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NCERT Solutions Class 12 Chemistry Chapter 11 Alcohols, Phenols, and Ethers have discussed important topics such as alcohols, phenols, and other related chemicals. NCERT Solutions is perfectly articulated by the experts on the subject. NCERT Solutions Class 12 is the one-stop solution for all kinds of queries one might face.

  • Alcohols – such organic chemical compounds which have at least one hydroxyl functional group (-OH) are bound to the main carbon atom. Alcohol is originally referred to as alcohol such as the primary alcohol or the ethanol (ethyl alcohol). This is the main ingredient used in medicines and other alcoholic drinks. In the category of alcohol, the simplest alcohols are the methanols and ethanols.
  • Phenols – such class of organic compound in which the hydroxyl group is attached to the benzene ring. The simplest of this chemical is known as phenol. Phenols are often used in the production of resins and nylons. The phenolic compounds are further divided into phenols or polyphenols according to several phenol units attached to the molecule.
  • Ethers – in ethers there are different organic compounds attached to them. These are organic compounds or oxygen atom which is connected to two alkyl or the aryl groups. These ethers have a general formula of R-O-R’, in this, the R represents the alkyl group and the R’ represents the aryl group.
  • Nomenclature – IUPAC has proposed the rules which govern the naming of all kinds of chemical compounds which we find in our surroundings. Alcohols are mainly divided into three main types
    • Monohydric Alcohol – the general formula of monohydric alcohol is CnH2n+1OH, Here n = 1, 2, etc. these chemical compounds can also be described as the alkyl group. There are three main ways to describe the nomenclature of a monohydric alcohol.
      1. Common system
      2. Carbinol system
      3. IUPAC System
    • Dihydric Alcohol – the general formula of Dihydric alcohol has a general formula is (CH2)n(OH)2 in which n = 2, 3, 4 … etc.

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

1. Write IUPAC names of the following compounds:

Answer:

(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 – Methyl phenol

(vii) 2, 5 – Dimethylphenol

(viii) 2, 6 – Dimethylphenol

(ix) 1 – Methoxy – 2 – methyl propane

(x) Ethoxy benzene

(xi) 1 – Phenoxyheptane

(xii) 2 – Ethoxybutane

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) 4-Chloro-3-ethylbutan-1-ol.

Answer:

3. (i) Draw the structures of all isomeric alcohols of molecular formula C5H12O and give their IUPAC names.
(ii) Classify the isomers of alcohols in question.

Answer:

(i) The structures & IUPAC names of all isomeric alcohols with a molecular formula of C5H12O are shown below:

(a) CH3-CH2-CH2-CH2-CH2-OH

Pentan – 1 – ol (1°)

2 – Methylbutan – 2 – ol ( 3 ° )

(ii) Primary alcohol: Pentan – 1 – ol ; 2 – Methylbutan – 1 – ol ;

3 – Methylbutan – 1 – ol ; 2, 2 – Dimethylpropan – 1 – ol

Secondary alcohol: Pentan – 2 – ol ; 3 – Methylbutan – 2 – ol ;

Pentan – 3 – ol

Tertiary alcohol: 2 – methylbutan – 2 – ol

4. Explain why propanol has higher boiling point than that of the hydrocarbon, butane.

Answer:

The presence of – OH group makes Propanol undergo intermolecular H-bonding. Butane, while on the other side does not have the same privilege

Hence, additional energy would be required to break the intermolecular hydrogen bonds.

This is the reason why hydrocarbon butane has a lower boiling point than propanol.

5. Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact.

Answer:

Due to the presence of – OH group, alcohols form H – bonds with water.

As a result, alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses.

6. What is meant by the hydroboration-oxidation reaction? Illustrate it with an example.

Answer:

The hydroboration – oxidation reaction is the reaction where borane is added in order for the oxidation to take place. For example, propan – 1 – ol is formed by making propene undergo the hydroboration – oxidation reaction. In the above reaction, the reaction between propene and diborane ( BH3)2 takes place in order to form trialkyl borane which acts an additional product. This additional product is oxidized to alcohol by hydrogen peroxide in the presence of aqueous sodium hydroxide.

7. Give the structures and IUPAC names of monohydric phenols of molecular formula, C7H8O.

Answer:

8. While separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which will be steam volatile. Give reason.

Answer:

Intramolecular H – bonding is present in o – nitrophenol & p – nitrophenol. In p – nitrophenol, the molecules are strongly associated due to the existence of intermolecular bonding.

Therefore, o – nitrophenol is steam volatile.

9. Give the equations of reactions for the preparation of phenol from cumene.

Answer:

To synthesize phenol, cumene is initially oxidized in the presence of air of cumene hydroperoxide.

Followed by, treating cumene hydroxide with dilute acid to prepare phenol & acetone as byproducts.

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