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Chapter 9: Q73P (page 451)

Starting with cyclohexene, how could the following compounds be prepared?

a. methoxycyclohexane

b. cyclohexylmethylamine

c. dicyclohexyl ether

Short Answer

Expert verified

c.

Step by step solution

01

How SN1 reaction proceed

1. It is a two step reaction.

2. In this carbocation is formed.

3. In this negative ion will attack from the side where there is less number of bulkier groups.

4. As the carbocation intermediate is formed so the stability order 3掳 carbocation>2掳 carbocation>1掳 carbocation

02

Final answer

A large excess of methylamine will have to be used in the second step in order to minimize the formation of a tertiary Amine and a quarternary ammonium ion.

c.Half of the cyclohexene is converted to bromocyclohexane and half is converted to an alkoxide ion. The ether is formed from the reaction of bromocyclohexane with the alkoxide ion.

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Most popular questions from this chapter

cis-1-Bromo-4-tert-butylcyclohexane and trans-1-bromo-4-tert-butylcyclohexane both react with sodium ethoxide in ethanol to form 4-tert-butylcyclohexene. Explain why the cis isomer reacts much more rapidly than the trans isomer.

Draw the major product obtained when each of the following alkyl halides undergoes an E2 reaction:

a. Explain why the reaction of an alkyl halide with ammonia gives a low yield of primary amine.

b. Explain why a much better yield of primary amine is obtained from the reaction of an alkyl halide with an azide ion (-N3), followed by catalytic hydrogenation. (Hint: An alkyl azide is not nucleophilic.)

When the following compound undergoes solvolysis in ethanol, three products are obtained. Propose a mechanism to account for the formation of these products.

After a proton is removed from the OH group, which compound in each pair forms a cyclic ether more rapidly?
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