Download ClickerQs #8-14

11/27/2014
Consider the following boiling points. Why is there such a dramatic drop in boiling points across the series?
NaF: 1704°C MgF2: 2260°C AlF3: 1291°C SiF4: –86°C PF5: –85°C SF6: –64°C
a)
b)
c)
d)
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Which compound has the strongest intermolecular attractive forces?
a) BCl3
b) GeCl4
c) AsCl3
d) Cl2
The Si P and S compounds have larger molar masses
The Na Mg and Al compounds are ionic
The Si P and S compounds are polar
The Na Mg and Al compounds have fewer F atoms
NaF MgF2 AlF3 are ionic lattices of cations and anions, with very strong ion‐ion forces holding them together. SiF4 PF5 and SF6 are covalent compounds, with weaker intermolecular forces. It takes far less heat to separate the covalent molecules from each other.
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AX3E, pyramidal, polar, bp = 130°C
linear, non‐polar, bp = –34°C
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Which intermolecular forces are exhibited by methylamine, CH3NH2?
only dispersion forces
only dipole‐dipole forces
only hydrogen bonding
both dispersion and dipole‐dipole
all three of dispersion, dipole‐dipole, and hydrogen bonding
H2S is bent, so it is polar and will exhibit dipole‐
dipole interactions. All molecules have dispersion forces, but these will be less important for H2S. S‐H bonds are not polar enough for true intermolecular H‐bonding: N‐H, O‐H, or F‐H bonds are required.
AX4, tetrahedral, non‐polar, bp = 87°C
BCl3, SnCl4, and Cl2 are all non‐polar covalent compounds, and will exhibit only dispersion forces. AsCl3 is polar, and will exhibit stronger dipole‐dipole interactions as well. Which intermolecular forces are exhibited by H2S?
a)
b)
c)
d)
e)
AX3, trigonal planar, non‐polar, bp = 13°C
a)
b)
c)
d)
e)
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only dispersion forces
only dipole‐dipole forces
only hydrogen bonding
both dispersion and dipole‐dipole
all three of dispersion, dipole‐dipole, and hydrogen bonding
CH3NH2 is pyramidal at N, so it is polar and will exhibit dipole‐dipole interactions. It also exhibits hydrogen bonding, because it has N‐H bonds and an N lone pair. Hydrogen bonding will be the most significant interaction.
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11/27/2014
Wax consists of long‐chain hydrocarbons (C and H atoms). Which liquid will form droplets on wax, rather than spreading out in a film?
Which liquid should have the highest boiling point?
a) ethanol, CH3CH2OH
b) chloroethane, CH3CH2Cl
c) water, H2O
d) carbon disulfide, CS2
bp 78°C
a) octane, a chain of 8 C atoms: C8H18
b) benzene, a ring of 6 C atoms: C6H6
c) water, H2O
d) carbon disulfide, CS2
bp 12°C
bp 100°C
bp 46°C
High bp requires strong attractive forces, so that more temperature is required to cause the molecules to escape to the gas phase.
Water has the strongest IM forces, so water has the highest bp.
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Vitamins are categorized as fat‐soluble or water‐soluble. Which vitamin is water‐soluble?
One of the following compounds has a melting point far lower than the others. Which one, and why?
a)
b)
c)
d)
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Line structures do not show C or H atoms explicitly: C atoms occur where lines join, and enough H atoms are assumed present to give each C four bonds. MgO, it has the lowest molar mass
Al2O3, it has the weakest bonds
SiO2, it forms the only non‐polar compound
P4O6, it is a molecular compound, while the others are ionic or networks solids
MgO (2852°C) and Al2O3 (2072°C) are ionic complexes.
SiO2 (~1650°C) is a covalent network, with no individual molecules. P4O6 is a molecular compound, mp = 24°C.
Water has the strongest intermolecular forces: the others are all non‐polar compounds. The cohesive forces within water droplets (hydrogen bonding) are stronger than the adhesive forces between water and wax, so water doesn’t form a film.
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a) vitamin A, retinol
b) vitamin C, ascorbic acid
c) vitamin D3,
cholecalciferol
d) vitamin E,
α‐tocopherol
Vitamin C has many OH groups and O LPs capable of forming hydrogen bonds to water. The others are almost entirely non‐polar C‐C and C‐H bonds.
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11/27/2014
Amino acids are categorized as hydrophilic or hydrophobic, depending on whether their side chains interact strongly with water. Which amino acid is hydrophilic?
H
N
a) histidine
b) methionine
O
O
N
O
CH3S
d) valine
O
HN
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a) carvacol
O
NH
c) phenylalanine
Oil of oregano contains dozens of organic compounds. The three principal components are carvacol (36%), thymol (30%), and p‐cymene (24%). Which of these three is not extracted into hot water?
NH
O
c) p‐cymene
HN
Histidine has an N‐H bond and an N LP in its imidazole ring: it can form H‐bonds to water. The other sidechains all contain non‐polar groups, and so do not interact strongly with water.
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All three compounds are mostly non‐polar, with only weak dispersion forces. However, the two –OH compounds will have some polarity and hydrogen‐
bonding, so they will have some solubility in hot water.
Oil of oregano is an antioxidant, due to the –OH compounds (“phenols”). Which extract of oil of oregano will be the have the most potent antioxidant activity?
J. Food. Sci. 2011, 76, C512‐8.
TLC of spinach leaf extract separates various pigments. Stationary phase: SiO2. Mobile phase: low‐polarity organic solvent mixture. Why do the pigments separate?
a)
b)
c)
d)
e)
a) different colours correspond to molecules xanthophylls
of different sizes b) carotenes are organic, but xanthophylls are inorganic
c) xanthophylls contain more SiO2 than carotenes
d) carotenes are more polar than xanthophylls
e) carotenes are less polar than xanthophylls
pentane (CH3CH2CH2CH2CH3) extract cold water extract
hot water extract
all these extracts should have antioxidant activity
none of them, only the pure oil will be an antioxidant Pentane has only dispersion forces, so it will not easily the –OH compounds. Solubility can increase with higher temperature. Hot water extracts the greatest concentration of the active phenol compounds.
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b) thymol
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carotenes
chlorophylls
lutein
solvent
Carotenes are less polar, so they are less attracted to polar silica and move more quickly with the mobile solvent. Xanthophylls are more polar, and are more strongly attracted to polar silica, so they move more slowly.
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11/27/2014
The melting point of polyethylene plastic increases as the length of the polymer chains increases. This is because
The bond angles at the N atom in a polypeptide are all ~120°. What is the hybridization of atomic orbitals at the N atom?
a) longer chains have stronger bonds
b) the longer chains have heavier mass, reducing their mobility compared to lighter chains
c) longer chains have larger dispersions forces
d) longer chains have more C‐H hydrogen bonding
e) there are more C2H4 units in a longer chain, so it takes more heat to separate them all back to ethylene
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Polyethylene is made up repeating (‐CH2CH2‐) units, with only non‐polar C‐H bonds, so only dispersion forces are present. Longer chains have larger and more polarizable electron clouds, so the dispersion forces are stronger. It requires more heat to overcome the stronger attractive forces between longer chains, permitting them to slide by each other in the liquid phase.
a)
b)
c)
d)
e)
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bond to C
sp
E
pz
2
sp
sp3
sp2
there is no hybridization
three bonds
it depends on the resonance form H
N
C
O
Three 120° angles means a trigonal planar geometry of three σ
bonds, explained with sp2 hybrids, leaving two electrons in pz to form the π bond. Limiting resonance forms are not the real molecule! The N lone pair does not represent the real geometry!
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