GSEB Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers

GSEB Class 12 Chemistry Alcohols, Phenols and Ehers InText Questions and Answers

Question 1. Classify the following as primary, secondary and tertiary alcohols:
i. \( \text{CH}_3\text{-}\text{C(CH}_3\text{)-CH}_2\text{OH} \)
ii. \( \text{H}_2\text{C = CH - CHOH} \)
iii. \( \text{CH}_3 \text{- CH}_2 \text{- CH}_2 \text{- OH} \)
iv. \( \text{OH - CH(CH}_3\text{)} \)
v. \( \text{CH}_2 \text{- CH - CH}_3 \text{OH} \)
vi. \( \text{CH=CH-C(CH}_3\text{)-OH} \)
Answer:
The classification of the given alcohols is as follows:
Primary alcohols: Compounds (i), (ii), and (iii)
Secondary alcohols: Compounds (iv) and (v)
Tertiary alcohols: Compound (vi)
In simple words: Primary alcohols have the -OH group attached to a carbon that is bonded to only one other carbon, secondary to two, and tertiary to three other carbons. This distinction is crucial for understanding their reactivity.

🎯 Exam Tip: Accurately identifying the carbon atom directly bonded to the hydroxyl group and counting its alkyl substituents is key for correct classification.

Question 2. Identify allylic alcohols in the above examples.
Answer:
Allylic alcohols are compounds where the hydroxyl group (OH) is attached to a \( \text{sp}^3 \) hybridized carbon atom, which is adjacent to a carbon-carbon double bond. Based on this definition, the allylic alcohols from the provided examples are:
Allylic alcohols: (ii) and (vi)
In simple words: An allylic alcohol has its OH group on a carbon next to a double bond, but not directly on the double bond itself.

🎯 Exam Tip: Remember that the carbon atom bearing the -OH group in an allylic alcohol must be \( \text{sp}^3 \) hybridized, and this \( \text{sp}^3 \) carbon must be adjacent to a \( \text{sp}^2 \) hybridized carbon involved in a double bond.

Question 3. Name the following compounds according to IUPAC system.
i. \( \text{CH}_3 \text{- CH}_2 \text{- CH(CH}_2\text{Cl) - CH(CH}_3\text{) - CH}_3 \)
ii. \( \text{CH}_3 \text{- CH(OH) - CH}_2 \text{- CH(C}_2\text{H}_5\text{) - CH}_2 \text{- CH}_3 \)
iii.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक साइक्लोहेक्सेन वलय को दर्शाता है जिसमें एक ब्रोमीन परमाणु और एक हाइड्रॉक्सिल (OH) समूह जुड़ा हुआ है।
iv. \( \text{H}_2\text{C=CH-CH(OH)-CH}_2\text{-CH}_2\text{-CH}_3 \)
v. \( \text{CH}_3 \text{- CH(OH) - CH}_2 \text{- CH=C(Br) - CH}_3 \)
Answer:
The IUPAC names for the given compounds are:
i. 3-Chloromethyl-2-isopropylpentan-1-ol
ii. 2,5-Dimethylhexane-1,3-diol
iii. 3-Bromocyclohexanol
iv. Hex-1-en-3-ol
v. 2-Bromo-3-methylbut-2-en-1-ol
In simple words: IUPAC naming involves identifying the longest carbon chain with functional groups, numbering carbons to give functional groups the lowest possible numbers, and then listing substituents alphabetically.

🎯 Exam Tip: Pay close attention to numbering the parent chain correctly to assign the lowest possible locants to functional groups and substituents. Prioritize functional groups like -OH over alkyl or halo substituents.

Question 4. Show how are the following alcohols prepared by the reaction of a suitable Grignard reagent on methanal.
i. \( \text{CH}_3 \text{- CH(CH}_3\text{) - CH}_2\text{OH} \)
ii.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक साइक्लोहेक्सेन वलय को दर्शाता है जिसमें एक मेथिल हाइड्रॉक्सिल समूह (-CH2OH) जुड़ा हुआ है।
Answer:
The preparation of the given alcohols from a Grignard reagent and methanal is shown below:
i. \( \text{CH}_3 \text{- CH(CH}_3\text{)-MgBr + HCHO } \xrightarrow{} \text{ CH}_3 \text{- CH(CH}_3\text{)-CH}_2\text{OMgBr } \xrightarrow{\text{ H}_2\text{O}} \text{ CH}_3 \text{- CH(CH}_3\text{)-CH}_2\text{OH + Mg(OH)Br} \)
ii. \( \text{C}_6\text{H}_{11}\text{MgBr + HCHO } \xrightarrow{} \text{ C}_6\text{H}_{11}\text{CH}_2\text{OMgBr } \xrightarrow{\text{ H}_2\text{O}} \text{ C}_6\text{H}_{11}\text{CH}_2\text{OH} \)
In simple words: Primary alcohols can be synthesized by reacting a Grignard reagent with methanal (formaldehyde), followed by hydrolysis of the intermediate alkoxide.

🎯 Exam Tip: Grignard reagents reacting with formaldehyde always yield primary alcohols after hydrolysis. For other aldehydes, secondary alcohols are formed, and for ketones, tertiary alcohols.

Question 5. Write structures of the products of the following reactions:
i. \( \text{CH}_3 \text{- CH = CH}_2 \xrightarrow{\text{ H}_2\text{O/H}^+} \)
ii. \( \text{CH}_3 \text{- C(O) - OCH}_3 \xrightarrow{\text{ NaBH}_4} \)
iii. \( \text{CH}_3 \text{- CH}_2 \text{- CH(CH}_3\text{) - CHO } \xrightarrow{\text{ NaBH}_4} \)
Answer:
The structures of the products are:
i. \( \text{CH}_3 \text{- CH(OH) - CH}_3 \)
ii. \( \text{CH}_3 \text{- CH}_2 \text{- OCH}_3 \)
iii. \( \text{CH}_3 \text{- CH}_2 \text{- CH(CH}_3\text{) - CH}_2\text{OH} \)
In simple words: These reactions involve hydration of an alkene, reduction of an ester (to an ether and an alcohol via NaBH4 reducing the carbonyl), and reduction of an aldehyde to a primary alcohol using a mild reducing agent.

🎯 Exam Tip: Understand the reactivity of different functional groups under specific conditions. \( \text{H}_2\text{O/H}^+ \) adds water across a double bond (Markovnikov's rule). \( \text{NaBH}_4 \) is a selective reducing agent, typically reducing aldehydes and ketones to alcohols, and sometimes esters, but not usually carboxylic acids or nitriles.

Question 6. Give structures of the products you would expect when each of the following alcohol reacts with
(a) HCl - \( \text{ZnCl}_2 \)
(b) HBr
(c) \( \text{SOCl}_2 \)
(i) Butan-1-ol
(ii) 2-Methylbutan-2-ol
Answer:
(i) Butan-1-ol
a. \( \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH } \xrightarrow{\text{ HCl//ZnCl}_2} \text{ CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Cl} \) (1-chlorobutane)
b. \( \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH } \xrightarrow{\text{ HBr}} \text{ CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Br} \)
c. \( \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH } \xrightarrow{\text{ SOCl}_2} \text{ CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Cl} \)
(ii) 2-Methylbutan-2-ol
a. \( \text{CH}_3\text{-C(OH)(CH}_3\text{)-CH}_2\text{CH}_3 \xrightarrow{\text{ HCl//ZnCl}_2} \text{ CH}_3\text{-C(Cl)(CH}_3\text{)-CH}_2\text{CH}_3 \)
b. \( \text{CH}_3\text{-C(OH)(CH}_3\text{)-CH}_2\text{CH}_3 \xrightarrow{\text{ HBr}} \text{ CH}_3\text{-C(Br)(CH}_3\text{)-CH}_2\text{CH}_3 \)
c. \( \text{CH}_3\text{-C(OH)(CH}_3\text{)-CH}_2\text{CH}_3 \xrightarrow{\text{ SOCl}_2} \text{ CH}_3\text{-C(Cl)(CH}_3\text{)-CH}_2\text{CH}_3 \)
In simple words: These reactions demonstrate the conversion of alcohols into alkyl halides using various halogenating reagents. Primary alcohols often react via \( \text{S}_{\text{N}}2 \) pathways, while tertiary alcohols tend to follow \( \text{S}_{\text{N}}1 \) mechanisms due to stable carbocation intermediates.

🎯 Exam Tip: The reactivity of alcohols towards hydrogen halides and thionyl chloride follows the order: tertiary > secondary > primary. \( \text{HCl} \) with \( \text{ZnCl}_2 \) (Lucas reagent) is particularly effective for secondary and tertiary alcohols.

Question 7. Predict the major product of acid catalysed dehydration of
(i) 1-methyl cyclohexanol
(ii) butan-1-ol
Answer:
(i) 1-Methylcyclohexane
(ii) But-1-ene
In simple words: Acid-catalyzed dehydration removes a water molecule from an alcohol to form an alkene. The major product usually follows Zaitsev's rule, meaning the most substituted alkene is formed.

🎯 Exam Tip: For dehydration, remember Zaitsev's rule: the more substituted alkene (the one with more alkyl groups on the double bond carbons) is typically the major product. Carbocation rearrangements can also occur, leading to more stable intermediates and different products.

Question 8. Ortho and para nitrophenols are more acidic than phenol. Draw the resonance structures of the corresponding phenoxide ions.
Answer:
(i) Orthonitro phenol:
\[ \text{C}_6\text{H}_4(\text{NO}_2)\text{O}^- \leftrightarrow \text{O=N(O}^-)\text{-C}_6\text{H}_4\text{O}^- \leftrightarrow \text{O=N(O}^-)\text{=C}_6\text{H}_4\text{O}^- \leftrightarrow \text{O=N(O}^-)\text{=C}_6\text{H}_4\text{O}^- \]
ℹ️ चित्र व्याख्या (Diagram Explanation): यह ऑर्थो-नाइट्रोफेनॉक्साइड आयन की अनुनाद संरचनाओं को दर्शाता है। नकारात्मक चार्ज ऑक्सीजन परमाणु से बेंजीन वलय में स्थानांतरित होता है और नाइट्रो समूह के ऑक्सीजन परमाणुओं द्वारा स्थिरीकृत होता है, जिससे ऑर्थो और पैरा स्थितियों पर नकारात्मक चार्ज आता है।
(ii) Paranitro Phenol:
\[ \text{C}_6\text{H}_4(\text{NO}_2)\text{O}^- \leftrightarrow \text{O=N(O}^-)\text{-C}_6\text{H}_4\text{O}^- \leftrightarrow \text{O=N(O}^-)\text{=C}_6\text{H}_4\text{O}^- \leftrightarrow \text{O=N(O}^-)\text{=C}_6\text{H}_4\text{O}^- \]
ℹ️ चित्र व्याख्या (Diagram Explanation): यह पैरा-नाइट्रोफेनॉक्साइड आयन की अनुनाद संरचनाओं को दर्शाता है। नकारात्मक चार्ज ऑक्सीजन परमाणु से बेंजीन वलय में स्थानांतरित होता है और नाइट्रो समूह के ऑक्सीजन परमाणुओं द्वारा स्थिरीकृत होता है, जिससे मुख्य रूप से पैरा स्थिति पर नकारात्मक चार्ज आता है।
In simple words: The nitro group is an electron-withdrawing group that stabilizes the phenoxide ion through resonance, especially when at the ortho or para positions. This stabilization makes orthonitrophenol and paranitrophenol more acidic than phenol, as the conjugate base is more stable.

🎯 Exam Tip: Electron-withdrawing groups (like \( \text{-NO}_2 \)) stabilize conjugate bases by delocalizing negative charge, thereby increasing acidity. Conversely, electron-donating groups destabilize conjugate bases and decrease acidity.

Question 9. Write the equations involved in the following reactions:
**ii. Kolbe's reaction**
Answer:
i. For the synthesis of Salicylaldehyde from Phenol (Reimer-Tiemann reaction is shown, not explicitly asked here but from the image, this is the context):
\[ \text{C}_6\text{H}_5\text{OH + CHCl}_3\text{ + NaOH } \xrightarrow{} \text{ ONa-C}_6\text{H}_4\text{CHO } \xrightarrow{\text{ H}^+} \text{ OH-C}_6\text{H}_4\text{CHO} \]
ii. For Kolbe's reaction:
\[ \text{C}_6\text{H}_5\text{ONa } \xrightarrow{\text{ (i) CO}_2 \text{ (ii) H}^+} \text{ C}_6\text{H}_4\text{(OH)COOH} \]
In simple words: Kolbe's reaction involves treating sodium phenoxide with carbon dioxide under specific conditions, followed by acidification, to produce salicylic acid. This is an important reaction for synthesizing aromatic carboxylic acids.

🎯 Exam Tip: Remember the specific reagents and conditions for Kolbe's reaction: sodium phenoxide, carbon dioxide, heating under pressure, and subsequent acidification. The product is always salicylic acid (2-hydroxybenzoic acid).

Question 10. Write the reactions of Williamson synthesis of 2-ethoxy-3-methyl pentane starting from ethanol and 3-methylpentan-2-ol.
Answer:
The Williamson synthesis of 2-ethoxy-3-methylpentane from ethanol and 3-methylpentan-2-ol involves the following steps:
1. Convert 3-methylpentan-2-ol into its corresponding sodium alkoxide:
\( \text{CH}_3 \text{- CH}_2 \text{- CH(CH}_3\text{) - CH(OH) - CH}_3 \xrightarrow{\text{ Na}} \text{ CH}_3 \text{- CH}_2 \text{- CH(CH}_3\text{) - CH(ONa) - CH}_3 \)
2. Convert ethanol into an alkyl halide (bromoethane):
\( \text{C}_2\text{H}_5\text{OH } \xrightarrow{\text{ HBr}} \text{ C}_2\text{H}_5\text{Br} \)
3. React the sodium alkoxide with the alkyl halide:
\( \text{CH}_3 \text{- CH}_2 \text{- CH(CH}_3\text{) - CH(ONa) - CH}_3 \text{ + C}_2\text{H}_5\text{Br } \xrightarrow{} \text{ CH}_3 \text{- CH}_2 \text{- CH(CH}_3\text{) - CH(OC}_2\text{H}_5\text{) - CH}_3 \)
(2-Ethoxy-3-methylpentane)
In simple words: Williamson ether synthesis creates an ether by reacting an alkoxide ion (derived from an alcohol) with a primary alkyl halide. It's best when the alkyl halide is primary to avoid elimination reactions.

🎯 Exam Tip: For Williamson synthesis, it's crucial to use a primary alkyl halide to ensure the reaction proceeds via an \( \text{S}_{\text{N}}2 \) pathway and avoids side reactions like elimination (which would lead to alkenes, especially with secondary or tertiary alkyl halides).

Question 11. Which of the following is an appropriate set of reactants for the preparation of 1-methoxy-4-nitrobenzene and why?
**i.**

ℹ️ चित्र व्याख्या (Diagram Explanation): यह 4-नाइट्रोब्रोमोबेंजीन को सोडियम मेथोक्साइड के साथ दर्शाता है।
**ii.**

ℹ️ चित्र व्याख्या (Diagram Explanation): यह सोडियम 4-नाइट्रोफेनॉक्साइड को ब्रोमोमेथेन के साथ दर्शाता है।
Answer:
The appropriate set of reactants for the preparation of 1-methoxy-4-nitrobenzene is **ii**. This involves reacting sodium 4-nitrophenoxide with methyl bromide.
Aryl halides (like 4-nitro-1-bromobenzene in option i) have very low reactivity towards nucleophilic substitution reactions. This is due to the partial double bond character of the C-X bond caused by resonance with the benzene ring, and the electron density of the benzene ring. Therefore, reacting 4-nitro-1-bromobenzene with \( \text{CH}_3\text{ONa} \) would not readily yield the desired ether.
In contrast, option ii, which uses sodium 4-nitrophenoxide (an alkoxide) and methyl bromide (a primary alkyl halide), is suitable for Williamson ether synthesis. The phenoxide acts as a strong nucleophile and attacks the methyl group of methyl bromide, leading to the formation of the ether.
In simple words: To make an aryl alkyl ether, it's better to react a sodium phenoxide with a primary alkyl halide (Williamson synthesis) because aryl halides are very unreactive in nucleophilic substitution, making direct reaction with an alkoxide difficult.

🎯 Exam Tip: Williamson ether synthesis works best when the alkyl halide is primary and the alkoxide is derived from either a primary/secondary alcohol or a phenol. Avoid using aryl halides or tertiary alkyl halides to prevent elimination or no reaction.

**Question 12. Predict the products of the following reactions:**
**i. CH$_3$ - CH$_2$ - CH$_2$ - O - CH$_3$ + HBr \(\implies\)**
**ii.**

ℹ️ चित्र व्याख्या (Diagram Explanation): यह मेथॉक्सीबेंजीन (एनीसोल) को HBr के साथ दर्शाता है।
**iii.**

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एथॉक्सीबेंजीन को सांद्र \( \text{H}_2\text{SO}_4 \) और सांद्र \( \text{HNO}_3 \) के साथ दर्शाता है।
**iv. (CH$_3$)$_3$C-O-C$_2$H$_5$ + HI \(\implies\)**
Answer:
The products of the reactions are:
i. \( \text{CH}_3 \text{- CH}_2 \text{- CH}_2\text{OH + CH}_3\text{Br} \)
ii. \( \text{C}_6\text{H}_5\text{OH + CH}_3\text{Br} \)
iii.

ℹ️ चित्र व्याख्या (Diagram Explanation): एथॉक्सीबेंजीन की नाइट्रेशन प्रतिक्रिया से ऑर्थो-नाइट्रोएथॉक्सीबेंजीन और पैरा-नाइट्रोएथॉक्सीबेंजीन (प्रमुख) उत्पाद बनते हैं।
iv. \( \text{(CH}_3\text{)}_3\text{C-I + C}_2\text{H}_5\text{OH} \)
In simple words: The cleavage of ethers by HBr or HI depends on the nature of the alkyl groups. For simple ethers, the smaller alkyl group forms the alkyl halide. For aromatic ethers or tertiary alkyl ethers, the more stable carbocation intermediate dictates product formation. Aromatic ethers undergo electrophilic substitution reactions like nitration at ortho/para positions.

🎯 Exam Tip: When an ether reacts with \( \text{HBr} \) or \( \text{HI} \): if one alkyl group is tertiary, the tertiary group forms the halide ( \( \text{S}_{\text{N}}1 \) mechanism). If one group is aryl (phenol derivatives), the aryl-oxygen bond is stronger due to resonance, so the alkyl group forms the halide, and phenol is the other product. If both are primary/secondary, the smaller group forms the halide.

GSEB Class 12 Chemistry Alcohols, Phenols and Ethers Text Book Questions and Answers

Question 1. Write IUPAC names of the following compounds:
i. \( \text{CH}_3 \text{- CH(CH}_3\text{) - CH(OH) - C(CH}_3\text{)}_2 \text{- CH}_3 \)
ii. \( \text{H}_3\text{C - CH(OH) - CH}_2 \text{- CH(C}_2\text{H}_5\text{) - CH(OH) - CH}_2 \text{- CH}_3 \)
iii. \( \text{CH}_3 \text{- CH(OH) - CH(OH) - CH}_3 \)
iv. \( \text{HO - CH}_2 \text{- CH(OH) - CH}_2 \text{- OH} \)
v.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक बेंजीन वलय को दर्शाता है जिसमें एक मेथिल समूह और एक हाइड्रॉक्सिल (OH) समूह ऑर्थो स्थिति पर जुड़ा हुआ है।
vi.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक बेंजीन वलय को दर्शाता है जिसमें एक मेथिल समूह और एक हाइड्रॉक्सिल (OH) समूह पैरा स्थिति पर जुड़ा हुआ है।
vii.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक बेंजीन वलय को दर्शाता है जिसमें दो मेथिल समूह (एक 2-स्थिति पर, दूसरा 5-स्थिति पर) और एक हाइड्रॉक्सिल (OH) समूह जुड़ा हुआ है।
viii.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक बेंजीन वलय को दर्शाता है जिसमें दो मेथिल समूह (एक 2-स्थिति पर, दूसरा 6-स्थिति पर) और एक हाइड्रॉक्सिल (OH) समूह जुड़ा हुआ है।
ix. \( \text{CH}_3 \text{- O - CH}_2 \text{- CH(CH}_3\text{) - CH}_3 \)
x. \( \text{C}_6\text{H}_5 \text{- O - C}_2\text{H}_5 \)
xi. \( \text{C}_6\text{H}_5 \text{- O - C}_7\text{H}_{15} \text{(n-)} \)
xii. \( \text{CH}_3 \text{- CH}_2 \text{- O - CH(CH}_3\text{) - CH}_2 \text{- CH}_3 \)
Answer:
The IUPAC names are:
(i) 2,2,4-Trimethylpentan-3-ol
(ii) 5-Ethylheptane-2,4-diol
(iii) Butane-2,3-diol
(iv) Propane-1,2,3-triol
(v) 2-Methylphenol
(vi) 4-Methylphenol
(vii) 2,5-Dimethylphenol
(viii) 2,6-Dimethylphenol
(ix) 1-Methoxy-2-methylpropane
(x) Ethoxybenzene
(xi) 1-Phenoxyheptane
(xii) 3-Ethoxy-2-methylpentane
In simple words: IUPAC nomenclature ensures unique and systematic naming of chemical compounds. It involves identifying the longest parent chain, numbering substituents and functional groups, and arranging them alphabetically.

🎯 Exam Tip: Practice identifying the longest carbon chain containing the principal functional group. Assign numbers such that the functional group receives the lowest possible locant. For aromatic compounds, "phenol" is the parent name for benzene with an -OH group.

Question 2. Write structures of the compounds whose IUPAC names are as follows:
(i) 2-Methylbutan-2-ol
(ii) 1-Phenylpropan-2-ol
(iii) 3,5-Dimethylhexane-1,3,5-triol
(iv) 2,3-Diethylphenol
(v) 1-Ethoxypropane
(vi) 2-Ethoxy-3-methylpentane
(vii) Cyclohexylmethanol
(viii) 3-Cyclohexylpentan-3-ol
(ix) Cyclopent-3-en-1-ol
(x) 3-Chloromethylpentan-1-ol.
Answer:
The structures of the compounds are:
i. \( \text{CH}_3 \text{- C(OH)(CH}_3\text{) - CH}_2 \text{- CH}_3 \)
ii. \( \text{C}_6\text{H}_5 \text{- CH}_2 \text{- CH(OH) - CH}_3 \)
iii. \( \text{HO - CH}_2 \text{- CH}_2 \text{- C(OH)(CH}_3\text{) - CH}_2 \text{- C(OH)(CH}_3\text{) - CH}_3 \)
iv.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह 2,3-डाईएथिलफेनोल की संरचना को दर्शाता है, जिसमें एक बेंजीन वलय पर एक हाइड्रॉक्सिल समूह और 2- और 3- स्थितियों पर दो एथिल समूह जुड़े होते हैं।
v. \( \text{CH}_3 \text{- CH}_2 \text{- O - CH}_2 \text{- CH}_3 \)
vi. \( \text{CH}_3 \text{- CH}_2 \text{- O - CH(CH}_3\text{) - CH}_2 \text{- CH}_2 \text{- CH}_3 \)
vii. \( \text{C}_6\text{H}_{11} \text{- CH}_2 \text{- OH} \)
viii. \( \text{CH}_3 \text{- CH}_2 \text{- C(OH)(C}_6\text{H}_{11}\text{) - CH}_2 \text{- CH}_3 \)
ix.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह साइक्लोपेंट-3-एन-1-ओएल की संरचना को दर्शाता है, जिसमें एक साइक्लोपेंटेन वलय पर एक डबल बॉन्ड (3-स्थिति पर) और एक हाइड्रॉक्सिल समूह (1-स्थिति पर) जुड़ा होता है।
x. \( \text{ClCH}_2 \text{- CH}_2 \text{- CH(CH}_2\text{OH) - CH}_2 \text{- CH}_3 \)
In simple words: Drawing structures from IUPAC names requires breaking down the name into the parent chain, functional groups, and substituents. The numbering in the name indicates the position of each group.

🎯 Exam Tip: When drawing structures from IUPAC names, always start by drawing the parent carbon chain or ring. Then, place the main functional group, and finally add substituents at their specified positions, ensuring correct valencies.

Question 3. (i) Draw the structures of all isomeric alcohols of molecular formula C$_5$H$_{12}$O and give their
(ii) Classify them as primary, secondary and tertiary alcohols.
Answer:
(i) Isomeric alcohols of molecular formula \( \text{C}_5\text{H}_{12}\text{O} \):
a. \( \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{OH} \) (Pentan-1-ol)
b. \( \text{CH}_3\text{-CH(CH}_3\text{)-CH}_2\text{CH}_2\text{OH} \) (2-Methylbutan-1-ol)
c. \( \text{CH}_3\text{-C(CH}_3\text{)}_2\text{-CH}_2\text{OH} \) (2,2-Dimethylpropan-1-ol)
d. \( \text{CH}_3\text{CH}_2\text{CH}_2\text{CH(OH)CH}_3 \) (Pentan-2-ol)
e. \( \text{CH}_3\text{CH}_2\text{CH(CH}_3\text{)CH}_2\text{OH} \) (Pentan-3-ol)
f. \( \text{CH}_3\text{-CH(CH}_3\text{)-CH(OH)-CH}_3 \) (3-Methylbutan-2-ol)
g. \( \text{CH}_3\text{CH}_2\text{-C(OH)(CH}_3\text{)-CH}_3 \) (2-Methylbutan-2-ol)
(ii) Classification:
Primary alcohols: (a), (b), (c)
Secondary alcohols: (d), (e), (f)
Tertiary alcohol: (g)
In simple words: Isomers have the same molecular formula but different structural arrangements. For alcohols, these structural differences lead to different classifications (primary, secondary, tertiary) based on the carbon atom holding the -OH group.

🎯 Exam Tip: Systematically draw all possible carbon skeletons first, then place the -OH group on different carbons, being careful to avoid duplicates. For classification, identify how many alkyl groups are attached to the carbon bearing the hydroxyl group.

Question 4. Explain why propanol has a higher boiling point than that of hydrocarbon, butane?
Answer:
Propanol exhibits a higher boiling point compared to butane primarily due to the presence of intermolecular hydrogen bonding. In propanol (an alcohol), the hydroxyl (-OH) group allows hydrogen atoms bonded to oxygen to form strong hydrogen bonds with the oxygen atoms of neighboring propanol molecules. Butane, being a hydrocarbon, only experiences weaker London dispersion forces (van der Waals forces). These stronger hydrogen bonds in propanol require significantly more energy to overcome during boiling, resulting in a higher boiling point.
In simple words: Propanol has a higher boiling point than butane because propanol molecules form strong hydrogen bonds with each other, requiring more energy to boil, while butane only has weaker forces.

🎯 Exam Tip: Always relate boiling point differences between organic compounds to the types and strengths of intermolecular forces present. Hydrogen bonding (present in alcohols) is much stronger than dipole-dipole interactions or London dispersion forces (present in hydrocarbons).

Question 5. Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact.
Answer:
Alcohols are considerably more soluble in water than hydrocarbons of similar molecular masses because they can form hydrogen bonds with water molecules. The hydroxyl (-OH) group in alcohols is polar, with the oxygen atom being electronegative and the hydrogen atom being electropositive. This allows the alcohol molecules to interact favorably with polar water molecules through hydrogen bonding. Hydrocarbons, on the other hand, are non-polar and cannot form hydrogen bonds with water. Consequently, they are insoluble in water as they cannot disrupt the existing hydrogen bonding network of water molecules.
In simple words: Alcohols dissolve better in water than hydrocarbons because their -OH groups can form hydrogen bonds with water, while non-polar hydrocarbons cannot.

🎯 Exam Tip: Solubility in water is often determined by a molecule's ability to form hydrogen bonds with water. The "like dissolves like" principle applies, meaning polar solvents dissolve polar solutes, and non-polar solvents dissolve non-polar solutes.

Question 6. What is meant by hydroboration-oxidation reaction? Illustrate it with an example.
Answer:
Hydroboration-oxidation reaction is a two-step process used to convert alkenes into alcohols, specifically following an anti-Markovnikov addition and *syn* stereochemistry. In the first step, an alkene reacts with diborane \( \text{(B}_2\text{H}_6\text{)} \) or borane \( \text{(BH}_3\text{)} \) to form an alkylborane intermediate. Since borane is electron-deficient, it acts as an electrophile. The addition is anti-Markovnikov because the boron atom (electrophile) adds to the less substituted carbon of the double bond. In the second step, this alkylborane is oxidized with alkaline hydrogen peroxide \( \text{(H}_2\text{O}_2\text{/OH}^-\text{)} \) to yield an alcohol.
Example:
\( \text{CH}_3\text{CH = CH}_2 \text{ + B}_2\text{H}_6 \xrightarrow{} \text{ (CH}_3\text{CH}_2\text{CH}_2\text{)}_3\text{B} \)
\( \text{(CH}_3\text{CH}_2\text{CH}_2\text{)}_3\text{B } \xrightarrow{\text{ 3H}_2\text{O}_2\text{/OH}^-} \text{ 3CH}_3\text{CH}_2\text{CH}_2\text{OH + B(OH)}_3 \)
(Propan-1-ol)
In simple words: Hydroboration-oxidation is a way to turn alkenes into alcohols. It adds an -OH group to the less crowded carbon of the double bond (anti-Markovnikov) and does so in a concerted manner.

🎯 Exam Tip: The key features of hydroboration-oxidation are anti-Markovnikov addition of water and *syn* addition (both H and OH add to the same face of the double bond). This reaction is valuable for synthesizing primary alcohols from terminal alkenes.

Question 7. Give the structures and IUPAC names of monohydric phenols of molecular formula, C$_7$H$_8$O.
Answer:
The monohydric phenols with the molecular formula \( \text{C}_7\text{H}_8\text{O} \) are the cresols, which are methylphenols. There are three isomers:
1. Ortho-cresol (2-methylphenol)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह 2-मिथाइलफेनोल की संरचना को दर्शाता है, जिसमें एक बेंजीन वलय पर एक हाइड्रॉक्सिल (OH) समूह और ऑर्थो स्थिति पर एक मेथिल समूह जुड़ा होता है।
2. Meta-cresol (3-methylphenol)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह 3-मिथाइलफेनोल की संरचना को दर्शाता है, जिसमें एक बेंजीन वलय पर एक हाइड्रॉक्सिल (OH) समूह और मेटा स्थिति पर एक मेथिल समूह जुड़ा होता है।
3. Para-cresol (4-methylphenol)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह 4-मिथाइलफेनोल की संरचना को दर्शाता है, जिसमें एक बेंजीन वलय पर एक हाइड्रॉक्सिल (OH) समूह और पैरा स्थिति पर एक मेथिल समूह जुड़ा होता है।
In simple words: For the formula \( \text{C}_7\text{H}_8\text{O} \), monohydric phenols are variations of methyl-substituted phenols, commonly known as cresols, with the methyl group at ortho, meta, or para positions relative to the hydroxyl group.

🎯 Exam Tip: Remember that monohydric phenols contain only one hydroxyl group. For aromatic compounds, drawing all possible isomer positions for substituents (ortho, meta, para) is essential to identify all unique structures.

Question 8. While separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which will be steam volatile. Give reason.
Answer:
In separating a mixture of ortho and para nitrophenols by steam distillation, **o-nitrophenol** will be steam volatile. The reason for its steam volatility is the presence of **intramolecular hydrogen bonding**. In o-nitrophenol, the hydrogen atom of the hydroxyl group forms a hydrogen bond with the oxygen atom of the adjacent nitro group within the *same* molecule. This intramolecular hydrogen bonding reduces the intermolecular attractive forces between o-nitrophenol molecules, allowing them to vaporize more easily with steam. In contrast, p-nitrophenol exhibits intermolecular hydrogen bonding with other p-nitrophenol molecules, leading to association and a higher boiling point, making it less volatile with steam.
In simple words: Ortho-nitrophenol is steam volatile because it forms an internal hydrogen bond, which prevents it from strongly interacting with other molecules or water, unlike para-nitrophenol which forms external hydrogen bonds.

🎯 Exam Tip: Intramolecular hydrogen bonding leads to lower boiling points and higher volatility (like steam volatility) compared to intermolecular hydrogen bonding, which increases boiling points and reduces volatility.

Question 9. Give the equations of reactions for the preparation of phenol from cumene.
Answer:
Phenol can be prepared from cumene (isopropylbenzene) through a two-step process:
1. **Oxidation of Cumene:** Cumene is oxidized by air (oxygen) in the presence of a catalyst, forming cumene hydroperoxide.
\[ \text{C}_6\text{H}_5\text{-CH(CH}_3\text{)}_2 \text{ (Cumene) } \xrightarrow{\text{ O}_2 \text{, 400K, catalyst}} \text{ C}_6\text{H}_5\text{-C(CH}_3\text{)}_2\text{-OOH} \text{ (Cumene hydroperoxide)} \]
2. **Acidic Hydrolysis of Cumene Hydroperoxide:** Cumene hydroperoxide is then treated with dilute acid ( \( \text{H}^+ \) ) to hydrolyze it, yielding phenol and acetone as by-products.
\[ \text{C}_6\text{H}_5\text{-C(CH}_3\text{)}_2\text{-OOH } \xrightarrow{\text{ H}^+} \text{ C}_6\text{H}_5\text{OH (Phenol) + CH}_3\text{COCH}_3 \text{ (Acetone)} \]
In simple words: Phenol is made from cumene by first reacting cumene with oxygen to form cumene hydroperoxide, which is then broken down with acid to give phenol and acetone.

🎯 Exam Tip: The cumene process is an important industrial method for phenol synthesis, producing acetone as a valuable co-product. Remember the intermediate (cumene hydroperoxide) and the conditions for each step.

Question 10. Write chemical reaction for the preparation of phenol from chlorobenzene.
Answer:
Phenol can be prepared from chlorobenzene via the Dow process, which involves nucleophilic aromatic substitution under harsh conditions:
1. **Reaction with NaOH:** Chlorobenzene is heated with aqueous sodium hydroxide \( \text{(NaOH)} \) at high temperature (623 K) and high pressure (300 atm). This converts chlorobenzene into sodium phenoxide.
\[ \text{C}_6\text{H}_5\text{Cl (Chlorobenzene) + NaOH } \xrightarrow{\text{ 623K, 300 atm}} \text{ C}_6\text{H}_5\text{ONa (Sodium phenoxide) + NaCl} \]
2. **Acidification:** The resulting sodium phenoxide is then acidified with dilute acid \( \text{(H}^+\text{)} \) to regenerate phenol.
\[ \text{C}_6\text{H}_5\text{ONa } \xrightarrow{\text{ H}^+} \text{ C}_6\text{H}_5\text{OH (Phenol) + Na}^+ \]
In simple words: Phenol can be made from chlorobenzene by first reacting it with strong base at high heat and pressure to make sodium phenoxide, then adding acid to get phenol.

🎯 Exam Tip: The Dow process demonstrates a nucleophilic aromatic substitution reaction, which typically requires extreme conditions due to the unreactivity of aryl halides. Note the two-step process: formation of phenoxide, then acidification.

Question 11. Write the mechanism of hydration of ethene to yield ethanol.
Answer:
The mechanism of acid-catalyzed hydration of ethene to yield ethanol proceeds via the formation of a carbocation intermediate and can be described in three main steps:
1. **Protonation of Alkene (Formation of carbocation):** The ethene molecule acts as a nucleophile and attacks a proton \( \text{(H}^+\text{)} \) from the acid (e.g., \( \text{H}_3\text{O}^+ \) derived from \( \text{H}_2\text{SO}_4 \)). This leads to the formation of a primary carbocation.
\[ \text{CH}_2\text{=CH}_2 \text{ + H}^+ \leftrightarrow \text{CH}_3\text{-CH}_2^+ \text{ (Ethyl carbocation)} \]
2. **Nucleophilic attack by water:** The carbocation is then attacked by a water molecule (a nucleophile), forming a protonated alcohol.
\[ \text{CH}_3\text{-CH}_2^+ \text{ + H}_2\text{O } \leftrightarrow \text{CH}_3\text{-CH}_2\text{-OH}_2^+ \text{ (Protonated ethanol)} \]
3. **Deprotonation:** The protonated alcohol loses a proton to a water molecule (or the conjugate base of the acid), regenerating the acid catalyst and yielding the final ethanol product.
\[ \text{CH}_3\text{-CH}_2\text{-OH}_2^+ \text{ + H}_2\text{O } \leftrightarrow \text{CH}_3\text{-CH}_2\text{OH (Ethanol) + H}_3\text{O}^+ \]
*Example:* Hydration of alkenes can also occur by absorption in concentrated sulfuric acid, followed by hydrolysis of the alkyl sulfate.
\( \text{CH}_2 \text{= CH}_2 \text{ + H}_2\text{SO}_4 \xrightarrow{\text{ \(\Delta\)}} \text{ CH}_3\text{CH}_2\text{HSO}_4 \text{ (Ethyl hydrogen sulfate)} \)
\( \text{CH}_3\text{CH}_2\text{HSO}_4 \text{ + H}_2\text{O } \xrightarrow{} \text{ CH}_3\text{CH}_2\text{OH + H}_2\text{SO}_4 \)
For unsymmetrical alkenes, this addition follows Markovnikov's rule.
In simple words: Ethene hydration with acid involves three steps: first, ethene gains a proton to form a carbocation, then water attacks this carbocation, and finally, the protonated alcohol loses a proton to become ethanol.

🎯 Exam Tip: Remember the three key steps: protonation of the alkene to form a carbocation, nucleophilic attack by water, and deprotonation to yield the alcohol. Carbocation stability influences the regioselectivity (Markovnikov's rule) and reaction rate.

Question 12. You are given benzene, cone. H$_2$SO$_4$ and NaOH. Write the equations for the preparation of phenol using these reagents.
Answer:
Phenol can be prepared from benzene using concentrated \( \text{H}_2\text{SO}_4 \) and \( \text{NaOH} \) through a three-step process:
1. **Sulfonation of Benzene:** Benzene reacts with concentrated sulfuric acid to form benzenesulfonic acid.
\[ \text{C}_6\text{H}_6 \text{ (Benzene) + H}_2\text{SO}_4 \text{ (conc.) } \xrightarrow{\text{ 298K}} \text{ C}_6\text{H}_5\text{SO}_3\text{H (Benzenesulfonic acid) + H}_2\text{O} \]
2. **Fusion with NaOH:** Benzenesulfonic acid is then heated with molten sodium hydroxide \( \text{(NaOH)} \) (fusion) to form sodium phenoxide.
\[ \text{C}_6\text{H}_5\text{SO}_3\text{H + 2NaOH } \xrightarrow{\text{ fuse}} \text{ C}_6\text{H}_5\text{ONa (Sodium phenoxide) + Na}_2\text{SO}_3 \text{ + H}_2\text{O} \]
3. **Acidification:** Finally, the sodium phenoxide is treated with an acid (e.g., \( \text{H}^+ \)) to liberate phenol.
\[ \text{C}_6\text{H}_5\text{ONa } \xrightarrow{\text{ H}^+} \text{ C}_6\text{H}_5\text{OH (Phenol) + Na}^+ \]
In simple words: Phenol can be made from benzene by first sulfonating benzene to form benzenesulfonic acid, then fusing it with sodium hydroxide to get sodium phenoxide, and finally acidifying to produce phenol.

🎯 Exam Tip: This method involves electrophilic aromatic substitution (sulfonation) followed by a strong nucleophilic attack on the sulfonate group and subsequent acidification. Each step's reagents and conditions are crucial for a complete reaction.

Question 18. Explain the following with an example.
(i) Kolbe's reaction.
(ii) Reimer-Tiemann reaction
(iii) Williamson ether synthesis.
(iv) Unsymmetrical ether.
Answer:
(i) **Kolbe's reaction:** Kolbe's reaction involves heating sodium phenoxide with carbon dioxide at approximately 413 K under a pressure of 4-7 atmospheres. This process initially forms the sodium salt of salicylic acid, which upon subsequent acidification, produces salicylic acid (2-hydroxy benzoic acid).
\[ \text{C}_6\text{H}_5\text{OH} + \text{NaOH} \implies \text{C}_6\text{H}_5\text{ONa} + \text{H}_2\text{O} \]
\[ \text{C}_6\text{H}_5\text{ONa} + \text{CO}_2 \xrightarrow{\text{413K, 4-7 atm}} \text{C}_6\text{H}_4(\text{OH})\text{COONa} \]
\[ \text{C}_6\text{H}_4(\text{OH})\text{COONa} + \text{H}^+ \implies \text{C}_6\text{H}_4(\text{OH})\text{COOH} (\text{Salicylic acid}) \]
Salicylic acid is a precursor for manufacturing aspirin (2-acetoxy benzoic acid), which is known for its analgesic, anti-inflammatory, and antipyretic properties.
In simple words: Kolbe's reaction transforms phenol into salicylic acid by reacting sodium phenoxide with carbon dioxide, followed by acidification. This reaction is a key method for introducing a carboxyl group onto an activated aromatic ring.
🎯 Exam Tip: Remember the specific conditions (temperature, pressure) and the two-step nature (carboxylation then acidification) for Kolbe's reaction, as well as the structure of the salicylic acid product. This reaction is important for its industrial application in aspirin synthesis.

(ii) **Reimer-Tiemann reaction:** The Reimer-Tiemann reaction occurs when phenol is subjected to treatment with chloroform and an alkali (like NaOH) at 340 K. This reaction yields a phenolic aldehyde, specifically salicylaldehyde (o-hydroxy benzaldehyde), as the main product, with a minor amount of the para-isomer also forming. Conversely, if phenol is reacted with carbon tetrachloride in the presence of an aqueous alkali, o-hydroxybenzoic acid is obtained as the principal component.
\[ \text{C}_6\text{H}_5\text{OH} + \text{CHCl}_3 + \text{NaOH} \xrightarrow{\text{340 K}} \text{Salicylaldehyde (o-hydroxybenzaldehyde)} \]
\[ \text{C}_6\text{H}_5\text{OH} + \text{CCl}_4 + \text{NaOH} \implies \text{Salicylic acid (o-hydroxybenzoic acid)} \]
In simple words: The Reimer-Tiemann reaction allows phenol to be converted into an ortho-formylated product (salicylaldehyde) using chloroform and a base, or into salicylic acid using carbon tetrachloride and a base. It's a method to add an aldehyde or carboxyl group to the ortho position of a phenol.
🎯 Exam Tip: Distinguish between the reagents used (chloroform vs. carbon tetrachloride) and their respective products (salicylaldehyde vs. salicylic acid) in the Reimer-Tiemann reaction. Knowing the temperature and the role of alkali is also crucial.

(iii) **Williamson ether synthesis:** Williamson ether synthesis is a chemical reaction where an ether is formed by heating a sodium alkoxide with an alkyl halide. This synthetic approach is particularly effective for producing mixed ethers.
**Example:** For instance, ethyl methyl ether can be synthesized in two ways:
\[ \text{CH}_3\text{ONa} + \text{BrC}_2\text{H}_5 \implies \text{CH}_3\text{OC}_2\text{H}_5 + \text{NaBr} \]
**OR**
\[ \text{C}_2\text{H}_5\text{ONa} + \text{BrCH}_3 \implies \text{C}_2\text{H}_5\text{OCH}_3 + \text{NaBr} \]
In simple words: Williamson ether synthesis is a reaction where an alkoxide ion attacks an alkyl halide to form an ether. It's great for making ethers, especially mixed ones.
🎯 Exam Tip: Understand that Williamson synthesis works best with a primary alkyl halide and an alkoxide. For mixed ethers, choose the less hindered alkyl halide for better yields and to avoid elimination side reactions.

(iv) **Unsymmetrical ether:** Unsymmetrical ethers, also known as mixed ethers, are those where the two alkyl groups bonded to the oxygen atom are distinct. A common example is ethyl methyl ether (\( \text{C}_2\text{H}_5\text{OCH}_3 \)).
In simple words: Unsymmetrical ethers are simply ethers where the carbon groups on each side of the oxygen are different.
🎯 Exam Tip: Be able to identify and name unsymmetrical ethers based on their structure. Recognize that Williamson synthesis is particularly useful for preparing these types of ethers.

Question 19. Write the equations involved in the dehydration of ethanol to yield ethene.
Answer: The dehydration of ethanol to ethene proceeds via a three-step mechanism:
(i) **Formation of protonated alcohols:** The hydroxyl group of ethanol is protonated by an acid catalyst, forming a better leaving group (water). This step is fast.
\[ \text{CH}_3\text{CH}_2\text{OH} + \text{H}^+ \xrightarrow{\text{Fast}} \text{CH}_3\text{CH}_2\text{OH}_2^+ \]
(ii) **Formation of carbocation:** The protonated alcohol loses a water molecule, leading to the formation of a carbocation. This is the slowest step and therefore the rate-determining step of the reaction.
\[ \text{CH}_3\text{CH}_2\text{OH}_2^+ \xrightarrow{\text{Slow}} \text{CH}_3\text{CH}_2^+ + \text{H}_2\text{O} \]
(iii) **Loss of proton by the carbocation to form the alkene molecule:** The carbocation quickly loses a proton from an adjacent carbon atom, forming the stable alkene (ethene). Removing the alkene product helps shift the equilibrium towards product formation.
\[ \text{CH}_3\text{CH}_2^+ \xrightarrow{\text{Fast}} \text{CH}_2=\text{CH}_2 + \text{H}^+ \]
In simple words: Ethanol loses water when heated with acid to form ethene. The process involves protonating the alcohol, then losing water to form a carbocation, and finally losing a proton from the carbocation to create the double bond.
🎯 Exam Tip: When writing mechanisms, clearly indicate the movement of electrons with curved arrows, show all intermediates (like carbocations), and correctly label fast/slow (rate-determining) steps. For dehydration, recall that carbocation stability can lead to rearrangements.

Question 20. How are the following conversions carried out?
(i) Propene → Propan-2-ol.
(ii) Benzyl chloride → Benzyl alcohol.
(iii) Ethyl magnesium bromide → Propan-1-ol.
(iv) Methyl magnesium bromide → 2-Methylpropan-2-ol.
Answer:
(i) **Propene → Propan-2-ol:** Propene can be converted to propan-2-ol through hydration, typically an acid-catalyzed process (Markovnikov addition) or oxymercuration-demercuration. The latter involves reacting propene with water in the presence of mercuric acetate, followed by reduction, to yield propan-2-ol.
\[ \text{CH}_3\text{CH}=\text{CH}_2 + \text{H}_2\text{O} \xrightarrow{\text{Hg(OAc)}_2/\text{NaBH}_4} \text{CH}_3\text{CH(OH)}\text{CH}_3 (\text{Propan-2-ol}) \]
In simple words: Propene turns into propan-2-ol by adding water to its double bond, usually with acid or mercury-based reagents, following Markovnikov's rule.
🎯 Exam Tip: Remember Markovnikov's rule for hydration of unsymmetrical alkenes: hydrogen adds to the carbon with more hydrogens, and the hydroxyl group adds to the carbon with fewer hydrogens.

(ii) **Benzyl chloride → Benzyl alcohol:** Benzyl chloride is converted to benzyl alcohol through a nucleophilic substitution reaction, where the chlorine atom is replaced by a hydroxyl group using dilute aqueous sodium hydroxide.
\[ \text{C}_6\text{H}_5\text{CH}_2\text{Cl} + \text{NaOH(aq)} \implies \text{C}_6\text{H}_5\text{CH}_2\text{OH} (\text{Benzyl alcohol}) + \text{NaCl} \]
In simple words: Benzyl chloride becomes benzyl alcohol by replacing the chlorine with an -OH group using sodium hydroxide.
🎯 Exam Tip: Nucleophilic substitution reactions, especially \( \text{SN}_1 \) or \( \text{SN}_2 \) with aqueous NaOH, are standard ways to convert alkyl halides to alcohols.

(iii) **Ethyl magnesium bromide → Propan-1-ol:** Propan-1-ol is synthesized from ethyl magnesium bromide by first reacting the Grignard reagent with methanal (formaldehyde, \( \text{HCHO} \)) to form an intermediate adduct. Subsequent acidic hydrolysis of this adduct yields propan-1-ol.
\[ \text{C}_2\text{H}_5\text{MgBr} + \text{HCHO} \implies \text{C}_2\text{H}_5\text{CH}_2\text{OMgBr} \]
\[ \text{C}_2\text{H}_5\text{CH}_2\text{OMgBr} \xrightarrow{\text{H}_2\text{O}} \text{C}_2\text{H}_5\text{CH}_2\text{OH} (\text{Propan-1-ol}) + \text{Mg(OH)Br} \]
In simple words: To make propan-1-ol from ethyl magnesium bromide, react it with formaldehyde first, then add water to release the alcohol.
🎯 Exam Tip: Grignard reagents reacting with formaldehyde always produce primary alcohols after hydrolysis. This is a reliable method for chain extension to primary alcohols.

(iv) **Methyl magnesium bromide → 2-Methylpropan-2-ol:** 2-Methylpropan-2-ol can be prepared from methyl magnesium bromide by reacting it with propanone (acetone, \( \text{CH}_3\text{COCH}_3 \)) to form a tertiary alkoxide intermediate. Acidic hydrolysis of this intermediate then produces the tertiary alcohol, 2-methylpropan-2-ol.
\[ \text{CH}_3\text{MgBr} + \text{CH}_3\text{COCH}_3 \implies (\text{CH}_3)_3\text{COMgBr} \]
\[ (\text{CH}_3)_3\text{COMgBr} \xrightarrow{\text{H}_2\text{O}} (\text{CH}_3)_3\text{COH} (\text{2-Methylpropan-2-ol}) + \text{Mg(OH)Br} \]
In simple words: To get 2-methylpropan-2-ol from methyl magnesium bromide, react it with acetone, then add water to form the alcohol.
🎯 Exam Tip: Grignard reagents reacting with ketones (other than formaldehyde) yield tertiary alcohols after hydrolysis. Keep track of the carbon chain length and the type of alcohol formed.

Question 21. Name the reagents used in the following reactions:
1. Oxidation of a primary alcohol to carboxylic acid.
2. Oxidation of a primary alcohol to aldehyde.
3. Bromination of phenol to 2,4,6-tribromophenol.
4. Benzyl alcohol to benzoic acid.
5. Dehydration of propan-2-ol to propene.
6. Butan-2-one to butan-2-ol.
Answer:
1. **Oxidation of a primary alcohol to carboxylic acid:** Strong oxidizing agents like acidified potassium dichromate (\( \text{K}_2\text{Cr}_2\text{O}_7 \)) or neutral, acidic, or alkaline potassium permanganate (\( \text{KMnO}_4 \)) are typically used.
2. **Oxidation of a primary alcohol to aldehyde:** Pyridinium chlorochromate (PCC) in dichloromethane (\( \text{CH}_2\text{Cl}_2 \)) or copper metal at 573 K can be employed to achieve this selective oxidation.
3. **Bromination of phenol to 2,4,6-tribromophenol:** Bromine water (\( \text{Br}_2/\text{H}_2\text{O} \)) is used for this exhaustive bromination reaction.
4. **Benzyl alcohol to benzoic acid:** Acidified or alkaline potassium permanganate (\( \text{KMnO}_4 \)) serves as an effective oxidizing agent for this conversion.
5. **Dehydration of propan-2-ol to propene:** Concentrated sulfuric acid (\( \text{H}_2\text{SO}_4 \)) at 443 K or 85% phosphoric acid (\( \text{H}_3\text{PO}_4 \)) at 443 K are common dehydrating agents.
6. **Butan-2-one to butan-2-ol:** Reduction of the ketone to an alcohol is achieved using hydrogen gas with a nickel catalyst (\( \text{Ni}/\text{H}_2 \)), sodium borohydride (\( \text{NaBH}_4 \)), or lithium aluminum hydride (\( \text{LiAlH}_4 \)).
In simple words: Different reagents are chosen depending on the desired product: strong oxidizers for acids, mild oxidizers for aldehydes, bromine water for phenol bromination, and reducing agents for converting ketones to alcohols.
🎯 Exam Tip: Memorize the specific reagents for each type of transformation. Pay attention to the conditions (temperature, solvent) as they can significantly influence the product, especially in oxidations and dehydrations.

Question 22. Give reason for the higher boiling point of ethanol in comparison to methoxymethane.
Answer: Ethanol exhibits a higher boiling point compared to methoxymethane primarily because ethanol molecules form strong intermolecular hydrogen bonds with each other due to the presence of the highly polar O-H group. In contrast, methoxymethane, being an ether, lacks a hydrogen atom directly bonded to the oxygen, and thus cannot form intermolecular hydrogen bonds. Its intermolecular forces are limited to weaker dipole-dipole interactions and London dispersion forces, which require less energy to overcome.
In simple words: Ethanol boils at a higher temperature than methoxymethane because ethanol molecules can form strong hydrogen bonds, which are harder to break than the weaker forces in methoxymethane.
🎯 Exam Tip: Always consider intermolecular forces (hydrogen bonding, dipole-dipole, London dispersion) when comparing boiling points of organic compounds. Hydrogen bonding is a particularly strong force that significantly elevates boiling points.

Question 23. Give IUPAC names of the following ethers:
i. \( \text{C}_2\text{H}_5\text{OCH}_2\text{-CH(CH}_3)\text{-CH}_3 \)
ii. \( \text{CH}_3\text{OCH}_2\text{CH}_2\text{Cl} \)
iii. \( \text{O}_2\text{N - C}_6\text{H}_4\text{ - OCH}_3\text{(p)} \)
iv. \( \text{CH}_3\text{CH}_2\text{CH}_2\text{OCH}_3 \)
v.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक ईथर अणु को दर्शाता है जहाँ एक एथॉक्सी (\( -\text{OC}_2\text{H}_5 \)) समूह एक आइसोप्रोपाइल (\( -\text{CH(CH}_3)_2 \)) समूह से जुड़ा है. कार्बन परमाणु का एक समूह ऑक्सीजन से जुड़ा है और दूसरा समूह एथाइल समूह से जुड़ा है.
vi.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक ईथर अणु को दर्शाता है जिसमें एक बेंजीन वलय एक एथॉक्सी (\( -\text{OC}_2\text{H}_5 \)) समूह से जुड़ा है. यह एक एथॉक्सीबेंजीन या फेनेटोल अणु है.
Answer:
(i) 1-Ethoxy-2-methylpropane
(ii) 2-Chloro-1-methoxyethane (Correcting original "2-Chloro-l-dimethoxyethane" as there is only one methoxy group and one chloro group.)
(iii) 4-Nitroanisole (or 1-methoxy-4-nitrobenzene)
(iv) 1-Methoxypropane
(v) The diagram shows 2-ethoxypropane. Based on the source's provided answer for this part, which seems to be for a different compound: 1-Ethoxy-4,4-dimethyl cyclohexane.
(vi) Ethoxybenzene
In simple words: IUPAC naming of ethers involves identifying the larger alkyl group as the alkane parent and the smaller alkoxy group as a substituent. For cyclic or aromatic systems, the ether is named as an alkoxy derivative of the parent ring.
🎯 Exam Tip: When naming ethers, remember to identify the longest carbon chain for the parent alkane and treat the smaller alkyl-oxygen group as an alkoxy substituent. For complex structures or rings, prioritize functional groups or larger rings.

Question 24. Write the names of reagents and equations for the preparation of the following ethers by Williamson's synthesis:
(i) 1-Propoxypropane
(ii) Ethoxybenzene
(iii) 2-Methoxy-2-methylpropane
(iv) 1-Methoxyethane
Answer:
(i) **1-Propoxypropane:**
Reagents: 1-Bromopropane and sodium propoxide.
\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{Br} (\text{1-bromopropane}) + \text{CH}_3\text{CH}_2\text{CH}_2\text{ONa} (\text{sodium propoxide}) \implies \text{CH}_3\text{CH}_2\text{CH}_2\text{-O-CH}_2\text{CH}_2\text{CH}_3 (\text{1-propoxypropane}) + \text{NaBr} \]
(ii) **Ethoxybenzene:**
Reagents: Bromoethane and sodium phenoxide.
\[ \text{C}_6\text{H}_5\text{ONa} (\text{sodium phenoxide}) + \text{C}_2\text{H}_5\text{Br} (\text{bromoethane}) \implies \text{C}_6\text{H}_5\text{OC}_2\text{H}_5 (\text{ethoxybenzene}) + \text{NaBr} \]
(iii) **2-Methoxy-2-methylpropane:**
Reagents: Methyl bromide (bromomethane) and sodium 2-methylpropan-2-oxide (sodium tert-butoxide).
\[ (\text{CH}_3)_3\text{CONa} (\text{sodium tert-butoxide}) + \text{CH}_3\text{Br} (\text{bromomethane}) \implies (\text{CH}_3)_3\text{COCH}_3 (\text{2-methoxy-2-methylpropane}) + \text{NaBr} \]
(iv) **1-Methoxyethane:**
Reagents: Bromomethane and sodium ethoxide.
\[ \text{CH}_3\text{Br} (\text{bromomethane}) + \text{CH}_3\text{CH}_2\text{ONa} (\text{sodium ethoxide}) \implies \text{CH}_3\text{OCH}_2\text{CH}_3 (\text{1-methoxyethane}) + \text{NaBr} \]
In simple words: Williamson synthesis involves reacting an alkoxide (formed from an alcohol and sodium) with an alkyl halide to create an ether and a salt. This method is versatile for making various ethers.
🎯 Exam Tip: For Williamson synthesis, remember the general reaction: \( \text{R-ONa} + \text{R'-X} \implies \text{R-O-R'} + \text{NaX} \). Always use a primary alkyl halide to minimize elimination side reactions, especially if the alkoxide is bulky.

Question 25. Illustrate with examples the limitations of Williamson synthesis for the preparation of certain types of ethers.
Answer: The Williamson ether synthesis is highly effective but has limitations, particularly when attempting to prepare ethers involving secondary or tertiary alkyl halides. In such cases, elimination reactions (E2) often compete with the desired nucleophilic substitution (SN2) reaction, leading to alkene formation as the major product instead of the ether.
**Example:** If we attempt to synthesize tert-butyl methyl ether using methyl bromide and sodium tert-butoxide, the reaction proceeds smoothly:
\[ (\text{CH}_3)_3\text{CONa} + \text{CH}_3\text{Br} \implies (\text{CH}_3)_3\text{COCH}_3 (\text{tert-butyl methyl ether}) + \text{NaBr} \]
However, if we try to use tert-butyl bromide and sodium methoxide, the bulkiness of tert-butyl bromide and the basicity of methoxide favor an E2 elimination, yielding isobutylene (2-methylpropene) as the main product, with little to no ether:
\[ (\text{CH}_3)_3\text{CBr} + \text{CH}_3\text{ONa} \implies (\text{CH}_3)_2\text{C}=\text{CH}_2 (\text{2-methylpropene}) + \text{CH}_3\text{OH} + \text{NaBr} \]
This demonstrates that a primary alkyl halide and a bulky alkoxide generally give a good yield of ether, while a bulky alkyl halide with any alkoxide favors elimination.
In simple words: Williamson synthesis works best when one of the reactants is a primary alkyl halide. If you use a bulky (secondary or tertiary) alkyl halide, elimination (making alkenes) becomes the main reaction instead of ether formation.
🎯 Exam Tip: To avoid elimination in Williamson synthesis, always pair a less hindered alkyl halide (preferably primary) with the desired alkoxide. This ensures the SN2 mechanism predominates.

Question 26. How is 1-propoxypropane synthesised from propan-1-ol? Write mechanism of this reaction.
Answer: 1-Propoxypropane can be synthesized from propan-1-ol through a two-step process involving the formation of a sodium alkoxide and an alkyl halide, followed by Williamson ether synthesis.
**Synthesis:**
1. **Formation of sodium propoxide:** Propan-1-ol reacts with sodium to form sodium propoxide.
\[ 2\text{C}_3\text{H}_7\text{OH} (\text{Propan-1-ol}) + 2\text{Na} \implies 2\text{C}_3\text{H}_7\text{ONa} (\text{Sodium propoxide}) + \text{H}_2 \]
2. **Formation of 1-bromopropane:** Propan-1-ol reacts with hydrogen bromide to form 1-bromopropane.
\[ \text{C}_3\text{H}_7\text{OH} (\text{Propan-1-ol}) + \text{HBr} \implies \text{C}_3\text{H}_7\text{Br} (\text{1-bromopropane}) + \text{H}_2\text{O} \]
3. **Williamson synthesis:** Sodium propoxide reacts with 1-bromopropane to yield 1-propoxypropane.
\[ \text{C}_3\text{H}_7\text{ONa} (\text{Sodium propoxide}) + \text{C}_3\text{H}_7\text{Br} (\text{1-bromopropane}) \implies \text{C}_3\text{H}_7\text{OC}_3\text{H}_7 (\text{1-propoxypropane}) + \text{NaBr} \]
**Mechanism (of the Williamson synthesis step):**
The final step, the Williamson ether synthesis, proceeds via an \( \text{SN}_2 \) mechanism. The propoxide ion, a strong nucleophile, attacks the electrophilic carbon atom bearing the bromine in 1-bromopropane. This simultaneous attack and departure of the bromide ion forms the ether bond and releases sodium bromide.
In simple words: To make 1-propoxypropane from propan-1-ol, you first convert some of the propan-1-ol to sodium propoxide and some to 1-bromopropane. Then, these two react in a Williamson synthesis, where the alkoxide attacks the alkyl bromide, to form the ether.
🎯 Exam Tip: When asked for a mechanism, ensure you show the step-by-step electron movement using curved arrows for both bond breaking and bond forming. For Williamson synthesis, specifically highlight the nucleophilic attack in the \( \text{SN}_2 \) step.

Question 27. Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give reason.
Answer: Acid-catalyzed dehydration is not an appropriate method for synthesizing ethers from secondary or tertiary alcohols because these alcohols predominantly undergo elimination reactions (E1 or E2, depending on conditions) under acidic conditions, leading to the formation of alkenes rather than ethers. The more substituted carbocations formed during dehydration of secondary and tertiary alcohols are stable, favoring the loss of a proton to form an alkene over a nucleophilic attack by another alcohol molecule to form an ether.
In simple words: Using acid to dehydrate secondary or tertiary alcohols to make ethers doesn't work well because these alcohols prefer to lose water and form alkenes instead.
🎯 Exam Tip: Remember the competition between substitution and elimination reactions. For secondary and tertiary alcohols in acidic conditions, elimination to form alkenes is generally favored over ether formation, especially at higher temperatures.

Question 28. Write the equation of the reaction of hydrogen iodide with:
(i) 1-propoxypropane
(ii) methoxybenzene
(iii) benzyl ethyl ether.
Answer:
(i) **1-Propoxypropane:** When 1-propoxypropane reacts with hydrogen iodide, it undergoes cleavage to yield 1-iodopropane and propan-1-ol.
\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{-O-CH}_2\text{CH}_2\text{CH}_3 (\text{1-propoxypropane}) + \text{HI} \implies \text{CH}_3\text{CH}_2\text{CH}_2\text{I} (\text{1-iodopropane}) + \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} (\text{1-propanol}) \]
(ii) **Methoxybenzene:** Methoxybenzene (anisole) reacts with hydrogen iodide, cleaving the ether linkage to produce phenol and iodomethane.
\[ \text{C}_6\text{H}_5\text{OCH}_3 (\text{Methoxybenzene}) + \text{HI} \implies \text{C}_6\text{H}_5\text{OH} (\text{Phenol}) + \text{CH}_3\text{I} (\text{Iodomethane}) \]
(iii) **Benzyl ethyl ether:** Benzyl ethyl ether reacts with hydrogen iodide to form benzyl alcohol and iodoethane.
\[ \text{C}_6\text{H}_5\text{CH}_2\text{OC}_2\text{H}_5 (\text{Benzyl ethyl ether}) + \text{HI} \implies \text{C}_6\text{H}_5\text{CH}_2\text{OH} (\text{Benzyl alcohol}) + \text{C}_2\text{H}_5\text{I} (\text{Iodoethane}) \]
In simple words: Ethers react with hot hydrogen iodide to break the ether bond, forming an alkyl iodide and an alcohol or phenol, depending on the structure.
🎯 Exam Tip: For ether cleavage with HI, remember that the reaction often follows an \( \text{SN}_2 \) pathway, so the iodide typically attacks the less hindered carbon. However, for aryl alkyl ethers, the aryl-oxygen bond is stronger due to resonance and typically remains intact, forming phenol.

Question 29. Explain the fact that in aryl alkyl ethers (i) the alkoxy group activates the benzene ring towards electrophilic substitution and (ii) it directs the incoming substituents to ortho and para positions in benzene ring.
Answer:
(i) **Activation of the benzene ring:** In aryl alkyl ethers, the alkoxy group (\( \text{-OR} \)) functions as an electron-donating group through resonance. The lone pair of electrons on the oxygen atom can be delocalized into the benzene ring, increasing the overall electron density of the ring. This enhanced electron density makes the benzene ring more nucleophilic and thus more reactive towards electrophilic substitution reactions, hence it "activates" the ring.
(ii) **Ortho and para directing effect:** The resonance effect of the alkoxy group preferentially increases the electron density at the ortho and para positions of the benzene ring more than at the meta positions. This localized increase in electron density makes the ortho and para positions more susceptible to attack by incoming electrophiles. Therefore, the alkoxy group is an ortho- and para-directing group.
In simple words: The oxygen in an alkoxy group pushes electrons into the benzene ring, making the ring more reactive to attack by positive species. This electron push happens strongest at the ortho and para spots, so new groups attach there.
🎯 Exam Tip: Understanding resonance structures is key to explaining both the activating and directing effects of substituents on aromatic rings. Draw the resonance structures to visualize the electron delocalization and charge distribution at ortho and para positions.

Question 30. Show how would you synthesise the following alcohols from appropriate alkenes:
(i) 1-methyl cyclohexanol
(ii) butan-1-ol
(iii) pentan-2-ol
(iv) 2-cyclohexylbutan-2-ol
Answer:
(i) **1-Methylcyclohexanol:**
ℹ️ चित्र व्याख्या (Diagram Explanation): यह संश्लेषण 1-मिथाइलसाइक्लोहेक्सीन से 1-मिथाइलसाइक्लोहेक्सानोल के निर्माण को दर्शाता है. यह एक एसिड-उत्प्रेरित जलयोजन (hydration) प्रतिक्रिया है जहाँ पानी एल्कीन के द्विबंध पर मार्कोवनिकोव के नियम के अनुसार जुड़ता है.
1-Methylcyclohexanol can be synthesized by the acid-catalyzed hydration of 1-methylcyclohexene. The alkene reacts with water in the presence of an acid catalyst to yield the desired alcohol following Markovnikov's rule.
\[ \text{C}_7\text{H}_{12} (\text{1-methylcyclohexene}) + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{C}_7\text{H}_{14}\text{O} (\text{1-methylcyclohexanol}) \]
(ii) **Butan-1-ol:**
ℹ️ चित्र व्याख्या (Diagram Explanation): यह संश्लेषण 4-मिथाइलहेप्ट-3-ईन से 4-मिथाइलहेप्टान-4-ऑल के निर्माण को दर्शाता है. यह एक एसिड-उत्प्रेरित जलयोजन प्रतिक्रिया है जहाँ पानी एल्कीन के द्विबंध पर मार्कोवनिकोव के नियम के अनुसार जुड़ता है.
Butan-1-ol (a primary alcohol) can be synthesized from but-1-ene via hydroboration-oxidation, which is an anti-Markovnikov addition of water. Alternatively, for simple hydration to produce an alcohol, Markovnikov's rule generally applies. To specifically get butan-1-ol, hydroboration-oxidation of but-1-ene is preferred. For brevity, assuming a typical hydration from the source's answer pattern, which here, however, describes synthesis of 4-methylheptan-4-ol from 4-methylhept-3-ene, which is inconsistent with the product 'butan-1-ol' asked. The correct reaction for butan-1-ol from an alkene would be:
\[ \text{CH}_3\text{CH}_2\text{CH}=\text{CH}_2 (\text{But-1-ene}) \xrightarrow{\text{i) } (\text{BH}_3)_2/\text{THF} \text{ ii) } \text{H}_2\text{O}_2/\text{NaOH}} \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH} (\text{Butan-1-ol}) \]
(iii) **Pentan-2-ol:**
ℹ️ चित्र व्याख्या (Diagram Explanation): यह संश्लेषण पेंट-1-ईन से पेंटान-2-ऑल के निर्माण को दर्शाता है. यह एक एसिड-उत्प्रेरित जलयोजन प्रतिक्रिया है जहाँ पानी एल्कीन के द्विबंध पर मार्कोवनिकोव के नियम के अनुसार जुड़ता है.
Pentan-2-ol can be prepared by the acid-catalyzed hydration of pent-1-ene. This involves adding water across the double bond of pent-1-ene, following Markovnikov's rule, to form pentan-2-ol.
\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}=\text{CH}_2 (\text{Pent-1-ene}) + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{CH}_3\text{CH}_2\text{CH}_2\text{CH(OH)}\text{CH}_3 (\text{Pentan-2-ol}) \]
(iv) **2-Cyclohexylbutan-2-ol:**
ℹ️ चित्र व्याख्या (Diagram Explanation): यह संश्लेषण 2-साइक्लोहेक्सिलब्यूट-2-ईन से 2-साइक्लोहेक्सिलब्यूटान-2-ऑल के निर्माण को दर्शाता है. यह एक एसिड-उत्प्रेरित जलयोजन प्रतिक्रिया है जहाँ पानी एल्कीन के द्विबंध पर मार्कोवनिकोव के नियम के अनुसार जुड़ता है.
2-Cyclohexylbutan-2-ol is synthesized through the acid-catalyzed hydration of 2-cyclohexylbut-2-ene. Water adds to the alkene in the presence of an acid to yield the tertiary alcohol following Markovnikov's rule.
\[ \text{C}_6\text{H}_{11}\text{CH}=\text{C}(\text{CH}_3)\text{CH}_2\text{CH}_3 (\text{2-cyclohexylbut-2-ene}) + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{C}_6\text{H}_{11}\text{C(OH)}(\text{CH}_3)\text{CH}_2\text{CH}_3 (\text{2-cyclohexylbutan-2-ol}) \]
In simple words: Alcohols are generally made from alkenes by adding water to the double bond. Depending on the desired alcohol, you might use acid-catalyzed hydration (Markovnikov addition) or hydroboration-oxidation (anti-Markovnikov addition).
🎯 Exam Tip: Understand the difference between Markovnikov and anti-Markovnikov additions of water to alkenes. Acid-catalyzed hydration yields Markovnikov products, while hydroboration-oxidation yields anti-Markovnikov products. Select the appropriate method based on the desired alcohol structure.

Question 31. When 3-methylbutan-2-ol is treated with HBr, the following reaction takes place:
\[ \text{CH}_3\text{CH}(\text{CH}_3)\text{CH}(\text{OH})\text{CH}_3 + \text{HBr} \implies \text{CH}_3\text{C(Br)}(\text{CH}_3)\text{CH}_2\text{CH}_3 + \text{H}_2\text{O} \]
**Give a mechanism for this reaction.**
**(Hint: The secondary carbocation formed in step II rearranges to a more stable tertiary carbocation by a hydride ion shift from 3rd carbon atom.)**
Answer: The reaction of 3-methylbutan-2-ol with HBr to yield 2-bromo-2-methylbutane involves a carbocation rearrangement, following these steps:
**Mechanism:**
1. **Protonation of the alcohol:** The hydroxyl group of 3-methylbutan-2-ol is protonated by \( \text{H}^+ \) from HBr, forming a protonated alcohol with water as a good leaving group.
\[ \text{CH}_3\text{CH}(\text{CH}_3)\text{CH}(\text{OH})\text{CH}_3 + \text{H}^+ \implies \text{CH}_3\text{CH}(\text{CH}_3)\text{CH}(\text{OH}_2^+)\text{CH}_3 \]
2. **Loss of water to form a secondary carbocation:** The protonated alcohol loses a water molecule, generating a secondary carbocation at the C-2 position.
\[ \text{CH}_3\text{CH}(\text{CH}_3)\text{CH}(\text{OH}_2^+)\text{CH}_3 \implies \text{CH}_3\text{CH}(\text{CH}_3)\text{C}^+(\text{H})\text{CH}_3 + \text{H}_2\text{O} \]
3. **Hydride shift (1,2-shift) to form a more stable tertiary carbocation:** The secondary carbocation undergoes a 1,2-hydride shift from the adjacent tertiary carbon (C-3) to the secondary carbon (C-2), resulting in a more stable tertiary carbocation at the C-3 position.
\[ \text{CH}_3\text{CH}(\text{CH}_3)\text{C}^+(\text{H})\text{CH}_3 \xrightarrow{\text{1,2-Hydride shift}} \text{CH}_3\text{C}^+(\text{CH}_3)\text{CH}_2\text{CH}_3 \]
4. **Attack by bromide ion:** The stable tertiary carbocation is then attacked by the bromide ion (\( \text{Br}^- \)) acting as a nucleophile, forming the final product, 2-bromo-2-methylbutane.
\[ \text{CH}_3\text{C}^+(\text{CH}_3)\text{CH}_2\text{CH}_3 + \text{Br}^- \implies \text{CH}_3\text{C(Br)}(\text{CH}_3)\text{CH}_2\text{CH}_3 (\text{2-bromo-2-methylbutane}) \]
In simple words: When 3-methylbutan-2-ol reacts with HBr, the alcohol first gets protonated and loses water, forming a carbocation. This carbocation then rearranges itself by moving a hydrogen atom to form a more stable (tertiary) carbocation, which is then attacked by bromine to form the final product.
🎯 Exam Tip: Always look for possibilities of carbocation rearrangements (hydride or alkyl shifts) in reactions involving carbocation intermediates, especially when converting secondary to tertiary carbocations, as this leads to more stable products and is a common source of unexpected products.

GSEB Class 12 Chemistry Alcohols, Phenols and Ethers Additional Important Questions and Answers

Question 1. Methanol is known as 'wood alcohol' because it is produced by the destructive distillation of wood.
(i) How is methanol manufactured?
(ii) Methanol is usually used for denaturing ethanol. Give reason.
(iii) Suggest a method to convert methanol to ethanol.
Answer:
(i) **Methanol manufacturing:** Industrially, methanol is primarily produced from 'water gas', which is a mixture of carbon monoxide (\( \text{CO} \)) and hydrogen (\( \text{H}_2 \)). Water gas is combined with approximately half its volume of additional hydrogen and then passed over a catalyst composed of zinc oxide (\( \text{ZnO} \)), chromium oxide (\( \text{Cr}_2\text{O}_3 \)), and copper oxide (\( \text{CuO} \)). This reaction occurs under high pressure (around 200 atm) and at elevated temperatures (between 623-670 K), yielding methanol.
\[ \text{CO} (\text{Water gas}) + 2\text{H}_2 (\text{Water gas}) \xrightarrow{\text{Catalyst (ZnO, Cr}_2\text{O}_3, \text{CuO}), 200 atm, 623-670 K} \text{CH}_3\text{OH} (\text{Methanol}) \]
(ii) **Denaturing ethanol:** Methanol is employed to denature ethanol because it is a highly poisonous substance. Its addition renders ethanol unsuitable for consumption, thereby preventing its use as an alcoholic beverage by making it toxic.
(iii) **Conversion of methanol to ethanol:** Methanol can be converted to ethanol through a multi-step Grignard reaction sequence:
1. **Conversion to chloromethane:** Methanol is first converted to chloromethane using phosphorus pentachloride (\( \text{PCl}_5 \)).
\[ \text{CH}_3\text{OH} \xrightarrow{\text{PCl}_5} \text{CH}_3\text{Cl} \]
2. **Formation of methyl magnesium chloride:** Chloromethane then reacts with magnesium in ether to form methyl magnesium chloride (a Grignard reagent).
\[ \text{CH}_3\text{Cl} + \text{Mg} \xrightarrow{\text{in ether}} \text{CH}_3\text{MgCl} \]
3. **Reaction with formaldehyde:** This Grignard reagent is treated with formaldehyde (methanal, \( \text{HCHO} \)) to form an intermediate adduct.
\[ \text{CH}_3\text{MgCl} + \text{HCHO} \implies \text{CH}_3\text{CH}_2\text{OMgCl} \]
4. **Acidic hydrolysis:** The adduct undergoes acidic hydrolysis to produce ethanol.
\[ \text{CH}_3\text{CH}_2\text{OMgCl} \xrightarrow{\text{H}_3\text{O}^+/\text{H}^+} \text{CH}_3\text{CH}_2\text{OH} (\text{Ethanol}) + \text{Mg(OH)Cl} \]
In simple words: Methanol is made from water gas and hydrogen. It's added to ethanol to make it undrinkable because it's poisonous. To convert methanol to ethanol, you first turn it into chloromethane, then a Grignard reagent, react with formaldehyde, and finally hydrolyze it.
🎯 Exam Tip: For industrial processes, remember the raw materials and catalyst conditions. For Grignard reactions, pay attention to the choice of carbonyl compound (formaldehyde for primary alcohols) and the hydrolysis step. Always remember the toxicity of methanol.

Question 2. What is absolute alcohol? How is it prepared?
Answer: Absolute alcohol refers to 100% pure ethanol (\( \text{CH}_3\text{CH}_2\text{OH} \)). It is typically prepared from rectified spirit (approximately 95.6% ethanol) using azeotropic distillation. In this process, benzene is added to the rectified spirit to form a ternary azeotrope with water and ethanol, which distills off at a lower boiling point (e.g., 64.8°C for a benzene-ethanol-water azeotrope). Once all the water is removed as part of this ternary azeotrope, a binary azeotrope of benzene and ethanol distills. Finally, pure anhydrous ethanol (absolute alcohol) can be obtained by distilling off the remaining benzene at 78.1°C.
In simple words: Absolute alcohol is pure (100%) ethanol. It's made by removing water from less pure ethanol using a special distillation method with benzene.
🎯 Exam Tip: Understand that ethanol and water form an azeotrope, meaning they cannot be completely separated by simple distillation. Azeotropic distillation with a third component like benzene is necessary to obtain absolute alcohol.

Question 3. A compound (A) reacts with thionyl chloride to give compound (B). (B) reacts with magnesium to form a Grignard reagent which is treated with acetone and the product on hydrolysis gives 2- methyl-2-butanol. Identify (A) and (B).
Answer: The final product, 2-methyl-2-butanol, is a tertiary alcohol. Tertiary alcohols are typically formed when a Grignard reagent reacts with a ketone (like acetone). The structure of 2-methyl-2-butanol is \( \text{CH}_3\text{CH}_2\text{C(OH)(CH}_3)_2 \). Since acetone contributes two methyl groups and the carbonyl carbon, the Grignard reagent must have provided the ethyl group (\( \text{CH}_3\text{CH}_2\text{-} \)). Thus, the Grignard reagent is ethyl magnesium halide (\( \text{CH}_3\text{CH}_2\text{MgX} \)).
This Grignard reagent is formed from compound (B), so (B) must be an ethyl halide, specifically chloroethane (\( \text{C}_2\text{H}_5\text{Cl} \)).
Compound (B) is formed when compound (A) reacts with thionyl chloride (\( \text{SOCl}_2 \)). This suggests (A) is an alcohol. Therefore, (A) is ethanol (\( \text{CH}_3\text{CH}_2\text{OH} \)).
**Identification:**
* (A) = Ethanol (\( \text{CH}_3\text{CH}_2\text{OH} \))
* (B) = Chloroethane (\( \text{C}_2\text{H}_5\text{Cl} \))
**Reactions:**
\[ \text{CH}_3\text{CH}_2\text{OH (A)} + \text{SOCl}_2 \implies \text{C}_2\text{H}_5\text{Cl (B)} + \text{SO}_2 + \text{HCl} \]
\[ \text{C}_2\text{H}_5\text{Cl (B)} + \text{Mg} \implies \text{C}_2\text{H}_5\text{MgCl} (\text{Grignard reagent}) \]
\[ \text{CH}_3\text{COCH}_3 (\text{Acetone}) + \text{C}_2\text{H}_5\text{MgCl} \implies (\text{CH}_3)_2\text{C(OMgCl)}\text{C}_2\text{H}_5 (\text{Adduct}) \]
\[ (\text{CH}_3)_2\text{C(OMgCl)}\text{C}_2\text{H}_5 \xrightarrow{\text{H}_3\text{O}^+} (\text{CH}_3)_2\text{C(OH)}\text{C}_2\text{H}_5 (\text{2-methyl-2-butanol}) + \text{Mg(OH)Cl} \]
In simple words: The problem describes forming a tertiary alcohol (2-methyl-2-butanol) using a Grignard reagent and acetone. By working backward, we find that the Grignard reagent is made from chloroethane, which in turn comes from ethanol and thionyl chloride. So, (A) is ethanol and (B) is chloroethane.
🎯 Exam Tip: For synthesis problems, especially those involving Grignard reagents, work backward from the final product to identify intermediates and starting materials. Remember that Grignard reactions with ketones yield tertiary alcohols.

Question 4. Complete the following reactions.
Answer:
(i) **Reaction 1:** The catalytic hydrogenation of propanone in the presence of a palladium catalyst reduces the ketone to a secondary alcohol, propan-2-ol.
\[ \text{CH}_3\text{COCH}_3 (\text{Propanone}) + \text{H}_2 \xrightarrow{\text{Pd}} \text{CH}_3\text{CH(OH)}\text{CH}_3 (\text{Propan-2-ol}) \]
(ii) **Reaction 2:** This is a Grignard reaction where methyl magnesium bromide reacts with butanone (ethyl methyl ketone), followed by acidic hydrolysis. This process yields a tertiary alcohol, 2-methylbutan-2-ol.
\[ \text{CH}_3\text{MgBr} + \text{CH}_3\text{CH}_2\text{COCH}_3 (\text{Butanone}) \implies \text{CH}_3\text{CH}_2\text{C(OMgBr)}(\text{CH}_3)_2 \xrightarrow{\text{H}^+/\text{H}_2\text{O}} \text{CH}_3\text{CH}_2\text{C(OH)}(\text{CH}_3)_2 (\text{2-methylbutan-2-ol}) \]
In simple words: The first reaction turns acetone into propan-2-ol by adding hydrogen. The second reaction uses a Grignard reagent and a ketone (butanone) to create a more complex tertiary alcohol (2-methylbutan-2-ol).
🎯 Exam Tip: Recognize common reduction reactions for carbonyls (e.g., catalytic hydrogenation) and the Grignard reaction's ability to create new carbon-carbon bonds and form alcohols (primary, secondary, or tertiary depending on the carbonyl compound).

Question 5. Phenol is acidic in nature. Give reason.
Answer: Phenol exhibits acidic character primarily because, upon deprotonation, it forms a phenoxide ion (\( \text{C}_6\text{H}_5\text{O}^- \)). This phenoxide ion is highly stabilized by resonance, where the negative charge on the oxygen atom is delocalized over the entire aromatic ring. This delocalization distributes the charge, leading to a more stable conjugate base compared to the un-ionized phenol molecule. The enhanced stability of the conjugate base facilitates the release of a proton (\( \text{H}^+ \)), making phenol significantly more acidic than simple alcohols, whose corresponding alkoxide ions are not resonance stabilized.
\[ \text{C}_6\text{H}_5\text{OH} (\text{Phenol}) \implies \text{C}_6\text{H}_5\text{O}^- (\text{Phenoxide ion}) + \text{H}^+ \]
In simple words: Phenol is acidic because when it loses a proton, the resulting phenoxide ion is very stable due to electron sharing (resonance) within its benzene ring. This stability makes it easier for phenol to give up a proton.
🎯 Exam Tip: To explain acidity, focus on the stability of the conjugate base. For phenols, draw resonance structures of the phenoxide ion to illustrate how the negative charge is delocalized, which is crucial for its enhanced acidity.

Question 6. The teacher asked the students to conduct nitration of phenol. Ramu used dil.HNO3 while John used a mixture of conc.HNO3 and conc.H2SO4.
(i) What are the products obtained in each case?
(ii) Write chemical equations.
Answer:
(i) **Products obtained:**
* **Ramu's experiment (dilute \( \text{HNO}_3 \)):** When phenol is nitrated with dilute nitric acid, it yields a mixture of ortho-nitrophenol and para-nitrophenol. The hydroxyl group activates the ortho and para positions, but the dilute acid prevents multiple nitrations.
* **John's experiment (concentrated \( \text{HNO}_3 \) and \( \text{H}_2\text{SO}_4 \)):** When treated with a concentrated nitrating mixture (concentrated nitric acid and concentrated sulfuric acid), phenol undergoes extensive nitration to form picric acid (2,4,6-trinitrophenol).
(ii) **Chemical equations:**
* **Nitration with dilute nitric acid:**
\[ \text{C}_6\text{H}_5\text{OH} (\text{Phenol}) + \text{dil. HNO}_3 \implies \text{o-nitrophenol} + \text{p-nitrophenol} \]
* **Nitration with concentrated nitrating mixture:**
\[ \text{C}_6\text{H}_5\text{OH} (\text{Phenol}) + 3\text{HNO}_3 (\text{conc.}) \xrightarrow{\text{conc. H}_2\text{SO}_4} \text{C}_6\text{H}_2(\text{NO}_2)_3\text{OH} (\text{Picric acid}) + 3\text{H}_2\text{O} \]
In simple words: Phenol reacts with weak nitric acid to give nitrophenol at ortho and para spots. With strong nitric and sulfuric acid, it adds three nitro groups, forming picric acid.
🎯 Exam Tip: The reactivity of phenol's ring towards electrophilic substitution is very high. Remember that the concentration of the nitrating agent determines the extent of nitration: dilute for mono-nitration, concentrated for multi-nitration (specifically picric acid).

Question 7.
(i) Identify the functional isomer of ethanol from the following, a. Ethanol, b. Propanol, c. Dimethyl ether.
(ii) Suggest a method to prepare the functional isomer of ethanol.
(iii) How will you distinguish the isomers?
Answer:
(i) **Functional isomer of ethanol:** The functional isomer of ethanol (\( \text{C}_2\text{H}_5\text{OH} \)) among the given options is dimethyl ether (\( \text{CH}_3\text{OCH}_3 \)). Both have the same molecular formula \( \text{C}_2\text{H}_6\text{O} \) but different functional groups (alcohol vs. ether).
(ii) **Method to prepare dimethyl ether:** Dimethyl ether can be prepared through Williamson's synthesis by reacting sodium methoxide with chloromethane.
\[ \text{CH}_3\text{ONa} (\text{Sodium methoxide}) + \text{CH}_3\text{Cl} (\text{Chloromethane}) \xrightarrow{\text{Williamson's synthesis}} \text{CH}_3\text{OCH}_3 (\text{Dimethyl ether}) + \text{NaCl} \]
(iii) **Distinguishing the isomers (Ethanol and Dimethyl ether):** Ethanol and dimethyl ether can be distinguished using the iodoform test. Ethanol, possessing the \( \text{CH}_3\text{CH(OH)} \)-group, will give a positive iodoform test, forming a yellow precipitate of iodoform (\( \text{CHI}_3 \)) when treated with iodine (\( \text{I}_2 \)) and a base (like sodium hydroxide). Dimethyl ether, lacking this specific structural requirement, will not respond to the iodoform test.
In simple words: Dimethyl ether is the functional isomer of ethanol. It can be made using Williamson synthesis. You can tell them apart using the iodoform test: ethanol gives a yellow solid, while dimethyl ether does not.
🎯 Exam Tip: Remember that functional isomers have the same molecular formula but different functional groups. The iodoform test is a crucial diagnostic tool for compounds containing \( \text{CH}_3\text{CH(OH)}- \) or \( \text{CH}_3\text{CO}- \) groups. Always remember to specify the reagents for the iodoform test (I2 and NaOH).

Question 8. Suggest a method to distinguish the following set of compounds.
(i) Ethanol and methanol.
(ii) Ethanol and phenol.
(iii) 1-propanol and 2-propanol
**Also, explain the chemistry behind.**
Answer:
(i) **Ethanol and methanol:**
**Method:** The iodoform test can distinguish between ethanol and methanol.
**Chemistry behind:** Ethanol contains the \( \text{CH}_3\text{CH(OH)}- \) group, which undergoes the iodoform reaction when warmed with iodine and a base (e.g., \( \text{NaOH} \) or \( \text{Na}_2\text{CO}_3 \)), producing a characteristic yellow precipitate of iodoform (\( \text{CHI}_3 \)). Methanol, lacking this specific structural requirement, does not give a positive iodoform test.
\[ \text{CH}_3\text{CH}_2\text{OH} (\text{Ethanol}) + 4\text{I}_2 + 3\text{Na}_2\text{CO}_3 \implies \text{CHI}_3 \downarrow (\text{yellow ppt}) + \text{HCOONa} + 5\text{NaI} + 2\text{H}_2\text{O} + 3\text{CO}_2 \]
**Methanol:** No reaction with \( \text{I}_2/\text{NaOH} \).
(ii) **Ethanol and phenol:**
**Method 1 (Bromine water test):** Phenol and ethanol can be distinguished by reacting them with bromine water.
**Chemistry behind:** Phenol rapidly undergoes electrophilic substitution with bromine water to form a white (often appearing yellow) precipitate of 2,4,6-tribromophenol, causing the reddish-brown color of bromine water to disappear. Ethanol, being an alcohol, does not react with bromine water under these conditions.
\[ \text{C}_6\text{H}_5\text{OH} (\text{Phenol}) + 3\text{Br}_2\text{(aq)} \implies \text{C}_6\text{H}_2\text{Br}_3\text{OH} (\text{2,4,6-tribromophenol, yellow ppt}) + 3\text{HBr} \]
**Method 2 (Ferric chloride test):** Another distinguishing test for phenol is the ferric chloride (\( \text{FeCl}_3 \)) test.
**Chemistry behind:** Phenol reacts with neutral ferric chloride solution to produce a distinctively colored complex (typically violet, but can vary to red, green, or blue) due to the formation of an iron-phenoxide complex. Ethanol does not show such a reaction.
\[ 6\text{C}_6\text{H}_5\text{OH} (\text{Phenol}) + \text{FeCl}_3 \implies (\text{C}_6\text{H}_5\text{O})_6\text{Fe}^{3-} (\text{Colored complex}) + 3\text{HCl} + 3\text{H}^+ \]
(iii) **1-Propanol and 2-propanol:**
**Method:** The iodoform test can distinguish between 1-propanol and 2-propanol.
**Chemistry behind:** 2-Propanol, which is a secondary alcohol, contains the \( \text{CH}_3\text{CH(OH)}- \) group. It will give a positive iodoform test, producing a yellow precipitate of iodoform (\( \text{CHI}_3 \)) when treated with iodine and a base. 1-Propanol, a primary alcohol, lacks this specific structural feature and thus does not give a positive iodoform test.
\[ \text{CH}_3\text{CH(OH)}\text{CH}_3 (\text{2-Propanol}) + \text{I}_2/\text{NaOH} \implies \text{CHI}_3 \downarrow (\text{yellow ppt}) + \text{CH}_3\text{COONa} \]
**1-Propanol:** No reaction with \( \text{I}_2/\text{NaOH} \).
In simple words: To tell these compounds apart, you can use specific chemical tests. The iodoform test works for ethanol vs. methanol and 2-propanol vs. 1-propanol. For ethanol vs. phenol, bromine water or ferric chloride tests are effective. The chemistry depends on unique reactive groups in each compound.
🎯 Exam Tip: Familiarize yourself with common qualitative tests like the iodoform test, bromine water test, and ferric chloride test. Understand the specific functional groups that react in each test and the observable changes (e.g., precipitate formation, color change).