Chapter 11: Alcohols, Phenols & Ethers
Exhaustive Revision Guide - 50 Subjective Questions with Detailed Solutions
Click on any question to reveal its answer.
1. Define Alcohols and Phenols. How do they differ structurally?
Alcohols: These are hydroxyl ($-OH$) derivatives of aliphatic hydrocarbons. The $-OH$ group is attached to an $sp^3$ hybridized carbon atom of an alkyl group. Example: Ethanol ($CH_3CH_2OH$).
Phenols: These are hydroxyl derivatives of aromatic hydrocarbons. The $-OH$ group is directly attached to an $sp^2$ hybridized carbon atom of a benzene ring. Example: Phenol ($C_6H_5OH$).
2. Classify alcohols based on the number of hydroxyl groups. Give one example of each.
- Monohydric Alcohols: Contain one $-OH$ group. Example: $CH_3CH_2OH$ (Ethanol).
- Dihydric Alcohols: Contain two $-OH$ groups. Example: $CH_2(OH)-CH_2(OH)$ (Ethane-1,2-diol or Ethylene glycol).
- Trihydric Alcohols: Contain three $-OH$ groups. Example: $CH_2(OH)-CH(OH)-CH_2(OH)$ (Propane-1,2,3-triol or Glycerol).
3. Define Ethers. How are they classified?
Ethers: These are organic compounds in which an oxygen atom is bonded to two alkyl or aryl groups. Their general formula is $R-O-R'$. They are considered anhydrides of alcohols.
Classification:
- Simple / Symmetrical Ethers: Both alkyl/aryl groups attached to the oxygen are the same ($R = R'$). Example: $CH_3-O-CH_3$ (Dimethyl ether).
- Mixed / Unsymmetrical Ethers: The two alkyl/aryl groups are different ($R \neq R'$). Example: $CH_3-O-C_2H_5$ (Ethyl methyl ether).
4. What are allylic and benzylic alcohols? Give examples.
- Allylic Alcohols: The $-OH$ group is attached to an $sp^3$ hybridized carbon next to a carbon-carbon double bond. Example: $CH_2=CH-CH_2OH$ (Prop-2-en-1-ol or Allyl alcohol).
- Benzylic Alcohols: The $-OH$ group is attached to an $sp^3$ hybridized carbon next to an aromatic ring. Example: $C_6H_5-CH_2OH$ (Benzyl alcohol).
5. Explain the preparation of alcohols by acid-catalyzed hydration of alkenes. State the rule followed.
Alkenes react with water in the presence of a strong acid catalyst (like dilute $H_2SO_4$) to form alcohols. The addition of water across the double bond follows Markownikoff's rule.
$$CH_3-CH=CH_2 + H_2O \xrightarrow{H^+} CH_3-CH(OH)-CH_3$$
Propene yields Propan-2-ol as the major product.
6. Explain the hydroboration-oxidation reaction of alkenes. What rule does the final product appear to follow?
Alkenes react with diborane ($B_2H_6$ or $BH_3 \cdot THF$) to form trialkylboranes. These are then oxidized by hydrogen peroxide ($H_2O_2$) in an alkaline medium ($NaOH$) to yield alcohols.
$3CH_3-CH=CH_2 + \frac{1}{2}B_2H_6 \rightarrow (CH_3-CH_2-CH_2)_3B$
$(CH_3-CH_2-CH_2)_3B + 3H_2O_2 \xrightarrow{OH^-} 3CH_3-CH_2-CH_2OH + B(OH)_3$
The final product appears as if water has been added across the double bond according to Anti-Markownikoff's rule, yielding primarily primary alcohols from terminal alkenes.
7. How are primary and secondary alcohols prepared by the reduction of aldehydes and ketones?
Aldehydes and ketones can be reduced using reducing agents like $H_2/Ni$, $LiAlH_4$, or $NaBH_4$.
- Aldehydes on reduction give Primary ($1^\circ$) alcohols.
$R-CHO + H_2 \xrightarrow{Ni} R-CH_2OH$ - Ketones on reduction give Secondary ($2^\circ$) alcohols.
$R-CO-R' + H_2 \xrightarrow{Ni} R-CH(OH)-R'$
8. Why is $LiAlH_4$ not widely used for the commercial reduction of carboxylic acids to alcohols? What is the alternative?
$LiAlH_4$ is an excellent reagent that directly reduces carboxylic acids to primary alcohols. However, it is an expensive reagent, making the commercial process unviable.
Alternative: Commercially, carboxylic acids are first converted to their corresponding esters by reacting with an alcohol. The ester is then reduced to alcohols using catalytic hydrogenation ($H_2$ / catalyst), which is much cheaper.
$RCOOH + R'OH \xrightarrow{H^+} RCOOR' + H_2O$
$RCOOR' + 2H_2 \xrightarrow{Catalyst} RCH_2OH + R'OH$
9. Explain the general mechanism of the reaction of a Grignard reagent with a carbonyl compound.
Grignard reagents ($R^{\delta-}-Mg^{\delta+}X$) are highly nucleophilic at the alkyl carbon.
Step 1 (Nucleophilic addition): The nucleophilic alkyl group ($R^-$) attacks the electrophilic carbonyl carbon ($>C=O$), and the $MgX^+$ attaches to the oxygen, forming an adduct (magnesium alkoxide).
$>C=O + R-MgX \xrightarrow{\text{dry ether}} >C(R)-OMgX$
Step 2 (Hydrolysis): The adduct is hydrolyzed with dilute acid ($H_2O/H^+$) to yield an alcohol.
$>C(R)-OMgX + H_2O \xrightarrow{H^+} >C(R)-OH + Mg(OH)X$
10. How will you prepare a primary alcohol using a Grignard reagent?
A primary alcohol is prepared by reacting a Grignard reagent strictly with Formaldehyde (Methanal, HCHO).
$HCHO + CH_3MgBr \xrightarrow{\text{dry ether}} CH_3-CH_2-OMgBr$
$CH_3-CH_2-OMgBr + H_2O \xrightarrow{H^+} CH_3CH_2OH \text{ (Primary alcohol)} + Mg(OH)Br$
11. How will you prepare a secondary alcohol using a Grignard reagent?
A secondary alcohol is prepared by reacting a Grignard reagent with any aldehyde other than formaldehyde (e.g., Acetaldehyde, $CH_3CHO$).
$CH_3CHO + CH_3MgBr \xrightarrow{\text{dry ether}} CH_3-CH(OMgBr)-CH_3$
$CH_3-CH(OMgBr)-CH_3 + H_2O \xrightarrow{H^+} CH_3-CH(OH)-CH_3 \text{ (Propan-2-ol, } 2^\circ \text{ alcohol)}$
12. How will you prepare a tertiary alcohol using a Grignard reagent?
A tertiary alcohol is prepared by reacting a Grignard reagent with a ketone (e.g., Acetone, $CH_3COCH_3$).
$CH_3COCH_3 + CH_3MgBr \xrightarrow{\text{dry ether}} (CH_3)_3C-OMgBr$
$(CH_3)_3C-OMgBr + H_2O \xrightarrow{H^+} (CH_3)_3C-OH \text{ (2-Methylpropan-2-ol, } 3^\circ \text{ alcohol)}$
13. Describe the commercial preparation of phenol by the Cumene process.
This is the most important commercial method. Cumene (isopropylbenzene) is oxidized in the presence of air to form cumene hydroperoxide.
$C_6H_5-CH(CH_3)_2 + O_2 \xrightarrow{423K} C_6H_5-C(O-O-H)(CH_3)_2$
The cumene hydroperoxide is then treated with dilute acid (e.g., dil. $H_2SO_4$), which cleaves it to form Phenol and a very valuable by-product, Acetone.
$C_6H_5-C(O-O-H)(CH_3)_2 \xrightarrow{H^+} C_6H_5OH \text{ (Phenol)} + CH_3COCH_3 \text{ (Acetone)}$
14. Explain the Dow Process for the preparation of phenol.
In the Dow process, Chlorobenzene is fused with aqueous Sodium hydroxide (NaOH) at a high temperature (623 K) and high pressure (300 atm) to form Sodium phenoxide.
$C_6H_5Cl + 2NaOH \xrightarrow{623K, \ 300atm} C_6H_5ONa + NaCl + H_2O$
The sodium phenoxide is then acidified (with dilute HCl) to yield phenol.
$C_6H_5ONa + HCl \rightarrow C_6H_5OH + NaCl$
15. How is phenol prepared from benzene sulfonic acid?
Benzene is first sulfonated using oleum to get benzene sulfonic acid.
The acid is neutralized by NaOH to form sodium benzene sulfonate. This salt is then fused with solid NaOH at 573 K to yield sodium phenoxide.
$C_6H_5SO_3Na + 2NaOH \xrightarrow{573K} C_6H_5ONa + Na_2SO_3 + H_2O$
Finally, acidification of sodium phenoxide gives phenol.
$C_6H_5ONa + H^+ \rightarrow C_6H_5OH + Na^+$
16. How is phenol prepared from Aniline (Diazotization)?
Aniline is treated with nitrous acid (prepared in situ from $NaNO_2 + HCl$) at 273-278 K to form benzene diazonium chloride. This is called Diazotization.
$C_6H_5NH_2 + NaNO_2 + 2HCl \xrightarrow{273-278K} C_6H_5N_2^+Cl^- + NaCl + 2H_2O$
The diazonium salt solution is then warmed with water or dilute acids to yield phenol, releasing nitrogen gas.
$C_6H_5N_2^+Cl^- + H_2O \xrightarrow{\text{warm}} C_6H_5OH + N_2 \uparrow + HCl$
17. Describe the preparation of ethers by continuous etherification process.
Ethers can be prepared by intermolecular dehydration of primary alcohols in the presence of protic acids ($H_2SO_4$ or $H_3PO_4$) at a specific temperature (413 K).
$2CH_3CH_2OH \xrightarrow{H_2SO_4, \ 413K} CH_3CH_2-O-CH_2CH_3 \text{ (Diethyl ether)} + H_2O$
If the temperature is higher (443 K), intramolecular dehydration dominates, yielding an alkene (Ethene) instead.
18. Explain Williamson Synthesis for the preparation of ethers.
It is an important laboratory method for preparing both symmetrical and unsymmetrical ethers. An alkyl halide reacts with a sodium alkoxide or sodium phenoxide via an $S_N2$ mechanism.
$R-X + R'-O^-Na^+ \rightarrow R-O-R' + NaX$
Example: $CH_3Br + C_2H_5ONa \rightarrow CH_3-O-C_2H_5 \text{ (Methoxyethane)} + NaBr$
19. What is the limitation of Williamson synthesis regarding the choice of alkyl halides?
Because Williamson synthesis follows an $S_N2$ mechanism, the alkyl halide must be primary ($1^\circ$) to avoid steric hindrance.
If a secondary ($2^\circ$) or tertiary ($3^\circ$) alkyl halide is used, the strongly basic alkoxide ion will cause elimination rather than substitution, resulting in an alkene as the major product instead of an ether. Therefore, to make a mixed ether with a tertiary group, the alkoxide must be tertiary and the halide must be primary.
20. How will you prepare tert-butyl methyl ether using Williamson synthesis?
To prepare $(CH_3)_3C-O-CH_3$, the alkyl halide must be primary ($CH_3Br$) and the alkoxide must be tertiary ($(CH_3)_3C-ONa$).
$$CH_3-Br + (CH_3)_3C-O^-Na^+ \rightarrow (CH_3)_3C-O-CH_3 + NaBr$$
If we used tert-butyl bromide and sodium methoxide, elimination would occur, yielding 2-methylpropene.
21. Why do alcohols have exceptionally higher boiling points than alkanes or ethers of comparable molecular mass?
Alcohols possess highly polar $-OH$ groups. This allows alcohol molecules to form strong intermolecular hydrogen bonds with one another. A large amount of thermal energy is required to break these strong hydrogen bonds to vaporize the liquid, resulting in a higher boiling point.
Alkanes and ethers lack an $-OH$ group, so they cannot form intermolecular hydrogen bonds (ethers only have weaker dipole-dipole interactions).
22. Explain the solubility of lower alcohols and ethers in water.
Lower alcohols and lower ethers are highly soluble in water because the oxygen atom in their molecules can form hydrogen bonds with water molecules.
However, as the size of the hydrophobic (water-repelling) alkyl group increases in higher homologs, the solubility decreases rapidly.
23. Why are alcohols weakly acidic? What is the trend of acidity among $1^\circ, 2^\circ,$ and $3^\circ$ alcohols?
Alcohols are weakly acidic due to the polarity of the $O-H$ bond, allowing them to release an $H^+$ ion to strong bases (like Na metal).
Trend: $1^\circ > 2^\circ > 3^\circ$.
Reason: Alkyl groups are electron-donating (+I effect). In a $3^\circ$ alcohol, three alkyl groups push electrons towards the oxygen, increasing the electron density on oxygen. This strengthens the $O-H$ bond and destabilizes the resulting alkoxide ion, making it the least acidic.
24. Phenols are more acidic than alcohols. Justify.
Phenol loses an $H^+$ to form a phenoxide ion. This phenoxide ion is highly stabilized by resonance, as the negative charge on the oxygen is delocalized over the ortho and para positions of the benzene ring. This stabilization drives the equilibrium forward, releasing $H^+$.
In contrast, alcohols form alkoxide ions ($R-O^-$) which have no resonance stabilization and are destabilized by the +I effect of the alkyl group. Thus, phenols are much more acidic than alcohols.
25. How do electron-withdrawing groups (EWG) and electron-donating groups (EDG) affect the acidity of phenol?
- EWG (like $-NO_2, -X, -CN$): They withdraw electron density from the ring, which disperses the negative charge on the phenoxide ion, stabilizing it. Thus, they increase acidity (especially at ortho/para positions).
- EDG (like $-CH_3, -OCH_3$): They donate electron density to the ring, intensifying the negative charge on the phenoxide ion, destabilizing it. Thus, they decrease acidity.
26. Describe the esterification reaction of alcohols.
Alcohols react with carboxylic acids, acid chlorides, or acid anhydrides to form esters. The reaction with a carboxylic acid is reversible and catalyzed by a small amount of concentrated $H_2SO_4$.
$$R-COOH + R'-OH \xrightleftharpoons{H^+} R-COOR' \text{ (Ester)} + H_2O$$
The forward reaction is favored by removing water as it is formed.
27. What is Acetylation? Give an example involving phenol.
Acetylation: The introduction of an acetyl group ($-COCH_3$) into an alcohol or phenol is known as acetylation. It is usually done using acetyl chloride or acetic anhydride.
Example (Aspirin preparation): Salicylic acid reacts with acetic anhydride in the presence of an acid catalyst to form acetylsalicylic acid (Aspirin).
$$HOC_6H_4COOH + (CH_3CO)_2O \xrightarrow{H^+} CH_3COOC_6H_4COOH \text{ (Aspirin)} + CH_3COOH$$
28. Describe the dehydration of ethanol at 443 K. Write the mechanism.
Ethanol undergoes intramolecular dehydration when heated with concentrated $H_2SO_4$ at 443 K to yield ethene.
$$CH_3CH_2OH \xrightarrow{H_2SO_4, \ 443K} CH_2=CH_2 + H_2O$$
Mechanism (3 steps):
- Protonation: Formation of protonated alcohol. $CH_3CH_2OH + H^+ \rightleftharpoons CH_3CH_2OH_2^+$
- Loss of water (Slow, RDS): Formation of a carbocation. $CH_3CH_2OH_2^+ \rightarrow CH_3CH_2^+ + H_2O$
- Deprotonation: Loss of a proton to form ethene. $CH_3CH_2^+ \rightarrow CH_2=CH_2 + H^+$
29. Explain the oxidation of primary, secondary, and tertiary alcohols.
- Primary alcohols: Oxidized to aldehydes, which are easily further oxidized to carboxylic acids with strong oxidizing agents (like acidified $K_2Cr_2O_7$ or $KMnO_4$).
$RCH_2OH \rightarrow RCHO \rightarrow RCOOH$. - Secondary alcohols: Oxidized to ketones using $CrO_3$ or $K_2Cr_2O_7$. Ketones resist further oxidation.
$R-CH(OH)-R' \rightarrow R-CO-R'$. - Tertiary alcohols: Resist oxidation under normal conditions because they lack an $\alpha$-hydrogen. Under drastic conditions (strong acid + high heat), they undergo cleavage of $C-C$ bonds to form a mixture of carboxylic acids with fewer carbon atoms.
30. How can a primary alcohol be oxidized strictly to an aldehyde without forming a carboxylic acid?
To stop the oxidation at the aldehyde stage, a milder oxidizing agent such as PCC (Pyridinium chlorochromate) in an anhydrous solvent like dichloromethane ($CH_2Cl_2$) is used.
$$RCH_2OH \xrightarrow{PCC} RCHO$$
31. What happens when vapors of $1^\circ, 2^\circ,$ and $3^\circ$ alcohols are passed over heated Copper at 573 K?
- Primary ($1^\circ$): Undergoes dehydrogenation to form an Aldehyde. ($RCH_2OH \rightarrow RCHO + H_2$)
- Secondary ($2^\circ$): Undergoes dehydrogenation to form a Ketone. ($RCH(OH)R \rightarrow RCOR + H_2$)
- Tertiary ($3^\circ$): Does not undergo dehydrogenation. Instead, it undergoes dehydration to form an Alkene. ($(CH_3)_3COH \rightarrow CH_3-C(CH_3)=CH_2 + H_2O$)
32. Phenol undergoes electrophilic aromatic substitution much more readily than benzene. Why?
The $-OH$ group in phenol is a strongly activating group. Its oxygen atom donates a lone pair of electrons into the benzene ring via resonance (+R effect). This significantly increases the electron density of the ring, particularly at the ortho and para positions, making it highly attractive to incoming electrophiles.
33. Write the reaction of phenol with dilute $HNO_3$. How are the products separated?
Phenol reacts with dilute nitric acid at low temperature (298 K) to give a mixture of o-nitrophenol and p-nitrophenol.
$C_6H_5OH + \text{dil. } HNO_3 \rightarrow \text{o-Nitrophenol} + \text{p-Nitrophenol}$
Separation: They are separated by steam distillation. o-Nitrophenol is steam volatile due to intramolecular hydrogen bonding. p-Nitrophenol is less volatile because of intermolecular hydrogen bonding, which associates the molecules together.
34. What is Picric acid? How is it prepared from phenol?
Picric acid is 2,4,6-trinitrophenol. It is a strong acid due to the presence of three electron-withdrawing $-NO_2$ groups.
Preparation: It is prepared by treating phenol with a concentrated nitrating mixture (conc. $HNO_3$ + conc. $H_2SO_4$).
$C_6H_5OH + 3HNO_{3(conc)} \xrightarrow{H_2SO_4} C_6H_2(OH)(NO_2)_3 + 3H_2O$
35. Describe the bromination of phenol in (a) aqueous medium (b) non-polar solvent like $CS_2$.
- (a) Aqueous medium (Bromine water): Phenol is highly activated. It rapidly undergoes polyhalogenation to form a white precipitate of 2,4,6-tribromophenol.
- (b) In $CS_2$ or $CHCl_3$ at low temp: The non-polar solvent decreases the activating effect of the $-OH$ group. Monohalogenation occurs, giving a mixture of p-bromophenol (major) and o-bromophenol (minor).
36. Describe Kolbe's Reaction.
Kolbe's Reaction: Phenol is treated with NaOH to form sodium phenoxide (which is even more reactive than phenol). This is treated with carbon dioxide ($CO_2$, a weak electrophile) at 400 K and 4-7 atm pressure, followed by acidification, to yield Salicylic acid (2-Hydroxybenzoic acid).
1. $C_6H_5OH \xrightarrow{NaOH} C_6H_5ONa$
2. $C_6H_5ONa \xrightarrow{CO_2, \ 400K} \xrightarrow{H^+} HOC_6H_4COOH \text{ (Salicylic acid)}$
37. Describe the Reimer-Tiemann Reaction.
Reimer-Tiemann Reaction: When phenol is treated with chloroform ($CHCl_3$) in the presence of aqueous NaOH at 340 K, a $-CHO$ group is introduced at the ortho position. The intermediate substituted benzal chloride is hydrolyzed to yield Salicylaldehyde (2-Hydroxybenzaldehyde).
The electrophile in this reaction is dichlorocarbene ($:CCl_2$).
$C_6H_5OH \xrightarrow{CHCl_3 + \text{aq. NaOH}} \xrightarrow{H^+} HOC_6H_4CHO \text{ (Salicylaldehyde)}$
38. How is phenol oxidized? What is the product?
Phenol undergoes oxidation with strong oxidizing agents like chromic acid ($Na_2Cr_2O_7 + H_2SO_4$) to produce a conjugated diketone called p-benzoquinone.
In the presence of air, phenols slowly oxidize and turn dark due to the formation of this quinone and other colored polymeric mixtures.
39. What happens when phenol is heated with zinc dust?
When phenol is heated with zinc dust, it undergoes reduction. The zinc extracts the oxygen atom, reducing phenol to benzene.
$$C_6H_5OH + Zn \xrightarrow{\Delta} C_6H_6 \text{ (Benzene)} + ZnO$$
40. Ethers are generally quite unreactive. Why?
Ethers ($R-O-R$) are highly stable and relatively unreactive because the $C-O$ bonds are strong and difficult to cleave. They do not have an active hydrogen atom (like alcohols), nor are they electrophilic. They are inert towards bases, reducing agents, and active metals, which is why they are widely used as inert solvents.
41. Write the reaction for the cleavage of diethyl ether with cold concentrated Hydroiodic acid (HI).
With cold conc. HI, ether is cleaved to give one molecule of alkyl halide and one molecule of alcohol.
$$CH_3CH_2-O-CH_2CH_3 + HI \rightarrow CH_3CH_2-I \text{ (Ethyl iodide)} + CH_3CH_2OH \text{ (Ethanol)}$$
If an excess of hot HI is used, the alcohol formed further reacts to yield a second molecule of alkyl halide and water.
$CH_3CH_2-O-CH_2CH_3 + 2HI \xrightarrow{\Delta} 2CH_3CH_2I + H_2O$
42. In the cleavage of unsymmetrical ethers containing primary or secondary alkyl groups by HI, which alkyl group forms the iodide? Explain.
When the ether has $1^\circ$ or $2^\circ$ alkyl groups, the cleavage follows an $S_N2$ mechanism. The iodide ion ($I^-$) is a bulky nucleophile, so it attacks the smaller (less sterically hindered) alkyl group.
Example: $CH_3-O-CH_2CH_3 + HI \rightarrow CH_3-I \text{ (Methyl iodide)} + CH_3CH_2OH$
43. In the cleavage of an ether containing a tertiary alkyl group by HI, which group forms the iodide? Explain.
If one of the alkyl groups is tertiary ($3^\circ$), the cleavage follows an $S_N1$ mechanism. The reaction proceeds via the formation of a highly stable tertiary carbocation. Therefore, the iodide ion attaches to the tertiary alkyl group.
Example: $(CH_3)_3C-O-CH_3 + HI \rightarrow (CH_3)_3C-I \text{ (tert-Butyl iodide)} + CH_3OH$
44. Write the cleavage reaction of Anisole (Methoxybenzene) with HI. Which bonds break?
In anisole ($C_6H_5-O-CH_3$), the $O-CH_3$ bond is weaker than the $O-C_6H_5$ bond because the $O-Phenyl$ bond has partial double bond character due to resonance. Hence, the $O-CH_3$ bond breaks.
$$C_6H_5-O-CH_3 + HI \rightarrow C_6H_5OH \text{ (Phenol)} + CH_3I \text{ (Methyl iodide)}$$
Phenol cannot be further cleaved by HI because the $C-O$ bond is too strong.
45. What is the action of atmospheric oxygen on ethers?
When exposed to air and light for a long time, ethers slowly undergo auto-oxidation to form highly explosive peroxides.
$CH_3CH_2-O-CH_2CH_3 + O_2 \xrightarrow{light} CH_3-CH(OOH)-O-CH_2CH_3 \text{ (Ether peroxide)}$
This is why old bottles of ether must never be distilled to dryness, as the concentrated peroxides may explode.
46. How will you distinguish between 1-Propanol and 2-Propanol using a chemical test?
Use the Lucas Test ($ZnCl_2$ + conc. HCl):
- 2-Propanol (Secondary alcohol): Will react with Lucas reagent to produce cloudiness/turbidity within 5 minutes at room temperature.
- 1-Propanol (Primary alcohol): Will not produce turbidity at room temperature (remains clear).
47. How can you distinguish between Ethanol and Phenol chemically? (Give two tests).
- Ferric Chloride Test: Phenol reacts with neutral $FeCl_3$ to give a deep violet/purple colored complex. Ethanol does not react.
- Bromine Water Test: Phenol reacts with bromine water to give a white precipitate of 2,4,6-tribromophenol. Ethanol does not react.
48. What is denatured alcohol and why is it made?
Denatured Alcohol: Commercial ethanol is made unfit for drinking by adding toxic substances like methanol, pyridine, or copper sulfate. This process is called denaturation.
Reason: It is done to prevent the misuse of industrial alcohol for drinking purposes, ensuring it is used strictly for commercial/solvent applications.
49. Mention two uses of Methanol and Ethanol.
- Methanol ($CH_3OH$): Used as a solvent for paints and varnishes, and primarily for the manufacture of formaldehyde. (Note: Highly toxic).
- Ethanol ($C_2H_5OH$): Used as a solvent in the perfume, cosmetic, and pharmaceutical industries, and as an active ingredient in alcoholic beverages.
50. Mention two uses of Phenol.
- Used extensively in the manufacture of phenol-formaldehyde resins (like Bakelite).
- Used as a starting material for the synthesis of important drugs like Aspirin, and antiseptics like Dettol (which contains chloroxylenol, a phenol derivative). It is also used directly in dilute solutions as a disinfectant.
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