Questions: Alcohols and Ethers: Structure, Properties, and Nomenclature

Question 1

Ethanol (CH₃CH₂OH, MW = 46) has a boiling point of 78°C, while dimethyl ether (CH₃OCH₃, MW = 46) boils at −24°C. What best explains this ~100°C difference?

• A. Ethanol has a higher molecular weight and stronger dispersion forces
• B. Ethanol can both donate and accept hydrogen bonds, while dimethyl ether can only accept them
• C. Dimethyl ether is more symmetrical and therefore packs more tightly in the liquid phase
• D. Ethanol's O–H bond is longer, making it harder to break during vaporization

Answer: (B)

Both compounds have the same molecular weight (MW = 46), so dispersion forces are comparable. The decisive difference is hydrogen bonding: ethanol has an O–H group that can donate hydrogen bonds to other ethanol molecules (and accept them via the oxygen lone pairs), creating a strong intermolecular network. Dimethyl ether has no O–H, so it can only accept hydrogen bonds from other donors but cannot form the donor–acceptor network. This is a direct consequence of the structural difference: R–OH vs. R–O–R'.

Question 2

A tertiary alcohol is more acidic than a primary alcohol because the tertiary carbocation formed after deprotonation is more stable.

• T. True
• F. False

Answer: False

Acidity of an alcohol depends on the stability of the alkoxide anion (R–O⁻), not a carbocation. In fact, alkyl groups are electron-donating (inductive effect), which destabilizes a negative charge on oxygen. Tertiary alkoxides have more electron-donating alkyl groups around the oxygen, making them *less* stable and tertiary alcohols *less* acidic than primary ones. The misconception conflates carbocation stability (relevant to SN1) with alkoxide stability (relevant to acid–base equilibria).

Question 3

What structural feature of the alcohol determines whether it is classified as primary, secondary, or tertiary, and why does this classification matter for reactivity?

Model answer: The classification is based on how many carbon groups are attached to the carbon that bears the –OH group: one carbon = primary, two = secondary, three = tertiary. This governs oxidation (primary and secondary can be oxidized; tertiary cannot) and substitution pathways (tertiary favors SN1/E1 due to carbocation stability).

Degree of substitution at the carbon bonded to oxygen controls both the steric environment around that carbon and the stability of any carbocation intermediate formed if the –OH leaves. Tertiary carbons form stable 3° carbocations, enabling SN1 and E1 routes. Primary carbons resist ionization and react via SN2 instead. For oxidation, the carbon must have at least one C–H bond adjacent to –OH; a tertiary carbon bonded to three carbons has no such H, blocking oxidation to a carbonyl.