1, according to the molecular formula C6H 12, must be unsaturated hydrocarbon. It can make bromine water fade and be judged as olefin. Catalytic hydrogenation produces n-hexane, which indicates that this olefin is linear n-hexene. Only the position of double bond needs to be judged according to "acidic potassium permanganate oxidizes to produce a carboxylic acid". If you read this book, I'm not sure if it's at the end.

2. Obviously, this topic is to study isomers of unsaturated hydrocarbons. Because 3- methylpentane is produced, the position of double bond is inferred from this result. Because A has cis-trans isomerism, it means that the groups on both sides are different. After deducting A, look at the addition of HBr, then subtract HBr to get C, and finally the remaining isomer is B. The result should be li 1lulu:

A: 3- methyl -2- pentene

B:3- methyl-1- pentene

C:2- ethyl-1- butene

D:3- methyl -3- bromopentane

You can verify it.