Login Registrieren

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

Chapter 18: 18-12P (page 677)

a. Draw resonance structures for the carbocation formed after loss of a leaving group from phytyl diphosphate.

b. Draw the two-step mechanism for Friedel–Crafts alkylation of 1,2-dihydroxynaphthoic acid with this carbocation to form X.

Short Answer

Expert verified

Step by step solution

01

Friedel-Crafts reaction

The Friedel-Craft is of two types, alkylation, and acylation. The alkylation proceeds via theelectrophile generation and acylation proceeds via the direct attack on the carbonyl carbon of the acyl group.

02

Explanation

a. The elimination of the leaving group \(^ - {\rm{OPP}}\) generates an allyl carbocation which can form another resonance structure (tertiary carbocation) as:

b. The two-step mechanism is the substitution of carbocation of diphytyl diphosphate and the abstraction of the proton as:

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.

Start your free trial

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

Question: Friedel-Crafts alkylation of benzene with (R)-2-chlorobutane and AlCl₃ affords sec-butylbenzene.

How many stereogenic centers are present in the product?
Would you expect the product to exhibit optical activity? Explain, with reference to the mechanism.

Question: Draw a stepwise mechanism for the following reaction.

Benzyl bromide\({\bf{(}}{{\bf{C}}_{\bf{6}}}{{\bf{H}}_{\bf{5}}}{\bf{C}}{{\bf{H}}_{\bf{2}}}{\bf{Br)}}\) reacts rapidly with \({\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{OH}}\)to afford benzyl methyl ether\({\bf{(}}{{\bf{C}}_{\bf{6}}}{{\bf{H}}_{\bf{5}}}{\bf{C}}{{\bf{H}}_{\bf{2}}}{\bf{OC}}{{\bf{H}}_{\bf{3}}}{\bf{)}}\). Draw a stepwise mechanism for the reaction, and explain why this 1° alkyl halide reacts rapidly with a weak nucleophile under conditions that favor an\({{\bf{S}}_{\bf{N}}}{\bf{1}}\) mechanism. Would you expect the para-substituted benzylic halides \({\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{O}}{{\bf{C}}_{\bf{6}}}{{\bf{H}}_{\bf{4}}}{\bf{C}}{{\bf{H}}_{\bf{2}}}{\bf{Br}}\) and \({{\bf{O}}_{\bf{2}}}{\bf{N}}{{\bf{C}}_{\bf{6}}}{{\bf{H}}_{\bf{4}}}{\bf{C}}{{\bf{H}}_{\bf{2}}}{\bf{Br}}\) to each be more or less reactive than benzyl bromide in this reaction?Explain your reasoning.

Question: Explain the following observation. Ethyl 3-phenylpropanoate $(C_{6} H_{5} {CH}_{2} {CH}_{2} {CO}_{2} {CH}_{2} {CH}_{3})$ reacts with electrophiles to afford ortho- and para-disubstituted arenes, but ethyl 3-phenylprop-2-enoate $(C_{6} H_{5} = {CHCO}_{2} {CH}_{2} {CH}_{3})$ reacts with electrophiles to afford meta-disubstituted arenes.

Which halides are unreactive in a Friedel–Crafts alkylation reaction?

See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Chemical Analysis

Read Explanation

The Earths Atmosphere

Read Explanation

Making Measurements

Read Explanation

Physical Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

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