Chemistry Preparation of Ethers and Physical Properties

Topics Covered :

● Preparation of Ethers
● By Dehydration of Alcohols
● By Williamson Synthesis
● Physical Properties of Ethers

Preparation of Ethers :

Ethers are prepared by following methods :

By dehydration of alcohols :

`=>` Alcohols undergo dehydration in the presence of protic acids `color{red}(H_2SO_4, H_3PO_4)`.

`=>` The formation of the reaction product, alkene or ether depends on the reaction conditions.

`=>` Example : Ethanol is dehydrated to ethene in the presence of sulphuric acid at `443 K`. At `413 K`, ethoxyethane is the main product.

`color{red}(CH_3CH_2OH underset(443K) overset(H_2SO_4)→ CH_2 = CH_2)`

`color{red}(CH_3CH_2OH underset(413K) overset(H_2SO_4)→ C_2H_5OC_2H_5)`

`=>` The formation of ether is a nucleophilic bimolecular reaction `color{red}(S_N 2)` involving the attack of alcohol molecule on a protonated alcohol, as shown in fig.

`=>` Acidic dehydration of alcohols, to give an alkene is also associated with substitution reaction to give an ether.

`=>` This method is suitable for the preparation of ethers having primary alkyl groups only.

● The alkyl group should be unhindered and the temperature be kept low. Otherwise the reaction favours the formation of alkene.

`=>` The reaction follows `color{red}(S_N 1)` pathway when the alcohol is secondary or tertiary.

● However, the dehydration of secondary and tertiary alcohols to give corresponding ethers is unsuccessful as elimination competes over substitution and as a consequence, alkenes are easily formed.

Williamson synthesis :

`=>` It is an important laboratory method for the preparation of symmetrical and unsymmetrical ethers.

`=>` In this method, an alkyl halide is allowed to react with sodium alkoxide.

`color{red}(R-X + R' - underset(* *) overset(* *)O Na → R - underset(* * ) overset(* *)O- R' +NaX)`

`=>` Ethers containing substituted alkyl groups (secondary or tertiary) may also be prepared by this method. The reaction involves `color{red}(S_N 2)` attack of an alkoxide ion on primary alkyl halide. See fig.1.

`=>` Better results are obtained if the alkyl halide is primary.

● In case of secondary and tertiary alkyl halides, elimination competes over substitution.

● If a tertiary alkyl halide is used, an alkene is the only reaction product and no ether is formed.

● For example, the reaction of `color{red}(CH_3ONa)` with `color{red}((CH_3)_3C–Br)` gives exclusively 2-methylpropene.

`color{red}(CH_3 - underset(underset(CH_3)(|)) overset (overset(CH_3)(||))C-Br + overset(+)Na underset(* *) overset(* *)O - CH_3 → undersettext(2-Methylpropene)(CH_3 - underset ( underset (CH_3)(|))C = CH_2+NaBr+CH_3OH))`

● It is because alkoxides are not only nucleophiles but strong bases as well. They react with alkyl halides leading to elimination reactions.

`=>` Phenols are also converted to ethers by this method. In this, phenol is used as the phenoxide moiety. See fig.2.
Q 3052156934

The following is not an appropriate reaction for the preparation of t-butyl ethyl ether.

`C_2H_5ONa+CH_3- underset(underset(CH_3)(|)) overset(overset(CH_3)(|))C-Cl → CH_3 - underset(underset(CH_3)(|)) overset(overset(CH_3)(|))C- OC_2H_5`

(i) What would be the major product of this reaction ?
(ii) Write a suitable reaction for the preparation of t-butylethyl ether.

Solution:

(i) The major product of the given reaction is 2-methylprop-1-ene. It is because sodium ethoxide is a strong nucleophile as well as a strong base. Thus elimination reaction predominates over substitution.

(ii). `CH_3 - underset(underset(CH_3)(|)) overset( overset(CH_3)(|))C- underset(* *) overset( * *) O overset(+)Na + CH_3CH_2Cl → CH_3 - underset(underset(CH_3)(|)) overset(overset(CH_3)(|))C- OC_2H_5`

Physical Properties :

`=>` The `C-O` bonds in ethers are polar and thus, ethers have a net dipole moment.

`=>` The weak polarity of ethers do not appreciably affect their boiling points which are comparable to those of the alkanes of comparable molecular masses but are much lower than the boiling points of alcohols as shown in the following cases :

`color{red}(tt((text(Formula) , undersettext(n-Pentane)(CH_3(CH_2)_3CH_3), undersettext(Ethoxyethane)(C_2H_5-O-C_2H_5) , undersettext(Butan-1-ol)(CH_3(CH_2)_3-OH)), (b.p//K, 309.1, 307.6, 390)))`

`=>` The large difference in boiling points of alcohols and ethers is due to the presence of hydrogen bonding in alcohols.

`=>` The miscibility of ethers with water resembles those of alcohols of the same molecular mass.

● Both ethoxyethane and butan-1-ol are miscible to almost the same extent i.e., `7.5` and `9` g per `100` mL water, respectively while pentane is essentially immiscible with water.

● This is due to the fact that just like alcohols, oxygen of ether can also form hydrogen bonds with water molecule as shown in fig.

 
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