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Chapter 7: Q57Q (page 332)

Question:

Show how you would prepare cyclohexene from each compound.

(a) cyclohexanol

(b) cyclohexyl bromide

(c) cyclohexane (not by dehydrogenation)

Short Answer

Expert verified

Cyclohexene can be prepared by dehydration of cyclohexanolby phosphoric acid (acid catalyst). This is one of the most common methods of preparing alkenes.

Step by step solution

01

Step by step solution:Step 1: (a)

Cyclohexene can be prepared by dehydration of cyclohexanolby phosphoric acid (acid catalyst). This is one of the most common methods of preparing alkenes.

02

(b)

Cyclohexene can be prepared by dehydration of cyclohexyl bromide to produce cyclohexene, with water and KBr as by products

03

(c)

Cyclohexane can be prepared by the reaction with bromine to give cyclohexyl bromide, then elimination with a bulky base like potassium tert-butoxide gives cyclohexene

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Most popular questions from this chapter

Question: Give systematic (IUPAC) names of the following alkenes.

(a)

(b)

(c)

(d)

(e)

(f)

(g)

Finish Solved Problem 7-3 by showing how the rearranged carbocations give the four products shown in the problem. Be careful when using curved arrows to show deprotonation and/or nucleophilic attack by the solvent. The curved arrows always show movement of electrons, not movement of protons or other species.

Determine the number of elements of unsaturation in the molecular formula C3H4. Give all three possible structures having this formula. Remember that

a double bond = one element of unsaturation

a ring = one element of unsaturation

a triple bond = two elements of unsaturation

For each reaction, decide whether substitution or elimination (or both) is possible, and predict the product you expect. Label the major products.

(a) 1 - bromo - 1 - methylcyclohexane + NaOH in acetone

(b) 1 - bromo - 1 - methylcyclohexane + triethylamine (Et3N:)

(c) chlorocyclohexane + NaOCH3 in CH3OH

(d) chlorocyclohexane + NaOC(CH3) in (CH3)3COH

Show what happens in step-2 of the example if the solvent acts as a nucleophile (forming a bond to carbon) rather than as a base (removing a proton).

See all solutions

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