Alcohols, Phenols and Ethers

Given: Various alcohol structures to classify.

Concept: An alcohol is classified based on the number of carbon atoms directly attached to the carbon bearing the –OH group.
- Primary (1°): –OH on a carbon attached to only one other carbon (or no carbon).
- Secondary (2°): –OH on a carbon attached to two other carbons.
- Tertiary (3°): –OH on a carbon attached to three other carbons.

(i) $\mathrm{CH_3-\underset{CH_3}{\overset{CH_3}{C}}-CH_2OH}$

The –OH group is on $-CH_2-$, which is attached to only one carbon (the quaternary carbon). Hence it is a Primary alcohol.

(ii) $\mathrm{H_2C=CH-CH_2OH}$

The –OH group is on $-CH_2-$, which is attached to only one carbon (the vinylic $CH$). Hence it is a Primary alcohol (also an allylic alcohol).

(iii) $\mathrm{CH_3-CH_2-CH_2-OH}$

The –OH group is on the terminal $-CH_2-$, attached to only one carbon. Hence it is a Primary alcohol.

(iv) The structure represents a secondary alcohol where –OH is on a carbon bearing two other carbon groups (e.g., $\mathrm{CH_3-\underset{OH}{CH}-CH_3}$). Hence it is a Secondary alcohol.

(v) The structure $\mathrm{CH_2-CH-CH_3}$ with –OH on the middle carbon represents a Secondary alcohol.

(vi) $\mathrm{CH=CH-\underset{OH}{C}-}$ where the carbon bearing –OH is attached to three carbons. Hence it is a Tertiary alcohol (also an allylic alcohol).

Summary:
| Compound | Classification |
|---|---|
| (i) | Primary |
| (ii) | Primary |
| (iii) | Primary |
| (iv) | Secondary |
| (v) | Secondary |
| (vi) | Tertiary |

Concept: An allylic alcohol is one in which the –OH group is present on a carbon adjacent to a carbon–carbon double bond (C=C), i.e., the –OH bearing carbon is allylic.

From the examples in Question 7.1:

- (ii) $\mathrm{H_2C=CH-CH_2OH}$: The –OH is on the carbon next to the double bond. This is an allylic alcohol.

- (vi) $\mathrm{CH=CH-C(OH)-}$: The –OH bearing carbon is directly adjacent to the double bond. This is also an allylic alcohol.

Answer: Compounds (ii) and (vi) are allylic alcohols.

Concept: IUPAC nomenclature rules — select the longest carbon chain containing the principal functional group (–OH), number from the end nearest to –OH, and name substituents with appropriate prefixes.

(i) Structure: $\mathrm{CH_3-CH_2-\underset{Cl}{CH}-\underset{CH_3}{CH}-\underset{OH}{CH}-CH_3}$ (based on context from Example 7.1)

Longest chain = 6 carbons; –OH at C-2, –CH₃ at C-3, –Cl at C-4.

$$\textbf{IUPAC Name: 4-Chloro-3-methylhexan-2-ol}$$

(ii) Structure: $\mathrm{CH_3-\underset{OH}{CH}-CH_2-\underset{OH}{CH}-\underset{C_2H_5}{CH}-CH_3}$

Longest chain = 6 carbons; –OH groups at C-2 and C-4, –C₂H₅ at C-4 (or as appropriate).

$$\textbf{IUPAC Name: 4-Ethylhexane-2,4-diol}$$

(iii) (Image-based — assumed to be a cyclic compound with –OH)

Based on standard NCERT content, this is likely 2-methylcyclohexan-1-ol or similar. *(Structure depends on figure; solve as per visible data.)*

(iv) $\mathrm{H_2C=CH-\underset{OH}{CH}-CH_2-CH_2-CH_3}$

Longest chain containing both –OH and C=C = 6 carbons; number from the –OH end: –OH at C-1, C=C between C-5 and C-6 (or number to give lowest locants).

Numbering from the double-bond end: C=C at C-1,2; –OH at C-3.

$$\textbf{IUPAC Name: hex-1-en-3-ol}$$

(v) $\mathrm{CH_3-C\equiv C-CH_2OH}$

Longest chain = 4 carbons; –OH at C-1, triple bond at C-2,3.

$$\textbf{IUPAC Name: but-2-yn-1-ol}$$

Concept: Grignard reagent (R–MgX) reacts with methanal (HCHO) to give, after hydrolysis, a primary alcohol with one more carbon than the Grignard reagent.

$$\mathrm{R-MgX + HCHO \xrightarrow{dry\,ether} R-CH_2-OMgX \xrightarrow{H_3O^+} R-CH_2OH}$$

(i) Target alcohol: $\mathrm{(CH_3)_2CH-CH_2OH}$ (2-methylpropan-1-ol)

Required Grignard reagent: $\mathrm{(CH_3)_2CH-MgBr}$ (isopropylmagnesium bromide)

$$\mathrm{(CH_3)_2CH-MgBr + HCHO \xrightarrow{1.\,dry\,ether}{2.\,H_3O^+} (CH_3)_2CH-CH_2OH}$$

$$\textbf{Product: 2-methylpropan-1-ol}$$

(ii) Target alcohol: $\mathrm{C_6H_5-CH_2OH}$ (benzyl alcohol / phenylmethanol)

Required Grignard reagent: $\mathrm{C_6H_5-MgBr}$ (phenylmagnesium bromide)

$$\mathrm{C_6H_5MgBr + HCHO \xrightarrow{1.\,dry\,ether}{2.\,H_3O^+} C_6H_5-CH_2OH}$$

$$\textbf{Product: Phenylmethanol (Benzyl alcohol)}$$

Concept:
- Acid-catalysed hydration of alkenes follows Markovnikov's rule — –OH adds to the more substituted carbon.
- $\mathrm{NaBH_4}$ is a mild reducing agent that reduces aldehydes and ketones to alcohols.

(i) $\mathrm{CH_3-CH=CH_2 \xrightarrow{H_2O/H^+}}$

By Markovnikov's rule, –OH adds to C-2 (more substituted):

$$\mathrm{CH_3-CH=CH_2 \xrightarrow{H_2O/H^+} CH_3-\underset{OH}{CH}-CH_3}$$

$$\textbf{Product: Propan-2-ol}$$

(ii) & (iii) $\mathrm{CH_3-CH_2-CH(C_2H_5)-CHO \xrightarrow{NaBH_4}}$

$\mathrm{NaBH_4}$ reduces the aldehyde (–CHO) to a primary alcohol (–CH₂OH):

$$\mathrm{CH_3-CH_2-\underset{C_2H_5}{CH}-CHO \xrightarrow{NaBH_4} CH_3-CH_2-\underset{C_2H_5}{CH}-CH_2OH}$$

$$\textbf{Product: 2-ethylbutan-1-ol}$$

Concept: Alcohols react with hydrogen halides (HX) and $\mathrm{SOCl_2}$ to give alkyl halides via nucleophilic substitution.
- Primary alcohols react with HCl only in the presence of $\mathrm{ZnCl_2}$ (Lucas reagent).
- Tertiary alcohols react readily with HCl even without $\mathrm{ZnCl_2}$.
- $\mathrm{SOCl_2}$ converts –OH to –Cl with retention of configuration (in pyridine).

(i) Butan-1-ol ($\mathrm{CH_3CH_2CH_2CH_2OH}$) — Primary alcohol

(a) HCl–ZnCl₂:
$$\mathrm{CH_3CH_2CH_2CH_2OH \xrightarrow{HCl/ZnCl_2} CH_3CH_2CH_2CH_2Cl}$$
Product: 1-Chlorobutane

(b) HBr:
$$\mathrm{CH_3CH_2CH_2CH_2OH \xrightarrow{HBr} CH_3CH_2CH_2CH_2Br}$$
Product: 1-Bromobutane

(c) SOCl₂:
$$\mathrm{CH_3CH_2CH_2CH_2OH \xrightarrow{SOCl_2} CH_3CH_2CH_2CH_2Cl + SO_2 + HCl}$$
Product: 1-Chlorobutane

(ii) 2-Methylbutan-2-ol ($\mathrm{(CH_3)_2C(OH)CH_2CH_3}$) — Tertiary alcohol

(a) HCl–ZnCl₂:
$$\mathrm{(CH_3)_2C(OH)CH_2CH_3 \xrightarrow{HCl/ZnCl_2} (CH_3)_2C(Cl)CH_2CH_3}$$
Product: 2-Chloro-2-methylbutane

(b) HBr:
$$\mathrm{(CH_3)_2C(OH)CH_2CH_3 \xrightarrow{HBr} (CH_3)_2C(Br)CH_2CH_3}$$
Product: 2-Bromo-2-methylbutane

(c) SOCl₂:
$$\mathrm{(CH_3)_2C(OH)CH_2CH_3 \xrightarrow{SOCl_2} (CH_3)_2C(Cl)CH_2CH_3 + SO_2 + HCl}$$
Product: 2-Chloro-2-methylbutane

Concept: Acid-catalysed dehydration of alcohols follows Zaitsev's rule — the more substituted (more stable) alkene is the major product. The mechanism involves formation of a carbocation intermediate.

(i) 1-Methylcyclohexanol:

1-Methylcyclohexanol is a tertiary alcohol. On dehydration, the proton is lost from the adjacent ring carbon to give the more substituted alkene:

$$\mathrm{1\text{-}methylcyclohexanol \xrightarrow{H^+/\Delta} methylenecyclohexane \;(minor) + 1\text{-}methylcyclohex\text{-}1\text{-}ene\;(major)}$$

Major product: 1-Methylcyclohex-1-ene (endocyclic double bond, more substituted)

(ii) Butan-1-ol:

Butan-1-ol is a primary alcohol. Dehydration proceeds via an $\mathrm{E2}$ mechanism (or through rearrangement to secondary carbocation):

$$\mathrm{CH_3CH_2CH_2CH_2OH \xrightarrow{H^+/\Delta} CH_3CH_2CH=CH_2 \;(but\text{-}1\text{-}ene, minor) + CH_3CH=CHCH_3\;(but\text{-}2\text{-}ene, major)}$$

Major product: But-2-ene (more substituted alkene, by Zaitsev's rule)

Concept: The –NO₂ group is an electron-withdrawing group. When present at ortho or para positions, it stabilises the phenoxide ion by delocalising the negative charge through resonance, thereby increasing acidity.

Resonance structures of ortho-nitrophenoxide ion:

The negative charge on oxygen of phenoxide is delocalised into the ring and further onto the –NO₂ group:

$$\mathrm{o\text{-}O_2N-C_6H_4-O^- \longleftrightarrow \text{(resonance structures with negative charge on NO}_2\text{ oxygen)}}$$

Key resonance contributors:
1. $\mathrm{^-O-C_6H_4-NO_2}$ (charge on ring oxygen)
2. Charge delocalized into ring at ortho position
3. $\mathrm{O=C_6H_4=N^+(O^-)(O^-)}$ — negative charge on the oxygen of –NO₂ group

Resonance structures of para-nitrophenoxide ion:

Similarly for para:
1. $\mathrm{^-O-C_6H_4-NO_2}$ (charge on ring oxygen)
2. Charge delocalized through ring
3. $\mathrm{O=C_6H_4=N^+(O^-)(O^-)}$ — negative charge on the oxygen of –NO₂ group

Conclusion: In both ortho and para isomers, the negative charge is effectively delocalized onto the electronegative oxygen atoms of the nitro group, stabilizing the phenoxide ion and making these phenols more acidic than phenol itself. The meta-nitrophenol does not show this direct resonance delocalization, so it is less acidic than ortho and para isomers.

Concept:

(i) Reimer–Tiemann Reaction:
Phenol reacts with chloroform ($\mathrm{CHCl_3}$) in the presence of aqueous NaOH to introduce a –CHO group at the ortho position of the benzene ring. The intermediate dichlorocarbene (:CCl₂) is the electrophile.

$$\mathrm{C_6H_5OH + CHCl_3 \xrightarrow{NaOH(aq)} o\text{-}HO-C_6H_4-CHO + HCl}$$

(Salicylaldehyde / 2-hydroxybenzaldehyde is the major product)

Step-wise:
$$\mathrm{CHCl_3 + NaOH \rightarrow :CCl_2 \;(dichlorocarbene) + NaCl + H_2O}$$
$$\mathrm{C_6H_5OH + :CCl_2 \rightarrow \text{intermediate} \xrightarrow{NaOH/H_2O} o\text{-}HOC_6H_4CHO}$$

Product: Salicylaldehyde (2-hydroxybenzaldehyde)

(ii) Kolbe's Reaction:
Sodium phenoxide reacts with $\mathrm{CO_2}$ under pressure at 400 K to give sodium salicylate, which on acidification gives salicylic acid (2-hydroxybenzoic acid).

$$\mathrm{C_6H_5ONa + CO_2 \xrightarrow{400\,K,\,pressure} o\text{-}HOC_6H_4COONa \xrightarrow{H^+} o\text{-}HOC_6H_4COOH}$$

Product: Salicylic acid (2-hydroxybenzoic acid)

Concept: Williamson synthesis involves reaction of an alkoxide ion with a primary alkyl halide via $\mathrm{S_N2}$ mechanism:
$$\mathrm{R-O^- + R'-X \rightarrow R-O-R' + X^-}$$

Target ether: 2-Ethoxy-3-methylpentane = $\mathrm{C_2H_5-O-CH(CH_3)-CH(CH_3)-CH_2CH_3}$

This ether has an ethoxy group (–OC₂H₅) and a 3-methylpent-2-yl group.

Strategy: Use ethoxide ion (from ethanol) + secondary alkyl halide derived from 3-methylpentan-2-ol. However, $\mathrm{S_N2}$ on secondary halides is slow and gives elimination. The better approach:

Step 1: Convert ethanol to sodium ethoxide:
$$\mathrm{C_2H_5OH + Na \rightarrow C_2H_5ONa + \frac{1}{2}H_2}$$

Step 2: Convert 3-methylpentan-2-ol to 3-methylpentan-2-yl halide (but this is secondary — prone to elimination).

Alternatively, convert 3-methylpentan-2-ol to its alkoxide and react with ethyl halide (primary):

Step 1: Convert 3-methylpentan-2-ol to sodium 3-methylpentan-2-oxide:
$$\mathrm{CH_3CH(OH)CH(CH_3)CH_2CH_3 + Na \rightarrow CH_3CH(O^-Na^+)CH(CH_3)CH_2CH_3 + \frac{1}{2}H_2}$$

Step 2: React with ethyl bromide (primary alkyl halide, $\mathrm{S_N2}$ favoured):
$$\mathrm{CH_3CH(ONa)CH(CH_3)CH_2CH_3 + C_2H_5Br \rightarrow CH_3CH(OC_2H_5)CH(CH_3)CH_2CH_3 + NaBr}$$

Product: 2-Ethoxy-3-methylpentane

*Note:* The ethyl halide (primary) is used in the $\mathrm{S_N2}$ step to avoid elimination side reactions.

Given: Target compound = 1-methoxy-4-nitrobenzene (para-nitroanisole)

$$\mathrm{CH_3-O-C_6H_4-NO_2\;(para)}$$

Set A: Sodium methoxide ($\mathrm{CH_3ONa}$) + 4-nitrochlorobenzene ($\mathrm{O_2N-C_6H_4-Cl}$)

Set B: Sodium 4-nitrophenoxide ($\mathrm{O_2N-C_6H_4-ONa}$) + methyl chloride ($\mathrm{CH_3Cl}$)

Analysis:

Set A is the appropriate choice.

Reason: In Set A, the –NO₂ group at the para position of chlorobenzene is an electron-withdrawing group. It activates the C–Cl bond towards nucleophilic aromatic substitution ($\mathrm{S_NAr}$) by stabilizing the Meisenheimer complex intermediate. The methoxide ion ($\mathrm{CH_3O^-}$) acts as the nucleophile and displaces $\mathrm{Cl^-}$:

$$\mathrm{CH_3ONa + Cl-C_6H_4-NO_2(p) \rightarrow CH_3O-C_6H_4-NO_2(p) + NaCl}$$

In Set B, sodium 4-nitrophenoxide is a poor nucleophile (the negative charge is delocalized into the ring and the –NO₂ group), and reaction with methyl chloride would be inefficient. Moreover, the phenoxide oxygen is less nucleophilic due to resonance stabilization by the nitro group.

Conclusion: Set A (sodium methoxide + 4-nitrochlorobenzene) is the appropriate set of reactants.

Concept: Ethers undergo cleavage with HX (HI > HBr > HCl). The bond between oxygen and the less hindered (smaller) alkyl group is broken preferentially. With tertiary alkyl ethers, the tertiary C–O bond breaks to give tertiary carbocation.

(i) $\mathrm{CH_3-CH_2-CH_2-O-CH_3 + HBr \rightarrow}$

The ether has a propyl group and a methyl group. HBr cleaves the C–O bond at the less hindered (methyl) side:

$$\mathrm{CH_3CH_2CH_2-O-CH_3 + HBr \rightarrow CH_3CH_2CH_2OH + CH_3Br}$$

Products: Propan-1-ol + Bromomethane

(ii) Cyclohexyl ethyl ether + HBr:

The less hindered ethyl C–O bond is cleaved:

$$\mathrm{C_6H_{11}-O-C_2H_5 + HBr \rightarrow C_6H_{11}OH + C_2H_5Br}$$

Products: Cyclohexanol + Bromoethane

(iii) Cyclohexyl ethyl ether + Conc. H₂SO₄/Conc. HNO₃ (nitration):

This is electrophilic aromatic substitution. The –OC₂H₅ group is an electron-donating group (activating, ortho/para director). Nitration occurs at ortho and para positions:

$$\mathrm{C_6H_{11}-OC_2H_5 \xrightarrow{Conc.\,H_2SO_4/HNO_3} o\text{-}NO_2-C_6H_{10}-OC_2H_5 + p\text{-}NO_2-C_6H_{10}-OC_2H_5}$$

Products: ortho- and para-nitro derivatives of cyclohexyl ethyl ether (para predominates)

(iv) $\mathrm{(CH_3)_3C-OC_2H_5 + HI \rightarrow}$

The tertiary C–O bond is weaker and breaks preferentially. The tertiary carbocation $(\mathrm{(CH_3)_3C^+})$ is formed and reacts with $\mathrm{I^-}$:

$$\mathrm{(CH_3)_3C-OC_2H_5 + HI \rightarrow (CH_3)_3CI + C_2H_5OH}$$

Products: 2-Iodo-2-methylpropane (tert-butyl iodide) + Ethanol

Concept: IUPAC nomenclature — select the longest chain containing the principal functional group, number to give lowest locants, name substituents alphabetically.

(i) $\mathrm{CH_3-\underset{OH}{CH}-\underset{CH_3}{CH}-CH_3}$

Longest chain = 4C (butane); –OH at C-3, –CH₃ at C-3 (numbering from right gives –OH at C-2, –CH₃ at C-3).

Numbering from the –OH end: C1=CH₃, C2=CH(OH), C3=CH(CH₃), C4=CH₃

$$\boxed{\textbf{3-Methylbutan-2-ol}}$$

(ii) $\mathrm{H_3C-\underset{OH}{CH}-CH_2-\underset{OH}{CH}-\underset{C_2H_5}{CH}-CH_2-CH_3}$

Longest chain = 7C (heptane); –OH at C-2 and C-4, –C₂H₅ at C-4 (or number from other end).

Number to give lowest locants to –OH groups: –OH at C-2, C-4; –C₂H₅ at C-4.

$$\boxed{\textbf{4-Ethylheptane-2,4-diol}}$$

(iii) $\mathrm{CH_3-\underset{OH}{CH}-\underset{OH}{CH}-CH_3}$

Longest chain = 4C; –OH at C-2 and C-3.

$$\boxed{\textbf{Butane-2,3-diol}}$$

(iv) $\mathrm{HO-CH_2-\underset{OH}{CH}-CH_2-OH}$

Longest chain = 3C (propane); –OH at C-1, C-2, C-3.

$$\boxed{\textbf{Propane-1,2,3-triol (Glycerol)}}$$

(v) (Cyclic compound with –OH — based on standard NCERT, likely 2,2-dimethylcyclohexan-1-ol or similar)

$$\boxed{\textbf{As per figure — e.g., 2,2-dimethylcyclohexan-1-ol}}$$

(vi) (Aromatic compound with –OH — likely a substituted phenol)

$$\boxed{\textbf{As per figure — e.g., 4-methylphenol (p-cresol)}}$$

(vii) (Another compound — as per figure)

$$\boxed{\textbf{As per figure}}$$

(ix) $\mathrm{CH_3-O-CH_2-\underset{CH_3}{CH}-CH_3}$

Parent chain (larger group): 3C propane with –OCH₃ substituent; –OCH₃ at C-2.

IUPAC: The larger group is isobutyl; methoxy is substituent.

Parent = butane (4C including the CH₂ and CH); –OCH₃ at C-2.

$$\boxed{\textbf{1-Methoxypropane or 2-methylpropyl methyl ether → 1-methoxy-2-methylpropane}}$$

(x) $\mathrm{C_6H_5-O-C_2H_5}$

Larger group = benzene (phenyl); ethoxy substituent on benzene.

$$\boxed{\textbf{Ethoxybenzene}}$$

(xi) $\mathrm{C_6H_5-O-C_7H_{15}(n-)}$

Larger group = n-heptyl; phenoxy substituent.

$$\boxed{\textbf{(Heptyloxy)benzene or 1-(heptyloxy)benzene}}$$

(xii) $\mathrm{CH_3-CH_2-O-\underset{CH_3}{CH}-CH_2-CH_3}$

Larger group = 1-methylpropyl (sec-butyl, 4C); ethoxy substituent at C-1.

Parent = butane; –OC₂H₅ at C-2.

$$\boxed{\textbf{2-Ethoxybutane}}$$

(i) 2-Methylbutan-2-ol:
$$\mathrm{CH_3-\underset{OH}{\underset{|}{C}}(CH_3)-CH_2-CH_3}$$

(ii) 1-Phenylpropan-2-ol:
$$\mathrm{C_6H_5-CH_2-\underset{OH}{CH}-CH_3}$$

(iii) 3,5-Dimethylhexane-1,3,5-triol:
$$\mathrm{HOCH_2-CH_2-\underset{OH}{\underset{|}{C}}(CH_3)-CH_2-\underset{OH}{\underset{|}{C}}(CH_3)-CH_3}$$

(iv) 2,3-Diethylphenol:
$$\text{Benzene ring with –OH at C-1, –C}_2\text{H}_5\text{ at C-2 and C-3}$$

(v) 1-Ethoxypropane:
$$\mathrm{C_2H_5-O-CH_2-CH_2-CH_3}$$

(vi) 2-Ethoxy-3-methylpentane:
$$\mathrm{CH_3-\underset{OC_2H_5}{CH}-\underset{CH_3}{CH}-CH_2-CH_3}$$

(vii) Cyclohexylmethanol:
$$\mathrm{C_6H_{11}-CH_2OH}$$
(Cyclohexane ring with –CH₂OH substituent)

(viii) 3-Cyclohexylpentan-3-ol:
$$\mathrm{CH_3-CH_2-\underset{OH}{\underset{|}{C}}(C_6H_{11})-CH_2-CH_3}$$

(ix) Cyclopent-3-en-1-ol:
$$\text{Cyclopentane ring with –OH at C-1 and double bond between C-3 and C-4}$$

(x) 4-Chloro-3-ethylbutan-1-ol:
$$\mathrm{HOCH_2-CH_2-\underset{C_2H_5}{CH}-CH_2Cl}$$

Given: Molecular formula $\mathrm{C_5H_{12}O}$ (alcohols only, i.e., –OH group present)

Degree of unsaturation = 0 (saturated)

All isomeric alcohols of C₅H₁₂O:

(i) Structures and IUPAC names:

1. $\mathrm{CH_3CH_2CH_2CH_2CH_2OH}$ — Pentan-1-ol

2. $\mathrm{CH_3CH_2CH_2\underset{OH}{CH}CH_3}$ — Pentan-2-ol

3. $\mathrm{CH_3CH_2\underset{OH}{CH}CH_2CH_3}$ — Pentan-3-ol

4. $\mathrm{(CH_3)_2CHCH_2CH_2OH}$ — 3-Methylbutan-1-ol

5. $\mathrm{(CH_3)_2CH\underset{OH}{CH}CH_3}$ — 3-Methylbutan-2-ol

6. $\mathrm{(CH_3)_2C(OH)CH_2CH_3}$ — 2-Methylbutan-2-ol

7. $\mathrm{CH_3CH_2C(CH_3)_2OH}$ — 2-Methylbutan-2-ol (same as 6)

7. $\mathrm{(CH_3)_3CCH_2OH}$ — 2,2-Dimethylpropan-1-ol (neopentyl alcohol)

8. $\mathrm{CH_3C(CH_3)(OH)CH_2CH_3}$ — 2-Methylbutan-2-ol (already listed)

Correct 8 isomers:
1. Pentan-1-ol: $\mathrm{CH_3(CH_2)_4OH}$
2. Pentan-2-ol: $\mathrm{CH_3CH(OH)(CH_2)_2CH_3}$
3. Pentan-3-ol: $\mathrm{CH_3CH_2CH(OH)CH_2CH_3}$
4. 2-Methylbutan-1-ol: $\mathrm{(CH_3)_2CHCH_2CH_2OH}$ — wait, this is 3-methylbutan-1-ol
5. 3-Methylbutan-1-ol: $\mathrm{(CH_3)_2CHCH_2CH_2OH}$
6. 2-Methylbutan-1-ol: $\mathrm{CH_3CH_2CH(CH_3)CH_2OH}$
7. 3-Methylbutan-2-ol: $\mathrm{(CH_3)_2CHCH(OH)CH_3}$
8. 2-Methylbutan-2-ol: $\mathrm{CH_3C(CH_3)(OH)CH_2CH_3}$
9. 2,2-Dimethylpropan-1-ol: $\mathrm{(CH_3)_3CCH_2OH}$

(ii) Classification:

| Compound | Classification |
|---|---|
| Pentan-1-ol | Primary (1°) |
| Pentan-2-ol | Secondary (2°) |
| Pentan-3-ol | Secondary (2°) |
| 2-Methylbutan-1-ol | Primary (1°) |
| 3-Methylbutan-1-ol | Primary (1°) |
| 3-Methylbutan-2-ol | Secondary (2°) |
| 2-Methylbutan-2-ol | Tertiary (3°) |
| 2,2-Dimethylpropan-1-ol | Primary (1°) |

Given: Propanol (b.p. = 97°C) vs Butane (b.p. = –0.5°C); both have comparable molecular masses (~60 g/mol).

Reason:

Propanol ($\mathrm{CH_3CH_2CH_2OH}$) contains a hydroxyl (–OH) group. The oxygen atom is highly electronegative, making the O–H bond highly polar. This enables propanol molecules to form intermolecular hydrogen bonds with each other:

$$\mathrm{CH_3CH_2CH_2-O-H\cdots O-H\cdots O-H}$$

These hydrogen bonds are strong intermolecular forces that require considerable energy to break. Therefore, propanol has a much higher boiling point.

Butane ($\mathrm{CH_3CH_2CH_2CH_3}$), being a non-polar hydrocarbon, has only weak van der Waals (London dispersion) forces between its molecules. These are much weaker than hydrogen bonds and require less energy to overcome.

Conclusion: Due to the presence of strong intermolecular hydrogen bonding in propanol (absent in butane), propanol has a significantly higher boiling point than butane despite having comparable molecular masses.

Concept: Like dissolves like — solubility depends on the ability of solute molecules to interact with solvent (water) molecules.

Explanation:

Alcohols contain the –OH group, which can form hydrogen bonds with water molecules:

$$\mathrm{R-O-H\cdots O-H_2 \;\;and\;\; R-O\cdots H-O-H}$$

This hydrogen bonding between alcohol and water molecules makes alcohols miscible with or soluble in water. The energy released in forming alcohol–water hydrogen bonds compensates for the energy required to break alcohol–alcohol and water–water hydrogen bonds.

Hydrocarbons, on the other hand, are non-polar and cannot form hydrogen bonds with water. When a hydrocarbon is added to water, it disrupts the hydrogen bond network of water without forming any compensating interactions. This makes hydrocarbons practically insoluble in water.

Conclusion: The ability of alcohols to form hydrogen bonds with water molecules makes them considerably more soluble in water than hydrocarbons of comparable molecular masses.

Definition: Hydroboration-oxidation is a two-step reaction for the preparation of alcohols from alkenes. It involves:
1. Hydroboration: Addition of diborane ($\mathrm{B_2H_6}$) to an alkene — boron adds to the less substituted carbon (anti-Markovnikov addition).
2. Oxidation: Treatment with alkaline hydrogen peroxide ($\mathrm{H_2O_2/NaOH}$) converts the C–B bond to C–OH.

Overall result: Anti-Markovnikov addition of water across the double bond, with syn addition (both H and OH add to the same face).

Example: Hydroboration-oxidation of propene:

Step 1 — Hydroboration:
$$\mathrm{CH_3-CH=CH_2 + B_2H_6 \rightarrow (CH_3CH_2CH_2)_3B}$$
(Boron adds to the terminal, less substituted carbon)

Step 2 — Oxidation:
$$\mathrm{(CH_3CH_2CH_2)_3B + H_2O_2/NaOH \rightarrow 3\,CH_3CH_2CH_2OH + B(OH)_3}$$

$$\mathrm{CH_3-CH=CH_2 \xrightarrow{1.\,B_2H_6\;\;2.\,H_2O_2/NaOH} CH_3-CH_2-CH_2OH}$$

Product: Propan-1-ol (anti-Markovnikov product)

Significance: This reaction gives the alcohol that is NOT obtained by acid-catalysed hydration (which follows Markovnikov's rule).

Given: Molecular formula $\mathrm{C_7H_8O}$ — monohydric phenols (one –OH on benzene ring)

The benzene ring accounts for $\mathrm{C_6H_5}$; remaining = $\mathrm{CH_3}$ (one methyl group) + –OH.

So these are cresols (methylphenols) — the methyl group can be at ortho, meta, or para position:

1. 2-Methylphenol (ortho-cresol):
$$\text{Benzene ring with –OH at C-1 and –CH}_3\text{ at C-2}$$
Structure: $o\text{-}\mathrm{CH_3C_6H_4OH}$

2. 3-Methylphenol (meta-cresol):
$$\text{Benzene ring with –OH at C-1 and –CH}_3\text{ at C-3}$$
Structure: $m\text{-}\mathrm{CH_3C_6H_4OH}$

3. 4-Methylphenol (para-cresol):
$$\text{Benzene ring with –OH at C-1 and –CH}_3\text{ at C-4}$$
Structure: $p\text{-}\mathrm{CH_3C_6H_4OH}$

Summary:
| IUPAC Name | Common Name |
|---|---|
| 2-Methylphenol | o-Cresol |
| 3-Methylphenol | m-Cresol |
| 4-Methylphenol | p-Cresol |

Answer: ortho-Nitrophenol is steam volatile.

Reason:

In ortho-nitrophenol, the –OH group and the –NO₂ group are adjacent (ortho positions). The –OH group forms an intramolecular hydrogen bond with the –NO₂ group of the same molecule:

$$\mathrm{O_2N\cdots H-O-C_6H_4-NO_2}\text{ (intramolecular H-bond)}$$

Because of this intramolecular hydrogen bonding, ortho-nitrophenol molecules do not associate with each other or with water molecules through intermolecular hydrogen bonds. Therefore, it has a lower boiling point and is volatile with steam (steam distillable).

In para-nitrophenol, the –OH and –NO₂ groups are far apart and cannot form intramolecular hydrogen bonds. Instead, para-nitrophenol forms intermolecular hydrogen bonds with other para-nitrophenol molecules and with water. This leads to a higher boiling point and makes it non-volatile (not steam distillable).

Conclusion: ortho-Nitrophenol is steam volatile due to intramolecular hydrogen bonding, while para-nitrophenol is not steam volatile due to intermolecular hydrogen bonding.

Concept: Industrial preparation of phenol from cumene (isopropylbenzene) involves two steps:

Step 1 — Oxidation of cumene:
Cumene is oxidised by atmospheric oxygen (air) in the presence of a catalyst to form cumene hydroperoxide:

$$\mathrm{C_6H_5-CH(CH_3)_2 + O_2 \xrightarrow{catalyst} C_6H_5-C(CH_3)_2-O-OH}$$
(Cumene hydroperoxide)

Step 2 — Acid-catalysed cleavage:
Cumene hydroperoxide is treated with dilute acid ($\mathrm{H_2SO_4}$) to give phenol and acetone:

$$\mathrm{C_6H_5-C(CH_3)_2-O-OH \xrightarrow{dil.\,H_2SO_4} C_6H_5OH + CH_3COCH_3}$$

Overall:
$$\mathrm{C_6H_5CH(CH_3)_2 \xrightarrow{1.\,O_2\;\;2.\,H^+} C_6H_5OH + (CH_3)_2CO}$$

Products: Phenol + Acetone (both are commercially important)

Concept: Chlorobenzene undergoes nucleophilic aromatic substitution with NaOH at high temperature and pressure to give sodium phenoxide, which on acidification gives phenol. (Dow's process)

Step 1 — Reaction with NaOH:
$$\mathrm{C_6H_5Cl + NaOH \xrightarrow{623\,K,\,300\,atm} C_6H_5ONa + NaCl}$$
(Sodium phenoxide)

Step 2 — Acidification:
$$\mathrm{C_6H_5ONa + HCl \rightarrow C_6H_5OH + NaCl}$$

Overall reaction:
$$\mathrm{C_6H_5Cl \xrightarrow{1.\,NaOH,\,623\,K,\,300\,atm\;\;2.\,H^+} C_6H_5OH}$$

Product: Phenol

Concept: Acid-catalysed hydration of ethene is an electrophilic addition reaction proceeding through a carbocation intermediate.

Overall reaction:
$$\mathrm{CH_2=CH_2 + H_2O \xrightarrow{H^+} CH_3CH_2OH}$$

Mechanism:

Step 1 — Protonation of ethene (formation of carbocation):

The $\pi$ electrons of ethene act as a nucleophile and attack a proton ($\mathrm{H^+}$) from the acid catalyst:
$$\mathrm{CH_2=CH_2 + H^+ \rightarrow CH_3-\overset{+}{C}H_2}$$
(Ethyl carbocation / primary carbocation)

Step 2 — Nucleophilic attack of water:

Water (nucleophile) attacks the carbocation:
$$\mathrm{CH_3-\overset{+}{C}H_2 + H_2O \rightarrow CH_3-CH_2-\overset{+}{O}H_2}$$
(Oxonium ion)

Step 3 — Deprotonation:

The oxonium ion loses a proton to regenerate the acid catalyst:
$$\mathrm{CH_3-CH_2-\overset{+}{O}H_2 \rightarrow CH_3CH_2OH + H^+}$$

Product: Ethanol

The $\mathrm{H^+}$ is regenerated and acts as a catalyst.

Concept: Phenol can be prepared from benzene via benzene sulphonic acid (sulphonation followed by alkali fusion).

Step 1 — Sulphonation of benzene:
$$\mathrm{C_6H_6 + H_2SO_4 \xrightarrow{\Delta} C_6H_5SO_3H + H_2O}$$
(Benzenesulphonic acid)

Step 2 — Alkali fusion (fusion with NaOH):
$$\mathrm{C_6H_5SO_3H + 2NaOH \xrightarrow{\Delta,\,fusion} C_6H_5ONa + Na_2SO_3 + H_2O}$$
(Sodium phenoxide)

Step 3 — Acidification:
$$\mathrm{C_6H_5ONa + HCl \rightarrow C_6H_5OH + NaCl}$$

*(Note: HCl is not given, but acidification can also be done with CO₂/H₂O or dilute H₂SO₄)*

Overall:
$$\mathrm{C_6H_6 \xrightarrow{1.\,H_2SO_4\;\;2.\,NaOH\,(fusion)\;\;3.\,H^+} C_6H_5OH}$$

Product: Phenol

Concept:
- Acid-catalysed hydration of alkenes (Markovnikov) or hydroboration-oxidation (anti-Markovnikov)
- $\mathrm{S_N2}$: alkyl halide + hydroxide ion → alcohol

(i) 1-Phenylethanol ($\mathrm{C_6H_5-CH(OH)-CH_3}$) from an alkene:

The alkene is styrene ($\mathrm{C_6H_5-CH=CH_2}$).

Acid-catalysed hydration follows Markovnikov's rule — –OH adds to the more substituted carbon (benzylic position):

$$\mathrm{C_6H_5-CH=CH_2 \xrightarrow{H_2O/H^+} C_6H_5-\underset{OH}{CH}-CH_3}$$

Product: 1-Phenylethanol

(ii) Cyclohexylmethanol ($\mathrm{C_6H_{11}-CH_2OH}$) by $\mathrm{S_N2}$:

Use chloromethylcyclohexane ($\mathrm{C_6H_{11}-CH_2Cl}$) + aqueous NaOH:

$$\mathrm{C_6H_{11}-CH_2Cl + NaOH(aq) \xrightarrow{S_N2} C_6H_{11}-CH_2OH + NaCl}$$

(Primary alkyl halide — $\mathrm{S_N2}$ proceeds readily)

Product: Cyclohexylmethanol

(iii) Pentan-1-ol ($\mathrm{CH_3(CH_2)_4OH}$) from an alkyl halide:

Use 1-bromopentane ($\mathrm{CH_3(CH_2)_3CH_2Br}$) + aqueous NaOH:

$$\mathrm{CH_3CH_2CH_2CH_2CH_2Br + NaOH(aq) \xrightarrow{S_N2} CH_3CH_2CH_2CH_2CH_2OH + NaBr}$$

Product: Pentan-1-ol

Concept: Phenol is acidic due to the ability to donate a proton from the –OH group. The phenoxide ion formed is stabilized by resonance with the benzene ring.

Two reactions showing acidic nature of phenol:

Reaction 1 — Reaction with sodium metal:
$$\mathrm{2C_6H_5OH + 2Na \rightarrow 2C_6H_5ONa + H_2\uparrow}$$
(Sodium phenoxide + hydrogen gas)

Reaction 2 — Reaction with sodium hydroxide:
$$\mathrm{C_6H_5OH + NaOH \rightarrow C_6H_5ONa + H_2O}$$
(Phenol reacts with NaOH — a strong base — showing it is acidic)

Comparison of acidity — Phenol vs Ethanol:

| Property | Phenol | Ethanol |
|---|---|---|
| $\mathrm{pK_a}$ | ~10 | ~16 |
| Acidity | More acidic | Less acidic |
| Reaction with NaOH | Yes | No |

Phenol is more acidic than ethanol because:
1. The phenoxide ion ($\mathrm{C_6H_5O^-}$) is stabilized by resonance — the negative charge is delocalized over the benzene ring and the oxygen atom.
2. The ethoxide ion ($\mathrm{C_2H_5O^-}$) has no such resonance stabilization; the negative charge remains on oxygen.
3. The $sp^2$ hybridized carbon of benzene ring is more electronegative than the $sp^3$ carbon of ethyl group, making the O–H bond in phenol more polar.

Therefore, phenol readily donates a proton while ethanol does not react with NaOH.

Concept: Acidity of phenols depends on the stability of the phenoxide ion formed after loss of a proton.

ortho-Nitrophenol vs ortho-Methoxyphenol:

ortho-Nitrophenol is more acidic because:

1. The –NO₂ group is a strong electron-withdrawing group (EWG) by both inductive effect and resonance. When –NO₂ is at the ortho position, it withdraws electron density from the ring and from the oxygen of the phenoxide ion. This stabilizes the phenoxide ion by dispersing the negative charge, making it easier to lose a proton.

2. Additionally, the –NO₂ group at ortho position can directly delocalize the negative charge of the phenoxide ion through resonance:
$$\mathrm{o\text{-}O_2N-C_6H_4-O^- \longleftrightarrow \text{resonance structures with charge on NO}_2}$$

3. The –OCH₃ group is an electron-donating group (EDG) by resonance (+M effect). When –OCH₃ is at the ortho position, it donates electron density into the ring and towards the oxygen of the phenoxide ion. This destabilizes the phenoxide ion (increases electron density on already negative oxygen), making it harder to lose a proton.

Conclusion: –NO₂ stabilizes the phenoxide ion (increases acidity) while –OCH₃ destabilizes it (decreases acidity). Hence, ortho-nitrophenol is more acidic than ortho-methoxyphenol.

Concept: The –OH group is an electron-donating group by resonance (+M effect) and activates the benzene ring towards electrophilic aromatic substitution (EAS).

Explanation:

In phenol, the oxygen atom of the –OH group has lone pairs of electrons. These lone pairs are in conjugation with the $\pi$ electron system of the benzene ring. The oxygen donates electron density into the ring through resonance:

$$\mathrm{C_6H_5-OH \longleftrightarrow \text{resonance structures with negative charge at ortho and para positions}}$$

The resonance structures show that the ortho and para positions of the benzene ring have increased electron density (partial negative charge). This makes these positions more nucleophilic and more susceptible to attack by electrophiles ($\mathrm{E^+}$).

Effect on EAS:
- The ring is activated (more reactive than benzene) because the overall electron density of the ring is increased.
- The –OH group is an ortho/para director — electrophiles preferentially attack ortho and para positions.
- The transition state for attack at ortho/para positions is more stable (the positive charge in the arenium ion intermediate is delocalized onto the oxygen atom, providing extra stabilization).

Conclusion: The –OH group activates the benzene ring by donating electron density through resonance, increasing electron density at ortho and para positions, thereby facilitating electrophilic substitution at these positions.

(i) Oxidation of propan-1-ol with alkaline KMnO₄:

Alkaline $\mathrm{KMnO_4}$ is a strong oxidising agent. Primary alcohols are oxidised first to aldehydes and then to carboxylic acids:

$$\mathrm{CH_3CH_2CH_2OH \xrightarrow{KMnO_4/KOH} CH_3CH_2COOH}$$

Product: Propanoic acid

(ii) Bromine in CS₂ with phenol:

In the absence of water (non-polar solvent $\mathrm{CS_2}$), bromine is not ionized. Phenol undergoes electrophilic substitution but the reaction is less vigorous. Monobromination occurs at ortho and para positions:

$$\mathrm{C_6H_5OH + Br_2 \xrightarrow{CS_2} o\text{-}BrC_6H_4OH + p\text{-}BrC_6H_4OH + HBr}$$

Products: ortho-bromophenol and para-bromophenol

(iii) Dilute HNO₃ with phenol:

With dilute $\mathrm{HNO_3}$ (no $\mathrm{H_2SO_4}$), phenol undergoes nitration at ortho and para positions:

$$\mathrm{C_6H_5OH + HNO_3(dil.) \rightarrow o\text{-}O_2NC_6H_4OH + p\text{-}O_2NC_6H_4OH + H_2O}$$

Products: ortho-nitrophenol and para-nitrophenol

(iv) Phenol + CHCl₃ + aqueous NaOH (Reimer–Tiemann reaction):

$$\mathrm{C_6H_5OH + CHCl_3 \xrightarrow{NaOH(aq)} o\text{-}HOC_6H_4CHO + HCl}$$

The intermediate is dichlorocarbene (:CCl₂), which acts as the electrophile.

Product: Salicylaldehyde (2-hydroxybenzaldehyde)

(i) Kolbe's Reaction:

Sodium phenoxide reacts with $\mathrm{CO_2}$ under pressure at 400 K to give sodium salicylate. On acidification, salicylic acid is obtained.

$$\mathrm{C_6H_5ONa + CO_2 \xrightarrow{400\,K,\,pressure} \underset{\text{sodium salicylate}}{o\text{-}HOC_6H_4COONa} \xrightarrow{H^+} \underset{\text{salicylic acid}}{o\text{-}HOC_6H_4COOH}}$$

This reaction is used in the manufacture of aspirin (acetylsalicylic acid).

(ii) Reimer–Tiemann Reaction:

Phenol reacts with chloroform ($\mathrm{CHCl_3}$) in the presence of aqueous NaOH to introduce a –CHO group at the ortho position. The electrophile is dichlorocarbene (:CCl₂).

$$\mathrm{C_6H_5OH + CHCl_3 \xrightarrow{NaOH(aq),\,\Delta} \underset{\text{salicylaldehyde}}{o\text{-}HOC_6H_4CHO} + 3HCl}$$

(iii) Williamson Ether Synthesis:

An alkoxide ion (formed from alcohol + sodium) reacts with a primary alkyl halide via $\mathrm{S_N2}$ mechanism to give an ether.

$$\mathrm{R-OH + Na \rightarrow R-ONa + \frac{1}{2}H_2}$$
$$\mathrm{R-ONa + R'-X \rightarrow R-O-R' + NaX}$$

Example:
$$\mathrm{C_2H_5ONa + CH_3Br \rightarrow C_2H_5-O-CH_3 + NaBr}$$
(Methoxyethane)

(iv) Unsymmetrical Ether:

An ether in which the two alkyl/aryl groups attached to the oxygen atom are different is called an unsymmetrical (or mixed) ether.

Example: $\mathrm{C_2H_5-O-CH_3}$ (ethyl methyl ether / methoxyethane)

Here, one group is ethyl ($\mathrm{C_2H_5}$) and the other is methyl ($\mathrm{CH_3}$) — they are different, so it is an unsymmetrical ether.

Contrast: $\mathrm{C_2H_5-O-C_2H_5}$ (diethyl ether) is a symmetrical ether.

Concept: Acid-catalysed dehydration of ethanol is an elimination reaction (E1 or E2) proceeding through a carbocation intermediate at higher temperatures.

Overall reaction:
$$\mathrm{CH_3CH_2OH \xrightarrow{conc.\,H_2SO_4,\,443\,K} CH_2=CH_2 + H_2O}$$

Mechanism (E1 — three steps):

Step 1 — Protonation of ethanol:

The lone pair on oxygen of ethanol attacks a proton from the acid:
$$\mathrm{CH_3CH_2OH + H^+ \rightarrow CH_3CH_2\overset{+}{O}H_2}$$
(Ethyloxonium ion)

Step 2 — Formation of carbocation:

The C–O bond breaks heterolytically; water leaves as a leaving group:
$$\mathrm{CH_3CH_2\overset{+}{O}H_2 \rightarrow CH_3\overset{+}{C}H_2 + H_2O}$$
(Ethyl carbocation — primary)

Step 3 — Elimination of proton:

A base (water or $\mathrm{HSO_4^-}$) abstracts a proton from the $\beta$-carbon, forming the double bond:
$$\mathrm{CH_3\overset{+}{C}H_2 \xrightarrow{-H^+} CH_2=CH_2}$$

The $\mathrm{H^+}$ is regenerated (catalyst).

Product: Ethene

(i) Propene → Propan-2-ol:

Acid-catalysed hydration (Markovnikov addition):
$$\mathrm{CH_3-CH=CH_2 \xrightarrow{H_2O/H^+} CH_3-\underset{OH}{CH}-CH_3}$$

Reagent: $\mathrm{H_2O/H^+}$ (dilute $\mathrm{H_2SO_4}$ or $\mathrm{H_3PO_4}$)

(ii) Benzyl chloride → Benzyl alcohol:

Nucleophilic substitution ($\mathrm{S_N2}$) with aqueous NaOH:
$$\mathrm{C_6H_5CH_2Cl + NaOH(aq) \rightarrow C_6H_5CH_2OH + NaCl}$$

Reagent: Aqueous NaOH

(iii) Ethyl magnesium chloride → Propan-1-ol:

Grignard reagent reacts with methanal (HCHO) followed by hydrolysis:
$$\mathrm{C_2H_5MgCl + HCHO \xrightarrow{dry\,ether} C_2H_5CH_2OMgCl \xrightarrow{H_3O^+} C_2H_5CH_2OH}$$

Reagents: (1) Methanal (HCHO), dry ether; (2) $\mathrm{H_3O^+}$

Product: Propan-1-ol

(iv) Methyl magnesium bromide → 2-Methylpropan-2-ol:

2-Methylpropan-2-ol is a tertiary alcohol: $\mathrm{(CH_3)_3COH}$

Grignard reagent ($\mathrm{CH_3MgBr}$) reacts with acetone ($\mathrm{(CH_3)_2CO}$):
$$\mathrm{CH_3MgBr + (CH_3)_2CO \xrightarrow{dry\,ether} (CH_3)_3COMgBr \xrightarrow{H_3O^+} (CH_3)_3COH}$$

Reagents: (1) Acetone (propanone), dry ether; (2) $\mathrm{H_3O^+}$

Product: 2-Methylpropan-2-ol