Benzene and Aromatic Compounds

Organic Chemistry ยท 73 exercises

Q.51.

Question: Use the inscribed polygon method to show the pattern of molecular orbitals in cyclooctatetraene.


a. Label the MOs as bonding, antibonding, or nonbonding.

b. Indicate the arrangement of electrons in these orbitals for cyclooctatetraene, and explain why cyclooctatetraene is not aromatic.

c. Treatment of cyclooctatetraene with potassium forms a dianion. How many π electrons does this dianion contain?

d. How are the π electrons in this dianion arranged in the molecular orbitals?

e. Classify the dianion of cyclooctatetraene as aromatic, antiaromatic, or not aromatic, and explain why this is so.

3 step solution

Q.52.

Question: Use the inscribed polygon method to show the pattern of molecular orbitals in cyclonona-1,3,5,7-tetraene and use it to label its cation, radical, and anion as aromatic, antiaromatic, or not aromatic.


3 step solution

Q.53.

Question: How many C13 NMR signals does each compound exhibit?

3 step solution

Q.54.

Question: Which of the diethylbenzene isomers (ortho, meta, or para) corresponds to each set of C13 NMR spectral data?

[A] C13NMR signals: 16, 29, 125, 127.5, 128.4, and 144 ppm

[B] C13 NMR signals: 15, 26, 126, 128, and 142 ppm

[C] C13 NMR signals: 16, 29, 128, and 141 ppm

3 step solution

Q.55.

Question: Propose a structure consistent with each set of data.

a. C10H14 : IR absorptions at 3150-28501600 , and 1500 cm-1 

b. C9H12: C13NMR signals at 21, 127, and 138 ppm

c. C8H10 : IR absorptions at 3108-2875 cm-1 , 1606, and  1496 cm-1

3 step solution

Q.56.

Question: Propose a structure consistent with each set of data.

a. Compound A:

Molecular formula:  C8H10O

IR absorption at   3150-2850 cm-1

H1 NMR data: 1.4 (triplet, 3 H), 3.95 (quartet, 2 H), and 6.8–7.3 (multiplet, 5 H) ppm

 

b. Compound B:

Molecular formula:  C9H10O2

IR absorption at   1669 cm-1

 H1 NMR data: 2.5 (singlet, 3 H), 3.8 (singlet, 3 H), 6.9 (doublet, 2 H), and 7.9 (doublet, 2 H) ppm

3 step solution

Q.57.

Question: Thymol (molecular formula C10H14O ) is the major component of the oil of thyme. Thymol shows IR absorptions at 3500–3200, 3150–2850, 1621, and 1585 cm-1 . The  H1 NMR spectrum of thymol is given below. Propose a possible structure for thymol.


3 step solution

Q.58.

Question: You have a sample of a compound of molecular formula C11H15NO2 , which has a benzene ring substituted by two groups, (CH3)2N- and -CO2CH2CH3  , and exhibits the given C13  NMR. What disubstituted benzene isomer corresponds to these C13  data?

3 step solution

Q.59.

Question: Explain why tetrahydrofuran has a higher boiling point and is much more water soluble than furan, even though both compounds are cyclic ethers containing four carbons.


3 step solution

Q.60.


Question: Rizatriptan (trade name Maxalt) is a prescription drug used for the treatment of migraines. 

a. How many aromatic rings does rizatriptan contain? 

b. Determine the hybridization of each N atom. 

c. In what type of orbital does the lone pair on each N reside? 

d. Draw all the resonance structures for rizatriptan that contain only neutral atoms. 

e. Draw all reasonable resonance structures for the five-membered ring that contains three N atoms.

3 step solution

Q.61.

Question: Zolpidem (trade name Ambien) promotes the rapid onset of sleep, making it a widely prescribed drug for treating insomnia.

a. In what type of orbital does the lone pair on each N atom in the heterocycle reside? 

b. Explain why the bicyclic ring system that contains both N atoms is aromatic. 

c. Draw all reasonable resonance structures for the bicyclic ring system.

3 step solution

Q.62.

Question: Answer the following questions about curcumin, a yellow pigment isolated from turmeric, a tropical perennial in the ginger family and a principal ingredient in curry powder.

a. In Chapter 11 we learned that most enols, compounds that contain a hydroxy group bonded to a C=C, are unstable and tautomerize to carbonyl groups. Draw the keto form of the enol of curcumin, and explain why the enol is more stable than many other enols.

b. Explain why the enol O-H proton is more acidic than an alcohol O-H proton.

c. Why is curcumin colored?

d. Explain why curcumin is an antioxidant.

3 step solution

Q.63.

Question: Stanozolol is an anabolic steroid that promotes muscle growth. Although stanozolol has been used by athletes and bodybuilders, many physical and psychological problems result from prolonged use and it is banned in competitive sports.

a. Explain why the nitrogen heterocycle—a pyrazole ring—is aromatic.

b. In what type of orbital is the lone pair on each N atom contained?

c. Draw all reasonable resonance structures for stanozolol.

d. Explain why the  pKa of the N-H bond in the pyrazole ring is comparable to the pKa  of O-H bond, making it considerably more acidic than amines such as CH3NH2 ( pKa= 40).

3 step solution

Q.64.

Question: Explain why A is aromatic but B is not aromatic.

3 step solution

Q.65.

Question: Use the observed 1 H NMR data to decide whether C and its dianion are aromatic, antiaromatic, or not aromatic. C shows NMR signals at –4.25 (6 H) and 8.14–8.67 (10 H) ppm. The dianion of C shows NMR signals at –3 (10 H) and 21 (6 H) ppm. Why are the signals shifted upfield (or downfield) to such a large extent?

3 step solution

Q.66.

Question: Explain why compound A is much more stable than compound B.

3 step solution

Q.67.

Question: (R)-Carvone, the major component of the oil of spearmint, undergoes acid-catalyzed isomerization to carvacrol, a major component of the oil of thyme. Draw a stepwise mechanism and explain why this isomerization occurs.

3 step solution

Q.68.

Question: Explain why triphenylene resembles benzene in that it does not undergo addition reactions with Br2 , but phenanthrene reacts with Br2 to yield the addition product drawn. (Hint: Draw resonance structures for both triphenylene and phenanthrene, and use them to determine how delocalized each π bond is.)

3 step solution

Q.69.

Question: Although benzene itself absorbs at 128 ppm in its C13 NMR spectrum, the carbons of substituted benzenes absorb either upfield or downfield from this value depending on the substituent. Explain the observed values for the carbon ortho to the given substituent in the monosubstituted benzene derivatives X and Y.

3 step solution

Problem 17.1

Draw all possible resonance structures for the antihistamine diphenhydramine, the active ingredient in Benadryl. 


2 step solution

Problem 17.2

What orbitals are used to form the labeled C-C bonds in the following molecule? Of the labeled bonds, which is the shortest?


3 step solution

Problem 17.3

Give the IUPAC name for each compound.

(a)

(b)



(c)



(d) 


2 step solution

Problem 17.4

Draw the structure corresponding to each name


  1. isobutyl benzene 
  2. o-dichlorobenzene 
  3. cis-1,2-diphenylcyclohexane 
  4. m-bromoaniline
  5. 4-chloro-1,2-diethylbenzene  
  6. 3-tert-butyl-2-ethyltoluene

2 step solution

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