Answer to Question #286459 in General Chemistry for Gionny

Question

Topics:Further synthesis and analysis 

Q1

Describe a simple chemical test for distinguishing between the following pairs of compounds:

a) cyclohexane and cyclohexene (2 marks detailed answers with explanation for every step answers please)

b) propanol and propanone (2 marks detailed answers with explanation for every step please)

c) propan-1-ol and propan-2-ol (3 marks detailed answers with explanation for every step please)

Detailed answers please with explanation for every step please

Q2

Suggest how the following syntheses could be carried out in the laboratory:

a) chloroethane to ethanoic acid (5 marks detailed explanation with answers please)

b) chloroethane to 2-hydroxypropanenitrile (5 mark detailed answers with explanation for every step please)

c) chloroethane to aminoethane (1 mark with detailed answer with explanation for every step please)

Answer 1

Cyclic cyclohexane is 1° saturated hydrocarbon and cyclic cyclohexene is 2° unsaturated hydrocarbon as cyclohexane has one ring whereas cyclohexene has one ring with one double bond.

Now, we have to consider a test that can differentiate both the compounds easily.

So, let’s take option A first i.e, Bromine water which is an alkene test.

When cyclic cyclohexane is reacted with Bromine water then no reaction occurs but when cyclic cyclohexene reacts with bromine water then it gets reacted with bromine water and an alkene with -Br and -OH bond is formed.

Thus, the main difference with Br2/H2O is that cyclic cyclohexane will not react and cyclic cyclohexene will react with Bromine water.

Now, let’s consider with option B i.e, Bayer’s reagent (alkaline KMnO4 solution)

Similarly, with alkaline KMnO4 solution, cyclic cyclohexane will not react whereas; cyclic cyclohexene will react and form a compound with –OH and –OH bonds.

Thus, cyclohexane remains unreacted with both Bromine water and Baeyer's reagent. And cyclohexene reacts with both Bromine water and Baeyer's reagent.

Thus, the reaction with sodium hypoiodite is the distinguishing test between propanal and propanone. This test is known as the iodoform test. Thus, the chemical test to distinguish between propanal and propanone is the iodoform test.