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Nucleophilic Substitution in Alkyl Halides

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Question

One molecule of dialkyl ether produces how many molecules of alkyl halides with excess of halogen acid?

A

1

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B

2

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C

3

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D

4

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Solution

The correct option is B 2
The given reaction will proceeds through $S_{N} 2$ reaction mechanism. Dialkyl ethers when treated with excess halogen acid first forms an alkyl halide and an alcohol. The alcohol further reacts with the excess acid to form the respective alkyl halide and a water molecule. Thus, two alkyl halide molecules are produced in the end.

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Q. Carbon - oxygen bond in ethers can be cleaved by halogen acids, sulphuric acid, phosphorouspentachloride etc. The cleavage of

C - O

bond takes place under drastic conditions with an excess of hydrogen halide,

R - O - R + 2 H X \to 2 R X + H_{2} O

If hydrogen halide is not taken in excess, then one molecule each of alcohol and alkyl halide are formed,

R - O - R + H I 373 K \to R - O H + R - I

. The order of reactivity of hydrogen halides towards the above-mentioned reaction depends on the bond strength of halogenated acids. Weaker is the

H - X

bond, greater is its reactivity. In a case of unsymmetrical ethers, the site of cleavage is such that the halide is formed from the alkyl group which is tertiary so that the halide formed is tertiary halide.

Indicate whether the following statements / reactions are true or false.
Hydrogen fluoride is not effective in ether cleavage.

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