The 0.502 g sample of cholesterol and 5 mL of diethyl ether was dissolved over the steam bath. Added to this solution was 2.5 mL of 0.56 M Br2, and then cooled in an ice-water bath. After stirring for 10 minutes to ensure precipitation, a white crystalline paste was formed. The precipitate was collected using vacuum filtration and washed with cold diethyl ether-acetic acid then methanol. The product was recrystallized with ethanol and then vacuum filtered to obtain white crystals. The product formed was cholesterol dibromide from 2 bromine additions to the double bond on cholesterol.1

The mass of the cholesterol dibromide product was 0.429 g and the amount of moles was 7.8796 x 10-4 mol. The moles of the reagents were 1.2983 x 10-3 mol for …show more content…
In both cases the reagent was bromine and the substrates were alkenes and hexanes. The bromine reagent was a dark red-brown colour. Test tubes 1 and 2 with alkenes and hexanes respectively, the test tubes were wrapped in tin foil and kept at room temperature. The congo red indicator paper folded over the opening did not change colour, and the bromine alkenes solution became clear after adding the bromine. This is consistent with the formation of a colorless vicinal dibromide, formed by the addition of bromine to the carbon-carbon double bonds in the alkenes.1 No gas was formed in this reaction explaining why the indicator paper did not change colour. When bromine was added to hexanes and placed under a strong light, the congo red paper turned blue indicating that an acidic gas was evolved. This is consistent with theory since the reaction of alkanes and bromine produces hydrogen bromide gas when placed under light, which contains an acidic proton that would lower the pH so the indicator paper would change colour.1 The solution became light yellow, which indicates that the same reaction as with alkenes did not occur since there are no double bonds present on hexanes and no colourless liquid was