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Chapter 18: 18-10P (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

The Lewis acids are identified by their sextet formation (six electrons in the octet). Therefore, it adds anions that are eliminated as a leaving group.

Step by step solution

01

Lewis acid

The Lewis acids are identified by their sextet formation (six electrons in the octet). Therefore, it adds anions that are eliminated as a leaving group.

02

Generation of electrophile

The electrophiles are generated when the Lewis acid attracts the anions. An anion has lone pair of electrons which can be donated to form a bond.

03

Explanation

The stepwise mechanism for the given question is shown as:

\(\left( 1 \right)\) The first step involves the generation of electrophiles by the Lewis acid.

\(\left( 2 \right)\)The second step involves the electrophilic substitution on the electron-rich position.

\(\left( 3 \right)\)The third step involves the abstraction of the proton from the base.

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Most popular questions from this chapter

Question: Rank the aryl halides in each group in order of increasing reactivity in nucleophilic aromatic substitution by an addition–elimination mechanism.

  1. chlorobenzene, p-fluoronitrobenzene, m-fluoronitrobenzene
  2. 1-fluoro-2,4-dinitrobenzene, 1-fluoro-3,5-dinitrobenzene, 1-fluoro-3,4-dinitrobenzene
  3. 1-fluoro-2,4-dinitrobenzene, 4-chloro-3-nitrotoluene, 4-fluoro-3-nitrotoluene

Question: Although two products (A and B) are possible when naphthalene undergoes electrophilic aromatic substitution, only A is formed. Draw resonance structures for the intermediate carbocation to explain why this is observed.


Question: Draw a stepwise, detailed mechanism for the following intramolecular reaction.

Explain why \({{\rm{C}}_{\rm{6}}}{{\rm{H}}_{\rm{5}}}{\rm{C}}{{\rm{H}}_{\rm{2}}}{\rm{C}}{{\rm{H}}_{\rm{2}}}{\rm{Br}}\)is not formed during the radical bromination of \({{\rm{C}}_{\rm{6}}}{{\rm{H}}_{\rm{5}}}{\rm{C}}{{\rm{H}}_{\rm{2}}}{\rm{C}}{{\rm{H}}_{\rm{3}}}\).

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

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

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