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NCERT Questions for Class 12 Chemistry Chapter 7 – Alcohols, Phenols, and Ethers
NCERT Class 12 Chemistry Chapter 7 – Alcohols, Phenols, and Ethers Important Questions and detailed answers is a valuable resource for students preparing for their board exams. In this chapter students are introduced to a variety of organic compounds, including alcohols, phenols, and ethers. These three classes of compounds have unique properties and play important roles in various chemical reactions. These questions are a useful tool for focused revision because they have been carefully chosen to cover the main concepts covered in the chapter.
Important Questions with Solutions of Class 12 Chemistry Chapter 7 – Alcohols, Phenols, and Ethers
1) Explain the following statements:
(i) Nitrophenol is a stronger acid than o-methoxyphenol. Is it true?
Ans – Yes, ortho-nitrophenol is more acidic than ortho-methoxyphenol. The nitro group is electron withdrawing and in the ortho position; it will depress the electron density of the O-H bond. Thus, the resonance stabilizes the o-nitrophenoxide ion that results following the removal of protons.
(ii) Alcohols contain more extreme boiling points than ethers carrying relevant molecular masses.
Ans – Since the molecules of alcohol could form hydrogen bonds due to the existing presence of the OH group, this substance boils at a higher temperature than the ethers and alkanes of the same molar weight.
(iii) Why is the C-O-C bond angle in dimethyl ether 111.7°?
Ans: In this molecule, the 2 standalone pairs of electrons are trying to link with the oxygen atom. However, the repulsive nature of lone electron pairs might not compress the side groups as they are deemed heavy. Hence the bond angle remains at 111.7°.
(iv) Phenol is more reactive towards electrophilic substitution reaction than benzene.
Ans – Phenol reacts far more readily than benzene in electrophilic substitution reactions. The lone electron pairs donated to the oxygen atom into the ring system enhance electron density around the ring system. In comparison to benzene, this makes it substantially much more reactive. The intermediate carbocation is more stable through resonance.
(v) Phenols are extremely acidic over alcohol. Why?
Ans – Although phenols are stronger than alcohols, they are weak acids. Since the phenoxide ion is more highly stable than the alkoxide ion, phenol becomes acidic compared to cyclohexanol and acyclic alcohols. An alkoxide ion like the one created from cyclohexanol has a negative charge localized close to the oxygen atom.
2) Explain the below reactions with desired examples
(i) Reimer-Tiemann reaction.
Ans – The -CHO group is launched at the benzene ring’s ortho zone when phenol is mixed with chloroform during a sodium hydroxide combination. This is known as the Reimer-Tiemann reaction. Intermediate is hydrolyzed by the action of alkalis into salicylaldehyde.
(ii) Williamson synthesis
Ans – Williamson’s ether synthesis reacts alkyl group halides and sodium alkoxides inside the laboratory to produce symmetrical and unsymmetrical ethers. It also encompasses the SN² mechanism by which the alkoxide particle attacks the organic compound. When primary alkyl groups are the halides, extreme results are obtained. Meanwhile, if organic compounds are secondary/tertiary, elimination is least preferred as substitution will be highly sought after.
(iii) Acid-catalyzed dehydration at 443 Κ of alcohols.
Ans – The below reaction is an ethanol acid dehydration to yield ethene.
The first protonation occurs on the oxygen atom in the OH group. The second step is eliminating a water molecule to create the carbonium ion. Lastly, there is a deprotonation for building a carbon-carbon dual bond.
(iv) Kolbe’s reaction
Ans – Kolbe’s reaction: Salicylic acid is prepared by boiling a mixture of sodium phenoxide and carbon dioxide at 180-200°c under pressurized conditions.
(v) Friedel-Crafts acylation of Anisole.
Ans – A Friedel-Crafts acylation reaction converts an aromatic ring into an aryl ketone. Anisole reacts with acetyl chloride to become a ketone.
(vi) Hydroboration oxidation of alkenes
Ans – The hydroboration-oxidation sequence is a 2 step cycle for alcohol synthesis. The reaction starts with an Anti-Markovnikov model, where hydrogen (from BH3/BHR2) links to the most substituted carbon in the alkene dual bond. Meanwhile, the boron merges itself with the least substituted carbon.
3) Write down the reactions
(i) The reaction of ethanol with copper at a temperature of 573 K.
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(ii) Butanal to Butanol
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(iii) The reaction between the sodium salt of substituted phenol and methyl bromide.
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(iv) Reaction of methoxy proprane HBr
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(v) The reaction of Tertiary butyl ethyl ether with Hydrogen Iodide.
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