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Chapter 9: Problem 49

The following ethers can, in principle, be synthesized by two different combinations of haloalkane or halocycloalkane and metal alkoxide. Show one combination that forms ether bond (1) and another that forms ether bond (2). Which combination gives the higher yield of ether?

Short Answer

Expert verified
Answer: The combination of a primary haloalkane or halocycloalkane (R1-X) and metal alkoxide (R2'-O-M) forming ether bond (1) (-R1-O-R2'-) tends to give a higher yield. However, actual yields may vary due to practical factors and competing reactions.

Step by step solution

01

Identifying the ether structure and bond (1) and bond (2)

Looking carefully at the given ethers, we can identify the ether bond (1) (-R1-O-R2-) and ether bond (2) (-R2'-O-R1'-).
02

Determining possible combinations of haloalkane or halocycloalkane and metal alkoxide

In the Williamsons Ether Synthesis, the general reaction can be represented as follows: Primary Haloalkane or Halocycloalkane (R1-X) + Metal Alkoxide (R2'-O-M) -> Ether (R1-O-R2'-) Primary Haloalkane or Halocycloalkane (R2-X') + Metal Alkoxide (R1'-O-M') -> Ether (R2'-O-R1'-) Where X and X' are halogens, and M and M' are metal ions. Note that for higher yield, we want to choose primary haloalkanes because secondary and tertiary ones have higher tendencies for an elimination reaction.
03

Combination for Ether Bond (1)

Let R1 be a primary alkyl halide and M be a metal ion. Combining R1-X with R2'-O-M will give us the ether having bond (1): R1-X + R2'-O-M -> R1-O-R2'- For example, let R1-X be ethyl bromide (CH3CH2-Br) and R2'-O-M be potassium methoxide (CH3-O-K), the ether formed with bond (1) will be ethyl methyl ether (CH3CH2-O-CH3).
04

Combination for Ether Bond (2)

Similarly, let R2' be a primary alkyl halide and M' be a metal ion. Combining R2-X' with R1'-O-M' will give us the ether having bond (2): R2-X' + R1'-O-M' -> R2'-O-R1'- For example, let R2-X' be methyl bromide (CH3-Br) and R1'-O-M' be potassium ethoxide (CH3CH2-O-K), the ether formed with bond (2) will be ethyl methyl ether (CH3CH2-O-CH3).
05

Determining the Higher Yield Combination

Both combinations are similar in their theoretical yield; however, it is important to start with the primary alkyl halides when possible. The combination with R1-X being primary and starting with R2'-O-M results in a higher yield. So, the first combination that forms an ether bond (1) tends to give a higher yield of ether than the second combination. Remember that practical yields may differ due to various factors like reaction conditions, purity of reactants, and competing reactions like the elimination reaction.

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Most popular questions from this chapter

The reaction of bromomethane with azide ion \(\left(\mathrm{N}_{3}{ }^{-}\right)\)in methanol is a typical \(\mathrm{S}_{\mathrm{N}} 2\) reaction. What happens to the rate of the reaction if \(\left[\mathrm{N}_{3}{ }^{-}\right]\)is doubled?

Draw a structural formula for the most stable carbocation with each molecular formula. (a) \(\mathrm{C}_{4} \mathrm{H}_{9}{ }^{+}\) (b) \(\mathrm{C}_{3} \mathrm{H}_{7}^{+}\) (c) \(\mathrm{C}_{5} \mathrm{H}_{11}^{+}\) (d) \(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{O}^{+}\)

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