Question

The victor Meyers test, alcohol is reacted with---- and iodide to form the corresponding alkyl iodide.
This question has multiple correct options.
(a) concentrate HI
(b) concentrated hydrochloric acid
(c)sodium dichromate
(d)red phosphorus

Answer 1

The victor Meyers test is used to distinguish between the primary, secondary and tertiary alcohols and for this they are first converted into the alkyl iodide by making it react with a non -metal belonging to the nitrogen family and has the atomic number as 15 or a halogen halide. Identify it.

Complete answer:
Compounds containing -OH groups as the functional group are called as the alcohols so, they may also be regarded as the hydroxy derivatives of aliphatic hydrocarbons. Depending whether the number of oh groups present in a molecule is one, two , three or many they are classified as monohydric, hydridic, trihydric or polyhydric and so on respectively.
Examples: ethyl alcohol( contains one -OH group), glycerol ( contains three -OH groups) etc.
Now, let’s discuss the victor Meyers test, this test is performed to distinguish between the primary, secondary and tertiary alcohols. The alcohol in which the carbon atom consists of one alkyl group is called a primary alcohol, the carbon atom which consists of two alkyl groups is called secondary alcohol and which consists of three alkyl groups is called tertiary alcohol.
In this test, first the given alcohol is treated with HI or phosphorus and iodine to convert it into alkyl iodide and then , it is treated with silver nitrate to form the corresponding nitroalkane. Now, this nitroalkane is treated with nitrous acid and the then resulting solution is finally made to react with aqueous sodium hydroxide or potassium hydroxide.
If the red coloration appears, it indicates a primary alcohol.
If the blue coloration appears, it indicates a secondary alcohol.
If the solution remains colourless, it indicates a tertiary alcohol.
Hence, in the victor Meyers test, alcohol is reacted with the concentrated HI or phosphorus and iodide to form the corresponding alkyl iodide.

So, options (a) and (d) are correct.

Note:
It should be kept in mind that in each monohydric, dihydric and polyhydric each -OH group is attached to the different carbon atom because the presence of two or more hydroxyl groups on the same carbon atom makes the molecule very unstable and it loses a water molecule to form aldehyde or ketone or carboxylic acid.