How many alkoxyalkanes can be obtained from the molecular formula C$_4$H$_{10}$O?

Analyze the question: Alkoxyalkanes are ethers (R–O–R'). We need to count the ether isomers of C₄H₁₀O.

Concept: The two alkyl groups on either side of the oxygen must have a combined total of 4 carbon atoms. We list the possible combinations of alkyl groups:

• Methyl + propyl (1 C + 3 C)
• Ethyl + ethyl (2 C + 2 C)

Propyl can be n-propyl or isopropyl, so the methyl + propyl combination gives two different ethers.

Solution:

1. CH₃–O–CH₂CH₂CH₃ is 1-methoxypropane (methyl propyl ether).
2. CH₃–O–CH(CH₃)₂ is 2-methoxypropane (methyl isopropyl ether).
3. CH₃CH₂–O–CH₂CH₃ is ethoxyethane (diethyl ether).

Total = 3 alkoxyalkanes.

Check: a 1 C + 3 C split gives two isomers because of the two propyl forms, and a 2 C + 2 C split gives one. Therefore 2 + 1 = 3.

Answer: D (3)

Tip: For ether isomers, split the carbons between the two sides of the oxygen (1+3, 2+2) and count the branching possibilities for each alkyl group. Remember that propyl has 2 forms and butyl has 4. The other C₄H₁₀O isomers are alcohols (4 of them), which are a different functional group, so do not count them.