E/Z stereoisomerism with alcohols

Butan-2-ol is heated with an H2SO4 catalyst.
A mixture of three alkenes form B, C and D.

The alkenes B and C are stereoisomers.

Draw the two structures of the two stereoisomers B and C.

My answer (incorrect):
I draw the double bond between the carbons and put the alkyl group CH3 on both sides to represent the E and Z isomers..

But I put OH and H, because butan-2-ol is an alcohol.

The mark scheme requires the other side group to only be H and not OH and H, alongside the alkyl group CH3 which was correct.

Can someone tell me why that is....

Appreciate the help from anyone.

For EZ isomerism the double bond must have 2 different groups on each carbon.

If the double bond is between carbons 1 & 2 then the first carbon MUST have two hydrogen atoms and cannot form EZ isomers.
Hence, the double bond forms between 2 & 3 for B and C

After dehydration the OH has disappeared ....

CH₃CHOHCH₂CH₃ ==> CH₃CH=CHCH₃ + H₂O

So the OH is completely disregarded