another question too..
27 Which isomer of C4H10O forms three alkenes on dehydration?
A butan-1-ol
B butan-2-ol
C 2-methylpropan-1-ol
D 2-methylpropan-2-ol
Butan-1-ol will form only one alkene since the OH group is found on the first carbon atom. Hence upon dehydration the OH group and an adjacent H atom will condense to form one mole of water to form an alkene.
Butan-2-ol will form 2 alkenes since it has equal chance of producing an alkene with the hydrogen atom from adjacent H atoms from both sides(Either the first carbon atom or the third one). Its OH group is situated on the second carbon atom.
2-methylpropan-1-ol has one OH group on the first carbon atom and two methyl groups on the neighbouring carbon atom. However, since it is a primary alcohol, dehydration will result in the formation of only one alkene since the carbon atom containing the OH group is attached to only one other carbon atom.
2-methyl-propan-2-ol is a secondary alcohol. If you draw its structural formula, you'll realize that any alkene formed will be the same one due to different possible orientations. Hence it will form only one type of alkene.
For me, all the four answers given are inappropriate. But if I had to choose one, it will be
butan-2-ol since it can form the most alkenes possible (2).
NOTE : All primary alcohols are first eliminated since they can form only one type of alkene upon dehydration.
Though I can't find any other worth explanation