1
Alcohols react using both acid and base catalysis: please elaborate.
Choose one answer.
a. Acids catalyze hydrolysis and ether formation from alcohols. Bases polarize the OH bond, even to the point of forming an alkoxide salt.
b. Acids are used as catalysts in the Williamson ether synthesis and for hydrolysis of ethers. Bases dissolve in ethers for use in other reactions.
c. Acids and bases will react in alcohols to form salts that are useful for ether synthesis.
d. Acids catalyze alcohol dimerization to ethers while bases hydrolyse them.
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Question 2
The Williamson ether synthesis involves which of the following?
Choose one answer.
a. The two Williamson reagents normally used in combination with heat and light.
b. An alkyl alcohol heated up with acid to distill off water.
c. An alkoxide salt of an alcohol reacted with a primary or secondary halide.
d. An alkoxide salt of an alcohol reacted with the alcohol it was made from.
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Question 3
Epoxides are a special kind of ether that:
Choose one answer.
a. are reactive because of ring string (3 membered ring).
b. react faster with acid or base catalysis.
c. can react with amines, alcohols and acids.
d. all of the above.
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Question 4
Ethers can react with which of the following?
Choose one answer.
a. With base and water to give two alcohols.
b. With acid and another alcohol to give transetherification- another ether and a different alcohol from the starting one.
c. With ethyl bromide to give an ethyl ether and a new alkyl bromide.
d. With a carboxylic acid and heat to give an ester and an alcohol.
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Question 5
Alcohols are unique in forming ethers because:
Choose one answer.
a. they can react with themselves in the presence of heat and acid to give ethers.
b. they have an exchangeable hydrogen (the OH one).
c. they have higher boiling points than hydrocarbons or ethers with the same number of carbon atoms.
d. they are all of the above.
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Question 6
Ether alcohols (molecules with both an ether and a beta alcohol) can best be made by:
Choose one answer.
a. acid catalyzed reaction of an epoxide and an alcohol with maybe a little heat.
b. addition of an alcohol to an alkene with base catalysis and heat.
c. acid catalyzed addition of water to a vinyl ether (ether with a double bond next to the oxygen).
d. addition of an alcohol to a ketone in the presence of acid and heat.
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Question 7
There are many ways to make an ether including:
Choose one answer.
a. reaction of two alcohols with base and heat catalysis.
b. reaction of an epoxide with an alkene using acid catalysis.
c. reaction of an alkoxide and an alkyl bromide.
d. addition of water to an epoxide with base catalysis.
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Question 8
Pick the synthetic conditions that will make an ether.
Choose one answer.
a. Reaction of an epoxide with a carboxylic acid.
b. Conversion of an alcohol with base catalysis and heat, distilling off the ether as it forms.
c. Reaction of an alkyl halide with NaOH in water.
d. the conditions for a Williamson synthesis (alkoxide plus halide).
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Question 9
An epoxide reacts with an alkoxide to give an ether plus something else:
Choose one answer.
a. and the something else is another alkoxide.
b. and the something else is NaOH.
c. and the something else is another epoxide.
d. and there is nothing else formed.
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Question 10
Ethers are pretty good solvents for:
Choose one answer.
a. alkyl halides.
b. lithium alkyl and sodium alkoxide salts.
c. alcohols.
d. all of the above to some extent.
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Question 11
Why are thiols and their anionic conjugate bases more reactive than their oxygen analogs?
Choose one answer.
a. They aren't really, since alcohols and acids are more reactive.
b. Because the sulfur is big, and the bigger an atom is with unpaired outer shell electrons, the more reactive it is.
c. Because sulfur has a bigger, less tightly packed nucleus.
d. Because it's smaller than oxygen and has more localized outer shell electrons which are more nucleophilic.
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Question 12
Why does sulfur oxidize so easily to sulfoxides, sulfones, sulfonic and sulfuric acid?
Choose one answer.
a. No one knows, it's one of the mysterious of the universe.
b. Oxygen is more electron rich, and wants to share electrons with any atom it can bind to.
c. Oxidation is just "burning," and we know that sulfur burns well (high sulfur coal?).
d. Sulfur just loves to share its outer shell electrons, and the more sharing it does, the happier it is.
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Question 13
Do eletrophilic compounds always react at the sulfur, no matter what?
Choose one answer.
a. Yes, except when oxygens are attached to the sulfur, then the oxygens react first.
b. Yes, except when carbons are attached to the sulfur, then substitution occurs at the carbon.
c. Yes, except when bromine is involved.
d. Pretty much, yes, even when oxygens are attached to the sulfur.
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Question 14
The Swern oxidation involves using DMSO (remember this great solvent?) to do what?
Choose one answer.
a. React with oxygen to form dimethylsulfone, the di-oxygen derivative of sulfur.
b. React with primary or secondary alcohols to give aldehydes or ketones.
c. Convert a primary alcohol to a carboxylic acid using oxygen as the co-reactant.
d. Convert amines to oxides such as nitric oxide using peroxide as co-reactant.
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Question 15
What is the difference between a thiol and a mercaptan?
Choose one answer.
a. There is none; they're the same and they both have SH bonds.
b. A thiol is the sulfur analog of an ether; a mercaptan is like an alcohol (SH bond).
c. A mercaptan is like an alcohol (SH bond) while a thiol is like an ether.
d. There is none; they're both ether analogs (C-S-C bonds).
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Question 16
It's possible to make a polymer from a conjugated double bond diene. Pick the right diene below and the polymer that it forms.
Choose one answer.
a. 1, 5-hexadiene (gives a polymer with four carbon backbone and two-carbon pendent group).
b. 2-methyl-1, 3-butadiene (also called isoprene), gives polyisoprene like the natural rubber used in tires.
c. 2, 4-hexadiene (gives a polymer with four carbon backbone and two pendent methyl groups).
d. 1, 2-propadiene (gives a polymer with three carbon backbone).
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Question 17
Diels-Alder reactions usually involve:
Choose one answer.
a. some kind of diene with pendent ester groups reacting with some kind of alkene to give a substituted cyclohexene.
b. 1, 3-butadiene reacting with ethylene to give cyclohexene.
c. 1, 3-cyclohexadiene reacting with carbon monoxide to give a cyclic ether-alkene.
d. some kind of alkene with pendent ester groups (e.g., maleic acid diethyl ester) reacting with some kind of diene (like 1, 3-hexadiene) to give a substituted cyclohexene.
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Question 18
Dienes can react with bromine (the diatomic molecule, that is) to give a product; which combination below is correct?
Choose one answer.
a. 1, 3-butadiene plus 1 equivalent of bromine gas to give mostly 1, 3-dibromo-3-butene.
b. 1, 3-butadiene plus 1 equivalent of bromine gas to give mostly 1, 4-dibromo-2-butene.
c. 1, 5-hexadiene plus excess bromine to give mainly 1, 2-dibromo-5-hexene.
d. 1, 2-propadiene plus one equivalent of bromine to give 1, 3-dibromopropene.
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Question 19
Dienes will readily react with chlorine, especially under the right conditions: which is the best answer below?
Choose one answer.
a. 2, 4-hexadiene with chlorine gas under a strong uv light to give the dichlorohexene.
b. 1, 3-butadiene added dropwise to liquid chlorine to give 1, 2-dichloro-3-butene.
c. 1equivalent of frozen chlorine (solid) added to cold 1, 4-cyclohexadiene to give 2, 5-dichloro-1, 4-cyclohexadiene..
d. chlorine gas dissolved in cold water with 1, 3-butadiene added dropwise to give dichlorobutanediols.
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Question 20
Conjugated double bonds in dienes often lead to an especially stable intermediate during reactions; pick a good explanation below for why.
Choose one answer.
a. Any reactive species whether radical, cationic or anionic will give a resonance stabilized "allylic" intermediate.
b. The first step in the reaction of conjugated double bonds is the same as for isolated double bonds, and the radical, cation or anion formed will be stabilized by the alkyl substituent.
c. The intermediate will have both the 1- and the 4-carbons of the starting conjugated double bonds attached to the attacking electrophile.
d. Actually, the reactions are so fast that no intermediate is expected no matter what electrophile is involved.
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Question 21
Reaction of 1, 3-butadiene with maleic anhydride gives a single product that is:
Choose one answer.
a. 3-vinylcyclobutane-1, 2-dicarboxylic acid anhydride.
b. 3-(3-butenyl)succinic anhydride.
c. 4-cyclohexene-cis-dicarboxylic acid anhydride.
d. a 1:1 copolymer of the two reactants.
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Question 22
You unexpectedly are given hundreds of tons of 1, 3-cyclohexadiene. You decide to convert it to benzene by:
Choose one answer.
a. first reacting with chlorine gas and UV light followed by KOH in alcohol.
b. treating with NaOH in water to make the alcohol followed by distillation from concentrated sulfuric acid.
c. carefully adding excess chlorine followed by excess KOH in alcohol.
d. heating it in the presence of a hydrogen and a catalyst such as palladium.
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Question 23
You can make butene alcohol by:
Choose one answer.
a. simply heating 1, 3-butadiene with NaOH in water.
b. carefully adding one equivalent of HCl followed by NaOH in water.
c. carefully adding one equivalent of bromine followed by NaOH in water.
d. reacting with hydrogen peroxide and acetic acid.
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Question 24
How would you best make a conjugated diene?
Choose one answer.
a. Starting with a linear alkane with at least four carbons, first halogenate once and dehydrohalogenate to an alkene; do radical allylic halogenation to give an allyl halide; then dehydrohalogenate again with KOH in alcohol to the diene.
b. Take any 4-carbon alkene, add bromine across the double bond and dehydrohalogenate twice with KOH in alcohol.
c. React a four carbon (at least) linear alkane with two equivalents of chlorine under UV light, then dehydrohalogenate twice with KOH in alcohol.
d. First chlorinate and dehydrochlorinate butane, then deprotonate with butyl lithium or concentrated KOH in alcohol, and pour the mixture into dilute acid to neutralize the base.
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Question 25
You decide to make 1, 3-cyclopentadiene from cyclopentene you happen to have by:
Choose one answer.
a. first reacting with one equivalent of chlorine gas and uv light followed by treating with KOH in alcohol.
b. treating with concentrated NaOH in water followed by dehydration with concentrated sulfuric acid.
c. adding one equivalent of bromine and treating the product with two equivalents of KOH in alcohol.
d. reacting it with HOCl (bleach) to give the halohydrin, and dehydrating with concentrated sulfuric acid.
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Question 26
Please name the diene that is a cyclic six-carbon ring with two double bonds (conjugated).
Choose one answer.
a. 1, 2-cyclohexadiene.
b. 1, 4-cyclohexadiene.
c. 1, 5-cyclohexadiene
d. 1, 3-cyclohexadiene
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Question 27
Huckel was one of the first scientists to describe aromaticity, and he used which simple rule?
Choose one answer.
a. involving 4n + 2 carbon atoms in a cyclic array.
b. involving 4n + 2 electrons in a continuous cyclic array of pi-bonds and/or p-orbitals.
c. requiring conjugated carbon atoms in a cyclic array like benzene or naphthalene.
d. with a continuous cyclic array of pi-bonds.
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Question 28
Aromatic molecules include which of the following?
Choose one answer.
a. Benzene, naphthalene and higher analogs with 4n + 2 electrons in the conjugated pi-bonds.
b. Furan which has four carbons and one sp2 hybridized oxygen totalling five atoms and 6 pi or non-bonded electrons.
c. Cycloheptatriene possessing 7 carbon atoms with three conjugated pi-bonds containing 6 pi electrons.
d. Cyclopentadiene cation with 5 carbons, two pi-bonds and one sp2 carbon containing 0 electrons in a p-orbital.
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Question 29
Naphthalene (10 carbons with two fused benzene rings) is aromatic according to Huckel because:
Choose one answer.
a. there is a continuous array of p-orbitals from pi-bonds around the outside of the fused rings AND there are 4n + 2 electrons in them (10).
b. fused benzene rings are always aromatic.
c. 10 carbons are an even number and there are 5 double bonds with 8 pairs of electrons in them.
d. none of the above.
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Question 30
Cyclopropene has three carbons in a ring with one double bond: is it aromatic?
Choose one answer.
a. No because there's only 2 electrons in the pi-bond (2 p-orbitals).
b. No, because there's no continuous array of p-orbitals around the ring.
c. No, because there needs to be a positive charge somewhere in the ring.
d. No, because the ring is too small.
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Question 31
Substituents on aromatic molecules can have long-range effects: explain why.
Choose one answer.
a. Sigma bonds transmit electronic effects by electronic polarization effects.
b. Resonance causes spreading of the electronic effects of various substituents throughout the ring through sigma bonds.
c. Resonance of the electron donating or withdrawing affects of a substituent can stabilize or destabilize charge at ortho and para positions.
d. Through space interactions of a substituent with sigma bonds in the ring affect individual carbon electron densities.
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Question 32
A good explanation of aromatic stabilization involves which of the following?
Choose one answer.
a. A more even distribution of pi electrons occurs in aromatic molecules.
b. Rapid resonance of pi electrons spreads the electron density over more atoms.
c. The molecular orbitals that form from a conjugated cyclic array of p-orbitalswith just the right number of electrons in them are substantially lower in energy than for non-conjugated systems.
d. There is no good explanation that we know of, it's just a fact of nature.
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Question 33
Because of their unique resonance stabilization and molecular structures, aromatic molecules possess unique properties: select the best description below.
Choose one answer.
a. They are more stable and less reactive then non-aromatic molecules.
b. They react with unique mechanisms such as electrophilic aromatic substitution and nucleophilic aromatic substitution.
c. They do not undergo simple addition reactions that occur with simple alkanes.
d. All of the above.
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Question 34
Benzene possess a unique structure: select from those below.
Choose one answer.
a. It has six carbons like cyclohexane and therefore has both chair and boat confirmations.
b. It consists of two fused cyclobutadiene rings that are basically flat.
c. The six carbons in the ring all occupy the same plane with pi bond orbitals above and below that plane.
d. Benzene has carbons connected by six sigma and three pi bonds, with the pi bonds being significantly shorter than the ordinary sigma bonds in the benzene ring.
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Question 35
Heteroatom's and charges on atoms can affect aromaticity in what way?
Choose one answer.
a. Their orbitals can be hybridized in such a way as to participate in the cyclic array of p-orbitals needed for 4n+2 electrons.
b. They can certainly change the stability of an aromatic ring.
c. They modify the behavior of the ring in strange and mysterious ways.
d. They decrease the reactivity of the rings in which they are located.
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Question 36
Substituents on aromatic ring can affect each other: Pick the answer that best illustrates this.
Choose one answer.
a. Meta-nitrobenzoic acid is more acidic than the para-nitro isomer because the nitro is closer to the acid group.
b. 3, 4-Dinitrobenzoic acid is readily made by nitration of benzoic acid with excess fuming nitric acid.
c. The order of acidity of substituted benzoic acid increases for para-substituents nitro, chloro, hydroxy.
d. The more electron withdrawing groups on the ring, the more acidic is benzoic acid; i.e., trichloro more than dichloro more than monochlorobenzoic acid.
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Question 37
Phenol is an aromatic alcohol and differs from aliphatic alcohols in what ways?
Choose one answer.
a. The OH group is less acidic and less easily exchanged with deuterium because of resonance.
b. The OH Group has an enormous effect on reactivity and aliphatic alcohols but not in phenols.
c. Reactions on carbons alpha or beta to the OH in phenols is strictly controlled by electronegativity affects.
d. Because of resonance stabilization of intermediates, phenols are more easily brominated and nitrated.
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Question 38
The reactivities of substituents on aromatic rings are affected by the aromatic ring as illustrated by which of the following answers?
Choose one answer.
a. The methyl group of toluene readily undergoes free radical reaction and oxidation.
b. Benzoic acid is more acidic than acetic acid.
c. Styrene or vinylbenzene is a better free radical monomer then propene.
d. All of the above.
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Question 39
The acidity of phenol is greater than cyclohexanol because:
Choose one answer.
a. an aromatic ring is inherently electron withdrawing, pulling electrons away from the O-H bond.
b. resonance stabilization of the ring bond to oxygen makes the bond between O and H weaker.
c. resonance stabilization of the conjugate base (phenolate anion) is greater than is stabilization of the cyclohexanol alkoxide by the attached ring.
d. alkyl carbons are electron donating compared to an aromatic ring, baking the cyclohexanol OH bond stronger than that of phenol.
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Question 40
Compare the reactivity of the oxygen to acetylation with acetic anhydride for phenol versus cyclohexanol by picking the best answer below.
Choose one answer.
a. Electron donation by the cyclohexane ring increases nucleophilic attack by the oxygen of cyclohexanol versus phenol.
b. Resonance is greater for the phenol acetyl oxonium intermediate than for the cyclohexanol analog.
c. Base catalyzed formation of the phenolate is faster than for cyclohexanol due to resonance stabilization, and the anion formed is more reactive.
d. In fact, they both react at the same rate.
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Question 41
You find to your surprise that the reaction of phenol with acetic anhydride and aluminum trichloride gives a mixture of two disubstituted products. Pick the best reason for why.
Choose one answer.
a. The OH oxygen reacts twice to give the bisester derivative.
b. Ring acylation occurs twice in the ortho-ortho' and ortho-para positions.
c. The first acylation product is in the ortho and para positions which deactivates the ring and causes meta-substitution with the second acetyl group to give ortho-meta and meta-para diacetylphenols.
d. The first reaction is at the oxygen to give phenyl acetate followed by acetylation in the ortho and para positions to give a mixture of products.
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Question 42
EAS reactions involve attack by something seeking:
Choose one answer.
a. all of the below.
b. electrons, and those are almost always relatively "available" ones in pi-orbitals.
c. a nucleus to bond to, making the reaction also a "nucleophilic" attack.
d. a one-step reaction involving loss of leaving group at the same time the attacking group makes a new bond.
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Question 43
All EAS reactions involve a "tetrahedral" or Td intermediate. Which of the following is true of this statement?
Choose one answer.
a. Can't happen with an aromatic ring carbon which by definition is sp2 or trigonal planar.
b. There is no intermediate in EAS which simple involves attack at electrons.
c. No, the leaving group leaves at the same time the attacking group forms a bond.
d. Almost all do: an sp2 carbon becomes sp3 in the intermediate.
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Question 44
The best way to describe the "arrow" showing the mechanism of EAS is:
Choose one answer.
a. it goes from the attacking atom (electrophile) to the carbon being attacked.
b. it goes from the attacking atom to the new bond being formed.
c. it goes from the electron pair in the pi-bond on the ring to the new sigma bond being formed.
d. it involves a half-head arrow from the attacking species and a half-head arrow from the pi-bond to the new sigma bond to be formed.
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Question 45
In general, electrophilic aromatic substitution (EAS) happens:
Choose one answer.
a. between any aromatic molecule and an appropriate electrophile, assuming aromaticity follows the 2(n + 2) rule.
b. with any pi-bond containing molecule and an electrophile.
c. with chlorine or bromine and iron catalysis on any pi-bond species.
d. when an electrophile bonds to a nucleophile of some kind.
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Question 46
EAS with chlorine gas bubbled into chlorobenzene containing aluminum trichloride would:
Choose one answer.
a. not happen because chlorine is deactivating.
b. gives mostly meta-dichlorobenzene with mild heating.
c. gives mostly ortho-dichlorobenzene with mild heating.
d. generates HCl and mostly para-dichlorobenzene.
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Question 47
47. Fuming nitric acid converts phenol into:
Choose one answer.
a. pure trinitrophenol (also called "picric acid" which has a really low pKa for a phenolic compound).
b. 2, 4-dinitrophenol because of deactivation with each nitro group added.
c. para-nitrophenol since the OH group on an aromatic ring is activating and ortho-para directing.
d. a complex mixture of ortho- and para-substituted phenols with di- and trinitro derivatives predominating.
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Question 48
48. Friedel-Crafts acylation of ortho-nitrotoluene with acetyl chloride and aluminum trichloride gives:
Choose one answer.
a. a mixture of acetophenones with the nitro group always ortho or para to the carbonyl.
b. almost exclusively a single product, 3-nitro-4-methyl-acetophenone.
c. 2-methyl-acetophenone by nitro group displacement.
d. no reaction since the nitro group is so deactivating.
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Question 49
49. Common explosives developed by Alfred Nobel use trinitro-aromatics such as:
Choose one answer.
a. trinitrophenol made by low-temperature nitration of phenol.
b. trinitrobenzene made by heating benzene in fuming nitric acid.
c. 2, 5-dinitro-1, 4-dichlorobenzene made by chloronating the dinitrobenzene with ferric chloride as catalyst.
d. TNT or trinitrotoulene made by careful exhaustive nitration of toluene.
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Question 50
50. Friedel-Crafts alkylation of benzene with 1-chloro-2-methylpropane and aluminum trichloride gives:
Choose one answer.
a. isobutylbenzene (or 1-phenyl-2-methylpropane).
b. t-butylbenzene (or 1-phenyl-1, 1-dimethylethane).
c. dimethylstyrene (or phenyl-dimethylethene).
d. paramethylpropylbenzene.
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Question 51
51. You decide to make meta-nitrophenol by doing which of the following steps?
Choose one answer.
a. Reacting phenol with fuming nitric acid in an ice bath.
b. Reacting benzene with fuming nitric acid, then reacting the product with chlorine/iron, and last heating with NaOH in water.
c. Reacting benzene with chlorine/iron, then reacting the product with nitric acid followed by NaOH in water.
d. Reacting benzene with fuming nitric acid to make trinitrobenzene and then with NaOH in water.
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Question 52
52. You acquire a mixture of meta- and para-fluoronitrobenzene and want to react it with a nucleophile. What should you do first?
Choose one answer.
a. You pick sodium methoxide because it's more reactive than a sulfur anion, sodium hydrosulfide (NaSH).
b. You choose NaCl in ethanol since that's a pretty good nucleophile system.
c. You decide on sodium methoxide dissolved in water to make the para-nitrophenyl methyl ether.
d. You mix NaSH with ethanol and then add the aromatic molecule to make the aryl sulfide.
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Question 53
53. You want to make para-hydroxybenzoic acid, so you decide to do both EAS and NAS as follows:
Choose one answer.
a. You react para-chlorotoulene with hot KMnO4, isolate the product and then react it with hot methanol and HCl catalyst. The product of this you react with neat NaOH in the melt.
b. You react toluene with concentrated NaOH in water at reflux to do NAS formation of the hydroxide, and then oxidize the methyl group with hot permanganate to give the acid.
c. You first react toluene with hot permanganate to make benzoic acid, then react that with chlorine/iron to make the para-chlorobenzoic acid which you heat with NaOH in water.
d. You react benzoic acid with chlorine and iron catalyst, then treat with concentrated NaOH in water and heat.
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Question 54
54. There are two components to a nucleophilic aromatic substitution reaction with the order of reactivity of:
Choose one answer.
a. halide leaving groups on the aromatic ring being F>Cl>Br, I not reactive.
b. the nature of the attacking nucleophile being unimportant.
c. other activating groups that are NOT displaced but stay on the ring such as ester, nitro, sulfone and carbonyl.
d. all of the above.
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Question 55
55. The Kolbe and Riemer-Tieman reactions are special types of EAS reaction in which:
Choose one answer.
a. chloroform is the electrophile and it reacts with the sodium salt of phenol (sodium phenolate).
b. bubbling carbon dioxide through a solution of phenol in water generates the para-hydroxybenzoic acid.
c. sodium phenolate (sodium salt of phenol) reacts directly with carbon dioxide (high temperature and pressure) to give ortho- and para-hydroxybenzoic acids as their sodium salts.
d. phenol dissolved in ethanol with iron chloride reacts rapidly with carbonic acid to form the ortho- and para-hydroxybenzoic acids.
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Question 56
56. The key intermediate in electrophilic aromatic substitution (e.g., bromination of benzene) can be described best by which answer below?
Choose one answer.
a. Bezene-derived anion adduct with delocalization of the charge around the ring.
b. Bezenonium ion, the cation formed by attack of the electrophile (e.g., bromine cation) on the ring to give an sp3 hybridized carbon.
c. There is no intermediate, just a transition state with the H-atom leaving as the electrophile forms a bond to that carbon.
d. A complex salt with the catalyst bound to the pi-cloud of the ring while the electrophile is attached to one of the bezene hydrogens.
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Question 57
57. Almost all EAS reactions require a catalyst because (pick the best answer below):
Choose one answer.
a. EAS is a difficult reaction since your going uphill energetically from an aromatic molecule, and the catalyst helps do this by forming a more reactive electrophile.
b. The catalyst complexes with the aromatic ring to activate it for attack.
c. The catalyst brings the aromatic ring and the electrophile together so that they can react.
d. Actually, a catalyst is rarely used and this is a false statement.
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Question 58
58. Nucleophilic aromatic substitution (NAS) is very similar to Sn2 substitution of alkanes because:
Choose one answer.
a. Displacement occurs in a single step.
b. It is a process involving an attacking species and a leaving group, although for NAS there is also an intermediate.
c. In both cases, a highly active nucleophile is the attacking species.
d. Inversion of the stereochemistry at the substituted carbon occurs during reaction.
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Question 59
59. The NAS mechanism is special in what regard?
Choose one answer.
a. It requires a much more nucleophilic attacking species than in alkyl substitution reactions.
b. The aromatic nucleus being attacked must have electron donating groups attached.
c. A 3-center Intermediate exists involving the catalyst, nucleophile and aromatic ring.
d. It involves nucleophilic attack at an sp2 carbon which only occurs with strong electron withdrawing groups attached at or in resonance with the substituted carbon.
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Question 60
60. The two types of aromatic substitution reactions (electrophilic and nucleophilic) are complementary in that (pick the best answer):
Choose one answer.
a. The same types of substituents on the aromatic ring are activators in both cases.
b. The arrow showing electron motion in both cases always starts at one of the pi bonds on the ring.
c. EAS is enhanced by electron donating groups on the ring while NAS is enhanced by electron withdrawing groups.
d. Both use Lewis acid catalysts like iron chloride or aluminum chloride.
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Question 61
Based on the outer shell hybridization of the nitrogen, amines have what geometry?
Choose one answer.
a. Modified structure with three sp2 orbitals in a trigonal planar array and one p-orbital with 2 electrons.
b. Two electrons in the 2s orbital and three bonding pairs of electrons in the three 2p orbitals.
c. It is incorrect to equate hybridization with molecular geometry.
d. Tetrahedral, just like carbon and oxygen, with four energetically equivalent orbitals pointing at the apexes of a tetrahedron.
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Question 62
Amines are "basic" in a Bronstead scheme because of what?
Choose one answer.
a. They have a lone pair of electrons in an sp3 orbital just sitting there waiting for something to happen, like protonation.
b. They have a conjugated set of orbitals capable of forming dative bonds.
c. A proton can bind to one of the attached atoms through the excess electron density in the bonding orbital.
d. That's just the way they roll.
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Question 63
Amines can be made from aldehydes by doing what?
Choose one answer.
a. Displacing the oxygen through an Sn2 reaction.
b. Protonating the oxygen of the aldehyde and attacking with a neutral amine of some kind in an Sn2 reaction.
c. Mixing ammonia or a primary or secondary amine with the aldehyde, forming the imine and reducing with hydrogen and a nickel catalyst.
d. Reduce the aldehyde to the alcohol with hydrogen and a catalyst, then reacting with an amine or ammonia at high temperature in an Sn2 reaction.
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Question 64
How would you convert a carboxylic acid to an amine?
Choose one answer.
a. First reduce the acid with lithium aluminum hydride to the alcohol, convert to an alkyl chloride with thionyl chloride, then heat with excess amine.
b. Heat the acid with ammonium hydroxide under high pressure and temperature, isolate the amide and reduce it with lithium aluminum hydride.
c. Convert the acid to the acid chloride with thionyl chloride and heat, add sodium cyanide to form the nitrile and reduce with lithium aluminum hydride.
d. All three of the above but with careful consideration of conditions and structure of the reactants.
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Question 65
Aniline is an aromatic amine that can be made how?
Choose one answer.
a. Reacting benzene with chlorine gas and iron chloride catalyst, then nitrating with nitric acid in sulfuric acid, isolating the product and reducing with hydrogen using a palladium catalyst.
b. Heating benzene with ammonia under high heat and pressure; once the first amine adds, further reaction stops because the amine is such a strong electron donator.
c. Reacting phenol with thionyl chloride to form chlorobenzene, then heating with ammonia under high temperature and pressure.
d. By treating benzene with nitric acid in sulfuric acid, isolating the nitrobenzene and reducing with hydrogen and a palladium catalyst.
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Question 66
In terms of basicity, what is the order for alkyl amines?
Choose one answer.
a. Secondary > tertiary > primary > ammonia due to steric effects for the tertiary putting it "out of order."
b. Tertiary amine > secondary > primary > ammonia due to electron donating ability of alkyl substituents.
c. Ammonia > primary amine > secondary > tertiary due to electron withdrawing ability of any heteroatom.
d. It will depend on whether we're talking about Lewis or Bronstead acid-base chemistry.
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Question 67
Reaction of isopropyl bromide (2-bromopropane) with amines varies how with amine structure?
Choose one answer.
a. In a non-regular manner, with secondary amines often being more reactive than either primary or tertiary amines.
b. Only primary and secondary amines react due to steric inhibition with tertiary amines.
c. Because of steric inhibition caused by the huge isopropyl group, there is no reaction with any amines.
d. It doesn't vary at all since all amines and ammonia have the same basic reactivity.
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Question 68
Pyridine is an aromatic amine: what is its structure and is it basic?
Choose one answer.
a. Pyridine is a benzene ring with an NH2 group attached that is basic like any amine.
b. It is a seven-membered ring with the six sp2 hybridized carbons accounting for the aromaticity and the nitrogen in the ring providing basicity.
c. It is like benzene with one carbon replaced with nitrogen that also has a lone pair of electrons (basic) in an sp2 orbital within the plane of the ring.
d. Pyridine is one of the nucleic acid bases possessing two nitrogens in the aromatic (benzene-like) ring and one nitrogen NH2 attached.
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Question 69
Quanidine (CN3H5 with three nitrogens attached to the carbon) is a very strong base for protonation for what reason?
Choose one answer.
a. Each of the amines has a lone pair of electrons making it three times more likely that a proton can bond to one pair.
b. Once one of the nitrogens is protonated, hydrogen bonding with the other nitrogens stabilizes the charge.
c. The proton in the protonated quandinium ion sits on top of and bonded to all three nitrogens at the same time.
d. Mainly because the protonated form has a positive charge resonance delocalized onto all three nitrogens through the central carbon.
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Question 70
Nitrogen is similar to carbon and oxygen (adjacent 2nd row elements) in what unique way?
Choose one answer.
a. All three have two electron in the 1s orbital.
b. It combines outer shell s and p electrons into four sp3 orbitals that are tetrahedral in shape.
c. It can form three bonds to other atoms in a neutral molecule.
d. It has a pair of electrons in a non-bonding sp3 orbital.
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Question 71
Amines and acids both have:
Choose one answer.
a. basic characters- they make water have a higher pH value when dissolved in it.
b. acidic characters- they make water have a lower pH value when dissolved in it.
c. exchangable hydrogens that can be converted to deuterium with heavy water (D2O).
d. greater volatility than alkanes with the same number of carbons.
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Question 72
Amines can be made from:
Choose one answer.
a. the reaction of ammonia with alkyl chloride with maybe heat to speed it up.
b. reaction of ammonium hydroxide with an alkyl bromide.
c. two-step reaction involving imine formation (amine plus aldehyde) followed by hydrogen reduction (with catalyst).
d. All of the above, more or less.
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Question 73
There are three types of amines that include:
Choose one answer.
a. primary, secondary, and tertiary, referring to the type of carbon directly attached to the nitrogen.
b. primary, secondary, and tertiary, referring to how many alkyl carbons are directly attached to the nitrogen.
c. alkyl, aryl (aromatic), and ammonium salts.
d. substituted amines, unsubstituted amines, and amides.
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Question 74
An alkyl carboxylic acid can be made by:
Choose one answer.
a. heating a methyl ketone with acid and iodine.
b. treating an alkyl halide with hydrogen gas and platinum.
c. reacting an alcohol with ozone.
d. oxidation of a primary alcohol with hot permanganate or chromate.
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Question 75
Butyl amine and acetic acid will react:
Choose one answer.
a. with heat to drive off water and make N-butyl acetamide.
b. in ethanol to form an ester and amine salt.
c. with an butyl bromide to form dibutyl amine.
d. hydrochloric acid to form an amide.
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Question 76
The name of a tertiary amine with one butyl and two methyl groups is:
Choose one answer.
a. butyl amine dimethyl.
b. methyl butyl aminomethane.
c. dimethylaminobutane.
d. butyl dimethyl amine.
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Question 77
The carboxylic acid made by oxidation of n-butanol is called:
Choose one answer.
a. butanoic acid.
b. propyl carboxylic acid.
c. carbonic acid.
d. benzoic acid.
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Question 78
Reacting 2-butene with peracetic acid (a peroxy acid) would lead to:
Choose one answer.
a. 2, 3-butanediol.
b. tetrahydrofuran (four carbon cyclic ether).
c. 2-butanol.
d. 2, 3-butene epoxide or 2, 3-epoxybutane.
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Question 79
Carboxylic acids can be converted to primary amines by:
Choose one answer.
a. decarboxylation with HBr and peroxide, then reaction of the alkyl bromide with ammonia.
b. reduction of the acid to the alcohol with NaOH/formaldehyde, then reaction with ammonium chloride and heat.
c. a two-step conversion, first to the amide with ammonia and heat, and then by reduction with litium aluminum hydride or hydrogen plus catalyst.
d. using lactase fermentation in the presence of ammonia atmosphere.
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Question 80
Primary amines of primary alkanes can be converted to alcohols using the Hoffman-type reaction involving:
Choose one answer.
a. reaction first with the Hoffman reagent and then heat with NaOH.
b. Exhaustive methylation with methyl iodide followed by heat with NaOH in water.
c. Conversion of the amine to an acetamide with acetic anhydride followed by reduction with hydrogen and catalyst.
d. Exhaustive ethylation with ethyl bromide followed by heat with KOH in ethanol.
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Question 81
Why is a hydrogen attached to the carbon next to a carbonyl so reactive?
Choose one answer.
a. The carbonyl polarization pushes electrons onto the alpha-carbon and hydrogen through sigma bonds.
b. In the anionic intermediate formed, delocalization of charge onto the more electronegative oxygen occurs through pi-bonds.
c. Enol-keto tautomers equilibrate and the hydrogen isn
d. The double bond to oxygen from the alpha carbon is conjugated to the intermediate carbanion through a double bond to that carbon as well.
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Question 82
The acidity of a proton (hydrogen) alpha to most carbonyl compounds is around pKa = 25, but drops to what with two carbonyls attached to the same sp3 carbon with a hydrogen?
Choose one answer.
a. Less than 20 but just barely.
b. Less than 14 or so which means that aqueous base can deprotonate such compounds.
c. Less than 10 which makes them weak acids.
d. Less than 6 which makes them about as acidic as carboxylic acids.
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Question 83
What is the best way to draw (describe) the sodium salt of 2, 4-pentadione (two acetyls on the same methylene)?
Choose one answer.
a. Since only one hydrogen on the carbon is lost to form this, the sodium cation should be on the alpha carbon.
b. Since only one of the oxygens can undergo enol-keto tautomerism, the sodium cation should be on one oxygen.
c. Since both oxygens can equilibrate with the carbanion, why not put the sodium cation between them in a chelate?
d. Since the carbonyl oxygen lone pairs repel each other, one carbonyl will be twisted out of plane with the other, and will have the sodium attached.
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Question 84
Reaction of almost any aldehyde or ketone with ethylene glycol (1, 2-ethanediol) and acid catalyst with water distillation does what?
Choose one answer.
a. Reduces the carbonyl to an alcohol while oxidizing the ethylene glycol to acetaldehyde.
b. Dehydrates the glycol to vinyl alcohol which forms a C-C bond to the carbon alpha to the carbonyl.
c. Transfers one of the hydroxyl groups from the glycol to the carbonyl to form a hydrate (two OH groups on the same carbon).
d. Well, if you're careful, in the end you form the acetal or ketal with no change in oxidation state.
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Question 85
A fairly general synthesis of a ketone would involve what steps?
Choose one answer.
a. Reaction of an alkane with chlorine and UV light to form an alkyl chloride, conversion to an alcohol with NaOH and water, and oxidation with CrO3 in sulfuric acid.
b. Treating an alkane with peroxide and acid to give an alcohol, followed by in situ (one-pot) oxidation to the carbonyl with the same reagents.
c. Treating an alkene with potassium permanganate in refluxing aqueous acid.
d. Reacting an alkyl carboxylic acid with thionyl chloride followed by treating the product with an alcohol and aluminum trichloride.
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Question 86
A neat way to make a symmetrical dialdehyde from cyclohexene is done how?
Choose one answer.
a. Reacting the alkene with hydrogen peroxide in acetic acid followed by adding water plus heat to give a glycol that is then cleaved with more peroxide and heat.
b. Treating first with osmium tetroxide in pyridine, then adding hydrogen sulfide and extracting the product.
c. Reacting the alkene with ozone (O3) in chloroform (CHCl3) followed by treatment with peroxide.
d. All of the above, I hope.
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Question 87
In a different problem (somewhere here) you oxidized an alcohol to an aldehyde or ketone; now you want the alcohol back so you:
Choose one answer.
a. Add concentrated NaOH in water and heat to drive off the water, then extract with methylene chloride to obtain the alcohol.
b. Add to the carbonyl compound in ethanol concentrated sulfuric acid containing sodium borohydride, then neutralizing and extracting with ether.
c. Dissolve the ketone or aldehyde in ethanol (maybe with a little water and NaOH added) and then slowly adding sodium borohydride followed by neutralization and extraction.
d. Treat with hot potassium permanganate (KMnO4) in aqueous acid, then neutralize and extract.
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Question 88
The Claissen-Schmidt reaction is a simple reaction (cyclohexanone plus 1 equivalent formaldehyde and sodium ethoxide catalyst) that forms what product?
Choose one answer.
a. Since the formaldehyde is more reactive, it adds the ethoxide and the resulting carbanion adds to the ketone carbonyl followed by loss of water to give cyclohexenylmethanal.
b. Abstraction of proton alpha to the ketone followed by attack at the formaldehyde carbonyl followed by a two-step proton exchange and water loss to form the alpha methylene of cyclohexanone.
c. The cyclohexanone dimerizes and then formaldehyde donates a hydride to reduce the remaining ketone to give the diol.
d. The ethoxide attacks the formaldehyde first (more reactive carbonyl) and the adduct attacks the ketone to form the acetal.
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Question 89
Ketones and aldehydes react with HCN to give what?
Choose one answer.
a. Poisonous materials that are extremely dangerous due to reversible release of the HCN.
b. A member of the family of adducts called cyanohydrins, which can be hydrolyzed to the alpha-hydroxy acid.
c. The dicyano compound with both CN's attached to the carbonyl carbon and one equivalent of water released.
d. Reaction at the alpha-position of the enol form of the carbonyl to give a beta-cyano ketone or aldehyde.
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Question 90
The aldol reaction of acetaldehyde (2 carbons) with sodium ethoxide catalyst gives what?
Choose one answer.
a. Addition of two equivalents of the ethoxide to the carbonyl carbon to form the acetal.
b. No reaction- ethoxide is not a strong enough base to actually catalyze any reaction of acetaldehyde.
c. A dimer joined at what were the carbonyl carbons, the product named as 2, 3-butandiol.
d. A dimer of the aldehyde (kind of) that could be named 3-hydroxy-1-butanal.
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Question 91
Cyclopentanone reacts with the non-nucleophilc but very strong base NaH to form an anion that then reacts with methyl iodide to give what product?
Choose one answer.
a. The base removes an alpha proton to give the enolate anion which nucleophilically attacks MeI to displace the iodide and form the alpha-methylcyclopentanone.
b. The initial enolate anion formed adds the iodine to the alpha carbon while the methyl ends up on the oxygen.
c. Initial enolate anion abstracts a proton from the methyl iodide carbon, which then attacks the carbonyl to form alpha- iodomethyl-hydroxycyclopentane.
d. No reaction occurs- NaH is not a strong enough base to deprotonate a carbon alpha to a carbonyl.
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Question 92
You try to react 1, 5-cyclooctadione with chlorine using NaH as catalyst. You forget to weigh the NaH and end up using 2 equivalents which gives you a product with no chlorine but different from the starting material: what is it?
Choose one answer.
a. The first step goes fine to give 2, 6-dichloro-1, 5-cyclooctadione. The second NaH dehydrohalogenates to give a double bond conjugated to each of the ring carbonyls.
b. The NaH reduces both carbonyls to alcohols.
c. The first step gives 2-chloro-1, 5-cyclooctadione but the second NaH reduces off the chlorine to give back the starting material.
d. The first step works fine, with deprotonation alpha to one carbonyl followed by reaction with chlorine to give the alpha-chloro species. However, the second NaH dehydrochlorinates to give the cyclooctadione with a double bond conjugated to one of the ketones.
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Question 93
Why are carboxamides are such wonderful and useful derivatives of carboxylic acids?
Choose one answer.
a. Didn't know they were, but if you say so, it must be because they are so reactive.
b. Well, if nature uses them in peptides and proteins, they most offer something useful, like maybe hydrolytic stability and intermolecular hydrogen bonding?
c. The sp3 hybridized nitrogen is almost as basic as a simple amine but with the added contribution of the attached carbonyl to enhance metal binding.
d. There so easy to make that all living creatures can just ingest amines from food and convert them to amides in the stomach.
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Question 94
Which of the following statements is true about a carboxylic derivative or reaction?
Choose one answer.
a. Nitriles have a carbon-nitrogen "double bond" and are also called Shiff bases.
b. An imide is a carbonyl with two nitrogens attached (like ureas).
c. The key atoms of an amide bond (carbon and oxygen of the carbonyl plus the amide of the nitrogen and its hydrogen) all lie in the same plane making for a rigid, conformationally restrained segment (like in an enzyme).
d. Nature uses acid chlorides (acyl halides) in the body to make natural amides such as peptides.
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Question 95
Even without a topic on polymers, can you pick out which of the following is not true?
Choose one answer.
a. The common polymer cellulose is found in virtually all living organisms.
b. Polyester used in clothing and soda bottles is made from terephthalic acid and ethlene glycol (the same chemical used in antifreeze).
c. Lipids are long-chain alkyl esters (of fatty acids) found in many living organisms.
d. RNA and DNA are polyesters made from inorganic phosporic acid and substituted sugar molecules.
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Question 96
Carboxylic acids can react by:
Choose one answer.
a. electrophilic attack of an amine on the central carbon of the acid group.
b. nucleophilic attack of the polarized central carbon of the acid group on alcohols and amines.
c. conversion to acid chlorides with thionyl chloride, a high-energy and very reactive inorganic acid chloride.
d. rehybridization of the central sp2 carbon to an sp3 one.
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Question 97
Synthesis of a carboxylic acid can occur by:
Choose one answer.
a. reaction of an alcohol with a strong oxidizing agent like KMnO4.
b. reaction with oxygen in the air catalyzed by heat.
c. conversion of the alcohol OH to an alkyl halide with thionyl chloride.
d. reaction of an alkene with a peracid or peroxide plus acid catalyst.
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Question 98
Amides are derivatives of carboxylic acids that:
Choose one answer.
a. can be formed by simple mixing of an amine with an acid.
b. first reacting the amine with thionyl chloride and then adding the acid.
c. first reacting the acid with sodium hydroxide and then adding the amine.
d. first reacting the acid with thionyl chloride and then adding the amine.
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Question 99
Carboxylic acids have a unique combination of:
Choose one answer.
a. a central sp2 hybridized carbon, one sp2 hybridized oxygen and one sp3 hybridized oxygen.
b. an sp2 hybridized carbon that can accept a proton from another acid's sp3 hybridized OH moiety.
c. basicity and nucleophilicity to substitution reactions.
d. none of the above.
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Question 100
Carboxylic acids can react to:
Choose one answer.
a. form alcohols by oxidation with peracids.
b. form alkyl bromides by reaction with PBr3.
c. form amides by reaction with amines, heat and water removal by distillation.
d. form an ester by reaction with an ether.
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Question 101
A general synthesis of carboxylic acids involves:
Choose one answer.
a. first conversion of an alcohol to an alkyl iodide and then reaction with oxygen.
b. alcohol reaction with strong oxidizing agents like hydrogen gas and platinum catalyst.
c. conversion of an alkyl iodide to a Grignard reagent with magnesium followed by quenching with dry ice.
d. reaction of an alkyl bromide with sodium hydroxide in water.
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Question 102
Esters can be made from carboxylic acids by:
Choose one answer.
a. none of the answers below.
b. mixing the acid with excess primary alcohol and distilling off water.
c. mixing an acid with a tertiary alkyl bromide and KOH in t-butyl alcohol.
d. first converting to the acid chloride with thionyl chloride and then adding alcohol and dilute base.
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Question 103
Ketones can be made from carboxylic acids by:
Choose one answer.
a. first converting the acid to an acid chloride then mixing with toluene and iron chloride, and heating for a while.
b. adding a nail to a mixture of a tertiary bromide and benzene, and heating.
c. simple reduction with hydrogen and a transition metal catalyst.
d. reacting the acid with KOH and then heating with phenyl bromide.
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Question 104
Acid chlorides are highly reactive derivatives of carboxylic acids that can:
Choose one answer.
a. be made from alkyl alcohols by reaction with thionyl chloride.
b. react with themselves to form ketones.
c. be reduced to ketones with hydrogen and a platinum catalyst.
d. react with amines and alcohols to give amides and esters.
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Question 105
The basic order of stability of carboxylic acids is (pick one).
Choose one answer.
a. acids > acyl halides > esters > anhydrides > amides
b. amides >> acids ~ esters > anhydrides > acyl halides
c. esters > amides > acids >> anhydrides > acyl halides
d. amides >> acids > esters > acyl halides > anhydrides
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Question 106
Which is the best way to make an amide from a carboxylic acid derivative?
Choose one answer.
a. Use two equivalents of amine with one equipment of acid chloride.
b. React an ester with an amine through heat-induced transamidation.
c. Use an anhydride with two equivalents of amine, one for each of the acid equivalents (oh, and add heat).
d. Use the classic Shotten-Baumen conditions of one equivalent of acid chloride in an organic solvent and one equivalent of amine plus one equivalent of dilute inorganic base in water, stir rapidly, chill as needed.
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Question 107
Carboxylic acids form salts with amines that can be useful how?
Choose one answer.
a. They're more water soluble as the salt, and if one is a medication, this can help in delivery.
b. You can heat the salt up under vacuum and form an amide (usually).
c. You can extract impurities away using water plus an organic solvent.
d. All of the above, in a practical sense.
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Question 108
Under what conditions can hydrolysis of a carboxylic ester or amide lead back to the acid?
Choose one answer.
a. Simply mixing with water and stirring at room temperature.
b. Boiling with alcohol and acid catalyst.
c. Boiling with water and base catalyst.
d. Treating with NaH and ethanol.
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Question 109
Which acid derivative and what reaction conditions would give an aryl ketone?
Choose one answer.
a. Heating an anhydride with the aromatic in refluxing water containing a strong acid catalyst.
b. Slowly adding a carboxylic acid to a mixture of concentrated nitric acid and the aromatic compound.
c. Carefully adding an aryl Grignard reagent to a carboxylic acid dissolved in an organic solvent.
d. Reaction of an acid chloride with a reasonably activated aromatic using a strong Lewis acid catalyst.
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Question 110
What is the key intermediate(s) of the reactions of many acid derivatives?
Choose one answer.
a. That would have to be the Td or tetrahedral intermediate involved in the most common type, the addition-elimination mechanism.
b. Trick question: there is only a transition state in most acid reactions.
c. Acid reaction mechanisms are very much like an Sn2 reaction with simultaneous bond-formation and bond-breaking.
d. Acid reaction mechanisms are very much like an Sn1 mechanism with the loss of the leaving group occurring first to give an acylonium ion.
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Question 111
How would you most efficiently (key word) convert an acid derivative to a primary amine?
Choose one answer.
a. Reduce an acid to an alcohol, treat with thionyl chloride and add excess ammonia.
b. Make a primary-nitrogen amide and then reduce with excess hydrogen and a very reactive transition metal catalyst.
c. Mix ammonia and a carboxylic acid together with rapid stirring in water, then distill the water under vacuum.
d. Heat an alkyl ether or alcohol with ammonium hydroxide, then distill.
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Question 112
Carboxylic acids are different than alcohols because:
Choose one answer.
a. they have an exchangeable hydrogen on the OH.
b. their OH proton is easily removed by bases even as weak as water.
c. both oxygens attached to the acid carbon are sp2 hybridized.
d. all of the above.
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.
Question 113
The genetic code of all plants and animals is composed of what polymer?
Choose one answer.
a. Natural (what else?) polyamides called DNA and RNA which contain nucleic acid bases in the repeat units.
b. Natural polyamides making up collegen, proteins and peptides.
c. Lipid polymers that contain nucleic acid bases organized in three-unit sequences.
d. Polymers of nucleic acid bases with phophoric acid to give complex polyesters abbreviated RNA and DNA.
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Question 114
Sucrose is an excellent example of what kind of natural material?
Choose one answer.
a. Natural monosaccharides such as hexoses and pentoses, although sucrose is a dimer sugar with two different monosaccharides.
b. Oligomers and polymers of glucose, the most common of the sugars, with sucrose composed of two glucose monosaccharides.
c. An artificial sweetener made of glucose and fructose units with non-natural stereochemistry at the anomeric carbons.
d. The monosaccharide that forms part of the polymer chain in RNA and DNA.
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Question 115
Proteins are complex oligomers and polymers made up of what?
Choose one answer.
a. Sugar units (ribose with pendent nucleic bases) joined to each other with phosphate ester linkages to make RNA and DNA molecules.
b. Complex carbohydrates containing both saccharide units linked with amino acid groups.
c. Oligomers and polymers of the group of 23 or so natural amino acids linked through amide bonds.
d. Ester- and amide-linked units of various combinations of the natural materials we ingest in our diets such as citric acid and lysine.
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Question 116
Trees and cotton are made of what natural polymer?
Choose one answer.
a. The polysaccharide of glucose called starch; it's also found in legumes and potatoes.
b. The polysaccharide of glucose called cellulose, a polymer with strong intermolecular hydrogen bonds that give these materials such excellent physical properties.
c. Proteins such as collagen that are strong and have extensive intermolecular hydrogen bonding holding them together.
d. The polysaccharide with sucrose repeat units, which is why sucrose is so cheap and readily available.
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Question 117
Hair and skin are made of the same natural polymer as rhino horns: what is it?
Choose one answer.
a. A complex carbohydrate containing both saccharides and amino acids in the polymer repeat units.
b. A natural polymer called starch made up of glucose repeat units.
c. A complex mixture of inorganic phosphate and cellulose that has extremely strong intermolecular forces holding the chains together.
d. A natural polyamide or protein with a unique arrangement intra- and intermolecular hydrogen bonds along the chains.
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Question 118
If you wanted to make your own soap, how would you do it?
Choose one answer.
a. Take cotton fibers and mix with a strong base such as lye, add water, boil till the fibers dissolve, filter and evaporate the water.
b. Add fur or wool to water containing strong acid, boil till everything dissolves, filter and evaporate the water to give the slippery crude soap.
c. Start with animal fat or a plant oil, mix with lye (strong inorganic base) and water, heat till all the fat or oil dissolves, filter and evaporate the water to give the fatty acid salt (soap).
d. Clot cream with acetic acid, mix the curds you get with yeast and let ferment until the bubbling stops; evaporate the water to get soap.
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Question 119
There are 23 or so natural amino acids: name a few.
Choose one answer.
a. Lysine, alanine, glycine and aspartic acid for starters.
b. Serine, glutamic acid, thymine and adenine are the less common ones.
c. Steric acid, glucose, alanine and citric acid are the most common ones.
d. Lipoic acid, proline, lysergic acid and pyridine are the essential amino acids, meaning we must have them in our diet or we get sick.
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Question 120
RNA is fundamentally different from DNA in what ways?
Choose one answer.
a. RNA is the genetic material passed from parents to offspring while DNA transfers that genetic code into components of cells and human bodies.
b. RNA uses different nucleic acid bases (different than adenine and thymine, for example) to carry the genetic information to the enzymes that construct our bodies.
c. The sugar in RNA is ribose and the one in DNA is deoxyribose, but more important, DNA makes up our genes while RNA does other stuff.
d. There is no chemical difference in terms of nucleic acid bases and sugar units, although DNA is a much higher molecular weight polymer than RNA.
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Question 121
Of the 20 natural amino acids listed in the reading, 10 are called "essential", meaning what?
Choose one answer.
a. That without them, you would die a slow, painful death.
b. You must take vitamins to make sure you get enough of these essential amino acids to survive.
c. Your body doesn't really need or use the non-essential ones, while the essential ones make up all the proteins your body needs.
d. While your body can "manufacture" the non-essential ones from what you eat, it can't make the essential ones so you have to make sure you eat the right balance of foods to get them in your diet.
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Question 122
Peptides and proteins are polymers of amino acids, and what is their basic structure?
Choose one answer.
a. Peptides and proteins contain alpha-amino acids forming a two-carbon plus one-nitrogen polyamide with each repeat unit usually having a pendent or attached side chain of some kind.
b. The amine group of each amino acid is protonated by the acid group of the one next to it forming a long chain of ammonium-carboxylate salts.
c. The acid groups react with alcohol groups of other amino acids to form polyesters with amine-containing pendent groups along the chains.
d. The amines, alcohols and thiols of the pendent groups all react with the acid groups to form complex 3D structures that each have unique properties such as in enzymes.
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.
Question 123
The amide groups of peptides, proteins and enzymes have what unique property?
Choose one answer.
a. The nitrogen group is basic and can form ammonium salts with carboxylic acids of other amino acids.
b. Like ketones, they can tautomerize to switch the hydrogen on nitrogen over onto the carbonyl oxygen, leading to enhanced hydrogen bonding.
c. Because of the sp2 hybridization of the carbonyl and attached nitrogen, the amide group is "flat" and capable of strong hydrogen bonding with other amide groups.
d. The free rotation around the carbonyl bond to the nitrogen allows these amino acid polymers to flex and bend as needed in the body.
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.
Question 124
Lipids can best be described as what?
Choose one answer.
a. The main constiuents of natural oils (e.g., canola oil) that are polymers of long-chain hydroxyacids.
b. Triglycerides, or long-chain acid triesters of glycerine (1, 2, 3-propanetriol).
c. Unsaturated long-chain aliphatic alcohols that form organized structures such as cell walls in plants and animals.
d. The main components of animal fats (ours included), milk fats and butter.
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.
Question 125
Basically, all spectroscopic methods can be described as what?
Choose one answer.
a. Using the interaction of radio waves with molecules to understand molecular structure and properties.
b. Using energy to force molecules to change shape or structure in such a way that we can observe it with a microscope.
c. Using visible or ultraviolet light to illuminate molecules so we can see them with a microscope.
d. Exciting molecules with heat and light so that they emit vibrational energy that can be detected and analyzed with spectrometers.
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Question 126
The relative energies used in various kinds of spectroscopic methods vary how?
Choose one answer.
a. From low energy ultraviolet rays to infrared and then to microwaves used in NMR analysis and finally high energy electrons in mass spectrometry.
b. From high frequency radio waves to lower frequency electrons for NMR to infrared to UV to mass spectrometry.
c. You can't really compare the energies since each type of spectroscopy uses a unique mixture of frequencies.
d. It's not really the energy used, since this can vary independently of frequency or wavelength.
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Question 127
Mass spectroscopy can best be described as:
Choose one answer.
a. mass analysis of whole molecules that have been given a static electricity charge using a voltameter.
b. fragmentation of molecules with a magnetic field followed by analysis by charge.
c. breakdown and ionization of molecules with high energy electrons followed by magnetic separation and analysis of mass-to-charge ratios of the ions.
d. careful weighing of the molecules with super-sensitive nano-balances constructed with microlithography.
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Question 128
Fragmentation "patterns" in mass spectrometry tell us what about the molecules?
Choose one answer.
a. What kinds of impurities were present and how much of each was there.
b. How many atoms of each kind were present in the original molecule.
c. What functional groups were present so that we can determine the functionality of the original molecules.
d. The atomic compositions of segments of the molecule that we re-assemble virtually to determine the original structure.
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Question 129
Ultraviolet spectroscopy looks at what kinds of molecular segments?
Choose one answer.
a. All types of bonding from sigma bonds to pi bonds to conjugated and aromatic bonding.
b. Double bonds, aromatic rings and conjugated unsaturation, and non-bonded heteroatoms all absorb in the UV spectrum and can be detected.
c. We can see the fragments that give us broad peaks or humps in the spectrum and this tells us what kinds of bonds are involved.
d. It is most sensitive to carbonyl bonds and less to aromatic and conjugated double bonds.
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Question 130
UV spectroscopy generally shows a combination of broad peaks and sharper ones that tell us what?
Choose one answer.
a. What atoms are attached to a double bond or aromatic ring.
b. How many heteroatoms are in the molecule.
c. The degree of conjugation (pi electron delocalization) and something about substitution patterns.
d. What the elemental composition of the molecule is.
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Question 131
Infrared is used mainly for determining what about a molecule?
Choose one answer.
a. What kinds of functional groups like amides, ethers and alkene double bonds are present in the molecule.
b. Its purity and what kinds of impurities are present in the mixture.
c. The exact kind and number of each functional group comprising the molecular composition.
d. The total number of atoms and what kind of each are in the molecule.
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Question 132
Infrared was one of the first spectroscopic methods and is still widely used today to characterize molecules because:
Choose one answer.
a. You can sample solids, liquids or gases although the information obtained isn't useful without combining with other techniques of analysis.
b. It tells you more about the molecule than any other technique we've studied.
c. It is easy to do, takes very little sample and gives specific information about what functional groups are present in the molecule.
d. It used the least expensive analytical equipment available.
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Question 133
How is infrared spectroscopy is different from UV and NMR?
Choose one answer.
a. It uses a different energy range from UV but the same range differently from NMR.
b. Infrared overlaps with UV in the electromagnetic spectrum providing complimentary information on functional groups while NMR looks at electronic configuration and bonding.
c. They all see different parts of the molecule, and specifically, different atoms and how they're joined together.
d. Infrared detects vibrational motions of functional groups while UV "sees" electronic transitions and NMR looks at the nuclei and infers bonding based on reference tables.
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Question 134
NMR involves what kind of analysis?
Choose one answer.
a. Spinning nuclei in such a way that they can be seen with infrared radiation and functional groups determined from the frequency.
b. Using radio waves to detect nuclei present in a molecule (such as H or C) and based on extensive data bases of "chemical shifts, " determine what each type is bonded to.
c. Magnetizing the molecules so they can be separated by traveling through a vacuum down a long tube surrounded by static magnets.
d. Using high energy beams to excite the electrons so they can be detected by magnetic field fluctuations.
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Question 135
Proton NMR tells you what about a molecule?
Choose one answer.
a. What kind of hydrogens are present (what each kind is bonded to) and exactly how many there are relative to each other.
b. Whether there are any hydrogens present in the molecule and what their chemical shift is with respect to a standard (TMS).
c. Whether electron donating or withdrawing groups are present in the molecule.
d. What kind of heteratoms are near the hydrogens you observe in the spectrometer.
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Question 136
Carbon NMR is harder to do than proton NMR because:
Choose one answer.
a. It has a higher atomic mass which makes it slower to move and relax on the NMR time scale.
b. The mechanism of interaction of the isotopes of carbon (number 13) with a magnetic field is dependent on how many hydrogens are attached to it.
c. The actual atoms of carbon observed (the isotope we can detect) is of lower relative abundance than for hydrogen.
d. The gyromagnetic ratio (NMR sensitivity factor) is much less for carbon 13 than for hydrogen 1.
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Question 137
The chemical shift range for carbon 13 in NMR is different from proton NMR how?
Choose one answer.
a. It's about the same but the relative intensity of the peaks are less, making it much harder to detect.
b. Carbon covers about a 200 pm range while hydrogen has a range of only about 10 ppm and this makes carbon more sensitive to its chemical environment (groups present).
c. It's more sensitive to solvation and hydrogen bonding which provides key information about molecular motion and interactions.
d. It's actually much less, and thus provides less specific information about what kinds of functional groups are present in the molecule.
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