/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q 50. Question: What is the major prod... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Question: What is the major product of electrophilic addition of HBr to the following alkene? Explain your choice.

Short Answer

Expert verified

Answer

Step by step solution

01

Electrophilic addition reaction

The addition reaction initiated by the attack of an electrophile on a double or triple bond is termed the electrophilic addition reaction.

In an electrophilic addition reaction, the pi bond is broken, and two sigma bonds are formed.

02

Addition of HBr

Bromine (Br), being more electronegative than hydrogen, pulls the electron from the H-Br bond towards. This reduces the electron density on the hydrogen atom.

The electron-rich pi bond attacks H+ and forms a sigma bond with it, thereby generating a carbocation.

Br-attacks the carbocation and forms a bond with it.

03

Addition of HBr to the given alkene

The addition of HBr to the double bond proceeds as shown below;

Addition of to the double-bonded carbon

The carbocation is formed adjacent to the electron-rich ring (ring containing the electron-donating OCH3 group) as the delocalization of electrons from the ring increases its stability.

Attack of Br-on the carbocation

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 91Ó°ÊÓ!

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: Explain the following observation. Ethyl 3-phenylpropanoate (C6H5CH2CH2CO2CH2CH3) reacts with electrophiles to afford ortho- and para-disubstituted arenes, but ethyl 3-phenylprop-2-enoate (C6H5=CHCO2CH2CH3) reacts with electrophiles to afford meta-disubstituted arenes.

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.

The 1 H NMR spectrum of phenol (\({C_6}{H_5}OH\)) shows three absorptions in the aromatic region: 6.70 (2 ortho H’s), 7.14 (2 meta H’s), and 6.80 (1 para H) ppm. Explain why the ortho and para absorptions occur at lower chemical shift than the meta absorption.

Label each compound as more or less reactive than benzene in electrophilic aromatic substitution.

Question: Draw a stepwise mechanism for the following reaction.

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.