1. Draw the following alcohols
a) heptan-2-ol → CH₃–CH(OH)–(CH₂)₄–CH₃
b) 3-methylhexan-1-ol → HO–CH₂–CH₂–CH(CH₃)–CH₂–CH₂–CH₃
c) cyclopropanol → cyclopropane ring with –OH on one carbon
d) 2,4,6-trichlorooctan-2-ol → CH₃–C(Cl)(OH)–CH₂–CH(Cl)–CH₂–CH(Cl)–CH₂–CH₃
e) pentan-1,4-diol → HO–CH₂–CH₂–CH(OH)–CH₂–CH₃
f) benzene-1,3-diol → benzene ring with –OH at positions 1 and 3
g) but-2-ene-1-ol → HO–CH₂–CH=CH–CH₃
h) 4-methylpent-2-yn-1-ol → HO–CH₂–C≡C–CH(CH₃)–CH₃
i) 3,4-dimethylcycloheptan-1-ol → cycloheptane ring with –OH at C1, methyl groups at C3 and C4
2. Name the following alcohols
a) 2-methylbutan-1-ol
b) hexane-2,5-diol
c) 2-chloropropan-1-ol
d) 2-methylpentan-1-ol
e) cyclohexanol
f) prop-2-en-1-ol (allyl alcohol)
g) phenol
h) 4-ethylhex-2-yn-1-ol
i) 3,3,4-trimethylhexan-2-ol
3. Explain why propane used as fuel in BBQ is gas at room temperature, but 2-propanol used as rubbing alcohol is liquid at room temperature.
Propane (C₃H₈) is nonpolar and has only weak London dispersion forces; its low molecular weight and lack of hydrogen bonding result in low boiling point (~−42°C), so it’s a gas at room temperature. 2-Propanol (C₃H₈O) has an –OH group capable of strong hydrogen bonding, which significantly increases intermolecular forces and boiling point (~82°C), making it a liquid at room temperature.
4. Draw the following ethers
a) 1-propoxypentane → CH₃CH₂CH₂–O–CH₂CH₂CH₂CH₂CH₃
b) 2-ethoxybutane → CH₃CH₂–O–CH(CH₃)CH₂CH₃
c) 1-methoxy-4-chlorohexane → CH₃–O–CH₂CH₂CH₂CH(Cl)CH₂CH₃
d) 3-butoxy-2,4-dimethyloctane → CH₃CH₂CH₂CH₂–O–CH(CH₃)CH₂CH(CH₃)CH₂CH₂CH₂CH₃
e) 2-propoxy-4-phenylheptane → CH₃CH(CH₃)–O–CH₂CH₂CH(C₆H₅)CH₂CH₂CH₃
f) 1-ethoxycyclopentane → cyclopentane ring with –OCH₂CH₃ attached to one carbon
5. Name the following ethers
a) ethyl propyl ether
b) 1-methoxy-2-methylpropane
c) 2-ethoxy-2-methylbutane
Parent Tip: Review the logic above to help your child master the concept of organic chemistry nomenclature practice worksheet.
