A tertiary alkanol has a molecular formula C$_4$H$_{10}$O. What is the structural formula of the compound?

Analyze the question: We need the structural formula of a tertiary alkanol with formula C₄H₁₀O.

Concept: Alkanols (alcohols) are classified by the carbon atom that carries the –OH group:

• Primary (1°): the C–OH carbon is attached to one other carbon (it has two H atoms).
• Secondary (2°): the C–OH carbon is attached to two other carbons (it has one H atom).
• Tertiary (3°): the C–OH carbon is attached to three other carbons (it has no H atoms).

Checking each option:

• A. (CH₃)₂CHCH₂OH is 2-methylpropan-1-ol. The C–OH carbon is CH₂, attached to one carbon, so it is primary.
• B. CH₃CH₂CH(OH)CH₃ is butan-2-ol. The C–OH carbon is CH, attached to two carbons, so it is secondary.
• C. (CH₃)₃COH is 2-methylpropan-2-ol. The C–OH carbon is bonded to three CH₃ groups and has no hydrogen, so it is tertiary. Its formula is C₄H₁₀O, which matches.
• D. CH₃CH₂CH₂CH₂OH is butan-1-ol. The C–OH carbon is CH₂, attached to one carbon, so it is primary.

Answer: Option C, (CH₃)₃COH (2-methylpropan-2-ol, also called tert-butyl alcohol).

Tip: Count how many carbons are attached to the carbon bearing the –OH group. One means primary, two means secondary, three means tertiary. Any alcohol written as R₃C–OH is tertiary, so look for a carbon with the OH and three carbon groups around it, such as (CH₃)₃COH.