Two liquid hydrocarbons A and B each C6H6 were used in an experiment. A decolourises bromine water rapidly whereas B reacted very slowly
- draw possible structures for A and B
- explain why B reacted very slowly.
2007-01-14 09:29:54 · 4 answers · asked by Stephanieee! 1 in Science & Mathematics ➔ Chemistry
B is benzene which does not react as an unsaturated compound.
A contains double or triple bonds and is some isotope of an aliphatic six carbon compoud with several double and triple bond possibilities.
2007-01-14 23:16:19 · answer #1 · answered by lykovetos 5 · 0⤊ 0⤋
Scott S has acceptable answers here. The reason the alkyne can readily decolorize the solution is that bromine can readily react with the sites of unsaturation (multiple bonds) in a typical addition reaction (in the presence of light). To add bromine to benzene, however, FeBr3 needs to also be present. This would explain why benzene, B, does not react very much if at all.
2007-01-14 10:05:20 · answer #2 · answered by damico105 3 · 0⤊ 0⤋
A = Benzene
The 6 carbons form a hexagon (so there is one covalent bond between them) and each carbon has one hydrogen attached to it. The last electron roves around and is generally drawn as the double bond between every other carbon.
A decolorises bromine due to the saturation of the double bonds:
C6H6 + 3Br2 --> C6H6Br6
This is an addition reaction and also a test for unsaturated double bonds.
B = CH3 - C -triple bond- C - C triple bond- C - CH3
Alkynes do not react as easily as double bonds.
2007-01-14 15:46:41 · answer #3 · answered by Kemmy 6 · 0⤊ 0⤋
One is defiantly benzene. The other is 1,5-Hexene-3yne
2007-01-14 09:48:22 · answer #4 · answered by Scott S 4 · 0⤊ 0⤋