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188 Cards in this Set
- Front
- Back
- 3rd side (hint)
name:
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ortho-ethylphenol or 2-ethylphenol
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15.1.a
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|
name
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meta-bromochlorobenzene or 3-bromochlorobenzene
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15.1.b
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name:
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meta-bromobenzaldehyde or 3-bromobenzaldehyde
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15.1.c
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Name:
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ortho-ethyltoluene or 2-ethyltoluene
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15.1.d
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|
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structure of para-toluidine
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.
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15.2.a
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structure of meta-cresol
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.
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15.2.b
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structure of para-xylene
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.
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15.2.c
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stucture of ortho-chlorobenzenesulfonic acid
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.
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15.2.d
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structure of m-chlorotouene
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.
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15.3.a
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structure of p-bromophenol
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.
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15.3.b
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structure of o-nitroaniline
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.
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15.3.c
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structure of m-chlorobenzonitrile
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15.3.d
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structure of 2-bromo-4-iodophenol
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.
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15.3.e
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structure of m-dichlorobenze
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.
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15.3.f
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stucture of 2,5-dinitrobenzaldehyde
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.
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15.3.g
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structure of 4-bromo-3-chloroaniline
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.
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15.3.h
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correct the name
2,4,6-tribromobenzene |
.1,3,5-tribromobenzene
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15.4.a
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|
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correct the name
3-hydroxynitrobenzene |
meta-nitrophenol or 3-nitrophenol
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15.4.b
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correct the name
para-methylbromobenzene |
para-bromotoluene or 4-bromotoluene
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15.4.c
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correct the name
1,6-dichlorobenzene |
ortho-dichlorobenzene or 1,2-dichlorobenzene
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15.4.d
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does this substituent..
withdraw electrons by inductively withdraw electrons by resonance donate electrons by hyperconjugation donate electrons by resonance |
donates electrons by resonance and withdraws electrons inductively
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12.5.a
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does this substituent..
withdraw electrons by inductively withdraw electrons by resonance donate electrons by hyperconjugation donate electrons by resonance |
donates electrons by hyperconjugation
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12.5.b
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does this substituent..
withdraw electrons by inductively withdraw electrons by resonance donate electrons by hyperconjugation donate electrons by resonance |
withdraws electrons by resonance and withdraws electrons inductively
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12.5.c
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does this substituent..
withdraw electrons by inductively withdraw electrons by resonance donate electrons by hyperconjugation donate electrons by resonance |
donates electrons by resonance and withdraws electrons inductively
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12.5.d
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does this substituent..
withdraw electrons by inductively withdraw electrons by resonance donate electrons by hyperconjugation donate electrons by resonance |
donates electrons by resonance and withdraws electrons inudctively
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12.5.e
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does this substituent..
withdraw electrons by inductively withdraw electrons by resonance donate electrons by hyperconjugation donate electrons by resonance |
withdraws electrons inductively
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12.5.f
|
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Is this substituent ortho-para or meta director?
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meta, this group withdraws electrons by resonance from the ring. The relatively electronegative nitrogen atom causes it to also withdraw electrons inductively from the ring
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15.10.a
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Is this substituent ortho-para or meta director?
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NO_2 withdraws electrons inductively and withdraws electrons by resonance.
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15.10.b
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Is this substituent ortho-para or meta director?
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meta, CH_2OH withdraws electrons inductively from the ring.
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15.10.c
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Is this substituent ortho-para or meta director?
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meta, COOH withdraws electrons inductively and withdraws electrons by resonance.
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15.10.d
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Is this substituent ortho-para or meta director?
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meta, CF_3 withdraws electrons inductively from the ring.
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15.10.e
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Is this substituent ortho-para or meta director?
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meta, N=O withdraws electrons inductively and withdraws electrons by resonance. You could draw resonance contributors for electron donation into the ring by resoance. However, the most stable resonace contributors are obtained by electron flow out of the benzene ring toward oxygen, the most electronegative atom in the compound.
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15.10.f
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Give the product(s) obtained from the reaction of the following with one equivalent of Br_2 and employing the catalyst, FeBr_3:
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.
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15.11.a
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Give the product(s) obtained from the reaction of the following with one equivalent of Br_2 and employing the catalyst, FeBr_3:
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15.11.b
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Give the product(s) obtained from the reaction of the following with one equivalent of Br_2 and employing the catalyst, FeBr_3:
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15.11.c
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Give the product(s) obtained from the reaction of the following with one equivalent of Br_2 and employing the catalyst, FeBr_3:
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15.11.d
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Which of the compounds is more acidic?
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15.12.a
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Which of the compounds is more acidic?
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.
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15.12.b
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Which of the compounds is more acidic?
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15.12.c
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Which of the compounds is more acidic?
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15.12.d
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Which of the compounds is more acidic?
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15.12.e
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Which of the compounds is more acidic?
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15.12.f
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Which of the compounds is more acidic?
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15.12.g
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Which of the compounds is more acidic?
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15.12.h
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explain why p-nitrophenol has a pKa of 7.14, whereas the pKa of m-nitrophenol is 8.39.
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When para-nitrophenol oses a proton, the electrons that held the proton can be delocalized by resonance onto the nitro substiutent. Therefore, the para nitro substituent decreases the pKa by resonance electron withdrawal and by inductive electron withdrawl.
When meta-nitrophnol loses a pront, the electrons that held the proton cannot be delocalized by resonance onto the nitro substituent. Therefore, the meta-nitro substituent can decrease the pKa only by inductive electron withdrawal. Therefore, the para isome has a lower pKa. |
15.13
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synthesize from benzene:
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15.14.a
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synthesize from benzene:
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15.14.b
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synthesize from benzene:
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15.14.c
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Give products:<br>benzonitrile + methyl chloride + AlCl3
|
no reaction
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15.15.a
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Give products:
aniline + 3Br2 |
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15.15.b
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Give products:<br>benzoic acid + CH3CH2Cl + AlCl3
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Ignore picture.
no reaction |
15.15.c
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Give products:<br>benzene +2 CH3Cl +AlCl3
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15.15.d
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synthesize p-chloroaniline from benzene
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15.16.a
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synthesize m-chloroaniline from benzene
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15.16.b
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synthesize p-nitrobenzoic acid from benzene
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15.16.c
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synthesize m-nitrobenzoic acid from benzene
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15.16.d
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synthesize m-bromopropylbenzene from benzene
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15.16.e
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synthesize o-bromopropylbenzene from benzene
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15.16.f
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synthesize 1-phenylp-2-propanol from benzene
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15.16.g
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synthesize 2-phenylpropene from benzene
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15.16.h
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major products of:
bromination of p-methylbenzoic acid |
COOH is a meta directo and Br is an ortho-para director, so both direct to the same position
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15.17.a
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|
|
major products of:
chlorination of o-benzenedicarboxylic acid |
COOH directs to the meta position. The same product will be obtained regardless of which COOH is the director.
Less of this compound will be obtained because of steric hindrance |
15.17.b
|
|
|
major products of:
cromination of p-chlorbenzoic acid |
COOH directs to its meta postion and Cl directs to its ortho postioin, so they both direct to the same position on the ring
|
15.17.c
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|
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major products of:
nitration of p-fluoroanisole |
A methoxy substituent is strongly activating whereas a fluorine substituent is deactivating so the methoxy substituent will do the directing.
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15.17.d
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major products of:
nitration of p-methoxybenzaldehyde |
The aldehyde group is a meta director and the methoxy group is an ortho-para director, so both direct to the same position
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15.17.e
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major products of:
nitration of p-tert-butyltoluene |
Less of this product will be obtained because of steric hindrance
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15.17.f
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How many products are obtained from the chlorination of o-xylene
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15.18a
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How many products are obtained from the chlorination of p-xylene
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15.18.b
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How many products are obtained from the chlorination of m-xylene
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15.18.c
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When phenol is reated with Br2, a mixture of monobromo-, dibromo-, and tribromophenols is obtained. Design a synthesis that would convert phenol primarily to ortho-bromophenol.
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the bulky sulfonic acid group will add preferentially to the para position. Both the OH and SO3H groups will direct bromine to the position ortho to the OH group. Heating in dilute acid removes the sulfonic acid group. the use of a sulfonic acid group to bloc the para position is a common strategy for synthesizing high yields of ortho-substituted compounds
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15.19
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explain why a diazonium group on a benzene ring cannot be used to direct an incoming substituent to the meta position.
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because a diazonium ion is electron withdrawing, it deactivates the benzene ring. a deactivated benzene ring would be too unreactive to undergo an electrophillic substitution reaction at the ocld temperature necessary to keep the benzenediazonium from decomposing.
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15.20
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Explain why HBr should be used to generate the benzenediazonium salt if bromobenzene is the desired product of the Sandmeyer reaction, whereas HCL should be used if chlrobenzene is the desired product.
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More of the desired bromo-substituted compound will be obtained if the halide ion in solution is br- than if it is br- can cl-
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15.21
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Why is FeBr3 not used as a catalyst in:
|
FeBr3 would complex with the amino group, converting it into a meta director.
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15.22
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write the sequence of steps required fot hte conversion of benzene into benediazonium chloride
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15.23
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show how m-dibromobenzene could be synthesized from benzene:
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15.24.a
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show how o-chlorophenol could be synthesized from benzene:
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15.24.c
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show how m-nitrotoluene could be synthesized from benzene:
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15.24.d
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show how p-metabenzonitrile could be synthesized from benzene:
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15.24.e
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show how m-chlorobenzaldehyde could be synthesized from benzene:
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15.24.f
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show how m-bromophenol could be synthesized from benzene:
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15.24.b
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In the mechanism for electrophilic aromatic substitution with a diazonium ion as the electrophile, why does nucleophilic attack occur on the terminal nitrogen atom of the diazonium ion rather than on the nitrogen atom bonded to the benzene ring?
|
You can see why nucleophilic attack occurs on the neutral nitrogen if you compare the products of nucleophilic attack on the two nitrogens. Nucleophilic attack on the neutral nitrogen forms a stable product, whereas nucleophilic attack on the positively charged nitrogen would gorm an unstable compound with two charged nitrogen atoms.
The terminal nitrogen is electrophilic because of electron withdrawal by the positively charged nitrogen. If you draw the resonance contributors, you can see that the "neutral" nitrogen is electron deficient. |
15.25
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Give the structure of the activated benzene ring and the diazonium ion used in the synthesis of butter yellow
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15.26.a
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Give the structure of the activated benzene ring and the diazonium ion used in the synthesis of methyl orange
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15.26.b
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What products would be formed from the reaction of isopropylamine with sodium nitrite and aqueous HCl?
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The reaction of an amine with sodium nitrite and HCl converts the amino gropu into an excellent leaving group. Substitution and elimination reactions can then occur by both SN1/E1 and SN2/E2 pathways. The same products are obtained by both pathways. Because the reaction is carried out in an aqueous solution, the nucleophiles are Cl- and H2O.
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15.27
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Diazomethane can be used to convert a carboxylic acid into a methyl ester. Propose a mechanism for this reaction.
|
The first step in the reaction is formation of the methyldiazonium ion as a result of removal of a proton from the carboxylic acid by diazomethane. Diazomethane is both explosive and toxic, so it should be synthesized only in small amounts by experienced laboratory workers. In the second step of the reaction, the carboxylate ion displaces nitrogen from the methyldiazonium ion. High yields are obtained, since the only side product is the N2 gas.
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15.28
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List the following compounds in order of decreasing tendency to undergo nucleophilic aromatic substitution:
Chlorobenzene 1-chloro-2,4-dinitrobenzene p-chloronitrobenzene |
1-chloro-2,4-dinitrobenzene > p-chloronitrobenzene > Chlorobenzene
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15.30.a
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List the following compounds in order of decreasing tendency to undergo electrophilic aromatic substitution:
Chlorobenzene 1-chloro-2,4-dinitrobenzene p-chloronitrobenzene |
Chlorobenzene > p-chloronitrobenzene > 1-chloro-2,4-dinitrobenzene
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15.30.b
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Show how o-nitrophenol could be synthesized from benzene.
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15.31.a
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Show how p-nitroaniline could be synthesized from benzene.
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15.31.b
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Show how p-bromoanisole could be synthesized from benzene.
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15.31.c
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Show how p-bromoanisole could be synthesized from benzene. EDIT
|
. EDIT
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15.31.d
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Propose a mechanism to account for the fact that when a basic solution of p-chlorotoluene is heated with sodium amide in liquid ammonia, both p-cresol and m-cresol are formed.
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15.32
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Give the product when this compound is reacted with sodium amide in liquid ammonia
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15.33.a
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Give the product when this compound is reacted with sodium amide in liquid ammonia
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15.33.b
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Give the product when this compound is reacted with sodium amide in liquid ammonia
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15.33.c
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Give the product when this compound is reacted with sodium amide in liquid ammonia
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15.33.d
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structure m-ethylphenol
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15.34.a
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structure p-bitrobenzenesulfonic acid
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15.34.b
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structure (E)-2-phenyl-2-pentene
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15.34.c
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structure o-bromoaniline
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15.34.d
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structure 2-chloroanthracene
anthracene = 3 rings |
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15.34.e
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structure m-chlorostyrene
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15.34.f
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structure o-nitroanisole
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15.34.g
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structure 2,4-dichlorotoluene
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15.34.h
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name:
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m-bromobenzoic acid
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15.35.a
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name:
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1,2,4-tribromobenzene
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15.35.b
|
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name:
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2,6-dimethylphenol
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15.35.c
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name:
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p-nitrostyrene
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15.35.d
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name:
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m-ethylanisole
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15.35.e
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name:
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3,5-dichlorobenzenesulfonic acid
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15.35.f
|
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name:
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o-bromotoluene
|
15.35.g
|
|
name:
|
p-cyclhexyltoluene
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15.35.h
|
|
name:
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2-chloro-4-ethylazobenzene
|
15.35.i
|
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products
|
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15.38.a
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products
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15.38.b
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products
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15.38.c
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products
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15.38.d
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products
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15.38.e
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products
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15.38.f
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products
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15.38.g
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products
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15.38.h
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Rank the following substituted anilines in order of decreasing basicity:
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15.39
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Explain why anisole is nitrated more rapidly than thioanisole under the same ocnditions.
|
EDIT
|
15.40
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The compound with the following HNMR spectrum is known to be highly reactive toward electrophilic aromatic substitution. Identify the compount.
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15.41
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Show how m-chlorobenzenesulfonic acid could be synthesized from benzene
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15.42.a
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Show how m-chloroethylbenzene could be synthesized from benzene
|
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15.42.b
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Show how benzyl alcohol could be synthesized from benzene
|
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15.42.c
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Show how m-bromobenzonitrile could be synthesized from benzene
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15.42.d
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Show how 1-phenylpentane could be synthesized from benzene
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15.42.e
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Show how m-hydroxybenzoic acid could be synthesized from benzene
|
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15.42.f
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Show how m-bromobenzoic acid could be synthesized from benzene
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15.42.g
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|
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Show how p-cresol could be synthesized from benzene
|
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15.42.h
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a) the most reactive in an electrophilic aromatic substitution
b)the least reactive in an electrophilic aromatic substitution c)highest yield of the meta product |
|
15.43.a
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a) the most reactive in an electrophilic aromatic substitution
b)the least reactive in an electrophilic aromatic substitution c)highest yield of the meta product |
|
15.43.b
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a) the most reactive in an electrophilic aromatic substitution
b)the least reactive in an electrophilic aromatic substitution c)highest yield of the meta product |
|
15.43.c
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|
|
arrange in decreasing reactivity towards EAS
benzene, ethylbenzene, chlorobenzene, nitrobenzen, anisole |
anisole > ehtylbenzene > benzene > chlorobenzene > nitrobenzene
|
15.44.a
|
|
|
arrange in decreasing reactivity towards EAS
1-chloro-2,4-dinitrobenzene, 2,4-dinitrophnol, 2,4-dinitrotoluene |
2,4-dinitrophenol > 2,4-dinitrotoluene >1-chloro-2,4-dinitrbobenzene
|
15.44.b
|
|
|
arrange in decreasing reactivity towards EAS
toluene, p-cresol, benzene, p-xylene |
p-cresol > p-xylene >toleuene >benzene
|
15.44.c
|
|
|
arrange in decreasing reactivity towards EAS
benzene, benzoic acid, pheno, propylbenzene |
phenol > propylbenzene > benzene benzoic acid
|
15.44.d
|
|
|
arrange in decreasing reactivity towards EAS
p-bitrotoluene, 2-chloro-4-nitrotoluene, p-chlorotoluene |
p-chlorotoluene > p-nitrotoluene > 2-chloro-4-nitrotoluene > 2,4-dinitrotoluene
|
15.44.e
|
|
|
arrange in decreasing reactivity towards EAS
bromobenzene, chlorobenzene, fluorobenzene, iodobenzene |
fluorobenzene > chlorobenzene > bromobenzene > iodobenzene
|
15.44.f
|
|
products:
|
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15.45.a
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products:
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15.45.b
|
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products:
|
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15.45.c
|
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products:
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15.45.d
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products:
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15.45.e
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products:
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15.45.f
|
|
|
describe two ways to prepare anisole from benzene
|
|
15.47
|
|
indicate which carbon would be nitrated
|
|
15.48.a
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|
indicate which carbon would be nitrated
|
|
15.48.b
|
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indicate which carbon would be nitrated
|
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15.48.c
|
|
indicate which carbon would be nitrated
|
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15.48.d
|
|
indicate which carbon would be nitrated
|
|
15.48.e
|
|
indicate which carbon would be nitrated
|
|
15.48.f
|
|
indicate which carbon would be nitrated
|
|
15.48.g
|
|
indicate which carbon would be nitrated
|
|
15.48.h
|
|
|
show how N,N,N-trimethylanilinium iodide could be synth from benzene
|
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15.49.a
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|
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show how benzyl methyl ether could be synth from benzene
|
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15.49.b
|
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show how p-benzylchlorobenzene could be synth from benzene
|
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15.49.c
|
|
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show how 2-methyl-4-nitrophenol could be synth from benzene
|
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15.49.d
|
|
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show how p-nitroaniline could be synth from benzene
|
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15.49.e
|
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show how m-bromoiodobenzene could be synth from benzene
|
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15.49.f
|
|
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show how p-dideuteriobenzene could be synth from benzene
|
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15.49.g
|
|
|
show how --nitro-N-methylaniline could be synth from benzene
|
|
15.49.h
|
|
Which of the following compounds will react with HBr more rapidly?
|
The compound with the methoxy substiuent is the more reactive b/c it forms the more stable carbocation intermediate. The carbocation intermediate is stabilized by resonance electron donation.
|
15.50
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|
|
Would m-xylene or p-xylene react more rapidly with Cl2 +FeCl3? explian
|
m-xylene will react more rapidly. In m-xylene both methyl groups activate the same position, whereas in p-xylene each methyl group activates a different position
|
15.51
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|
What products would be obtained after reacting with chromate and H+
|
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15.52.a
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What products would be obtained after reacting with chromate and H+
|
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15.52.b
|
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What products would be obtained after reacting with chromate and H+
|
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15.52.c
|
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A student had prepared three ethyl-substuted benzaldehydes, but had neglected to label them. The premed student at the next bench said they could be identified by brominating a sample of each and determining how many bromo-substitued products were formed. Is the premed student's advice.
|
yes, the advice is sound. the para isomer will form one product, because the formyl and ethyl groups both direct to the same position and both positions result in the same product. The ortho isomer will form two products, b/c the formyl and ethyl groups both direct to the same positions but different products are obtained from each position. The meta iosmer will form as many as four products, because the formyl and ethyl group direct to four different positions and a different prodcut is obtained from each position.
|
15.53
|
|
When heated with acidic solution of sodium dichromate, compound A forms benzoic acid. Identify compound A from it HNMR spectrum
|
The spectrum indicates that the benzene ring has a substituent with two different kinds of hydrogens. The doublet and multiplet indicate that the substituent is an isopropyl group. Therefore, compound A is isopropylbenzene.
|
15.57
|
|
Which is more stable?
|
The hydroxy substituted carbocation intermediate is more stable because the positive charge cab be stabilized by resonance electron donation from the OH group
|
15.59.a
|
|
Which is more stable?
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The carbanion with the negative charge meta to the nitro group is more stable because a negative charge in the meta position can be delocalized onto the nitro group but a negative charge in the ortho position cannot.
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15.59.b
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Describe two synthetic routes for the preparation of p-methoxyaniline with benzene as the starting material.
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15.58
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if phenol is allowed to sit in D2O that contains a small amout of D2SO4, what products will be formed?
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15.60
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synth from benzene:
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15.61.a
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synth from benzene:
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15.61.b
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synth from benzene:
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15.61.c
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An unknown compound reacts with ethyl chloride and AlCl3 to form a compound that has the following HNMR spectrum. Give the structure.
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the unsplit signal at 7.1 ppm suggests that all the hydrogens of the benzene ring in the product are chemically equivalent.
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15.62
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p-Fluoronitrobenzene is mors reactive than p-chloronitrobenzene toward hydroxide ion. 'what does this tell you about the rate-determining step for aromatic substitution
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a flouro substituent is more electronegative than a chloro substituent. therefore, nucleophillic attack on the karbon bearing the fluoro substituent will be easier.
a fluoro substituent is a stronger base than a chloro substituent , so eliminition of the halogen in the second step of the reaction will be harder for a fluoro substituted benzene the fact teat the fluoro substituted compound is the more reactive tells you that attack of the nucleophile on the aromatic ring is the rate-determining step. |
15.65
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show how meta-aminoanisole could be synth from benzene,involving a benzyne intermediate
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15.66.a
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show how meta-aminoanisole could be synth from benzene, not involving a benzyne intermediate
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15.66.b
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explain why the following reaction leads to the products shown:
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the alkyl diazonium ion is very unstable. loss of N2 and a 1,2 hydride shift forms a tert butyl carbocation, which can undergo either substitution or elimination.
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15.67.a
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What product would bee obtained from the following reaction
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the cation formed from the diazonium ion wemm undergo a pinacol rearrangement.
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15.67.b
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describe how mescaline could be synth from benzene
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15.68
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propose a mechanism for the following reaction that explains why the configuration of the asymmetric center in the reactant is retained in the product:
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the configuration of the asymmetric center in the reactant will be retained only if the asymmetric center undergoes two successive SN2 reactions.
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15.69
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explain why hydroxide ion catalyzes the reaction of piperidine with 2,4 nitroanisole, but has no effect on the reaction of piperidine with 1-chloro-2,4-dinitrobenzene.
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a chloro group is a better elaving group thatn the ammonium group, so the product is formed without hydroxide ion catalysis.
A methoxy group is a poorer leaving group than the ammonium group, so the ammonium group is eliminate, reforming starting materials. If hydroxide is added to the solution, hydroxide ion converts the ammonium group into an amino group. Since the amino group is a poorer leaving group than the methoxy group, the methoxy group is eliminated |
15.70
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show how o-chlorophenol could be synthesized from benzene:
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15.24.c.b
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