The product of the reaction has to be identified and the stereochemistry has to be specified. Concept introduction: S N 1 reaction: The alcohol is reaction with acids like hydrochloric acid or hydrobromic acid which yield the corresponding carbocation intermediate, this carbocation intermediate undergoes substitution reaction which yields the corresponding substitution product. Tertiary alcohols undergo substitution very fast than the secondary alcohols because tertiary carbocation is more stable than the secondary carbocation than the primary carbocation. Primary alcohol is less stable therefore it won't undergo S N 1substitution reaction. E 1 elimination: Alkyl halide forms carbocation by the removal of bromine followed by the abstraction of the proton from the β-carbon atom in two steps which leads to the product as an alkene . Given information: The given compound is shown below,

Question

2. Name the following hydrocarbons. (9 marks) a) HHHHHHHH H-C-C- H-O-S b) HCEC-CH3 H H H H H d) c) H C=C- H H H e) CH3 CH3 CH2CH=CH-CH=CHCH3 HHHH H-C-C-C-C-H H HH H f) large CH2CH3 pola H3C section lovels tower, able ocart firs g) Tower H3C-CH2 then in H3C-CH-CH-CH3 enblbano bne noitsidab Copyright © 2008. Durham Continuing Education CH3

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Question

Chapter 11.SE, Problem 67AP

Interpretation Introduction

Interpretation:

The product of the reaction has to be identified and the stereochemistry has to be specified.

Concept introduction:

S_N1 reaction:

The alcohol is reaction with acids like hydrochloric acid or hydrobromic acid which yield the corresponding carbocation intermediate, this carbocation intermediate undergoes substitution reaction which yields the corresponding substitution product.

Tertiary alcohols undergo substitution very fast than the secondary alcohols because tertiary carbocation is more stable than the secondary carbocation than the primary carbocation. Primary alcohol is less stable therefore it won't undergo S_N1substitution reaction.

E₁ elimination:

Alkyl halide forms carbocation by the removal of bromine followed by the abstraction of the proton from the β-carbon atom in two steps which leads to the product as an alkene.

Given information:

The given compound is shown below,

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