The molecular formula for this compound is (C11H16O). Because the number of carbons is odd, the molecular formula and emperical formula are the same.
The compound is a primary alcohol (a primary carbon is bonded to the OH group).
The compound contains one tertiary center, that is, one sp3 carbon which is bonded to three other carbons.
All of the carbons in the benzene ring are sp2 centers.
Does the molecule shown above:
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