Question #c179b

1 Answer
Jul 6, 2017

How about these?

Explanation:

A). Tetrachloromethane and butane

Look at them.

#"CCl"_4# is a colourless liquid with a characteristic "sweet" smell.

Butane is a gas at room temperature and has an odour like natural gas.

B). 1-propanol and 2-methylpropan-2-ol

Do the Lucas test.

To the alcohol add a few drops of an equimolar mixture of concentrated #"HCl"# and anhydrous #"ZnCl"_2#.

The acid protonates the alcohol, and the nucleophile #"Cl"^"-"#, which is present in excess, displaces the #"H"_2"O"#.

#"R-OH + H"^"+" → "R-"stackrelcolor(blue)("+")("O"}"H"_2#

#"R-"stackrelcolor(blue)("+")("O"}"H"_2 + "Cl"^"-" → underbrace("R-Cl")_color(blue)("oily layer") +"H"_2"O"#

The 3 ° alcohol is much more reactive than the 1 ° alcohol, and the alkyl halide is insoluble in water.

Thus, a 3 ° alcohol forms an oily layer immediately.

A 1 ° alcohol forms an oily layer only on heating.

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C). Pentanamine and pentane

The odour alone should be sufficient.

Pentanamine has an ammonia-like odour, while pentane has a gasoline-like (petrol-like) odour.

If you need a chemical test, just add a few drops if each to a test-tube containing hydrochloric acid.

The basic amine will immediately be neutralized to form the soluble ammonium salt, while pentane is insoluble in water.

#"R-NH"_2 + "HCl" → underbrace("R-NH"_3^"+", "Cl"^"-")_color(red)("soluble")#

D). Pent-1-ene and benzene

To each, add a few drops of #"Br"_2"/CCl"_4# solution.

The alkene will decolourize the bromine immediately; the benzene will not react.

#"R-CH=CH"_2 + underbrace("Br"_2)_color(red)("brown") → underbrace("R-CHBr-CH"_2"Br")_color(red)("colourless")#