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86 Cards in this Set
- Front
- Back
- 3rd side (hint)
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what is the value of n in huckel's rule when a compound has nine pairs or pi electrons? Is such a compound aromatic?
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In he case of 9 pairs of pi electrons, there are 18 electrons. Therefore, 4n+2=18 and n=4. Because it has an odd number of pairs of pi electrons, it will be aromatic if it is cyclic, planar, and if every atom in the ring has a p orbital.
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14.0.1
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which compound is aromatic?
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EDIT..it is cyclic, planar every ring atom has a p orbitaland it has one pair of pi electrons. The first compound is not aromatic b/c one of the atoms is sp^3 hybridized and, therefore does not have a p orbital. The third compound is not aromatic b/c it has two pairs of pi electrons. This is the only one that is aromatic
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14.0.3
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which compound is aromatic?
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a & b, no, two pairs of pi e and some atoms don't have p orbitals
c, no, two pairs of pi e d, yes, cyclic, planar, seven pairs of pi e e, yes, cyclic, planar, three pairs of pi e f, no, not cyclic |
14.0.4
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How many monobromonaphthalenes are there?
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two, text
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14.0.5
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How many monobromophenanthrenes are there? (three ringed)
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five
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14.0.5
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The [10]- and [12]-annulenes have been synthesized, and neither has been found to be aromatic. Explain.
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To be aromatic, a compound must be cyclic and planar, every atom in the ring must have a p orbital, and the pi cloud must contain an odd number of pairs of pi electrons.
[10]-annulene is cyclic, every atom in the ring has a p orbital, and it has the correct number of pi electrons to be aromatic (five pairs). Knowing it is not aromatic, we can conclude that it is not planar. [12]-annulene has an even number of pairs of pi electrons (six pairs) in its pi cloud, so it cannot be aromatic. |
14.0.6
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Draw arrows to show the movement of electrons in going from one resonance contributor to the next in pyrrole.
How many ring atoms share the negative charge? |
four ring atoms share the negative charge
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14.0.7
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What orbitals contain the electrons represented as lone pairs in the structures of quinoline, indole, imidazole, purine, and pyrimidine?
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14.0.8
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What orbitals contain the electrons represented as lone pairs in the structures of quinoline, indole, imidazole, purine, and pyrimidine?
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Cyclopentadiene has a lower pK_a. When cyclopentadiene loses a proton, a relatively stable aromatic compound is formed. When cycloheptatriene loses a proton, an unstable antiaromatic compound is formed. It is, therefore, easier to lose a proton from cyclopentadiene.
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14.0.10
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In what direction is the dipole moment in fulvene? explain.
In what direction is the dipole moment in calicene? explain. |
In fulvene, the electrons move toward the five-membered ring because the resonance contributor that has a negative charge on a ring carbon is aromatic.
In calicene, the electrons move towared the five-membered ring because both rings are aromatic in the resonance contributor that has a negative charge on a carbon of the five-membered ring and a positive charge on a carbon of the three-membered ring. |
14.0.12
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Predict the relative pK_a values of cyclopropene and cyclopropane
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Cyclopropane has a lower pK_a because an antiaromatic compound is formed when cyclopropene loses a proton.
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14.0.13
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Which is more soluble in water, 3-bromocyclopropene or bromocyclopropene or bromocyclopropane?
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3-Bromocyclopropene is more soluble in water, because it is more apt to ionize since heterolytic cleavage of its carbon-bromine bond forms an aromatic compound.
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14.0.13
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How many bonding, nonbonding, and antibonding pi molecular orbitals does cyclobutadiene have? In which molecular orbitals are the pi electrons?
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Cyclobutadiene has 1 bonding molecular orbital, 2 nonbonding M/O, and 1 antibonding M/O
Because each of its four atoms has a p atomic orbital, it has 4 pi electrons: 2 pi electrons are in the bonding pi M/O, and each of the two nonbonding M/O contains one pi electron. |
14.15
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can a radical be aromatic?
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No. To be aromatic, all the bonding M/O msut be filled and there must be no partially filled orbitals. Each orbital contains two electrons. Since a radical has an unpaired electron, there will be an orbital with a single electron (a partially filled M/O). A radical, therefore, cannot be aromatic. (A radical can, however, be attached to an aromatic compound. For example, the phenyl radical has a methylene radical attached to an aromatic benzene ring.)
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14.16
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structure of 2-phenylhexane
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14.18.a
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structure of benzyl alcohol
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14.18.b
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structure of 3-benzylpentane
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14.18.c
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structure of bromomethylbenzene
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14.18.d
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if electrophilic addition to benzene is an endergonic reaction overall, how can electrophilic addition to an alkene be an exergonic reaction overall?
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Because benzene is an aromatic molecule it is particularly stable. Therefore, electrophilic addition to benzene is an endergonic reaction because benzene is more stable than the non-aromatic addition product.An alkene is much less stable than benzene because it does not have any delocalized electrons. Electrophilic addition to an alkene is an exergonic reaction because an alkene is less stable than the addition product (sigma bonds pi bonds yad yada)
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14.19
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why does hydration inactivate FeBr_3?
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Ferric bromide activates Br_2 for nucleophilic attack by accepting a pair of electrons from it. Hydrated ferric bromide cannot do this because it has already accepted a pair of electrons from water.
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14.20
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Show the mechanism for the generation of the acylium ion if an acid anhydride is used instead of an acyl chloride in a Friedel-Crafts acylation reaction.
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14.0.23
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14.0.24
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Show the mechanism for the alkylation of benzene by 2-butene +HF.
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14.0.25
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12.0.26
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synthesis:
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14.0.27
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14.027.b
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give the product:
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14.0.28.a
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give the product:
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14.0.28.b
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give the product:
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14.0.28.c
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give the product:
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14.0.28.d
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show how benzaldehyde could be prepared from benzene:
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Gatterman-Koch or this:
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14.29.a
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show how styrene could be prepared from benzene:
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.
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14.29.b
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show how 1-bromo-2phenylethane could be prepared from benzene:
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14.29.c
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show how 2-phenyl-1-ethanol could be prepared from benzene:
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14.29.d
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show how aniline could be prepared from benzene:
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.
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14.29.e
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Show how benzoic acid could be prepared from benzene
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14.29.f
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structure of phenol
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14.e.30.a
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structure of benzyl phenyl ether
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14.e.30.b
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structure of benzonitrile
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14.e.30.c
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structure of benzaldehyde
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14.e.30.d
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structure of anisole
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14.e.30.e
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structure of styrene
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14.e.30.f
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structure of toluene
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14.e.30.g
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structure of tert-butylbenzene
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14.e.30.h
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structure of benzyl chloride
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14.e.30.i
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aromatic, antiaromatic, nonaromatic
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AA
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14.31.a
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aromatic, antiaromatic, nonaromatic
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A
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14.31.b
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aromatic, antiaromatic, nonaromatic
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AA
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14.31.c
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aromatic, antiaromatic, nonaromatic
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A
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14.31.d
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aromatic, antiaromatic, nonaromatic
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nonA
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14.31.e
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aromatic, antiaromatic, nonaromatic
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A
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14.31.f
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aromatic, antiaromatic, nonaromatic
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A?
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14.31.g
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aromatic, antiaromatic, nonaromatic
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A
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14.31.h
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aromatic, antiaromatic, nonaromatic
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A
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14.31.i
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aromatic, antiaromatic, nonaromatic
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A
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14.31.j
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aromatic, antiaromatic, nonaromatic
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A
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14.31.k
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aromatic, antiaromatic, or nonaromatic
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A
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14.31.l
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aromatic, antiaromatic, or nonaromatic
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A
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14.31.m
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aromatic, antiaromatic, or nonaromatic
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nonA
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14.31.n
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aromatic, antiaromatic, or nonaromatic
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nonA
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14.31.o
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aromatic, antiaromatic, or nonaromatic
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AA
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14.31.p
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Give the product of the reaction of excess benzene with iosbutyl chloride +AlCl3
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14.32.a
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Give the product of the reaction of excess benzene with propene +HF
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14.32.b
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Give the product of the reaction of excess benzene with 1-chloro-2,2-dimethylpropane +AlCl3
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14.32.c
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Give the product of the reaction of excess benzene with dichloromethane + AlCl3
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14.32.d
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Which ion is more stable and why:
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aromatic
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14.33.a
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Which ion is more stable and why:
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aromatic
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14.33.b
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Which ion is more stable and why:
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aromatic
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14.33.c
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Which ion is more stable and why:
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not antiaromatic
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14.33.d
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How could you prepare the following compound with benzene as one of the starting materials?
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14.35.a
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How could you prepare the following compound with benzene as one of the starting materials?
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14.35.b
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Benzene underwent a Friedel-Crafts acylation reaciton followed by a Glemmensen reduction. The product gave the following HNMR spectrum. What acyl chloride was used in the Friedel-Crafts acylation reaction?
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The HNMR spectrum is the spectrum of 1-phenylbutane: the benzene ring protons show a signal at ~7.2ppm. The two triplets and two multiplets indicate a butyl substituent. Therefore, the acyl chloride has a straight chain prpoyl group and a varbonyl group that will be reduced to a methylene group. pic
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14.36
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product
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14.37.a
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product
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14.37.b
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Which compound is a stronger base? why?
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left one has a resonance contributor that is aromatic. Therefor the right one is the stronger base
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14.38
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Which compound undergoes an SN1 reaction more rapidly?
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The right one forms a resonance contributor of the carbocation that is aromatic.
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14.39
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Purine. Which nitrogen is most apt to be protonated? least?
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14.40
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Professor orbie tal isolated an aromatic compound with mol formula C6H4Br2. He treated this compound with nitric acid and sulfuric acid (conditions that replace an H with an NO2 group) and isolated three different isomer with molecular formula C6H3Br2NO2. What was the structure of the original compound?
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The compound is 1,3-dibromobenzene. 1,4-dibromobenzene would form only one product and 1,2-dibromobenzene would form only two.
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14.41
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Propose a mechanism
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14.42.a
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Propose a mechanism
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14.42.b
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give product:
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14.43.a
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give product:
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14.43.b
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Which of the following compounds is the strongest acid?
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The following compound is the strongest acid because it is the only one that forms a base that is aromatic. Recall that the more stable (weaker) the base, the stronger is its conjugate acid.
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14.44
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Propose a mechanism
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14.46.a
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give product
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14.46.b
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In a reaction called the Birch reduction, benzene can be partially reduced to 1,4-cyclohexadiene by an alkali metal (Na, Li, or K) in liquid ammonia and a low-molecular-weight alcohol. Propose a mechanism for this reaction
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14.47
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