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15.5.2: Chapter 2

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    Problem 2-1
    (a) H
    (b) Br
    (c) Cl
    (d) C
    Problem 2-2
    (a) The chemical structure of chloromethane with a delta plus charge on carbon and a delta minus charge on chlorine.
    (b) The chemical structure of methylamine with a delta plus charge on carbon and a delta minus on nitrogen.
    (c) The chemical structure of ammonia with a delta minus charge on nitrogen and a delta plus charge on hydrogen.
    (d) The chemical structure of methanethiol.
    (e) The chemical structure of methyl magnesium bromide with a delta minus charge on carbon and a delta plus charge on magnesium.
    (f) The chemical structure of fluoromethane with a delta plus charge on carbon and a delta minus charge on fluorine.
    Problem 2-3 H3C−OH < H3C−MgBr < H3C−Li = H3C−F < H3C−K
    Problem 2-4

    The nitrogen is electron-rich, and the carbon is electron-poor.

    The chemical structure of methylamine where the direction of the dipole moment points from carbon toward nitrogen.
    Problem 2-5

    The two C–O dipoles cancel because of the symmetry of the molecule:

    The chemical structure of ethylene glycol, where the direction of both dipole moments points from carbon toward the hydroxyl group in opposite directions.
    Problem 2-6
    (a) The chemical structure of ethylene, where the two carbon atoms are connected by a double bond and this structure has no dipole moment.
    (b) The chemical structure of trichloromethane, where the direction of the resulting dipole moment points from hydrogen toward chlorine.
    (c) The chemical structure of dichloromethane, where the direction of the resulting dipole moment points from carbon toward chlorine.
    (d) The chemical structure of 1,2 - dichloroethene where the direction of the resulting dipole moment points from carbon toward chlorine.
    Problem 2-7
    (a) For carbon: FC = 4 − 8/2 − 0 = 0 For the middle nitrogen: FC = 5 − 8/2 − 0 = +1 For the end nitrogen: FC = 5 − 4/2 − 4 = −1
    (b) For nitrogen: FC = 5 − 8/2 − 0 = +1 For oxygen: FC = 6 − 2/2 − 6 = −1
    (c) For nitrogen: FC = 5 − 8/2 − 0 = +1 For the triply bonded carbon: FC = 4 − 6/2 − 2 = −1
    Problem 2-9 The structures in (a) are resonance forms.
    Problem 2-10
    (a) Three different resonance forms of methyl phosphate anion, featuring a phosphorus atom at the center and separated by double-headed arrows.
    (b) Three different resonance forms of nitrate anion featuring a nitrogen atom at the center and separated by two double-headed arrows.
    (c) Two different resonance forms of allyl cation separated by one double-headed arrow.
    (d) Two different resonance forms of benzoate, resulting in the migration of a negative charge from one oxygen to the other.
    Problem 2-12 Phenylalanine is stronger.
    Problem 2-13 Water is a stronger acid.
    Problem 2-14 Neither reaction will take place.
    Problem 2-15 Reaction will take place.
    Problem 2-16 Ka = 4.9 × 10–10
    Problem 2-17
    (a) Three reversible reactions show ethanol, dimethylamine, and trimethylphosphine each reacting with hydrogen chloride to form protonated products.
    (b) Three reversible reactions show methyl cation, trimethyl borane, and magnesium bromide each reacting with hydroxide ion to form products.
    Problem 2-18
    (a) The structure of imidazole shows N1 labeled more basic (red) and the hydrogen bonded to N3 labeled most acidic (blue).
    (b) Imidazole reversibly reacts with acid at N1 to form two resonance structures of protonated product. Imidazole reversibly reacts with base at the N3 hydrogen, forming two resonance structures of deprotonated product.
    Problem 2-19 Vitamin C is water-soluble (hydrophilic); vitamin A is fat-soluble (hydrophilic).

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