Aromatic Compounds: Reaction

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Aromatic Compounds: Reaction - Quiz

Review the different reactions that are encountered when adding groups to aromatic rings when you can no longer force yourself to do more practice problems!


Questions and Answers
  • 1. 

    What is EAS? Why does it occur?

    • A.

      EAS is electron aromatics sequestering. It occurs because regular addition would destroy the aromaticity of a ring.

    • B.

      EAS is electrophilic aromatic substitution. It occurs because addition to pi bonds will always substitute, preserving the pi bond.

    • C.

      EAS is electrophilic aromatic substitution. It occurs because regular addition would destroy the aromaticity of a ring.

    Correct Answer
    C. EAS is electrophilic aromatic substitution. It occurs because regular addition would destroy the aromaticity of a ring.
    Explanation
    EAS stands for electrophilic aromatic substitution. It occurs because regular addition would destroy the aromaticity of a ring. In electrophilic aromatic substitution, a strong electrophile replaces a hydrogen atom on an aromatic ring. This reaction is preferred over regular addition because it maintains the stability and aromaticity of the ring. If regular addition were to occur, the aromaticity of the ring would be lost, resulting in a less stable compound. Therefore, EAS is a key reaction in organic chemistry to introduce functional groups onto aromatic compounds while preserving their aromatic character.

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  • 2. 

    The first step of the EAS reaction mechanism sees an _________ (electrophile/nucleophile) add to the ring in the addition step, which is the ______ step of the reaction. The ________ ion intermediate that is formed reacts _______ (slowly/quickly) with a __________(electrophile/nucleophile), which is a base,  in the _________ (RDS/ elimination) step. Separate answers with a comma.

    Correct Answer
    electrophile, RDS, arenium, quickly, nucleophile, elimination, electrophile, rate determining step, arenium, quickly, nucleophile, elimination
    Explanation
    In the first step of the EAS reaction mechanism, an electrophile adds to the ring in the addition step, which is the rate determining step of the reaction. The arenium ion intermediate that is formed reacts quickly with a nucleophile, which is a base, in the elimination step.

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  • 3. 

    True or false: An arenium ion is not stabilized by any conjugation.

    • A.

      True

    • B.

      False

    Correct Answer
    B. False
    Explanation
    An arenium ion is stabilized by conjugation. Conjugation refers to the delocalization of electrons in a molecule, which can stabilize charged species like ions. In the case of an arenium ion, the positive charge is delocalized over the aromatic ring through resonance, which provides stability to the ion. Therefore, the statement that an arenium ion is not stabilized by any conjugation is false.

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  • 4. 

    Halogenation of benzene cannot occur with only the instantaneous dipole of the helogen gas- it requires a catalyst. Which of the following reagents-catalyst combination would be typically used in halogenation of benzene?

    • A.

      Cl2 and FeCl3.

    • B.

      Br2 and AlBr3

    • C.

      F2 and FeF3

    Correct Answer
    A. Cl2 and FeCl3.
    Explanation
    Halogenation of benzene involves replacing one or more hydrogen atoms on the benzene ring with a halogen atom. This reaction is not spontaneous and requires a catalyst to proceed. In the given options, the combination of Cl2 (chlorine gas) and FeCl3 (iron(III) chloride) is typically used as a catalyst for the halogenation of benzene. The FeCl3 acts as a Lewis acid, accepting a lone pair of electrons from the chlorine molecule, which activates the chlorine for the reaction. Therefore, Cl2 and FeCl3 is the correct reagent-catalyst combination for the halogenation of benzene.

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  • 5. 

    If I2 was used, what catalyst would allow for the best addition to benzene in a halogenation reaction? (Subscripts and superscripts are not supported on some computers for these quizzes- if they are needed type as regular number)A: __________

    Correct Answer
    Cu2+
    Cu+2
  • 6. 

    Freidel Crafts reactions can be useful to add what functional groups to aromatic rings?

    • A.

      -R and -OR

    • B.

      -R and -C(O)R,

    • C.

      -R and -OH

    Correct Answer
    B. -R and -C(O)R,
    Explanation
    Freidel Crafts reactions are commonly used to add alkyl (-R) and acyl (-C(O)R) groups to aromatic rings. This is achieved by the electrophilic substitution of a hydrogen atom on the aromatic ring with the alkyl or acyl group. The reaction involves the use of a Lewis acid catalyst, such as aluminum chloride, to activate the aromatic ring and facilitate the substitution reaction. The resulting product is an aromatic compound with the desired functional group attached to the ring.

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  • 7. 

    What catalyst should be used to add ClCH2CH3 to benzene?

    • A.

      FeCl3

    • B.

      Na2Cr2O7 and water

    • C.

      AlCl3

    • D.

      HCl

    Correct Answer
    C. AlCl3
    Explanation
    AlCl3 should be used as a catalyst to add ClCH2CH3 to benzene. AlCl3 is a Lewis acid catalyst that can facilitate the addition of the ClCH2CH3 (ethyl chloride) to the benzene ring through electrophilic aromatic substitution. It can generate a strong electrophile, CH3CH2+ (ethyl cation), which can attack the benzene ring and replace one of its hydrogen atoms. This reaction is commonly known as Friedel-Crafts alkylation. FeCl3 is also a Lewis acid catalyst that can be used for this reaction, but AlCl3 is more commonly used due to its higher reactivity. Na2Cr2O7 and water are not suitable catalysts for this reaction, and HCl is not a catalyst but a reagent that can react with benzene to form chlorobenzene.

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  • 8. 

    At its core, a Friedel-Crafts reaction requires a ______ to be produced to allow the addition of an alkly group.

    Correct Answer
    Carbocation
    C+
    Explanation
    A Friedel-Crafts reaction involves the addition of an alkyl group to a molecule. This addition is facilitated by the presence of a carbocation, which is a positively charged carbon atom. The carbocation acts as an electrophile, attracting the alkyl group and allowing it to bond to the molecule. Therefore, a carbocation is necessary for the production of the alkyl group in a Friedel-Crafts reaction.

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  • 9. 

    If a long chain alkyl group is to be added to a aromatic ring , a variation of Friedel-Crafts alkylation can be used, called Friedel Crafts _______. This produces an _________ ion intermediate, which stops the problem of ___________, because of _________ in the intermediate.Separate answers with commas.

    Correct Answer
    Acylation, acylium, rearrangement, resonance
    Explanation
    Friedel-Crafts acylation is a variation of Friedel-Crafts alkylation where a long chain alkyl group is added to an aromatic ring. This reaction produces an acylium ion intermediate, which prevents the problem of rearrangement that can occur in Friedel-Crafts alkylation reactions. This is because the acylium ion is stabilized by resonance in the intermediate.

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  • 10. 

    Match the following Friedel Crafts related reactions and reagents:1. R-C(O)-Cl and AlCl32. CO and HCl and ZnCl23. FC acylation product and HCl and Zn(Hg)4. Aldehyde formation5. Clemenson Reduction6. Friedel Crafts Acylation

    • A.

      1-4, 2-6, 3-5

    • B.

      1-5, 2-4, 3-6

    • C.

      1-5, 2-6, 3-4

    • D.

      1-6, 2-4, 3-5

    • E.

      1-6, 2-5, 3-4

    Correct Answer
    D. 1-6, 2-4, 3-5
    Explanation
    The correct answer is 1-6, 2-4, 3-5. In the Friedel Crafts acylation reaction, R-C(O)-Cl (acyl chloride) and AlCl3 (aluminum chloride) are used as the reagents. This reaction leads to the formation of an aldehyde (choice 4). Clemmensen reduction (choice 5) involves the use of HCl and Zn(Hg) to convert a ketone or aldehyde into an alkane. Therefore, choice 1-6 matches the Friedel Crafts acylation reaction, choice 2-4 matches aldehyde formation, and choice 3-5 matches Clemmensen reduction.

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  • 11. 

    Nitrate and sulfate both add to benzene to produce ____ directors. Which requires the other in a catalytic role?

    • A.

      Meta. Nitration requires H2SO4 as a catalyst.

    • B.

      Ortho-para. Nitration requires H2SO4 as a catalyst

    • C.

      Meta. Sulfonation requires HNO3 as a catalyst

    • D.

      Ortho-para. Sulfonation requires HNO3 as a catalyst

    Correct Answer
    A. Meta. Nitration requires H2SO4 as a catalyst.
    Explanation
    Nitrate and sulfate both add to benzene to produce meta directors. In the case of nitration, H2SO4 is required as a catalyst to facilitate the reaction. This is because H2SO4 can donate a proton to the nitrate ion, making it a better electrophile and enhancing its reactivity towards benzene. Sulfonation, on the other hand, requires HNO3 as a catalyst. Therefore, the correct answer is that meta directors are produced when nitration is catalyzed by H2SO4.

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  • 12. 

    The rate of subsequent EAS reactions can be affected by the groups already attached. Activating groups _____ (increase/decrease) the rate of reaction by ______  (donating/withdrawing) electrons to the pi cloud, making it _______ (less/more) nucleophilic and ______ (increase/decrease) the stability of the carbocation intermediate. Deactivating groups _________ (increase/decrease) the rate of a reaction by _______ (donating/withdrawing) electrons from the pi cloud, and ______ (increase/decrease) the stability of the carbocation intermediate.Separate answers with commas.

    Correct Answer
    Incease, donating, more, increase, decrease, withdrawing, decrease
    Explanation
    Activating groups increase the rate of reaction by donating electrons to the pi cloud, making it more nucleophilic and increasing the stability of the carbocation intermediate. Deactivating groups decrease the rate of reaction by withdrawing electrons from the pi cloud, and decreasing the stability of the carbocation intermediate.

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  • 13. 

    Which of the following are ortho-para directors?

    • A.

      -NH2

    • B.

      -NO2

    • C.

      -C(O)OH

    • D.

      --OH

    • E.

      -CH3

    Correct Answer(s)
    A. -NH2
    D. --OH
    E. -CH3
    Explanation
    The compounds -NH2, --OH, and -CH3 are ortho-para directors. Ortho-para directors are groups that donate electron density to the benzene ring, making it more reactive towards electrophilic substitution reactions at the ortho and para positions. These groups have lone pairs of electrons or are electron-donating groups, which stabilize the intermediate carbocation formed during the reaction. In contrast, NO2 and C(O)OH are meta directors, which withdraw electron density from the benzene ring, making it less reactive towards electrophilic substitution reactions.

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  • 14. 

    Which of the following are meta directors?

    • A.

      -NH-R

    • B.

      -C(O)R

    • C.

      -O-C(O)R

    • D.

      -C(O)-NH2

    • E.

      -CF3

    Correct Answer(s)
    B. -C(O)R
    D. -C(O)-NH2
    E. -CF3
    Explanation
    The compounds -C(O)R, -C(O)-NH2, and -CF3 are all examples of meta directors. This means that they direct incoming groups to the meta position in an electrophilic aromatic substitution reaction. The presence of a carbonyl group (-C=O) or a cyano group (-CN) in a molecule can result in meta directing behavior. Additionally, the presence of a strong electron-withdrawing group like -CF3 can also lead to meta directing behavior.

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  • 15. 

    A halide substituent is:

    • A.

      A meta directing activator.

    • B.

      An ortho-para directing activator

    • C.

      A meta direction deactivator.

    • D.

      An ortho-para directing deactivator.

    Correct Answer
    D. An ortho-para directing deactivator.
    Explanation
    A halide substituent is an ortho-para directing deactivator because it withdraws electron density from the benzene ring, making it less reactive towards electrophilic substitution reactions. This means that the halide substituent prefers to direct incoming electrophiles to the ortho and para positions of the ring rather than the meta position. Additionally, the halide substituent's electron-withdrawing nature reduces the overall reactivity of the benzene ring, further deactivating it towards electrophilic substitution reactions.

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  • 16. 

    What group, located para on a phenol, would make the least acidic compound?

    • A.

      -CH3

    • B.

      -O-CH3

    • C.

      -C(O)OH

    • D.

      -NH2

    • E.

      -NO2

    Correct Answer
    B. -O-CH3
    Explanation
    The -O-CH3 group, also known as a methoxy group, would make the least acidic compound. This is because the presence of the electron-donating methoxy group (-OCH3) would stabilize the negative charge that forms on the phenol molecule when it loses a proton, making it less likely to dissociate and release a hydrogen ion. Consequently, the compound would have a higher pKa value, indicating lower acidity.

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  • 17. 

    True or false: A better electron donating group makes a better acid.

    • A.

      True

    • B.

      False

    Correct Answer
    B. False
    Explanation
    A better electron donating group does not make a better acid. In fact, it is the opposite. A better electron donating group tends to stabilize the negative charge on the conjugate base, making it less acidic. This is because the electron donating group donates electron density to the conjugate base, making it more stable and less likely to donate a proton. Therefore, a better electron donating group actually makes a weaker acid.

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  • 18. 

    Diazonium Salts aer very useful in multistep synthesis. Which of the following groups are only/best added by first making a diazonium salt?

    • A.

      -Cl

    • B.

      -F

    • C.

      -CH2CH3

    • D.

      -CN

    • E.

      -C(O)R

    Correct Answer(s)
    B. -F
    D. -CN
    Explanation
    Diazonium salts are very useful in multistep synthesis because they can be used to introduce specific functional groups into a molecule. In this case, the groups that are only/best added by first making a diazonium salt are -F and -CN. This is because the diazonium salt can react with these groups to form the corresponding fluorinated or cyano compounds, respectively. The other groups listed (-Cl, -CH2CH3, -C(O)R) can also be added using other methods, but they may not be as efficient or selective as the diazonium salt method.

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  • 19. 

    What reagents are needed to get a diazonium salt from a aniline?Check all that are required.

    • A.

      H2SO4

    • B.

      HNO3

    • C.

      NaNO2,

    • D.

      Cu2O and Cu(NO3)2

    • E.

      HCl, 0 degrees C

    Correct Answer(s)
    C. NaNO2,
    E. HCl, 0 degrees C
    Explanation
    You already have an NH2 group, since this is alinine not benzene. So no nitration is necessary.

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  • 20. 

    What is the name of the following compound?

    • A.

      Phenol

    • B.

      Styrene

    • C.

      Anisole

    • D.

      Benzonitrile

    Correct Answer
    B. Styrene
    Explanation
    Styrene is the correct answer because it is the name of the compound depicted in the question. It is an organic compound that is commonly used in the production of plastics, rubber, and resins. Styrene is a colorless liquid that has a distinctive sweet smell. It is widely used in various industries due to its versatility and ability to polymerize into polystyrene, which is a widely used plastic material.

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  • 21. 

    What substituent is on anisole?

    • A.

      -NH2

    • B.

      -CH3

    • C.

      -NO2

    • D.

      -OCH3

    • E.

      -CN

    Correct Answer
    D. -OCH3
    Explanation
    Anisole is a compound that consists of a benzene ring with a methoxy group (-OCH3) as a substituent. This means that the -OCH3 group is attached to the benzene ring in anisole.

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  • Current Version
  • Mar 21, 2023
    Quiz Edited by
    ProProfs Editorial Team
  • Nov 22, 2009
    Quiz Created by
    Ochem_ownsmylife
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