Chem2O6 - 1997/98

1. The digitoxin derivative shown below is a metal chelating agent that is specific for ATPase and has potential use in radio-imaging.

(i) Identify the functional groups present in this compound.
(ii) Identify a methyl group, and a primary, secondary, or tertiary carbon (if there are any present).
(iii) Identify a primary, secondary, or tertiary alcohol (if any are present).
(iv) Identify a primary, secondary, or tertiary amide (if any are present).

Answer:

2. Consider the following three organic molecules:

(a) Give the hybridization of each of the atoms (excluding hydrogens) in each molecule.

Answer: The lone pairs on EACH of the NH nitrogens in all three structures can overlap with either a C=O, C=C, or C=N system, and are therefore sp² hybridized. The only atom which is sp³ hybridized is the methyl carbon in Thymine.

(b) In which direction do the lone pairs on the N atoms in the three structures point relative to the plane of the rings?

Answer:

(c) A cyclic array of 6 electrons (or 6 electrons in overlapping 2p orbitals is said to be aromatic:

    (i) Would you expect thymine, adenine, or imidazole to be aromatic? If so, what resonance structures would you have to write in order to obtain 6 electrons in a cycle?

Answer: While only adenine has a six-membered ring with 3 conjugated double bonds (like benzene), all three compounds have some degree of aromatic character. If we consider the Lewis structures shown below, and count the lone pairs on N atoms which are pointing perpendicular to the planes of the rings as contributing to the cyclic array of 2p electrons, we see that both thymine and imidazole are aromatic. Adenine is like naphthalene, an aromatic compound with 10 electrons in a cyclic array of 2p orbitals. The general rule for aromatic compounds is Huckel's Rule, which states that any cyclic array of 4n+2 (where n is any integer) electrons in p type atomic orbitals will have special stability.

    (ii) The first sites of protonation of adenine by a strong acid (e.g. HCl) are N(1), N(3), and N(7), NOT the NH₂ group as one might expect. Write one or more resonance structures that would account for this.

Answer: The numbering system in adenine was omitted (sorry!); it's given below:

As shown by the two charge-separated resonance structures above, the lone pair on the NH₂ nitrogen can delocalize into the ring, giving it slight positive charge and N1 and N3 slight negative charge. This means that the NH₂ lone pair is less available for protonation by an acid than it would otherwise be. The same arguments can be applied to the lone pair on N9 (try it).

3. Show the direction (LHS or RHS) that you would expect the following equilibria to lie:

(i) CH₃CN + C₆H₅^- [CH₂CN]^- + C₆H₆

Answer: RHS. Acetonitrile (pK_a 25) is a stronger acid than benzene (pK_a 43) and will therefore protonate the benzene conjugate base, phenyl anion.

(ii) CH₂CH₂ + OH^- CH₂CH^- + H₂O

Answer: LHS. Water (pK_a 115.7) is a much stronger acid than ethene (pK_a 36) and will protonate the conjugate base, ethenyl anion.

(iii) OH^- + CH₃NH₂ CH₃NH^- + H₂O

Answer: LHS. Water (pK_a 15.7) is a much stronger acid than the N-H proton of methylamine ((pK_a ~35) and will therefore protonate the methylamide anion.

(iv) CH₃SH + CH₃^- CH₃S^- + CH₄

Answer: RHS. Mercaptomethane (CH₃SH; pK_a ~10) is a very much stronger acid than methane (pK_a ~50), and will thus protonate the methyl anion.

(v) CH₃CH₂O^- + C₆H₅COOH CH₃CH₂OH + C₆H₅COO^-

Answer: RHS. Benzoic acid (pK_a 4.2) is stronger acid than ethanol (pK_a 15.9), and will protonate ethoxide anion.

(vi) (C₂H₅)₂NH₂⁺ + HCO₂^- (C₂H₅)₂NH + HCO₂H

Answer: LHS. Formic acid (pK_a 3.75) is a stronger acid than the diethylammonium cation (pK_a ~11), and will thus protonate diethylamine.

4. The following equilibria all lie on the right hand side of the equation:

    NH₂^- + CH₃COOCH₃ NH₃ + ^-CH₂COOCH₃
    ^-CH₂COOCH₃ + CH₃COCH₃ CH₃COOCH₃ + ^-CH₂COCH₃
    ^-CH₂COCH₃ + CH₃OH CH₃COCH₃ + CH₃O^-
    CH₃O^- + CH₃NO₂ CH₃OH + ^-CH₂NO₂

(i) Identify which compounds are weaker acids than methanol.

Answer: Methanol is able to donate a proton to acetone anion (CH₃COCH₂-) and is therefore a stronger acid than acetone (2-propanone). Likewise, it is a stronger acid than methyl acetate (because acetone is a stronger acid than methyl acetate) and ammonia.

(ii) Identify which compounds are weaker bases than methoxide ion.

Answer: Methoxide ion is involved in only one equilibrium which lies on the right (the last one); thus methoxide anion is a stronger base than nitromethane anion (^-CH₂NO₂). All the others are stronger bases than methoxide anion.

(iii) Identify which compounds are weaker acids than acetone (2-propanone).

Answer: From the arguments in (i) above, 2-propanone is a stronger acid than methyl acetate and ammonia.

(iv) Identify which compounds are weaker bases than acetone anion.

Answer: 2-propanone anion is a stronger base than methoxide anion and nitromethane anion.

(v) Identify which is the strongest acid and which is the strongest base.

Answer: The strongest acid is nitromethane. The strongest base is amide anion (NH₂^-).

02oct97; wjl