atom has 1 lone pair, 3 bond pairs; electron geometry tetrahedral, ionic geometry trigonal pyramidal.

       

      Qn2. Since 1 mole of disulfur dichloride undergoes hydrolysis to give 1.5 moles of S(s), or 1.5/8 moles of S8(s), the coefficients of the balanced equation are : 2, 3, 1, 4, 3/8

      For disulfur dichloride, the 2 central S atoms have 2 lone pairs and 2 bond pairs; electron geometry tetrahedral, molecular geometry v-shape.

      For sulfuric(IV) acid {stock name} or sulfurous acid {latin name}, the central S atom has 1 lone pair, 4 bond pairs (S can expand its octet using its vacant and energetically accessible 3d orbitals); electron geometry tetrahedral, molecular geometry trigonal pyramidal.

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