Question

Compound A (C10H8) decolorizes a solution of bromine in CCl4. Independent analysis shows that compound A does not possess any phenyl rings. In the presence of Pt and excess H2 compound A undergoes conversion to compound B (C10H16). How many rings does compound A possess?

Answer 1

Given that;

Compound A (C10H8)

compound B (C10H16)

DBE = C - (H/2) + (N/2) +1

Compound A (C10H8)

here C = number of carbon atoms, H = number of hydrogen and halogen atoms, and N = number of nitrogen atoms.

DBE = 10- (8/2) + (0/2) +1

DBE = 10-4+0+1

DBE = 7

Compound B (C10H16)

here C = number of carbon atoms, H = number of hydrogen and halogen atoms, and N = number of nitrogen atoms.

DBE = 10- (16/2) + (0/2) +1

DBE = 10-8+0+1

DBE = 3

One DBE = one ring or one double bond here DBE is 2 means one and one double bond.

Compound B posses 3 DBE means three rings which also present in the compound A because all double and triple bonds are saturated with Pt/H2.