JAMB Chemistry 2019 Paper

Concept: Alkoxyalkanes are ethers with the general formula $R-O-R'$, so count the distinct ways to split 4 carbon atoms between $R$ and $R'$.Explanation:The formula $C_4H_{10}O$ has no double bonds or rings, so it represents a saturated ether.The two alkyl groups attached to oxygen must together contain exactly 4 carbon atoms.Case 1: methyl ($1$ carbon) combined with a propyl group ($3$ carbons).The propyl group can be either n-propyl or isopropyl, giving two different alkoxyalkanes: methoxypropane and methoxyisopropane.Case 2: ethyl ($2$ carbons) combined with another ethyl group ($2$ carbons).This gives only one alkoxyalkane: ethoxyethane.No other combination of alkyl groups has a total of 4 carbon atoms.Therefore, the total number of possible alkoxyalkanes is $2 + 1 = 3$.Answer: D. 3