Top 50 Most Important
Subjective Questions
Alcohols, Phenols, and Ethers. Master Name Reactions, Mechanisms, Acidity comparisons, and key conversions.
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Preparation of Alcohols
Alkenes react with water in the presence of an acid (like $H_2SO_4$) as a catalyst to form alcohols. In the case of unsymmetrical alkenes, the addition reaction strictly follows Markovnikov's rule, where the $OH$ group attaches to the more highly substituted carbon.
$CH_3-CH=CH_2 + H_2O \xrightarrow{H^+} CH_3-CH(OH)-CH_3 \text{ (Propan-2-ol)}$
Diborane ($B_2H_6$) adds to an alkene to form a trialkylborane. This is subsequently oxidized by hydrogen peroxide ($H_2O_2$) in an alkaline medium to form an alcohol.
Advantage: The net result is the addition of water to the alkene in a direction strictly opposite to Markovnikov's rule (Anti-Markovnikov hydration), yielding high amounts of primary alcohols.
Grignard reagents ($RMgX$) react with carbonyl compounds followed by hydrolysis:
- Methanal (Formaldehyde) yields a Primary ($1^\circ$) alcohol.
- Other Aldehydes yield Secondary ($2^\circ$) alcohols.
- Ketones yield Tertiary ($3^\circ$) alcohols.
$LiAlH_4$ is a very strong and highly reactive reducing agent; it reduces aldehydes, ketones, esters, and carboxylic acids. $NaBH_4$ is a milder, selective reducing agent. It strictly reduces only aldehydes and ketones into alcohols, leaving esters and carboxylic acid groups completely untouched.
It is manufactured via fermentation. The enzyme invertase (from yeast) converts sucrose in molasses into glucose and fructose. Then, the enzyme zymase converts glucose/fructose into ethanol and $CO_2$.
$C_6H_{12}O_6 \xrightarrow{\text{Zymase}} 2C_2H_5OH + 2CO_2$
To prevent the misuse of industrial-grade ethanol for drinking purposes, it is made unfit for human consumption by adding toxic substances like methanol, pyridine, or copper sulphate (which gives it a foul smell and a blue color). This process is called the denaturation of alcohol.
Because of the highly polar -OH group, alcohol molecules form strong intermolecular hydrogen bonds with each other. Breaking these strong intermolecular forces requires a significantly higher amount of thermal energy compared to the weak Van der Waals forces in alkanes or dipole-dipole forces in haloalkanes.
Lower alcohols are highly soluble in water because they can easily form hydrogen bonds with water molecules. However, as the molecular mass increases, the size of the hydrophobic (water-repelling) alkyl group becomes larger, dominating the molecule and hindering hydrogen bond formation. Thus, solubility steeply decreases.
Propan-1-ol has a higher boiling point. Propan-2-ol has a branched structure, making the molecule more compact and spherical. This decreases the overall surface area of contact, weakening the intermolecular Van der Waals forces compared to the straight-chain propan-1-ol.
An ethanol-water mixture forms a minimum-boiling azeotrope at approximately 95% ethanol concentration. Because an azeotrope boils at a constant temperature with the exact same composition in both the liquid and vapor phases, further separation by simple fractional distillation is physically impossible.
Chemical Reactions of Alcohols
Alcohols are weaker acids than water.
Reason: In an alcohol ($R-OH$), the alkyl group ($R$) has an electron-donating inductive effect (+I effect). It pushes electron density toward the oxygen atom, increasing the electron density on oxygen. This makes it harder for the oxygen to release the $H^+$ ion, and also destabilizes the resulting alkoxide ion ($RO^-$) compared to the hydroxide ion ($OH^-$) from water.
Order of acidity: $1^\circ > 2^\circ > 3^\circ$.
As the number of alkyl groups attached to the alpha-carbon increases from primary to tertiary, the cumulative +I effect increases massively. This heavily destabilizes the alkoxide ion, making tertiary alcohols the least acidic.
It is a nucleophilic acyl substitution.
1. The acid catalyst protonates the carbonyl oxygen of the carboxylic acid, making the carbonyl carbon highly electrophilic.
2. The alcohol acts as a nucleophile and attacks the carbonyl carbon.
3. After a proton transfer, a molecule of water is eliminated, yielding an ester.
Note: Isotopic labeling proves the oxygen in the water byproduct comes entirely from the carboxylic acid, not the alcohol.
The Lucas reagent is a mixture of concentrated HCl and anhydrous $ZnCl_2$. It converts soluble alcohols into insoluble alkyl chlorides, creating turbidity (cloudiness).
- $3^\circ$ Alcohols: Turbidity appears immediately (highly stable $3^\circ$ carbocation formed).
- $2^\circ$ Alcohols: Turbidity appears after 5 minutes.
- $1^\circ$ Alcohols: No turbidity at room temperature (only upon heating).
PCC (Pyridinium Chlorochromate) is a mild oxidizing agent. It selectively halts the oxidation of primary alcohols at the aldehyde stage.
Acidic $KMnO_4$ is a very strong oxidizing agent. It aggressively oxidizes the primary alcohol past the aldehyde stage all the way to a highly oxidized carboxylic acid.
While primary and secondary alcohols undergo dehydrogenation (oxidation) to form aldehydes and ketones respectively, tertiary alcohols lack an alpha-hydrogen. Therefore, they undergo dehydration over heated copper to form an Alkene.
- Protonation: The acid ($H^+$) protonates the oxygen of ethanol to form an oxonium ion ($CH_3CH_2OH_2^+$).
- Carbocation Formation: The oxonium ion loses a molecule of water to form a primary carbocation ($CH_3CH_2^+$). This is the rate-determining step.
- Elimination: A proton ($H^+$) is rapidly eliminated from the adjacent beta-carbon to form the double bond (Ethene).
Use the Iodoform Test.
Ethanol contains the required $CH_3-CH(OH)-$ group. When treated with $I_2$ and $NaOH$, it gives a prominent yellow precipitate of Iodoform ($CHI_3$).
Methanol does not contain this group and will not give a yellow precipitate.
In this reaction (Darzen's process), both byproducts formed ($SO_2$ and $HCl$) are gases. They escape naturally into the atmosphere, leaving behind pure liquid alkyl chloride. This completely avoids the need for complex separation techniques.
Secondary alcohols are oxidized by chromic anhydride ($CrO_3$) or acidic dichromate to strictly form ketones. They do not easily oxidize further.
$CH_3-CH(OH)-CH_3 \xrightarrow{CrO_3} CH_3-CO-CH_3 \text{ (Propanone/Acetone)}$
Phenols
Cumene (isopropylbenzene) is oxidized in the presence of air to form Cumene hydroperoxide. This is then treated with dilute acid to yield Phenol and a highly valuable byproduct, Acetone. It is the premier industrial method because both products are commercially important and separated easily.
When phenol loses a proton, it forms the phenoxide ion. The negative charge on the oxygen atom is highly stabilized by delocalizing into the aromatic benzene ring via resonance. In contrast, the alkoxide ion from an alcohol has no resonance stabilization. Greater stability of the conjugate base translates to stronger acidity.
Electron-withdrawing groups increase the acidity of phenol. They pull electron density out of the ring, helping to further disperse and stabilize the negative charge on the phenoxide ion. This effect is most pronounced when the EWG is at the ortho or para position.
p-Cresol < Phenol < p-Nitrophenol < 2,4,6-Trinitrophenol (Picric Acid).
p-Cresol has an electron-donating methyl group (decreases acidity). Phenol is neutral. p-Nitrophenol has one EWG (increases acidity). Picric acid has three powerful EWGs, making it exceptionally acidic.
Sodium phenoxide is heated with Carbon Dioxide ($CO_2$, a weak electrophile) under pressure, followed by acidic hydrolysis. This introduces a carboxylic acid group at the ortho position. The major product is Salicylic acid (2-Hydroxybenzoic acid).
Phenol is treated with Chloroform ($CHCl_3$) in the presence of aqueous $NaOH$, followed by acidification. An aldehyde group ($-CHO$) is introduced at the ortho position, forming Salicylaldehyde.
Intermediate Electrophile: Dichlorocarbene ($:CCl_2$).
- In Bromine water (highly polar): Phenol ionizes to highly reactive phenoxide. It undergoes tri-substitution immediately, giving a white precipitate of 2,4,6-Tribromophenol.
- In $CS_2$ (low polarity at low temp): Ionization is suppressed. It undergoes mono-substitution, yielding a mixture of o-bromophenol and p-bromophenol.
Phenol undergoes aggressive oxidation with chromic acid ($Na_2Cr_2O_7 / H_2SO_4$) to produce a conjugated diketone called p-benzoquinone. In the presence of air, phenol slowly oxidizes to dark-colored mixtures containing quinones.
o-Nitrophenol forms intramolecular hydrogen bonds (within the same molecule), preventing it from linking tightly with adjacent molecules. Therefore, it is volatile and has a lower boiling point.
p-Nitrophenol forms strong intermolecular hydrogen bonds between adjacent molecules, linking them into large chains, significantly raising its boiling point.
Use the Neutral Ferric Chloride ($FeCl_3$) Test.
Phenol reacts with neutral $FeCl_3$ to form a distinct violet/purple colored complex.
Ethanol does not react and gives no such coloration.
Ethers
It is the best method for preparing both symmetrical and unsymmetrical ethers. An alkyl halide is reacted with a sodium alkoxide. The alkoxide ion acts as a strong nucleophile, displacing the halide ion via an $S_N2$ mechanism.
$R-ONa + R'-X \rightarrow R-O-R' + NaX$
The reaction follows an $S_N2$ pathway. If a secondary or tertiary alkyl halide is used, the strong, bulky alkoxide base will preferentially trigger an elimination reaction (E2) rather than substitution, resulting in an alkene as the major product instead of an ether.
To avoid elimination, the bulky group must be the alkoxide, and the alkyl halide must be primary. Therefore, react Sodium tert-butoxide with Ethyl bromide (or chloride).
$(CH_3)_3C-O^- Na^+ + CH_3CH_2-Br \rightarrow (CH_3)_3C-O-CH_2CH_3 + NaBr$
The methoxy group ($-OCH_3$) has a strong +R (resonance) effect. The lone pairs on the oxygen atom are delocalized into the benzene ring, significantly increasing electron density strictly at the ortho and para positions. Thus, electrophiles preferentially attack these highly negative sites.
Anisole reacts with HI to yield Phenol and Methyl Iodide.
Reason: The $O-CH_3$ bond breaks, not the $O-Phenyl$ bond. The $O-Phenyl$ bond possesses partial double bond character due to resonance, making it much stronger and harder to break than the single $O-CH_3$ bond. Thus, the $I^-$ attacks the methyl group.
When the ether has primary or secondary alkyl groups, the cleavage follows an $S_N2$ mechanism. The bulky iodine nucleophile ($I^-$) prefers to attack the less sterically hindered carbon. Thus, the smaller alkyl group forms the alkyl iodide (Methyl iodide), and the larger group forms the alcohol (Ethanol).
If one of the alkyl groups is tertiary, the mechanism switches entirely to $S_N1$. The oxonium ion cleaves to form a highly stable tertiary carbocation. The $I^-$ ion then attacks this carbocation. Thus, the bulky tertiary group forms the iodide (tert-Butyl iodide), and the smaller group forms the alcohol (Methanol).
Ethers do not have a hydrogen atom directly bonded to an electronegative oxygen atom. Therefore, unlike alcohols, ether molecules cannot form intermolecular hydrogen bonds with each other. They only possess weak dipole-dipole forces, resulting in much lower boiling points.
Lower ethers (like dimethyl ether) are somewhat soluble in water. Although they cannot hydrogen bond with themselves, the oxygen atom in the ether can form hydrogen bonds with the hydrogen atoms of water molecules. However, solubility drops sharply for higher ethers as the bulky hydrophobic alkyl groups block interaction.
Anisole undergoes electrophilic substitution to give a mixture of p-bromoanisole (major product due to less steric hindrance) and o-bromoanisole (minor product). Due to the highly activating nature of the methoxy group, no Lewis acid catalyst (like $FeBr_3$) is required for this reaction.
Mechanisms & Conversions
- Protonation: The alkene's $\pi$-electrons attack a hydronium ion ($H_3O^+$), forming a stable carbocation intermediate (follows Markovnikov).
- Nucleophilic Attack: A water molecule uses its lone pairs to attack the electrophilic carbocation.
- Deprotonation: The positively charged oxygen in the intermediate loses a proton to another water molecule to yield the neutral alcohol.
This is a bimolecular nucleophilic substitution ($S_N2$) pathway:
1. The alcohol is protonated to form an oxonium ion.
2. A second molecule of unprotonated alcohol acts as a nucleophile, attacking the carbon of the oxonium ion and displacing water.
3. The resulting protonated ether loses a proton to form the final dialkyl ether. (Occurs at 413 K; higher temps like 443 K favor elimination to alkenes).
Step 1 (Kolbe's Reaction): React phenol with $NaOH$ to get sodium phenoxide, then heat with $CO_2$ under pressure, followed by acidification to get Salicylic acid.
Step 2 (Acetylation): Treat salicylic acid with acetic anhydride in the presence of a few drops of conc. $H_2SO_4$. The $-OH$ group is acetylated to form Acetylsalicylic acid (Aspirin).
1. Oxidize Ethanol using PCC to get Ethanal ($CH_3CHO$).
2. React Ethanal with Methylmagnesium bromide ($CH_3MgBr$) followed by hydrolysis to get a secondary alcohol, Propan-2-ol.
3. Oxidize Propan-2-ol with $CrO_3$ or acidic $K_2Cr_2O_7$ to yield Propanone.
Phenol reduces upon heating with Zinc dust to form Benzene.
When benzene is treated with $CH_3Cl$ and anhydrous $AlCl_3$ (Friedel-Crafts Alkylation), it undergoes electrophilic substitution to form Toluene (Methylbenzene).
In chlorobenzene, the lone pairs of chlorine undergo resonance with the benzene ring, giving the C-Cl bond partial double bond character. This makes the bond exceptionally strong. Normal nucleophilic substitution fails; extreme conditions (Dow's Process: 623 K and 300 atm) are required to force the substitution to yield phenol.
Use the Lucas Test.
2-Methyl-2-propanol is a tertiary alcohol and will produce immediate turbidity (cloudiness) upon addition of Lucas Reagent at room temperature.
1-Propanol is a primary alcohol and will show no reaction (no turbidity) at room temperature.
Phenol reacts with concentrated $HNO_3$ in the presence of concentrated $H_2SO_4$ to undergo aggressive tri-nitration. The product is 2,4,6-Trinitrophenol, commonly known as Picric acid.
Ethene is absorbed in cold concentrated $H_2SO_4$ to form ethyl hydrogen sulphate. This intermediate is then boiled with water (hydrolyzed) to yield Ethanol.
1. $CH_2=CH_2 + H_2SO_4 \rightarrow CH_3-CH_2-OSO_3H$
2. $CH_3-CH_2-OSO_3H + H_2O \xrightarrow{\Delta} CH_3CH_2OH + H_2SO_4$
It is a classical chemical test to definitively distinguish between $1^\circ, 2^\circ,$ and $3^\circ$ alcohols using a sequence of reagents (P/I2, AgNO2, HNO2, NaOH).
Final Colors:
- Primary ($1^\circ$): Blood Red color.
- Secondary ($2^\circ$): Blue color.
- Tertiary ($3^\circ$): Colorless (does not react with $HNO_2$).
Chapter 7 Mastered!
You have just conquered the 50 most critical subjective questions for Class 12 Chemistry, Chapter 7: Alcohols, Phenols, and Ethers. Your organic mechanisms are solid!
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