Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Draw a structural formula for each compound. (a) 1-Nitronaphthalene (b) 1,6 -Dichloronaphthalene (c) 9-Bromoanthracene (d) 2-Methylphenanthrene

Short Answer

Expert verified
Question: Draw the structural formula for 1,6-Dichloronaphthalene. Answer: 1,6-Dichloronaphthalene consists of a naphthalene parent structure with two chloro (-Cl) groups at carbon atom 1 and carbon atom 6. The structural formula for 1,6-Dichloronaphthalene is: ![1,6-Dichloronaphthalene](https://www.sigmaaldrich.com/content/dam/sigma-aldrich/structure2/156/mfcd00000693-2.eps/_jcr_content/renditions/mfcd00000693-2-large.png)

Step by step solution

01

Identify the parent structure

For each compound, we need to identify the base skeleton, also called the parent structure, before adding substituents. The parent structures are: (a) Naphthalene: a fused-ring system with two benzene rings (10 carbon atoms in total) (b) Naphthalene: same as (a) (c) Anthracene: a fused-ring system with three benzene rings (14 carbon atoms) (d) Phenanthrene: a fused-ring system with three benzene rings, similar to anthracene but with a different arrangement of double bonds.
02

Determine the position of substituents

Now that we have identified the parent structures, we need to add the substituents in the correct positions as indicated by the compound names: (a) 1-Nitronaphthalene: has a nitro (-NO2) group at the first carbon atom of the naphthalene structure. (b) 1,6-Dichloronaphthalene: has two chloro (-Cl) groups at carbon atom 1 and carbon atom 6 of the naphthalene structure. (c) 9-Bromoanthracene: has a bromo (-Br) group at the ninth carbon atom of the anthracene structure. (d) 2-Methylphenanthrene: has a methyl (-CH3) group at the second carbon atom of the phenanthrene structure.
03

Draw the structural formulas

Finally, we will draw the structural formulas for each compound, showing the parent structures with their substituents in the correct positions. (a) 1-Nitronaphthalene: ![1-Nitronaphthalene](https://www.sigmaaldrich.com/content/dam/sigma-aldrich/structure2/163/mfcd00002684-2.eps/_jcr_content/renditions/mfcd00002684-2-large.png) (b) 1,6-Dichloronaphthalene: ![1,6-Dichloronaphthalene](https://www.sigmaaldrich.com/content/dam/sigma-aldrich/structure2/156/mfcd00000693-2.eps/_jcr_content/renditions/mfcd00000693-2-large.png) (c) 9-Bromoanthracene: ![9-Bromoanthracene](https://www.sigmaaldrich.com/content/dam/sigma-aldrich/structure1/155/mfcd00011286-2.eps/_jcr_content/renditions/mfcd00011286-2-large.png) (d) 2-Methylphenanthrene: ![2-Methylphenanthrene](https://www.sigmaaldrich.com/content/dam/sigma-aldrich/structure2/084/mfcd00001215-2.eps/_jcr_content/renditions/mfcd00001215-2-large.png)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Following each name is the number of Kekulé structures that can be drawn for it. Draw these Kekulé structures and show, using curved arrows, how the first contributing structure for each molecule is converted to the second and so forth. (a) Naphthalene (3) (b) Phenanthrene (5)

Following is a synthesis for toremifene, a nonsteroidal estrogen antagonist whose structure is closely related to that of tamoxifen. (a) This synthesis makes use of two blocking groups, the benzyl (Bn) group and the tetrahydropyranyl (THP) group. Draw a structural formula of each group and describe the experimental conditions under which it is attached and removed. (b) Discuss the chemical logic behind the use of each blocking group in this synthesis. (c) Propose a mechanism for the conversion of \(\mathrm{D}\) to \(\mathrm{E}\). (d) Propose a mechanism for the conversion of \(F\) to toremifene. (e) Is toremifene chiral? If so, which of the possible stereoisomers are formed in this synthesis?

Propose a structural formula for each compound given these NMR data. (a) \(\mathrm{C}_{9} \mathrm{H}_{9} \mathrm{BrO}\) (b) \(\mathrm{C}_{\mathrm{g}} \mathrm{H}_{9} \mathrm{NO}\) (c) \(\mathrm{C}_{9} \mathrm{H}_{9} \mathrm{NO}_{3}\) $$ \begin{array}{lc} \hline{ }^{1} \text { H-NMR } & { }^{13} \text { C-NMR } \\ \hline 1.39(\mathrm{t}, 3 \mathrm{H}) & 165.73 \\ 4.38(\mathrm{q}, 2 \mathrm{H}) & 131.56 \\ 7.57(\mathrm{~d}, 2 \mathrm{H}) & 131.01 \\ 7.90(\mathrm{~d}, 2 \mathrm{H}) & 129.84 \\ & 127.81 \\ & 61.18 \\ & 14.18 \\ \hline \end{array} $$ $$ \begin{array}{lc} \hline{ }^{1} \text { H-NMR } & { }^{13} \text { C-NMR } \\ \hline 2.06(\mathrm{~s}, 3 \mathrm{H}) & 168.14 \\ 7.01(\mathrm{t}, 1 \mathrm{H}) & 139.24 \\ 7.30(\mathrm{~m}, 2 \mathrm{H}) & 128.51 \\ 7.59(\mathrm{~d}, 2 \mathrm{H}) & 122.83 \\ 9.90(\mathrm{~s}, 1 \mathrm{H}) & 118.90 \\ & 23.93 \\ \hline \end{array} $$ $$ \begin{array}{cc} \hline{ }^{1} \text { H-NMR } & { }^{13} \text { C-NMR } \\ \hline 2.10(\mathrm{~s}, 3 \mathrm{H}) & 168.74 \\ 7.72(\mathrm{~d}, 2 \mathrm{H}) & 166.85 \\ 7.91(\mathrm{~d}, 2 \mathrm{H}) & 143.23 \\ 10.3(\mathrm{~s}, 1 \mathrm{H}) & 130.28 \\ 12.7(\mathrm{~s}, 1 \mathrm{H}) & 124.80 \\ & 118.09 \\ & 24.09 \\ \hline \end{array} $$

Account for the fact that \(p\)-nitrophenol \(\left(K_{\mathrm{a}} 7.0 \times 10^{-8}\right)\) is a stronger acid than is phenol \(\left(K_{\mathrm{a}} 1.1 \times 10^{-10}\right)\).

Arrange these compounds in order of increasing acidity: 2,4-dichlorophenol, phenol, cyclohexanol.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free