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Chapter 10: Q10-25E-b (page 308)

How would you prepare the following compounds, starting with cyclopentene and any other reagents needed? (b) Methylcyclopentane.

Short Answer

Expert verified

Methylcyclopentane can be formed in the following way:

Step by step solution

01

Step-by-step solutionStep 1: Chlorination

It means substitution of one hydrogen from the reactant, so that halogen atom can be added there.

02

Alkyl halide

Alkyl halides are also known as haloalkanes or halogenoalkanes. They contains one or more halogen substituents i.e. 鈥揃r, -Cl,-I,-F.

03

Preparation of methylcyclopentane

The desired product methylcyclopentane can be prepared by brominating cyclopantene and then reducing bromocyclopantane to methylcyclopentane.

The reaction is as follows:

Preparation of methylcyclopentane

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Most popular questions from this chapter

Alkyl halides can be reduced to alkanes by a radical reaction with tributyltinhydride, (C4H9)3SnH in the presence of light (hv). Propose aradical chain mechanism by which the reaction might occur. The initiationstep is the light-induced homolytic cleavage of the Sn-H bondto yield a tributyltin radical.

R-X+(C4H9)3SnH鈫�hvR-H+(C4H9)3SnX

Draw structures corresponding to the following IUPAC names: (c) 3-Bromo-3-ethylpentane.

Sort the radicals below from most stable to least stable.

Carboxylic acids (RCOOH; pKa=5) are approximately 1011 times more

acidic than alcohols (ROH;pKa=16). In other words, a carboxylate ion(RCO2-)

is more stable than an alkoxide ion (RO-). Explain, using

resonance.

Draw structures corresponding to the following IUPAC names: (c) 3-Iodo-2,2,4,4-tetramethylpentane
See all solutions

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