Draw the structures for the ten names in Exercise 7.2.
Steps only
(i) (CH3)2C(OH)CH2CH3. (ii) C6H5CH2CH(OH)CH3. (iii) HOCH2CH2C(OH)(CH3)CH2CH(OH)(CH3)CH3. (iv) 2,3-diethylphenol: –OH at C1, ethyl at C2 and C3. (v) CH3CH2CH2OC2H5. (vi) CH3CH(OC2H5)CH(CH3)CH2CH3. (vii) C6H11CH2OH. (viii) CH3CH2C(OH)(C6H11)CH2CH3. (ix) cyclopent-3-en-1-ol. (x) HOCH2CH(C2H5)CH2CH2Cl.
Answer
(i) (CH3)2C(OH)CH2CH3.
(ii) C6H5CH2CH(OH)CH3.
(iii) HOCH2CH2C(OH)(CH3)CH2CH(OH)(CH3)CH3.
(iv) 2,3-diethylphenol: –OH at C1, ethyl at C2 and C3.
(v) CH3CH2CH2OC2H5.
(vi) CH3CH(OC2H5)CH(CH3)CH2CH3.
(vii) C6H11CH2OH.
(viii) CH3CH2C(OH)(C6H11)CH2CH3.
(ix) cyclopent-3-en-1-ol.
(x) HOCH2CH(C2H5)CH2CH2Cl.
Question 7.3
List and classify all alcohol isomers with formula C5H12O.
Steps only
Primary: pentan-1-ol, 2-methylbutan-1-ol, 3-methylbutan-1-ol, 2,2-dimethylpropan-1-ol. Secondary: pentan-2-ol, pentan-3-ol, 3-methylbutan-2-ol. Tertiary: 2-methylbutan-2-ol. There are eight in total.
Answer
Primary: pentan-1-ol, 2-methylbutan-1-ol, 3-methylbutan-1-ol, 2,2-dimethylpropan-1-ol. Secondary: pentan-2-ol, pentan-3-ol, 3-methylbutan-2-ol. Tertiary: 2-methylbutan-2-ol. There are eight in total.
Question 7.4
Why does propanol boil at a higher temperature than butane despite similar molar masses?
Steps only
Propanol molecules form hydrogen bonds through –OH groups. Butane molecules have only weak dispersion forces, so less energy is needed to separate them.
Answer
Propanol molecules form hydrogen bonds through –OH groups. Butane molecules have only weak dispersion forces, so less energy is needed to separate them.
Question 7.5
Why are alcohols more soluble in water than hydrocarbons of similar size?
Steps only
The –OH group of an alcohol forms hydrogen bonds with water. Hydrocarbons cannot form these strong interactions and are poorly soluble.
Answer
The –OH group of an alcohol forms hydrogen bonds with water. Hydrocarbons cannot form these strong interactions and are poorly soluble.
Question 7.6
How can propan-1-ol and 2-methylpropan-2-ol be distinguished?
Steps only
Use Lucas reagent (concentrated HCl and anhydrous ZnCl2). The tertiary alcohol turns cloudy quickly at room temperature; the primary alcohol remains clear at first.
Answer
Use Lucas reagent (concentrated HCl and anhydrous ZnCl2). The tertiary alcohol turns cloudy quickly at room temperature; the primary alcohol remains clear at first.
Question 7.7
Name the monohydric phenols with formula C7H8O.
Steps only
2-methylphenol, 3-methylphenol and 4-methylphenol (the o-, m- and p-cresols). Anisole has the same formula but is an ether, not a phenol.
Answer
2-methylphenol, 3-methylphenol and 4-methylphenol (the o-, m- and p-cresols). Anisole has the same formula but is an ether, not a phenol.
Question 7.8
Why is o-nitrophenol steam volatile while p-nitrophenol is not?
Steps only
o-Nitrophenol forms an intramolecular hydrogen bond, so its molecules associate less with one another. p-Nitrophenol forms intermolecular hydrogen bonds, making its molecules associate strongly and raising its boiling point.
Answer
o-Nitrophenol forms an intramolecular hydrogen bond, so its molecules associate less with one another. p-Nitrophenol forms intermolecular hydrogen bonds, making its molecules associate strongly and raising its boiling point.
Question 7.9
How is phenol prepared from cumene?
Steps only
Oxidise cumene with oxygen to cumene hydroperoxide, then split it using dilute acid. The products are phenol and propanone (acetone).
Answer
Oxidise cumene with oxygen to cumene hydroperoxide, then split it using dilute acid. The products are phenol and propanone (acetone).
Question 7.10
Write the conversion of chlorobenzene to phenol.
Steps only
Heat chlorobenzene with aqueous NaOH at about 623 K and 300 atm to form sodium phenoxide; acidify it to obtain phenol.
Answer
Heat chlorobenzene with aqueous NaOH at about 623 K and 300 atm to form sodium phenoxide; acidify it to obtain phenol.
Question 7.11
Describe acid-catalysed hydration of ethene to ethanol.
Steps only
(1) H+ adds to ethene to form an ethyl carbocation. (2) Water attacks to make protonated ethanol. (3) Loss of H+ gives ethanol and restores the acid catalyst.
Answer
(1) H+ adds to ethene to form an ethyl carbocation. (2) Water attacks to make protonated ethanol. (3) Loss of H+ gives ethanol and restores the acid catalyst.
Question 7.12
Show the benzenesulphonic acid route from benzene to phenol.
Steps only
Benzene + fuming H2SO4 → benzenesulphonic acid. Fuse with NaOH to form sodium phenoxide; acidify to give phenol.
Answer
Benzene + fuming H2SO4 → benzenesulphonic acid. Fuse with NaOH to form sodium phenoxide; acidify to give phenol.
Question 7.13
Give routes to (i) 1-phenylethanol from an alkene, (ii) cyclohexylmethanol from a halide and (iii) pentan-1-ol from a halide.
(ii) Cyclohexylmethyl chloride + aqueous KOH → cyclohexylmethanol.
(iii) 1-bromopentane + aqueous KOH → pentan-1-ol.
Question 7.14
Write the reaction of ethanoic acid with propan-1-ol. Why is phenol more acidic than ethanol?
Steps only
CH3COOH + CH3CH2CH2OH ⇌ CH3COOCH2CH2CH3 + H2O (acid catalyst). Phenol is more acidic because its phenoxide ion is stabilised by resonance; ethoxide has no such stabilisation.
Answer
CH3COOH + CH3CH2CH2OH ⇌ CH3COOCH2CH2CH3 + H2O (acid catalyst). Phenol is more acidic because its phenoxide ion is stabilised by resonance; ethoxide has no such stabilisation.
Question 7.15
Explain the acidity order o-nitrophenol > phenol > o-methoxyphenol.
Steps only
The nitro group withdraws electrons and stabilises the phenoxide ion, increasing acidity. The methoxy group donates electrons and destabilises phenoxide, decreasing acidity.
Answer
The nitro group withdraws electrons and stabilises the phenoxide ion, increasing acidity. The methoxy group donates electrons and destabilises phenoxide, decreasing acidity.
Question 7.16
Why is phenol resistant to nucleophilic replacement of –OH but reactive toward ring electrophilic substitution?
Steps only
Resonance gives the aryl C–O bond partial double-bond character, making it difficult to break. The same oxygen lone-pair donation increases electron density at ortho and para ring positions, activating the ring toward electrophiles.
Answer
Resonance gives the aryl C–O bond partial double-bond character, making it difficult to break. The same oxygen lone-pair donation increases electron density at ortho and para ring positions, activating the ring toward electrophiles.
Question 7.17
Write equations or general reactions for Reimer–Tiemann, Kolbe, primary-alcohol oxidation and alcohol dehydration.
(ii) Sodium phenoxide + CO2, pressure/heat; then acid → 2-hydroxybenzoic acid.
(iii) RCH2OH + [O] → RCOOH.
(iv) RCH2CH2OH, conc. H2SO4/heat → alkene + H2O.
Question 7.18
Explain Kolbe, Reimer–Tiemann, Williamson synthesis and unsymmetrical ethers.
Steps only
Kolbe: sodium phenoxide + CO2, then acid → salicylic acid. Reimer–Tiemann: phenol + CHCl3/NaOH, then acid → salicylaldehyde. Williamson: RONa + R′X (primary) → ROR′ + NaX by SN2. An unsymmetrical ether has different groups on oxygen, e.g. CH3OC2H5.
Answer
Kolbe: sodium phenoxide + CO2, then acid → salicylic acid. Reimer–Tiemann: phenol + CHCl3/NaOH, then acid → salicylaldehyde. Williamson: RONa + R′X (primary) → ROR′ + NaX by SN2. An unsymmetrical ether has different groups on oxygen, e.g. CH3OC2H5.
Question 7.19
Describe HI cleavage of methoxymethane and dehydration of ethanol to ethene.
Steps only
For methoxymethane: oxygen is protonated, then I− attacks a methyl group to give CH3I + CH3OH; excess HI converts methanol to another CH3I + H2O. For ethanol: protonate –OH, remove water, then lose a neighbouring H to form ethene.
Answer
For methoxymethane: oxygen is protonated, then I− attacks a methyl group to give CH3I + CH3OH; excess HI converts methanol to another CH3I + H2O. For ethanol: protonate –OH, remove water, then lose a neighbouring H to form ethene.
Question 7.20
Give reagents for propene→propan-2-ol, benzyl chloride→benzyl alcohol, ethylmagnesium chloride→propan-1-ol and methylmagnesium bromide→2-methylpropan-2-ol.
Steps only
(i) H2O/H+. (ii) Aqueous KOH. (iii) Add HCHO, then H3O+. (iv) Add propanone, then H3O+.
Answer
(i) H2O/H+.
(ii) Aqueous KOH.
(iii) Add HCHO, then H3O+.
(iv) Add propanone, then H3O+.
Question 7.21
Name suitable reagents for the six transformations in Exercise 7.21.
Steps only
(i) Excess acidified KMnO4 or K2Cr2O7. (ii) PCC. (iii) Bromine water. (iv) Acidified KMnO4. (v) Concentrated H2SO4 and heat, or hot Al2O3. (vi) NaBH4 or LiAlH4.
Answer
(i) Excess acidified KMnO4 or K2Cr2O7.
(ii) PCC.
(iii) Bromine water.
(iv) Acidified KMnO4.
(v) Concentrated H2SO4 and heat, or hot Al2O3.
(vi) NaBH4 or LiAlH4.
Question 7.22
Why does ethanol boil at a much higher temperature than methoxymethane?
Steps only
Ethanol molecules form intermolecular hydrogen bonds through O–H. Methoxymethane has no O–H bond, so its molecules attract each other less strongly and boil at a lower temperature.
Answer
Ethanol molecules form intermolecular hydrogen bonds through O–H. Methoxymethane has no O–H bond, so its molecules attract each other less strongly and boil at a lower temperature.
Describe the Williamson ether mechanism and explain why tertiary halides are poor substrates.
Steps only
An alkoxide attacks the carbon bearing halogen from the back in a single SN2 step, displacing the halide. A tertiary halide is too crowded for this attack; the strong alkoxide base instead causes elimination to an alkene.
Answer
An alkoxide attacks the carbon bearing halogen from the back in a single SN2 step, displacing the halide. A tertiary halide is too crowded for this attack; the strong alkoxide base instead causes elimination to an alkene.
Question 7.26
How can 1-propoxypropane be prepared from propan-1-ol? Give the mechanism.
Steps only
React propan-1-ol with sodium to form sodium propoxide. Then sodium propoxide + 1-bromopropane → 1-propoxypropane + NaBr by one-step SN2 attack.
Answer
React propan-1-ol with sodium to form sodium propoxide. Then sodium propoxide + 1-bromopropane → 1-propoxypropane + NaBr by one-step SN2 attack.
Question 7.27
Why pair a secondary or tertiary alkoxide with a primary halide when making an ether with bulky groups?
Steps only
The halide must be the less crowded partner so the required SN2 attack can occur. A secondary or tertiary halide would favour elimination with the strongly basic alkoxide.
Answer
The halide must be the less crowded partner so the required SN2 attack can occur. A secondary or tertiary halide would favour elimination with the strongly basic alkoxide.
Question 7.28
Write products when HI reacts with 1-propoxypropane, anisole and benzyl ethyl ether.
Steps only
(i) 1-Propoxypropane + HI → 1-iodopropane + propan-1-ol; excess HI can convert the alcohol to more 1-iodopropane. (ii) Anisole + HI → phenol + CH3I. (iii) Benzyl ethyl ether gives benzyl iodide + ethanol under suitable cleavage conditions.
Answer
(i) 1-Propoxypropane + HI → 1-iodopropane + propan-1-ol; excess HI can convert the alcohol to more 1-iodopropane.
Why is anisole ortho/para-directing and more reactive than benzene?
Steps only
Acid-catalysed hydration of ethene: (1) H+ adds to C=C, giving an ethyl carbocation. (2) Water attacks to form protonated ethanol. (3) Loss of H+ gives ethanol and regenerates the acid. In anisole, oxygen donates a lone pair into the ring; resonance increases electron density at ortho and para positions, so the ring is activated and directs substitution to those positions.
Answer
Acid-catalysed hydration of ethene: (1) H+ adds to C=C, giving an ethyl carbocation. (2) Water attacks to form protonated ethanol. (3) Loss of H+ gives ethanol and regenerates the acid. In anisole, oxygen donates a lone pair into the ring; resonance increases electron density at ortho and para positions, so the ring is activated and directs substitution to those positions.
Question 7.30
Describe the mechanism for HI reacting with methoxymethane.
Steps only
(1) HI protonates the ether oxygen. (2) I− attacks a methyl carbon by SN2 and breaks a C–O bond, giving CH3I and methanol. (3) With excess HI, methanol is converted to a second CH3I and water.
Answer
(1) HI protonates the ether oxygen. (2) I− attacks a methyl carbon by SN2 and breaks a C–O bond, giving CH3I and methanol. (3) With excess HI, methanol is converted to a second CH3I and water.
Question 7.31
Give main products for anisole alkylation, nitration, bromination and acetylation.
Steps only
Each reaction gives mainly para product with some ortho product: (i) p-methylanisole (plus o-isomer); (ii) p-nitroanisole (plus o-isomer); (iii) p-bromoanisole (plus o-isomer); (iv) p-methoxyacetophenone (plus o-isomer).
Answer
Each reaction gives mainly para product with some ortho product:
(i) p-methylanisole (plus o-isomer);
(ii) p-nitroanisole (plus o-isomer);
(iii) p-bromoanisole (plus o-isomer);
(iv) p-methoxyacetophenone (plus o-isomer).
Question 7.32
How do hydration, oxymercuration, hydroboration and cold alkaline KMnO4 convert alkenes?
Steps only
(i) H2O/H+ gives a Markovnikov alcohol and may allow rearrangement. (ii) Hg(OAc)2/H2O then NaBH4 gives a Markovnikov alcohol without rearrangement. (iii) BH3 then H2O2/OH− gives an anti-Markovnikov alcohol. (iv) Cold dilute alkaline KMnO4 adds two –OH groups across the double bond to form a vicinal diol.
Answer
(i) H2O/H+ gives a Markovnikov alcohol and may allow rearrangement.
(ii) Hg(OAc)2/H2O then NaBH4 gives a Markovnikov alcohol without rearrangement.
(iii) BH3 then H2O2/OH− gives an anti-Markovnikov alcohol.
(iv) Cold dilute alkaline KMnO4 adds two –OH groups across the double bond to form a vicinal diol.
Question 7.33
Explain why 3-methylbutan-2-ol gives 2-bromo-2-methylbutane with HBr.
Steps only
First, –OH is protonated and leaves as water, forming a secondary carbocation. A hydride shifts from the neighbouring carbon to make a more stable tertiary carbocation. Br− then attacks, producing 2-bromo-2-methylbutane.
Answer
First, –OH is protonated and leaves as water, forming a secondary carbocation. A hydride shifts from the neighbouring carbon to make a more stable tertiary carbocation. Br− then attacks, producing 2-bromo-2-methylbutane.