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NCERT Solutions for Class 12 Chemistry Chapter 16 is the best study material for all the students. It not only prepares them for the board exams but also strengthens their concepts for competitive exams such as NEET and JEE.

The solutions are provided by subject mater experts and are accurate. The diagrammatic representations make it even easier for the students to understand. The students appearing for UP board, MP board, Gujarat board, CBSE find the NCERT Solutions beneficial while preparing for the exams.

NCERT Solutions for Class 12 Chemistry Chapter 16 Chemistry in Every Day Life

This chapter deals with the principles of chemistry in everyday life. All the products such as soaps, detergents have an organic composition. All our daily activities are controlled by chemicals. This chapter gives some interesting facts about the products we use in our daily lives and how are they controlled by the chemicals.

The students are advised to go through the NCERT solutions for Class 12 Chemstry for better understanding of the concepts provided in the chapter.

NCERT INTEXT QUESTIONS

Question 1:
Sleeping pills are recommended by doctors to the patients suffering from sleeplessness but it is not advisable to take their doses without consultation with the doctor. Why?
Answer:
Sleeping pills contain drugs which may be tranquilizers or antidepressants. They affect the nervous system and induce sleep. However, if these doses are not properly controlled, they may create havoc. They even adversely affect the vital organs of the body. It is advisable to take these sleeping pills under the strict supervision of a doctor.

Question 2.
With refrence to which classification has the statement “ranitidine is an antacid”, been given?
Answer:
This statement refers to the classification of drugs according to pharmacological effect because any drug which will be used to neutralise the excess acid present in the stomach will be called an antacid.

Question 3.
Why do we require artificial sweetening agents?
Answer:
The commonly used sweetening agent i.e., sucrose is a carbohydrate with molecular formula C₁₂H₂₂O₁₁. Since it has high calorific value, it is not recommended to the patients, diabetics in particular which require low calorie diet. Most of the artificial sweeteners are better than sucrose but hardly provide any calories to the body. These are being used as substitutes of sugar.

Question 4.
Write chemical equations for preparing sodium soap from glyceryl oleate and glyceryl palmitate. Structural formulas of these compounds are given:
(i) (C₁₅H₃₁COO)₃C₃H₅ (Glyceryl palmitate)
(ii) (C₁₇H₃₃COO)₃ C₃H₅ (Glyceryl oleate)
Answer:

Question 5.
Label the hydrophilic and hydrophobic parts in the following molecule which is a detergent. Also identify the functional group(s) present.

Answer:

NCERT Exercise

Question 1.
Why do we need to classify the drugs in different ways?
Answer:
Drugs have been classified in different ways depending

• upon their pharmacological effect
• upon their action on a particular biochemical process
• on the basis of their chemical structure
• on the basis of molecular targets.

For example, the classification of the drugs based on pharmacological effect is useful for doctors. The classification of drugs based on molecular targets is the most useful classification for medicinal chemists. Thus, drugs are classified in different ways to serve different purposes.

Question 2.
Explain the term, target molecules or drug targets as used in medicinal chemistry.
Answer:
In medicinal chemistry, drug targets refer to the key molecules involve in certain metabolic pathways that result in specific diseases Carbohydrates, proteins, lipids, and nucleic acids are examples of drug targets.
Drugs are chemical agents designed to inhibit these target molecules by binding with the active sites of the key molecules.

Question 3.
Name the macromolecules that are chosen as drug targets.
Answer:
The different macromolecules or biomolecules which are used as drug targets are carbohydrates, proteins, enzymes, nucleic acids. Out of these, enzymes are the most significant because their deficiency leads to many disorders in the body.

Question 4.
Why should not medicines be taken without consulting doctors?
Answer:
A medicine can bind to more than one receptor site, thus a medicine may be toxic for some receptor site Further, in most cases, medicines cause harmful effects when taken in a higher dose than recommended As a result, medicines may be poisonous in such cases Hence, medicines should not be taken without consulting doctors.

Question 5.
Define the term chemotherapy. (C.B.S.E. Delhi 2008)
Answer:
The branch of chemistry which deals with the treatment of diseases using chemicals is called chemotherapy.

Question 6.
Which forces are involved in holding the drugs to the active sites of enzymes?
Answer:
These are different intermolecular forces like dipolar forces, hydrogen bonding, van der Waals’ forces etc. The receptor targets have specific roles to perform. They help in transferring message from messengers to the cell. The messengers are in fact chemical compounds which are received by the active sites of the receptor proteins that project out of the surface. In order to accommodate these, the receptors may undergo a change in shape. The receptors are held by the active sites also called binding sites. Once the message is transferred to the cells.

Question 7.
While antacids and antiallergic drugs interfere with the function of histamines, why do these not interfere with the function of each other?
Answer:
They do not interfere with the functioning of each other because they work on different receptors in the body. Secretion of histamine causes allergy and acidity while antacid removes only acidity.

Question 8.
Low level of noradrenaline is the cause of depression. What type of drugs are needed to cure this problem ? Name two drugs (C.B.S.E. Outside Delhi 2008 Supp.)
Answer:
Low level of noradrenaline which acts as a neurotransmitter reduces the signal sending ability to the nerves and the patient suffers from depression. Antidepressants are needed to give relief from depression. These are also called tranquilizers or neurologically active drugs. The two specific drugs are iproniazid and phenelzine.

Question 9.
What is meant by the term ‘broad-spectrum antibiotic? Explain.
Answer:
Broad-spectrum antibiotics are the drugs which are effective against a large number of harmful micro-organisms causing diseases.
Chloramphenicol It is a broad-spectrum antibiotic, isolated in 1947. It is rapidly absorbed from the gastrointestinal tract and hence can be given orally. It is very effective against typhoid, dysentery, acute fever, certain form of urinary infections, meningitis and pneumonia.

Chloramphenicol is quite easy to synthesise. Therefore, most of the chloramphenicol available in the market is synthetic.

Question 10.
How do antiseptics differ from disinfectants? Give one example of each.
Answer:

Antiseptics and disinfectants are effective against micro-organisms However antiseptics are applied to the living tissues such as wounds, cuts, ulcers, and diseased skin surfaces, while disinfectants are applied to inanimate objects such as floors, drainage system, instruments, etc Disinfectants are harmful to the living tissues.

Iodine is an example of a strong antiseptic Tincture of iodine (2-3 percent of the solution of iodine in the alcohol-water mixture) is applied to wounds 1 percent solution of phenol is used as a disinfectant.

Question 11.
Why are cimetidine and ranitidine better antacids than sodium bicarbonate or magnesium or aluminium hydroxides?
Answer:
Both sodium bicarbonate and hydroxides of magnesium or aluminium are very good antacids since they neutralise the acidity in the stomach. But their prolonged use can cause the secretion of excessive acid in the stomach. This may be quite harmful and may lead to the formation of ulcers. Both cimetidine and ranitidine are better salts without any side effects.

Question 12.
Name a substance which can be used as an antiseptic as well as a disinfectant.
Answer:
Phenol can be used as an antiseptic as well as a disinfectant 0.2 percent solution of phenol is used as an antiseptic, while 1 percent of its solution is used as a disinfectant.

Question 13.
What are the main constituents of Dettol?
Answer:
The main constituents of antiseptic Dettol are chloroxylenol and terpenol.

Question 14.
What is the tincture of iodine? What is its use?
Answer:
Tincture of iodine is a 2-3 percent solution of iodine in an alcohol-water mixture It is applied to wounds as an antiseptic.

Question 15.
What are food preservatives?
Answer:
Chemical substances which are used to protect food against bacteria, yeasts and moulds are called preservatives. For example, sodium benzoate and sodium metabisulphite.

Question 16.
Why is the use of aspartame restricted to cold foods and drinks?
Answer:
Aspartame becomes unstable at cooking temperature This is the reason why its use is limited to cold foods and drinks.

Question 17.
What are artificial sweetening agents? Give two examples.
Answer:
Carbohydrates in the form of sugar (sucrose) are the traditional sweeteners and are the essential constituents of our diet. In the present lifestyle, people lack physical activities and exercise and it becomes rather difficult to burn the extra calories that are produced by the carbohydrates. Chemists have provided certain chemicals known as artificial sweeteners which provide the desired sweet taste io the food articles but hardly affect the calorie intake by the body. The most popular among the artificial sweeteners is saccharin which is nearly 550 times more sweet than the cane sugar. It is a boon for diabetic patients who don’t want to take carbohydrates (sugar) which is likely to increase the calories. It is in fact, a life saviour for these patients and is in the form of sodium or calcium salt which is water-soluble. These days, a number of other sweeteners are also available, e.g.. Aspartame, Alitame, Sucrolose etc.

(a) Aspartame: It is a very successful and commonly used artificial sweetener. As stated above, it is nearly 100 times as
sweet as cane sugar. However, it can be used in soft drinks and cold foods only since it decomposes upon heating. Chemically
aspartame is the methyl ester of dipeptide formed by the action of aspartic acid with phenylalanine.
(b) Sucrolose: The artificial sweetener as the name suggests is a trichioroderivative of sucrose. It is better, than
aspartame in the sense that it can be used in hot food at the cooking temperature, since it does not decompose on heating.
Moreover, it does not provide calories.

Question 18.
Name the sweetening agent used in the preparation of sweets for a diabetic patient.
Answer:
Artificial sweetening agents such as saccharin, alitame, and aspartame can be used in preparing sweets for diabetic patients.

Question 19.
What problem arises by using alitame as an artificial sweetener?
Answer:
Alitame is no doubt, a very potent sweetener. Its sweetening capacity is more than 2000 times as compared to ordinary cane sugar or sucrose. But sometimes, it becomes quite difficult to control the sweetness level in the food which is actually desired.

Question 20.
Why are detergents called soapless soaps?
Answer:
Soaps work in soft water, they are not effective in hard water In contrast synthetic detergents work both in soft water and hard water Therefore, synthetic detergents are better than soaps.

Question 21.
Explain the following terms with suitable examples.
(a) Cationic detergents
(b) Anionic detergents
(c) Neutral detergents.
Answer:
1. Anionic Detergents: These detergents contain anionic hydrophilic groups. These are generally made from long-chain alcohols which are reacted with concentrated sulphuric acid to form alkyl hydrogen sulphates. These are then neutralised with alkali to give water-soluble salts.
A few examples are listed below :
Sodium Alkyl Sulphates: These are the sodium salts of sulphonic acid esters of long-chain aliphatic alcohols which normally contain 10 to 15 carbon atoms. The alcohols are formed from fats and oils as a result of hydrogenolysis.

Sodium Alkyl Benzene Suiphonates: A common detergent belonging to this class is Sod – p – dodecyl benzene sulphonate. It is obtained from benzene by reacting with dodecyl chloride in the presence of anhydrous AlCl₃ acting as a catalyst.
The different steps involved are as follows :

2. Cationic Detergents: In these detergents, the hydrophilic group is of cationic nature. These are generally acetates, chlorides or bromides of quaternary ammonium CH CH CH B salts. The cationic part enclosed in the bracket contains a long hydrocarbon chain. These  detergents have germicidal qualities and are quite expensive as well, The cationic CH₃ detergents are present in hair conditioners. Cetyl trimethyl ammonium bromide

3. Non-ionic or neutral Detergent: These detergents are simply long-chain organic compounds and are esters in nature. For example, stearic acid and polyethylene glycol react to form a non-ionic detergent.

Non-ionic detergents contain polar groups and form hydrogen bonds with water. Some dishwashing liquids contain non-ionic detergents.
The field of detergents is very vast because of their immense utility. Companies engaged in their manufacture are spending huge amounts of money to bring products of better quality.

Question 22.
What are biodegradable and non-biodegradable detergents? Give an example of each. (C.B.S.E. Delhi 2008, 2009)
Answer:
Bio-degradable detergents are degraded by bacteria. In them, the hydrocarbon chain is unbranded. They do not cause water pollution and are bitter. Example: Sodium lauryl sulphate.

Non-biodegradable detergents possess highly branched hydrocarbon chain so bacteria cannot degrade them easily. They cause water pollution. Example: Sodium 4-(l, 3, 5, 7-tetramethyl-actyl) benzene sulphonate.

Question 23.
Why do soaps not work in hard water?
Answer:
Hard water contains calcium and magnesium salts. Therefore, in hard water soaps get precipitated as calcium and magnesium soaps which being insoluble stick to the clothes as gummy mass.

Question 24.
Can you use soaps and synthetic detergents to check the hardness of water?
Answer:
Soaps can be used to check hardness of water as they will form insoluble precipitates of calcium and magnesium salts on reacting with hard water. Since detergents do not form any precipitate, they cannot check hardness of water.

Question 25.
Explain the cleansing action of soaps.
Answer:
In order to understand the cleansing action of soaps let us try to analyse how the clothes become dirty. They first become oily because of the perspiration coming out of the skin and also from the organic matter dispersed in the atmosphere. Dust particles stick to oil drops and the clothes become dirty. In order to wash these, they are dipped in water and soap is applied.

In solution, it dissociates to give carboxylate ions (RCOO^–) and the cations (Na⁺). The alkyl portion which contains a long chain of hydrocarbons is a tail pointing towards the oil drops while the COO portion is the head directed towards water. This is quite evident from the figure where the solid circles (.) represent the polar groups and the wavy lines represent the alkyl portions. This formation is known as micelle and helps in forming a stable emulsion of oil and water by acting as a bridge between the two. The oil droplets along with the particles of the dirt get detached from the fibres of the clothes and pass into the emulsion. In this manner, the clothes become free from dust or dirt. The cleansing action of the soap is depicted in the Fig. 5.15.

Question 26.
If the water contains dissolved calcium bicarbonate, out of soaps and synthetic detergents, which one will you use for cleaning clothes?
Answer:
Synthetic detergents are preferred for cleaning clothes When soaps are dissolved in water containing calcium ions, these ions form insoluble salts that are of no further use, however when synthetic detergents are dissolved in water containing calcium ions, these ions form soluble salts that act as cleaning agents.

Question 27.
Label the hydrophilic and hydrophobic parts in the following compounds.
(a) CH₃(CH₂)₁₀CH₂OSO₃Na⁺
(b) CH₃(CH₂)₁₅-N⁺(CH₃)₃Br^–
(c) CH₃(CH₂)₁₆-COO(CH₂CH₂O)_nCH₂CH₂OH
Answer:

We hope the NCERT Solutions for Class 12 Chemistry Chapter 16 Chemistry in Every Day Life help you. If you have any query regarding NCERT Solutions for Class 12 Chemistry Chapter 16 Chemistry in Every Day Life, drop a comment below and we will get back to you at the earliest.

NCERT Class 12 Chemistry Solutions for Chapter 11 provides an insight into the various concepts related to alcohols, phenols and ethers. This is an important chapter and hence requires an indepth knowledge of the topics. The subject experts have provided accurate explanations and step wise solutions for the questions provided in the textbook. This will help the students prepare well during the exams.

NCERT Solutions not only help the students appearing for UP board, MP board, CBSE, Maharashtra board, Gujarat board, etc. but also prepares them for competitive exams such as JEE and NEET. The students should refer to the NCERT Solutions to score well in the examinations.

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ehers

Alcohols, phenols, and ethers is an important chapter from the examination perspective. The chapter explains the structure, properties, and applications of alcohols, phenols, and ethers. It also explains the correlation and differences between the three.

NCERT Solutions for Class 12 Chemistry Chapter 11 gives a better idea of the related concepts. The students can refer to these for better practice.

NCERT INTEXT QUESTIONS

Question 1.
Classify the following into primary, secondary, and tertiary alcohols

Answer:
(i) Primary alcohol
(ii) Primary alcohol
(iii) Primary alcohol
(iv) Secondary alcohol
(v) Secondary alcohol
(vi) Tertiary alcohol

Question 2.
Identify allylic alcohol in the above examples.
Answer:
allylic alcohols:

Question 3.
Give the IUPAC names of the following compounds : (C.B.S.E 2008)

Answer:

Question 4.
Show how the following alcohols can be prepared by the action of suitable Grignard reagent on methanal ?

Answer:
(i) The structure of alcohol suggests that the Grignard reagent that reacts with methanal is isopropyl magnesium halide.

(ii) The structure of alcohol suggests that the Grignard reagent that reacts with methanal is cyclohexyl magnesium halide.

Question 5.
Write the structures of the products of the following reactions :

Answer:
(i) The acidic hydration of propene gives propan-2-ol (isopropyl alcohol)

(ii) NaBH₄ is a weak reducing agent. It brings about the reduction of the ketonic group present in cyclohexane ring to secondary alcoholic group. However, it does not affect ester group.

(iii) NaBH₄ reduces aldehydic group to a primary alcoholic group.

Question 6.
Give structures of the products you would expect when each of the following alcohols reacts with (a) HCl/ZnCl₂ (b) HBr (c) SOCl₂ :
(i) Butan-l-ol
(ii) 2-Methylbutan-2-ol
Answer:
(i) Reactions of Butan-1-ol

The reaction of butan-l-ol (primary alcohol) can take place only upon heating. At room temperature, the reaction does not occur.

(ii) Reactions of 2-Methylbutan-2-ol

The alcohol being tertiary in nature reacts immediately with Lucas Reagent at room temperature.

Question 7.
Predict the major product of add catalysed dehydration of :
(i) 1-Methylcyclohexanol
(ii) Butan-1-ol
Answer:
(i) Acid catalysed dehydration of 1-methylcyclohexanol can give two products. However, 1-methylcyclohexene will be  preferably formed according to Satyzeff s rule since it is more substituted.

(ii) Butan-1-o1 upon acid dehydration will give but-2-ene as the major product along with but-l-ene as the minor product. Actually, primary carbocation formed can either lose a H⁺ to form but-1-ene or may undergo rearrangement and shift to secondary carbocation, which is more stable. The latter then forms but-2-ene by losing a H⁺.

Question 8.
Ortho and para nitrophenols are more acidic than phenol. Draw the resonating structures of the corresponding phenoxide ions.
Answer:
We know that the nitrophenols are more acidic than phenol. Their acidic strength can be compared in terms of the relative stabilities of the corresponding phenoxide ions based on resonance. For example,
(i) Phenoxide ion :


(ii) p-Nitrophenoxide ion :


(iii) p-Nitrophenoxide ion :

In case of nitrosubstituted phenoxides, the contributing structures that are enclosed in boxes have negative charge on the carbon atom to which die electron withdrawing nitro group is attached. They therefore, contribute more towards the acidic character than the rest of the contributing structures. Consequently, both ortho and para nitrophenol are stronger acids than phenol.

Question 9.
Write the equations involved in the following reactions :
(i) Reimer Tiemann Reaction
(ii) Kolbe’s Reaction.
Answer:
(i) In this reaction phenol is heated with chloroform alongwith aqueous NaOH (or KOH) to about 340 K. This is followed by acidification with dilute HCl when 2 – hydroxyhenzaldehyde (salicylaldehyde) is formed as the major product.

A small amount of para isomer is also formed in the reaction. In case, chLoroform is replaced by carbon tetrachioride, then 2—hydroxybenzoic acid (salicylic acid) is formed as the main product.

(ii) In this reaction, CO₂ gas is passed through sodium phenate at 400 K under a pressure of 4 to 7 atmospheres. This is followed by acidification with dilute HCI when salicylic acid is formed. This method is commonly used for the commercial preparation of saucy lic acid.

Question 10.
Write the reactions of Williamson’s synthesis of 2-ethoxy-3-methoxypentane starting from ethanol and 3-methylpentan- 2-ol.
Answer:
In the Williamson’s synthesis, the reactants are alkyl halide and sodium salt of an alcohol. In order to avoid the formation of alkene during the reaction, the alkyl halide should be primary while sodium salt must be of branched chain alcohol. In the present case, alkyl halide must be derived from ethanol upon heating with halogen acid (e.g HBr).

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

Answer:
The second set of reactants is more appropriate to give the products i.e., l-methoxy-4-nitrobenzene. In the first set, cleavage of C – Br bond is involved. It is rather difficult since the carbon atom is sp² hybridised and the bond has partial double bond character as well. The product is formed as a result of Williamson’s synthesis.

Question 12.
Predict the products of the following reactions :

Answer:
(i) The reaction involves the cleavage of C – 0 bond. The Br atom of HBr is to combine with the smaller alkyl group to give the following products.

For more details and explanation, consult section 12.24.
(ii) This reaction also proceeds in the same manner. The Br atom of HBr is expected to combine with ethyl group (smaller in size) and not with phenyl group (bigger in size).

(iii) Nitrating mixture brings about the nitration of benzene ring. The ethoxy group (OC₂H₅) is an activating group and increases the electron density at the ortho and para positions due to +M effect. As a result, a mixture of o-nitro and p-nitro derivatives is formed. Out of these, the p-isomer is in excess since there is less steric hindrance due to OC₂H₅ group at the para position than at the ortho position in the ring.

(iv) In this reaction, the ether is initially protonated by H⁺ ion of the acid HI. to accomodate I^– ion (nucleophile). The reaction follows S_n1 mechanism.

NCERT EXERCISE

Question 1.
Write IUPAC names of the following compounds :

Answer:
(i) 2, 2, 4 – Trimethylpentan-3-ol
(ii) 5 – Ethylheptan-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 – Dimethyiphenol
(ix) 1 -Methoxy – 2 – methy lpropane
(x) Ethoxybenzene
(xi) 1 – Phenoxyheptane
(xii) 2 – Ethoxybutane.

Question 2.
Write the structure of the compounds, whose IUPAC names are as follows
(i) 2-Methylbutan – 2 – ol
(ii) I-Phenylpropan – 2 – ol
(iii) 3-Phenylhexane – l, 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 –  – ol
(x) 3 – Chloromethvlpentan – l – ol
Answer:

Question 3.
(i) Draw the structures of all isomeric alcohols of molecular formula C₅HI₂0 and give their IUPAC names.
(ii) Classify the isomers of alcohols in question 11.3 (i)as primary, secondary, and tertiary alcohols.
Answer:
Eight isomers are possible. These are:

Question 4.
Explain why is propanol higher boiling than butane?
Answer:
Propanol (Propan-l-ol) and butane are of comparable molecular masses 60m and 58u respectively but the boiling point of propanol is higher (391 K) because of the presence of intermolecular hydrogen bonding in the molecules. However, it is not present in butane due to the absence of polar OH group. The only attractive forces are weak van der Waals forces. Therefore, the boiling point of propanol is more than that of butane (309 K).

Question 5:
Explain why are alcohols more soluble in water than the corresponding hydrocarbons?
Answer:
The solubility of alcohols in water may be attributed to two factors :
(i) Both of them are of polar nature.
(ii) Molecules of both of them are involved in the intermolecular hydrogen bonding.
However, hydrocarbons are non-polar and are also not involved in any hydrogen bonding with alcohols. Alcohols readily dissolve in a water while the hydrocarbons are almost insoluble.

Question 6.
What is meant by hydroboration oxidation reaction ? Illustrate with an example.
Answer:
By hydroboration- oxidation of alkenes. Indirect hydration of alkenes can also bedone by hvdroboration-oxidation which is completed in two steps. In the first step. alkene reacts ith diborane (B’1{6) as boron hydride (BH3) dissolved in tetrahydrofuran (THF) to form an alkyl horane. In fact. the boron atom along i th the hydrogen gets attached to the double
bonded carbon atom with more number of hydrogen atoms less sterically hindered side). One hydrogen is then transferred to the other carbon atom. In this manner, all the three hydrogen atoms of boron are transferred to alkene molecule to form
trialkyl borane as the product. In the next step. the alkyl borane is oxidised by alkaline 11202 to form an alcohol. The indirect hydration proceeds according to Antimarkownikov s rule. For example.

Question 7.
Give the structures and IUPAC names of the phenols of molecular formula C₇H₈O.
Answer:

Question 8.
In separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which is steam volatile? Give reason.
Answer:
o-nitrophenol is steam volatile while p-nitrophenol is not. This is on account of intramolecular hydrogen bonding in the molecules of o-nitrophenol. As a result, its boiling point is less than that of p-nitrophenol in which the molecules are linked by intermolecular hydrogen bonding.
It is interesting to note that in the substituted phenols, the nature and position of the substituent influences the boiling point of phenol.
For example .o-nitrophenol is steam volatile while p-nitrophenol is not. This is supported by the fact that the boiling point temperature of o-nitrophenol (100°C) is less than that of p-nitrophenol, (279°C). In o-nitrophenol, there is intramolecular hydrogen bonding in OH and NO₂ groups placed in a adjacent positions. However, these are linked by intermolecular hydrogen bonding in the p-isomers. It is quite obvious that extra energy is needed to cleave the hydrogen bonds in the p-isomer. Consequently, its boiling point is more.

o-nitrophenol with lower boiling point is steam volatile while p-nitrophenol is not. This helps in the separation of the two isomers present in the liquid mixture. On passing steam, o-nitrophenol volatilises, and its vapours rise alongwith steam and after condensation, collect in the receiver p-nitrophenol is left behind in the distillation flask. e-nkrophenol p-nnrophenol.

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

Question 10.
Write chemical equations involved in the preparation of phenol from chlorobenzene.
Answer:
From chlorobenzene, phenol is prepared by Dow’s process.In this method, chlorobenzene is heated with 6 to 8% solution of NaOH to about 623 K under a pressure of 300 atmospheres to form sodium phenate which upon acidification with dilute HCI gives phenol as follows:

Question 11.
Write the mechanism of hydration of ethene to yield ethanol.
Answer:
Ethene does not react with water as such. Water being little polar, is not in a position to provide H⁺ ion for initial electrophilic attack on ethene. The reaction is carried in the presence of H₂S0₄. The acid provides proton (H⁺) for the initial electrophilic attack.

In the second step, H₂0 attacks the carbocation in preference to HS0₄ ion as a nucleophile

Question 12.
You are given benzene, cone. H₂S0₄ and NaOH. Write equations for the preparation of phenol using these reagents.
Answer:

Question 13.
Show how well you synthesise:
(i) 1-phenyl ethanol from a suitable alkene
(ii) Cyclohexylmethanol using an alkyl halide by S_N² reaction
(iii) Pentan-1-ol using a suitable alkyl halide.
Answer:

Question 14.
Give two reactions to show the acidic nature of phenol. Compare the acidity of phenol with that of ethanol.
Answer:
Acidic nature of phenols. Phenols are weakly acidic in nature. The liquid form of phenol containing about 5 percent water is known as carbolic acid. The dissociation constant (K_a) for phenol is 10^-10 at 298 K (room temperature). The corresponding pK_a* value is 10.0. The acidic character is exhibited by the following properties:
(i) Reaction with active metals. Phenols evolve hydrogen with active metals such as sodium and potassium.

(ii) Reaction with alkalies. Phenols neutralise caustic alkalies such as NaOH or KOH to form salt and water.

In addition to these, phenols turn blue litmus red which is the characteristic property or acids. However, phenols do not
react with either alkali metal carbonates or bicarbonates since these are quite weak acids.

Question 15.
Explain why is ortho-nitrophenol more acidic than ortho-methoxy phenol?
Answer:
Due to strong -R and – I-effect of the -NO₂ group, electron density of the O – H bond decreases and hence the loss of a proton becomes easy.

Question 16.
Explain how does -OH group attached to a carbon atom of benzene ring activates it towards electrophilic substitution ?
Answer:
The OH group exerts a +M (or + R) effect on the ring under the influence of attacking electrophile.

As a result, there is an increase in the electron density in the ring particularly at the ortho and para positions. The electrophilic substitution readily takes place at these positions when electrophile attacks.

Question 17.
Give equations for the following chemical reactions :
(i) Oxidation of propan-1-ol with alkaline KMnO₄
(ii) Reaction of bromine in CS₂ with phenol
(iii) Action of dilute HNO₃ on phenol
(iv) Treating phenol with chloroform in the presence of aqueous NaOH at 343 K.
Answer:

Question 18.
Write short notes on (i) Kolbe reaction (ii) Reimer-Tiemann reaction.
Answer:
(i) In this reaction phenol is heated with chloroform alongwith aqueous NaOH (or KOH) to about 340 K. This is followed by acidification with dilute HCl when 2 – hydroxyhenzaldehyde (salicylaldehyde) is formed as the major product.

A small amount of para isomer is also formed in the reaction. In case, chLoroform is replaced by carbon tetrachloride, then 2—hydroxybenzoic acid (salicylic acid) is formed as the main product.

(ii) In this reaction, CO₂ gas is passed through sodium phenate at 400 K under a pressure of 4 to 7 atmospheres. This is followed by acidification with dilute HCl when salicylic acid is formed. This method is commonly used for the commercial preparation of salicylic acid.

Question 19.
Write a mechanism for the acid dehydration of ethanol to ethene.
Answer:
Mechanism of dehydration. The mechanism is illustrated with ethanol which is a primary alcohol.

The relative ease of dehydration of different alcohols i.e. primary, secondary and tertiary can be further justified on the basis of the relative stabilities of the carbocations formed in the slow step. Since tertiary carbocation is maximum stable while primary is the least, the tertiary alcohols are maximum reactive while the primary are the least reactive in nature.
In other words, greater the stability of carbocation formed, more is the reactivity of the alcohol.

Question 20.
How are the following conversions carried out?
(i) Propene → Propan -2-ol
(ii) Benzyl chloride → Benzyl alcohol
(iii) Ethyl magnesium chloride → Propan-1-ol
(iv) Methyl magnesium bromide → 2-Methylpropan-2-ol
Answer:

Question 21.
Name the reagents used in the following reactions :
(i) Oxidation of primary alcohol to carboxylic acid
(ii) Oxidation of primary alcohol to an aldehyde
(iii) Bromination of phenol to 2, 4, 6-tribromophenol
(iv) Benzyl alcohol to benzoic acid
(v) Dehydration of propan-2-ol to propene
(vi) Butan-2-one to butan-2-ol.
Answer:
(i) Acidified K₂Cr₂O₇
(ii) Pyridine chlorochromate (PCC)
(iii) Bromine water (Br₂/H₂O)
(iv) Alkaline KMnO₄
(v) 60% H₂S0₄ at 373 K
(vi) LiAlH₄ or NaBH₄.

Question 22.
Give a reason for the higher boiling point of ethanol in comparison to methoxymethane.
Answer:
Ethers are isomeric with alcohols but their boiling points are comparatively low due to the lack of hydrogen bonding. For example, boiling points of isomeric n – butyl alcohol (nC₄H₉OH) and diethyl ether (C₂H₅ – O – C₂H₅) are, 390 K and 308 K respectively.

Question 23:
Give the IUPAC names of the following ethers :


Answer:
(i) 1-Methoxy-2-methylpropane
(ii) 1-Chloro-2-methoxy ethane
(iii) 4-Nitroanisole
(iv) 1-Methoxypropane
(v) 4-Ethoxy-1, 1-dimethyl cyclohexane
(vi) Ethoxybenzene

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:

Question 25.
Illustrate with examples the limitations of Williamson’s synthesis for the preparation of certain types of ethers.
Answer:
Preparation from Alkyl Halides
From alkyl halides, ethers can be prepared by the following methods
By Williamson’s synthesis. It is the best method for the laboratory preparation of both simple and mixed ethers and involves the action of sodium alkoxide (formed by reaction between alcohol and sodium metal) on a suitable alkyl halide.

Limitations of the reaction. In the preparation of unsymmetrical ethers, a proper choice of the reactants is necessary.
Elimination leading to alkene can take place since alkoxide ion can also abstract one of the 3—hydrogen atom alongwith acting as a nucleophile. Thus, in order ro prepare ethyl tertiary butyl ether, we must use ethyl halide (primary) and sodium tertiary
butoxide.

In case, the alkyl halide is tertiary and sodium ethoxide is employed, then C2H5O ion will cause the elimination of alkyl halide to form an alkene as the main product.

Since secondary and tertiary alkyl halides prefer to undergo elimination rather substitution, symmetrical ethers containing secondary and tertiary alkyl groups are obtained only in poor yields by Willamson’s synthesis. For example,

This method is also not successful for preparing aryl alkyl ethers by reacting sod. alkoxide with aryl halides because the cleavage of C – X bond is not so easy due to partial double bond character. In such cases, we must react sodium phenoxide with
alkyl halide as follows:

However, diaryl ethers (both the groups are aryl or phenyl groups) cannot be prepared since aryl halides fail to participate in the nucleophilic substitution reactions.

Question 26.
How is 1-propoxypropane synthesised from propan-1-ol? Write the mechanism of this reaction.
Answer:
(a) Williamson’s synthesis

Question 27.
Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give reason.
Answer:

Question 28.
Write the equation for the reaction of HI with :
(i) 1-Propoxypropane
(ii) Methoxybenzene
(iii) Benzyl ethyl ether.
Answer:

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:
In aryl alkyl ethers, the +R-effect of the alkoxy (OR) group increases the electron density in the benzene ring, thereby activating the benzene ring towards electrophilic substitution reaction.

Since the electron density increases more at the two ortho and one para position as compared to meta position therefore, electrophilic substitution reactions mainly occur at o-and -positions.

Question 30.
Write the mechanism of the reaction of Hl with methoxymethane.
Answer:

Question 31.
Write equations for the following reactions :
(i) Friedel Crafts reaction-alkylation in anisole.
(ii) Nitration of anisole
(iii) Bromination of anisole in ethanolic medium
(iv) Friedel Crafts acetylation of anisole.
Answer:
(i) The halogenation of the benzene ring occurs at the ortho and para positions. However, the para isomer is
formed in excess. For example, the bromination of anisole in ethanoic acid gives nearly 90 percent p-bromoanisole.

(ii)  The nitration of anisole carried with a nitrating mixture of conc. UNO3 and conc. H2SO4 upon heating gives a
a mixture of ortho and para nitro derivatives.

(iii) Sulphonation: Anisole upon sulphonation gives a mixture of isomeric sulphonic acid derivatives.

(iv) Friedel Craft’s reaction: Both alkylation and acylation are carried in the presence of anhydrous Aid3 catalyst which
behaves as a Lewis acid.

Question 32.
Show how would you synthesise the following alcohols from appropriate alkanes?

Answer:

Question 33.
When 3-methylbutan-2-ol is treated with HBr, the following reaction takes place :

Give the mechanism of the reaction.
Answer:
The mechanism is explained as follows :

We hope the NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ehers help you. If you have any query regarding NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ehers, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 12 Chemistry Chapter 15 provides answers to the questions provided in the textbook. The answers are accompanied by diagrammatic representations for better understanding. Also, the solutions are explained in such a language that the students find easy to understand.

NCERT Solutions are beneficial for the students appearing in UP board, MP board, CBSE, Gujarat board, etc. Also the students appearing for competitive exams such as NEET and JEE will find the NCERT Solutions for Class 12 Chemistry beneficial.

NCERT Solutions for Class 12 Chemistry Chapter 15 Polymers

Polymers are very large molecules having high molecular mass. These are also known as macromolecules. This chapter explains the mechanical properties and applications of polymers. The classification of polymers is also described in detail.

The chapter gives an overview of different polymerization reactions. The chapter explains that the polymers are the backbone of four major industries, plastic, elastomers, fibres and paints.

NCERT INTEXT QUESTIONS

Question 1.
What are polymers?
Answer:
The word polymer has a Greek origin where poly means many and mers or meros stands for unit or part. Thus, the polymer may be defined as a substance of a high molecular mass (10³ – 10⁷ u) formed by the combination of a large number of simple
molecules called monomers by chemical bonds. The process by which monomers are converted into polymers is called polymerization.

Question 2.
How are polymers classified on the basis of structure?
Answer:
On the basis of structure, polymers are classified as:
(i) Linear polymers in which the monomers are joined together to form long straight chains of polymer molecules. Forex: HDPE, PVC, nylons, etc.
(ii) Branched-chain polymers in which the monomers not only join in a linear fashion but also form branches of different lengths along the main chain. For ex: LDPE, glycogen, etc.
(iii) Cross-linked polymers in which the initially formed linear polymer chains join together to form a 3D network structure. For ex: bakelite, Urea-formaldehyde resin, etc.

Question 3.
Write the names of the monomers of the following polymers:

Answer:
(i) The polymer is Nylon 66. Its monomer units are hexamethylenediamine and adipic acid.
(ii) The polymer is Nylon 6. Its monomer units are of caproilactum.
(iii) The polymer is Teflon. Its monomer units are of tetrafluoroethylene.

Question 4.
Classify the following as addition and condensation polymers:
(i) Terylene
(ii) Bakelite
(iii) Polyvinyl chloride
(iv) Polythene
Answer:
(i) Condensation polymer
(ii) Condensation polymer
(iii) Addition polymer
(iv) Addition polymer

Question 5.
Explain the difference between Buna-N and Buna-S.
Answer:
Both Buna-N and Buna-S are synthetic rubber and are co-polymers in nature. They differ in their constituents.
Buna-N: Constituents are : buta-1, 3-diene and acrylonitrile.
Buna-S: Constituents are : buta-1, 3-diene, and styrene. They condense in the presence of Na.

Buna – S: It is a co—polymer of 1. 3 – butadiene and styrene and is prepared by the polymerisation of these components in the
ratio of 3 : 1 in the presence of sodium.

Buna-N (Nitrile rubber):  h is a co-polymer of buta-1. 3-diene and acrylonitrile. It is formed as follows:

Question 6.
Arrange the following polymers in increasing order of their intermolecular forces :
(i) Nylon-66, Buna-S, Polythene
(ii) Nylon-6, Neoprene, Polyvinyl chloride
Answer:
We have studied the classification of polymers based upon intermolecular forces. These follow the order :
Elastomers < Plastics < Fibres.

The polymers listed may be arranged in increasing order of intermolecular forces as follows :
(i) Buna-S (Elastomer) < Polythene (Plastic) < Nylon-66 (Fibre)
(ii) Neoprene (Elastomer) < Polyvinyl chloride (Plastic) < Nylon-6 (Fibre).

NCERT Exercise

Question 1.
Explain the terms polymers and monomers.
Answer:
Polymers are a high molecular mass of macromolecules composed of repeating structural units derived from monomers. Polymers have a high molecular mass (10³-10⁷u). In a polymer, various monomer units are joined by strong covalent bonds. Polymers can be natural as well as synthetic. Polythene, rubber, and nylon 6,6 are examples of polymers.

Monomers are simple, reactive molecules that combine with each other in large numbers through covalent bonds to give rise to polymers. For example ethene, propene, styrene, vinyl chloride.

Question 2.
What are natural and synthetic polymers? Give two examples of each.
Answer:
Natural polymers:
Natural polymers are high molecular mass macromolecules and are found in nature mainly in plants and animals. For example, protein, nucleic acids, starch, cellulose, etc.

Synthetic polymer:
Synthetic polymers are man-made high molecular mass macromolecules. For example, plastics (Polyethene, P.V.C.), synthetic fibers (Polyesters, Nylon-6,6), synthetic rubber (Neoprene, Buna-S), etc.

Question 3.
Distinguish between the terms homopolymer and copolymer and give one example of each. (C.B.S.E. Outside Delhi 2008, Pb. Board 2008)
Answer:
A polymer in which the monomers are the same is called a homopolymer. For example, polystyrene is made from styrene only.
A polymer in which the monomers are different is known as a copolymer. For example, in Nylon-66 the monomers are adipic acid and hexamethylenediamine.

Question 4.
How do you explain the functionality of a monomer?
Answer:
The functionality of a monomer is the number of binding sites that is/are present in that monomer.
For example, the functionality of monomers such as ethene and propene is one and that of 1, 3-butadiene, and adipic acid is two.

Question 5.
Define the term polymerisation. (C.B.S.E. Delhi 2008)
Answer:
A very widely used polymer is polyethylene which is formed by the polymerisation of ethylene molecules (monomers) by heating under pressure in the presence of oxygen.

It may be noted that all the monomer units in a polymer may or may not be the same, In case, they are the same as in case of polyethylene, then the polymer is called homopolymer. On the other hand, if they happen to be different, then the polymer is known as copolymer or mixed polymer. For example, the monomer units in terylene are ethylene glycol and terephthalic acid. Similarly, hexamethylenediamine and adipic acid are the monomer units of nylon-66. Both are co-polymers or mixed polymers. We shall discuss these at a later stage in the present unit.

Question 6.

Answer:
It is a homopolymer in nature. It has only one type of monomer units i.e., NH₂ – CHR – COOH. These area-amino acids.

Question 7:
In which classes, are the polymers classified on the basis of molecular forces?
Answer:
Polymers are classified into four classes on the basis of molecular forces. These are:
elastomers, fibres, thermoplastic polymers, and thermosetting polymers.

1. Elastomers: In these polymers, the intermolecular forces are the weakest. As a result, they can be readily stretched by applying small stress and regain their original shape when the stress is removed. The elasticity can be further increased by introducing some cross-links in the polymer chains. Natural rubber is the most popular example of elastomers. A few more examples are of: buna-S, buna-N and neoprene.

2. Fibres: Fibres represent a class of polymers which are thread-like and can be woven into fabrics in a number of ways. These are widely used for making clothes, nets, ropes, gauzes etc. Fibres possess high tensile strength because the chains possess strong intermolecular forces such as hydrogen bonding. These forces are also responsible for close packing of the chains. As a result, the fibres are crystalline in nature and have also sharp melting points. A few common polymers belonging to this class are nylon – 66, terylene and polyacrylonitrile etc.

3. Thermoplastics: These are linear polymers and have weak van der Waals forces acting in the various chains and are intermediate of the forces present in the elastomers and in the fibres. When heated, they melt and form a fluid which sets into a hard mass on cooling, Thus, they can be cast into different shapes by using suitable moulds. A few common examples are polyethylene and polystyrene polyvinyls etc. These can be used for making toys, buckets, telephone apparatus, television cabinets etc.

4. Thermosetting plastics: These are normally semifluid substances with low molecular masses. When heated, they become hard and infusible due to the cross-linking between the polymer chains. As a result, they also become three-dimensional in nature. They do not melt when heated. A few common thermosetting polymers are bakelite, melamine-formaldehyde, urea-formaldehyde and polyurethane etc.

Question 8:
How can you distinguish between addition and condensation polymerisation?
Answer:
In addition polymerization: The monomers are unsaturated in nature i.e., they have atleast one double or triple bond in their molecules. They represent the functionality of the monomers which combine with each other at these sites. The addition polymerisation is generally chain growth polymerisation in nature. For example, polythene, polystyrene, PVC, Teflon etc. are all formed as a result of additional polymerisation.

In condensation polymerization: The monomer units have specific functional groups present which represent their functionality. The monomers combine through these functional groups and the polymerisation is step growth polymerisation in nature. For example, terylene, nylon-6, nylon-66, bakelite etc. are all formed as a result of condensation polymerisation.

Question 9:
Explain the term co-polymerization and give two examples.
Answer:
Co-polymerisation is a process in which a mixture of more than one different monomeric units are allowed to polymerise. The copolymer thus formed contains multiple units of each monomer in the chain. The formation of buna-S copolymer in which the monomers are : buta-1, 3-diene, styrene and sodium is an example of co­polymerisation.
Another Example of co-polymer: Buna-S is a co-polymer of 1 : 3 – butadiene and styrene in the presence of sodium metal.

Thus, as a result of addition polymerisation, the double bonds change to single bonds. A large number of monomer units can be linked in this way.
The addition polymerisation is normally a chain reaction in which the chain is built up by the successive addition of the monomer units. It involves a number of intermediates such as free radicals, carbocations or carbanions. Let us illustrate the chain growth polymerisation based upon different mechanisms.

Question 10:
Write the free radical mechanism for the polymerisation of ethene.
Answer:
The polymerisation of ethene to polythene consists of heating or exposing to light a mixture of ethene with a small amount of benzoyl peroxide initiator The process starts with the addition of phenyl free radical formed by the peroxide to the ethene double bond thus regenerating a new and larger free radical. This step is called chain initiating step. As this radical reacts with another molecule of ethene, another bigger sized radical is formed. The repetition of this sequence with new and bigger radicals carries the reaction forward and the step is termed as chain propagating step. Ultimately; at some stage the product radical thus formed reacts with another radical to form the polymerised product. This step is called the chain terminating step. The sequence of steps may be depicted as follows :
Chain initiating steps:

Chain terminating step:
For termination of the long-chain, these radicals can combine in different ways to form polythene. One mode of termination of the chain is shown as under:

Question 11:
Define thermosetting and thermoplastic polymers with two examples of each.
Answer:
Thermoplastics are polymers which can be easily softened repeatedly on heating and hardened on cooling. Therefore, it can be used again and again. For example, polyethylene and polyvinyl chloride. Thermosetting polymers are those which undergo permanent change on heating. They become hard and infusible on heating and cannot be softened again. For example, Bakelite, and Melamine formaldehyde.

Question 12:
Write the monomers used for getting the following polymers :

1. Polyvinyl chloride
2. Teflon
3. Bakelite

Answer:

1. vinyl chloride
2. tetrafluoroethylene
3. phenol and formaldehyde.

Question 13:
Write the name and structure of one of the common initiators used in free radical addition polymerisation.
Answer:

Question 14:
How does the presence of double bonds in rubber molecules influence their structure and reactivity?
Answer:
Chemically natural rubber is polyisoprene in which monomer units are of isoprene i.e., 2-methyl-i, 3-butadiene. It is in fact, 1, 4 polymer in which monomers are linked through CH₂ groups located at 1.4 positions. The residual double bonds are located at C₂ and C₃ positions in the isoprene units. All the double bonds have cis configurations. Thus, natural rubber is cis polyisoprene.

The high elasticity of natural rubber is due to the absence of polar groups and the cis-configurations about double bonds do I not allow the polymer chains to come closer. Therefore, only weak van der Waals’ forces are present. Since the chains are not linear, they can be stretched just like springs and exhibit elastic properties.

Question 15:
Discuss the main purpose of vulcanisation of rubber.
Answer:
The main purpose of vulcanization of rubber is to improve the following draw-back of natural rubber:

• At high temperature (T >335K) natural rubber becomes soft.
• At low temperature (T< 283K) natural rubber becomes brittle.
• Natural rubber is soluble in non-polar solvents.
• It is non-resistant to attack by oxidizing agents.

Question 16:
What are the monomers repeating units of Nylon 6 and Nylon 66?
Answer:
The monomeric repeating unit of nylon 6 is [NH – (CH₂)₅ – CO], which is derived from caprolactam.
The monomeric repeating unit of nylon 6, 6 is [NH – (CH₂)₆ – NH – CO – (CH₂)₄ – CO], which is derived from hexamethylene diamine and adipic acid.

Question 17:
Write the names and structures of the monomers of the following polymers:
(i) Buna-S
(ii) Buna-N
(iii) Dacron
(iv) Neoprene (C.B.S.E. Delhi 2008)
Answer:

Question 18:
Identify the monomers in the following polymeric structures:

Answer:
(i) Decanedioic acid: HOOC(CH₂)₈COOH
Octamethylenediamine: H₂N(CH₂)₈NH₂

(ii)

Question 19:
How is dacron obtained from ethylene glycol and terephthalic acid?
Answer:

Question 20:
What is a biodegradable polymer? Give an example of an aliphatic biodegradable polyester.
Answer:
A polymer that can be decomposed by bacteria is called a biodegradable polymer. Poly – β- hydroxybutyrate – Co -β – hydroxy valerate (PHBV) is a biodegradable aliphatic polyester.

We hope the NCERT Solutions for Class 12 Chemistry Chapter 15 Polymers help you. If you have any query regarding NCERT Solutions for Class 12 Chemistry Chapter 15 Polymers, drop a comment below and we will get back to you at the earliest.

NCERT Class 12 Chemistry Solutions for Chapter 12 provides answers for the questions provided in the textbook. The answers are accurate to the best of our knowledge and are provided by subject experts. The students can refer to these or sure shot success.

The students appearing for various boards and competitive exams can find these solutions helpful for practice. Most of the questions asked in UP board, MP board, Gujarat board, CBSE, etc. are asked from these. To score well in the examinations, the students should go through these solutions atleast once after finishing the entire syllabus.

NCERT Solutions for Class 12 Chemistry Chapter 12 Aldehydes, Ketones and Carboxylic Acids

This chapter explains the structure, physical and chemical properties and applications of aldehydes, ketones and carboxylic acid. It also explains a correlation between the three. It also explains the mechanism of different reactions of aldehydes, ketones and carboxylic acids. The factors affecting the acidity of carboxylic acids and their reactions help in understanding the advanced concepts related to this chapter.

NCERT Solutions for Class 12 Chapter 12 will help you revise the chapter during the examination. It also helps in clearing doubts if any.

NCERT INTEXT QUESTIONS

Question 1:
Write the structures of the following compoimds :
(i) α -Methoxypropionaldehyde
(ii) 3-Hydroxybutanal
(iii) 2-Hydroxycyclopentanecarbaldehyde
(iv) 4-Oxopentanal
(v) Di-sec butylketone
(vi) 4-Fluoroacetophenone
Answer:

Question 2.
Write the structures of products of following reactions:

Answer:

Question 3.
Arrange the following compounds in increasing order of their boiling points :
CH₃CHO,  CH₃CH₂OH,  CH₃OCH₃, CH₃CH₂CH₃
Answer:
The increasing order of boiling points of all these compounds of comparable molecular masses is :

Explanation: We know that the boiling points of liquids are directly related to the magnitude of the intermolecular forces of attraction.

1. Hydrocarbons (alkanes) are completely non-polar. The only attractive forces in their molecules are Van der Waals forces which are quite weak. That is why propane (CH₃CH₂CH₃) has the least boiling point. It is a gas at room temperature.
2. Ethers have bent structures and are also polar. However, there is no hydrogen bonding in their molecules. The only attractive forces are dipolar forces. Therefore, boiling point of dimethyl ether (CH₃OCH₃) is higher than that of propane. However, it is also a gas at room temperature.
3. Aldehydes contain polar carbonyl group and have strong dipolar interactions in their molecules. It is more than in ethers. Therefore, the boiling point of acetaldehyde (CH₃CHO) is more than that of dimethyl ether.
   
4. Out of all the families listed, alcohols have maximum intermolecular forces in the form of hydrogen bonding
   
   As a result, ethyl alcohol (C₂H₅OH) has the maximum boiling point.

Question 4.
Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions
(i) Ehtanal, propanaL, propanone, butanone
(ii) Benzaldehyde,p-Tolualdehyde, p-Nitrobenzaldehyde, acetophenone.
Ans: (i) Butanone < Propanone < Propanal < Ethanal .This is because as the no. of alkyl groups attached to carbonyl carbon increases, +I-effect increases. As a result, e^– density

Question 5.
Predict the products of the following reactions :

Answer:

Question 6.
Give the IUPAC names of the following compounds:

Answer:

Question 7.
Show how each of the following compounds can be converted into benzoic acid?
(i) Ethylbenzene (C.B.S.E. Outside Delhi 2017)   
(ii) Acetophenone  (C.B.S.E. Outside Delhi 2017)
(iii) Bromobenzene
(iv) Phenylethene (Styrene)
Answer:

Question 8.
Which acid of each pair would you expect to be stronger?
(i) CH₃CO₂H or FCH₂CO₂H
(ii) FCH₂CO₂H or ClCH₂CO₂H
(iii) FCH₂CH2CH₂CO₂H or CH₃CH(F)CH₂CO₂H

Answer:

NCERT EXERCISE

 Question 1.
What is meant by the following terms? Give an example in each case.
(a) Cyanohydrin
(b) Semicarbazone
(c) Acetal
(d) Oxime
(e) Cyanohydrin
(f) Ketal
(g) Aldol
(h) Schiff’s base
(i) 2, 4-D.N.P.
(ii) Imine
Answer:

Question 2.
Name the following compounds according to IUPAC system of nomenclature.
(i) CH₃CH(CH₃)CH₂CH₂CHO
(ii) CH₃CH=CHCHO
(iii) CH₃CH(CH₃)CH₂C(CH₃)₂COCH₃
(iv) OHCC₆H₄CHO(-p)
(v) CH₃CH₂COCH(C₂H₅)CH₂CH₂Cl
(vi) CH₃COCH₂COCH₃
(vii) (CH₃)₃CCH₂COOH
(viii) (CH₃)₂CHCH(CH₃)COCl
Answer:
(i) 4-Methylpentanal
(ii) But-2-enal
(iii) 3 3 5-Trimethylhexan-2-one
(iv) Benzene- 1, 4-dicarbaldehyde
(v) 6-Chloro-4-ethylhexan-3-one
(vi) Pentane-2, 4-dione
(vii) 3, 3-Dimethylbutanoic acid
(viii) 2, 3-Dimethylbutanoyl chloride

Question 3.
Draw the structures of the following compounds:
(i) 3-Methylbutanal
(ii) p-Nitropropiophenone
(iii) p-Methylbenzaldehyde
(iv) 4-Methylpent-3-en-2-one
(v) 4-Chloropentan-2-one
(vi) 3-Bromo-4-phenylpentanoic acid
(vii) pp’-Dihydroxybenzophenone
(viii) Hex-2-en-4-ynoic acid
Answer:

Question 4.
Write the IUPAC names of the following aldehydes and ketones. Also give the common names wherever possible.

Answer:

Answer:

Question 7.
Which of the following will undergo Aldol condensation, which Cannizzaro’s reaction, and which neither of these? Write the structures of the expected products in each case
(i) Methanal
(ii) 2-Methylpentanal
(iii) Benzaldehyde
(iv) Benzophenone
(v) Cyclohexanone
(vi) 1-Phenylpropanone
(vii) Phenylacetaldehyde
(viii) Butan-1-ol
(ix) 2, 2-Dimethyl butanal
Answer:
(i) Methanal (HCHO): It will give Cannizzaro’s reaction since the α-hydrogen atom is absent.

(ii) 2-Methylpentanal [CH₃CH₂CH₂CH (CH₃)CHO]: It will give Aldol condensation since the α-hydrogen atom is present.

(iii) Benzaldehyde (CgH₅CHO): It will give Cannizzaro’s reaction since a-hydrogen is not present.

(iv) Benzophenone (C₆H₅COC₆H₅): It will not give any of the two reactions. Being ketone, does not take part in Cannizzaro’s reaction. Without a-hydrogen, it fails to participate in Aldol condensation.

(vi) 1-Phenylpropanone (C₆H₅COCH₂CH₃): It will undergo Aldol condensation since the α-hydrogen atom is present.

(vii) Phenylacetaldehyde (C₆H₅CH₂CHO): It will give Aldol condensation since the α-hydrogen atom is present.

(viii) Butan-1-ol: It will not give any of the reactions.

Question 8.
How will you convert ethanal to the following compounds?
(i) Butane-1, 3-diol
(ii) But-2-enal
(iii) But-2-enoic acid.
Answer:
(i) Ethanal to butane -1, 3-diol

(ii) Ethanal to but-2-enal

(iii) Ethanal into but-2-enoic acid

Question 9.
Write the structural formulae and names of four possible aldol condensation products from propanal and butanal. In each case, indicate which aldehyde serves as a nucleophile and which as an electrophile.
Answer:
Both propanal and butanal have a-hydrogen atoms present. These can undergo self aldol condensation as well as cross aldol condensation to give four compounds as follows:
(i) Condensation involving propanal: It is a case of a self aldol condensation.

(ii) Condensation involving butanal: It is self aldol condensation.

(iii) Condensation involving butanal (electrophile) and propanal (nucleophile):  It is cross-aldol condensation.

(iv) Condensation involving propanal (electrophile) and butanal (nucleophile): It is cross-aldol condensation.

Question 10.
An organic compound with molecular formula CgHjoO forms 2, 4-DNP derivative, reduces Tollen’s reagent, and undergoes Cannizzaro’s reaction. On vigorous oxidation, it gives 1, 2 benzene dicarboxylic acid. Identify the compound.      (C.B.S.E. Outside Delhi 2012; Haryana Board 2013)
Answer:
Since the compound forms 2, 4-DNP derivative on reacting with 2, 4-DNP, it is a carbonyl compound. As the compound reduces Tollen’s reagent and undergoes Cannizzaro’s reaction, it is an aldehyde and not a ketone. The data further reveals that the compound on vigorous oxidation gives 1, 2-benzene dicarboxylic acid. This clearly shows that in the compound which is of aromatic nature, CHO group is present at position-1 and C₂H₅ side chain at position-2. The given compound is 2-ethyl benzaldehyde. The reactions involved are given below :

Question 11.
An organic compound [A] with molecular formula C₈H₁₆O₂ was hydrolysed with dilute sulphuric acid to give a carboxylic acid [B] and alcohol [C]. Oxidation of [C] with chromic acid produced [B]. The alcohol [C] on dehydration gave but-1-ene. Write equations for the reactions involved. (C.B.S.E. 2008 Supp., C.B.S.E. 2009)
Answer:
(i) The available data shows that the compound [A] upon hydrolysis gave a carboxylic acid [B] and alcohol [C]. It must be an ester.
(ii) Since the alcohol [C] upon oxidation with chromic acid gave back the carboxylic acid [B], both the acid and alcohol must have the same number of carbon atoms (four each).
(iii) The alcohol [C] upon dehydration gave an alkene.
The equations for the reactions are given:

Question 12.
Arrange the following compounds in increasing order of their property as indicated:
(i) Acetaldehyde, Acetone, Di-tert butyl ketone, Methyl tert-butyl ketone (reactivity towards HCN)
(ii) CH₃CH₂CH(Br)COOH, CH₃CH(Br)CH₂COOH, (CH₃)₂CH COOH, CH₃CH₂CH₂COOH (acid strength).
(iii) Benzoic acid, 4-Nitrobenzoic acid, 3,4-Dinitrobenzoic add, 4-Methoxybenzok acid (acid strength).
Answer:
(i) The reactivity of aldehydes and ketones towards HCN addition decreases as the +1 – effect of the alkyl groups increases. Secondly, it decreases with increase in steric hindrance to the nucleophilic attack by CN^– at the carbonyl carbon. Thus the decreasing order of reactivity towards HCN is,

(ii) We know that the + I-effect decreases while -I-effect increases the acidic strength of carboxylic acids. Since + I-effect of isopropyl group is more than that of propyl group, therefore, (CH₃)₂CHCOOH is a weaker acid than CH₃CH₂CH₂COOH. Further since -I-effect decreases with distance, therefore CH₃CH₂CHBrCOOH is a stronger acid than CH₃CHBrCH₂COOH. Thus, the overall acid strength increases in the order:

(iii) Since electron-donating groups decrease the acidic strength, therefore, 4-methoxy benzoic acid is a weaker acid than benzoic acid. Further, since electron-withdrawing groups increase the acidic strength, therefore, both 4-nitrobenzoic acid and 3,4-dinitrobenzoic acid are stronger acids than benzoic acid. Further due to the presence of an additional -NO₂ group at /w-position with respect to -COOH group, 3,4-dinitrobenzoic acid is a stronger acid than 4-nitrobenzoic acid. Thus, the overall acidic strength increases in the order:4-methoxy benzoic acid < benzoic acid < 4-nitrobenzoic acid < 3,4-dinitrobenzoic acid.

Question 13.
Give chemical tests to distinguish between the following pairs of compounds :
(i) Propanal and propanone  (C.B.S.E. Delhi 2011, 2012)
(ii) Phenol and benzoic acid
(iii) Acetophenone and benzophenone
(iv) Benzoic acid and ethyl benzoate  (C.B.S.E. Outside Delhi 2009, 2011)
(v) Pentan-2-one and pentane-3-one
(vi) Benzaldehyde and acetophenone (C.B.S.E. Outside Delhi 2015)
(vii) Ethanal and propanal (C.B.S.E. Outside Delhi 2009, 2011, 2012)
Answer:
(i) Propanal and propanone:  Propanal will give a silver mirror upon heating with Tollen’s reagent but propanone will not respond.
(ii) Phenol and benzoic acid: Benzoic acid will give brisk effervescence with sodium hydrogen carbonate (NaHC0₃) but phenol will not respond.
(iii) Acetophenone and benzophenone: Acetophenone is a methyl ketone. It will give a yellow precipitate upon heating with I₂ and NaOH. Benzophenone will not respond.
(iv) Benzoic acid and ethyl benzoate: Benzoic acid will give brisk effervescence with sodium hydrogen carbonate (NaHC0₃) but ethyl benzoate (ester) will not respond.
(v) Pentan-2-one and pentan-3-one: Pentan-3-one is a methyl ketone and will give a yellow precipitate upon heating with I₂ and NaOH. Pentan-3-one will not respond.
(vi) Benzaldehyde and acetophenone: The distinction can also be made by iodoform test. Acetophenone will give yellow precipitate while benzaldehyde will not react.
(vii) Ethanal and propanal: Ethanal will respond to iodoform test and give yellow precipitate. Propanal will not react.

Question 14.
How will you prepare the following compounds from benzene? You may use any inorganic reagent and any organic reagent having not more than one carbon atom.
(i) Methylbenzoate
(ii) m-Nitrobenzoic acid
(iii) p-Nitrobenzoic acid
(iv) Phenylacetic acid
(v) p-nitrobenzaldehyde
Answer:
(i) Benzene to methylbenzoate

(ii) Benzene to m-nitrobenzoic acid


(iii) Benzene to p-nitrobenzoic acid


(vi) Benzene to phenylacetic acid


(v) Benzene to p-nitrobenzaldehyde

Question 15.
How will you bring about the following conversions in not more than two steps?
(i) Propanone to propene   (C.B.S.E. Delhi 2009, Uttarakhand Board 2009)
(ii) Propanal to butanone
(iii) Ethanol to 3-hydroxybutanal (C.B.S.E. Outside Delhi 2012)
(iv) Benzaldehyde to benzophenone (C.B.S.E. Outside Delhi 2012)
(v) Benzaldehyde to 3-PhenyIpropan-1-ol
(vi) Benzaldehyde to a-Hydroxyphenylacetic acid
(vii) Benzoic acid to benzaldehyde (C.B.S.E. Delhi 2009, Outside Delhi 2017)
(viii) Benzene to m-nitroacetophenone
(ix) Benzoic acid to /n-nitrobenzyl alcohol.      (C.B.S.E. Delhi 2012)
Answer:
(i) Propanone to propene

(ii) Propanal to butanone

(iii) Ethanol to 3-hydroxybutanal

(iv) Benzaldehyde to benzophenone

(v) Benzaldehyde to 3-phenylpropan-1-ol

(vi) Benzaldehyde to a-Hydroxyphenylacetic acid

(vii) Benzoic acid to benzaldehyde

(viii) Benzene to m-nitroacetophenone

(ix) Benzoic acid to m-nitrobenzyl alcohol

Question 16:
Describe the following:
(i) Acylation
(ii) Cross-aldol condensation
(iii) Cannizzaro’s reaction
(iv) Decarboxylation.
Answer:

Question 17:
Complete each synthesis by giving missing starting material, reagent, or products.


Answer:

Question 18:
Give a plausible explanation for each of the following:
(i) Cyclohexanone forms cyanohydrin in good yield but 2, 4, 6-trimethylcyclohexanone does not.
(ii) There are two – NH₂ groups in semicarbazide. However, only one is involved in the formation of semicarbazone.
(iii) During the preparation of esters from carboxylic acid and alcohol in the presence of an acid catalyst, the water or the ester should be removed as fast as it is formed.
Answer:

In cyclohexanone, the attack of CN“ ion (nucleophile) can easily take place at the carbonyl carbon atom. However, in 2, 4, 6-trimethylcyclohexanone, the three CH₃ groups being electron releasing in nature (+ I effect) will considerably increase the electron density on the carbonyl carbon atom and the nucleophile attack does not seem to be feasible. Moreover, the two —CH₃ substituents at the ortho positions will also hinder the attack of nucleophile CN^– ion on the carbonyl group.

(ii) The structural formula of semi-carbazide is NH₂NHCONH₂. Although both the amino groups have lone electron pairs, one of these is in conjugation with the electron-withdrawing carbonyl group and acquires a positive charge. Therefore, it is not in a position to act as the nucleophile, and only one -NH₂ group is involved in the formation of semicarbazone.

(iii) The esterification carried in the presence of acid is of reversible nature and the reverse reaction is called ester hydrolysis.

In order that the reaction may proceed in the forward direction, ester or water formed in the reaction must be removed. Sulphuric acid added in esterification helps in removing molecules of H₂0 as it is a dehydrating agent.

Question 19:
An organic compound contains 69-77% carbon, 11-63% hydrogen and the rest is oxygen. The molecular mass of the compound is 86. It does not reduce Tollen’s reagent but forms an addition compound with sodium hydrogen sulphite and gives a positive iodoform test. On vigorous oxidation, it gives ethanoic acid and propanoic acid. Write the possible structure of the compound. (C.B.S.E. Delhi 2008, 2009, Uttarakhand Board 2015)
Answer:
Step I: Calculation of molecular formula of the compound
Percentage of oxygen = 100 – (% C + % H) = 100 – (69.77 + 11.63) = 18.6%

Step II. Predicting the structure of the compound

• Since the compound forms an addition compound with NaHS0₃, it must be a carbonyl compound.
• As the compound does not reduce Tollen’s reagent but gives a positive iodoform test, it must contain in it a methyl
  

Keeping in view these characteristics, the compound is CH₃CH₂CH₂COCH₃ (Pentan-2-one).
All the reactions in which pentan-2-one participates, are given for the benefit of the students.

Question 20.
Although phenoxide ion has more number of resonating structures than carboxylate ion, carboxylic acid is a stronger acid than phenol. Why?
Answer:

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