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PSEB 12th Class Chemistry Important Questions Chapter 16 Chemistry in Everyday Life

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Punjab State Board PSEB 12th Class Chemistry Important Questions Chapter 16 Chemistry in Everyday Life Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 16 Chemistry in Everyday Life

Very Short Answer Type Questions

Question 1.
Where are receptors located? [NCERT Exemplar]
Answer:
Receptors are embedded in cell membrane.

Question 2.
Which site of an enzyme is called allosteric site? [NCERT Exemplar]
Answer:
Sites different from active site of enzyme where a molecule can bind and affect the active site is called allosteric site.

Question 3.
What is the harmful effect of hyperacidity? [NCERT Exemplar]
Answer:
Ulcer development in stomach.

Question 4.
Write the name of an antacid which is often used as a medicine.
Answer:
Ranitidine (Zantac).

PSEB 12th Class Chemistry Important Questions Chapter 16 Chemistry in Everyday Life

Question 5.
What is the medicinal use of narcotic drugs? [NCERT Exemplar]
Answer:
Since narcotic drugs relieve pain and produce sleep, these are chiefly used for the relief of post-operative pain, cardiac pain and pain of terminal cancer and in childbirth.

Question 6.
Which type of drugs come under antimicrobial drugs? [NCERT Exemplar]
Answer:
Antiseptics, antibiotics and disinfectants.

Question 7.
What is the mode of action of antimicrobial drugs? [NCERT Exemplar]
Answer:
Antimicrobial drugs can kill the microorganism such as bacteria, virus, fungi or other parasites. They can, alternatively, inhibit the pathogenic action of microbes.

Question 8.
Which one of the following drugs is an antibiotic? Morphine, Equanil, Chloramphenicol, Aspirin
Answer:
Chloramphenicol.

Question 9.
What is meant by ‘narrow-spectrum antibiotics’?
Answer:
Antibiotics which are mainly effective against Gram-positive or Gram-negative bacteria are known as narrow-spectrum antibiotics. For , example, penicillin G.

Question 10.
Define the limited spectrum antibiotics.
Answer:
Antibiotics which are mainly effective against a single organism or disease, are called as limited spectrum antibiotics.

Short answer type questions

Question 1.
Why are certain drugs called enzyme inhibitors?[NCERT Exemplar]
Answer:
Enzymes have active sites that bind the substrate for effective and quick chemical reactions. The functional groups present at the active site of enzyme interact with functional groups of substrate via ionic bonding, hydrogen bonding, van der Waal’s interaction, etc. Some drugs interfere with this interaction by blocking the binding site of enzyme and prevent the binding of actual substrate with enzyme. This inhibits the catalytic activity of the enzyme, therefore, these are called inhibitors.

Question 2.
Sodium salts of some acids are very useful as food preservatives. Suggest a few such acids. [NCERT Exemplar]
Answer:
A preservative is naturally occurring or synthetically produced substance that is added to foods to prevent decomposition by microbial growth or by undesirable chemical changes. Sodium salts of some acids are very useful as food preservatives.

Some examples of such acids are as follows :

  • Benzoic acid in the form of its sodium salts constitutes one of the most common food preservatives. Sodium benzoate is a common preservative in acid or acidified foods such as fruit, juices, pickles etc. Yeasts are inhibited by benzoate to a greater extent than are moulds and bacteria.
  • Sorbic acid and its salts (sodium, potassium, and calcium) also have preservative activities but the applications of -sodium sorbate (C6H7NaO2) are limited compared to that for potassium salt.
  • Sodium erythorbate (C6H7NaO6) is a food additive used predominate in meats, poultry and soft drinks.
  • Sodium propanoate[Na(C2H5COO)] is used in bakery products as mould inhibitor.

Question 3.
What is the side product of soap industry? Give reactions showing soap formation. [NCERT Exemplar]
Answer:
Soaps are sodium or potassium salts of long-chain fatty acids such as stearic acid, oleic acid and palmitic acid. Soaps containing sodium salts are formed by heating fat (i. e., glyceryl ester of fatty acid) with aqueous sodium hydroxide solution.
PSEB 12th Class Chemistry Important Questions Chapter 16 Chemistry in Everyday Life 1
This reaction is known as saponification. In this reaction, esters of fatty acids are hydrolyzed and the soap obtained remains in colloidal form. It is precipitated from the solution by adding NaCl. The solution left after removing the soap contains glycerol as side product.

PSEB 12th Class Chemistry Important Questions Chapter 16 Chemistry in Everyday Life

Question 4.
Hair shampoos belong to which class of synthetic detergent? [NCERT Exemplar]
Answer:
Hair shampoos are made up of cationic detergents. These are quaternary ammonium salts of amines with acetates, chlorides or bromides as anions, e. g., cetyltrimethylammonium bromide.
PSEB 12th Class Chemistry Important Questions Chapter 16 Chemistry in Everyday Life 2

Question 5.
Explain why sometimes foaming is seen in river water near the place where sewage water is poured after treatment?
[NCERT Exemplar]
Answer:
Detergents have long hydrocarbon chains. If their hydrocarbon chain is highly branched, then bacteria cannot degrade this easily. Such detergents are non-biodegradable. Slow degradation of detergents leads to their accumulation.

These non-biodegradable detergents persist in water even after sewage treatment and cause foaming in rivers, ponds and their water get polluted. In order to overcome this issue branching of the hydrocarbon chain is controlled and kept to a minimum.

Long answer type questions

Question 1.
What are enzyme inhibitors? Classify them on the basis of their mode of attachment on the active site of enzymes. With the help of diagrams explain how do inhibitors inhibit the enzymatic activity? [NCERT Exemplar]
Answer:
Enzymes are responsible to hold the substrate molecule for a chemical reaction and they provide functional groups which will attack the substrate to carry out the chemical reaction. Drugs which inhibit any of the two activities of enzymes are called enzyme inhibitors.

Enzyme inhibitors can block the binding site thereby preventing the binding of the substrate to the active site and hence inhibiting the catalytic activity of the enzyme.
Drugs inhibit the attachment of natural substrate on the active site of enzymes in two different ways as explained below :
(i) Drugs which compete with natural substrate for their attachment on the active sites of enzymes are called competitive inhibitors.
PSEB 12th Class Chemistry Important Questions Chapter 16 Chemistry in Everyday Life 3
(ii) Some drugs, however, do not bind to the active site but bind to a different site of the enzyme which is called allosteric site. This binding of the drug at allosteric site changes the shape of the active site of the enzyme in such a way that the natural substrate cannot recognize it. Such enzymes are called non-competitive inhibitors.
PSEB 12th Class Chemistry Important Questions Chapter 16 Chemistry in Everyday Life 4

Question 2.
In what respect to prontosil and salvarsan resemble? Is there any resemblance between azo dye and prontosil? Explain. [NCERT Exemplar]
Answer:
Prontosil, also called sulfamide chrysoidine, trade name of the first synthetic drug used in the treatment of general bacterial infections in humans. Prontosil resulted from research, directed by German chemist and pathologist Gerhard Domagk, on the antibacterial action of azo dyes. A red azo dye of low toxicity, prontosil was shown by Domagk to prevent mortality in mice infected with Streptococcus bacteria.

The dye was also effective in controlling staphylococcus infections in rabbits. Within a relatively short period, it was demonstrated that prontosil was effective not only in combating experimental infections in animals but also against Streptococcal disease in humans, including meningitis and puerperal sepsis. Structural formula of prontosil is as follows :
PSEB 12th Class Chemistry Important Questions Chapter 16 Chemistry in Everyday Life 5
From the structure of prontosil, it is very clear that it has -N = N- linkage. It was discovered that the part of the structure of prontosil molecule shown inbox, i.e., p-amino benzene sulphonamide has antibacterial activity.
Salvarsan is also known as arsphenamine. It was introduced at the beginning of 1910s as the first effective treatment for syphilis. It is an organoarsenic molecule and has -As = As- double bond.
PSEB 12th Class Chemistry Important Questions Chapter 16 Chemistry in Everyday Life 6
Salvarsan and prontosil show similarity in their structure. Both of these drugs are antimicrobials. Salvarsan contains -As = As- linkage whereas prontosil has —N = N— linkage.
PSEB 12th Class Chemistry Important Questions Chapter 16 Chemistry in Everyday Life 7
Prontosil (a red azo dye) and azo dye both have -N = N- linkage.
PSEB 12th Class Chemistry Important Questions Chapter 16 Chemistry in Everyday Life 8

PSEB 12th Class Chemistry Important Questions Chapter 16 Chemistry in Everyday Life

Question 3.
Ashwin observed that his friend Shubhain was staying aloof, not playing with friends, and becoming easily irritable for some weeks. Ashwin told his teacher about this, who, in turn, called Shubham’s parents and advised them to consult a doctor. The doctor after examining Shubham prescribed antidepressant drugs for him.
After reading the above passage, answer the following questions:
(i) Name two antidepressant drugs.
(ii) Mention the values shown by Ashwin.
(iii) How should Shubham’s family help him other than providing medicine?
(iv) What is the scientific explanation for the feeling of depression?
Answer:
(i) Equanil, Iproniazid, phenelzine (any two)
(ii) Empathetic, caring, sensitive.
(iii) They should talk to him, be a patient listener, can discuss the matter with the psychologist.
(iv) If the level of noradrenaline is low, then the signal sending activity becomes low and the person suffers from depression.

PSEB 12th Class Chemistry Solutions Chapter 16 Chemistry in Everyday Life

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Punjab State Board PSEB 12th Class Chemistry Book Solutions Chapter 16 Chemistry in Everyday Life Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Chemistry Chapter 16 Chemistry in Everyday Life

PSEB 12th Class Chemistry Guide Chemistry in Everyday Life InText Questions and Answers

Question 1.
Why do we need to classify drugs in different ways?
Answer:
Different ways of classification of drugs and the usefulness of such classification are as follows :

  1. Classification on the basis of pharmacological effect is useful for doctors because it provides them the whole range of drugs available for the treatment of a particular type of problem.
  2. Classification on the basis of drug action on a particular biochemical process is useful for choosing the correct lead compound for designing the synthesis of a desired drug.
  3. Classification on the basis of molecular targets is useful for medicinal chemists so that they can design a drug which is most effective for a particular receptor site.
  4. Classification on the basis of chemical structure is useful for doctors to design the synthesis of a number of structurally similar compounds having different substituents and then choosing the drug having the least toxicity.

Question 2.
Explain the term target molecules or drug targets as used in medicinal chemistry.
Answer:
Drugs interact with macromolecules such as proteins, carbohydrates, lipids, enzymes and nucleic acids. Hence, these are called drug targets. Drugs possessing some common structural features may have the same mechanism of action on targets.

Question 3.
Name the macro-molecules that are chosen as drug targets.
Answer:
Nucleic acids, proteins, carbohydrates, lipids, enzymes are chosen as drug targets.

Question 4.
Why should not medicines be taken without consulting doctors?
Answer:
Side effects are caused when a drug binds to more than one receptor site. So, a doctor must be consulted to choose the right drug which has the maximum affinity for a particular receptor site to have the desired effect. The dose of the drug is also crucial because some drugs like opiates in higher doses act as poisons and may cause death.

PSEB 12th Class Chemistry Solutions Chapter 16 Chemistry in Everyday Life

Question 5.
Define the term chemotherapy.
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 site of enzymes?
Answer:
Ionic bonding, hydrogen bonding, van der Waals’ interaction, dipole-dipole interaction etc., are involved in holding the drugs to the active site of enzymes.

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:
These (antacids and antiallergic drugs) do not interfere with the function of each other because they work on different receptors in the body.

Question 8.
Low level of noradrenaline is the cause of depression. What type of drugs are needed to cure this problem? Name two drugs.
Answer:
In event of low level of neurotransmitters, noradrenaline, antidepressant drugs are required. These drugs inhibit the enzymes which catalyse the degradation of noradrenaline. If the enzyme is inhibited, noradrenaline is slowly metabolised and thus activates its receptor for longer periods of time thereby reducing depression. Two important drugs are iproniazid and phenelzine.

Question 9.
What is meant by the term ‘broad-spectrum antibiotics? Explain.
Answer:
Broad-spectrum antibiotics are effective against several different types of harmful bacteria. Examples are tetracycline, ofloxacin, chloramphenicol, etc. Chloramphenicol can be used in case of typhoid, acute fever, dysentery, urinary infections, meningitis and pneumonia.

Question 10.
How do antiseptics differ from disinfectants? Give one example of each.
Answer:
Differences between antiseptics and disinfectants are as follows :
Antiseptics

  • Antiseptics are chemical substances which prevent the growth of microorganisms and may even kill them but are not harmful to living tissues.
  • Antiseptics are generally applied to living tissues such as wounds, cuts, ulcers and diseased skin surfaces.
  • Dettol, furnace, soframicine are antiseptics.

Disinfectants

  • Disinfectants are chemical substances which kill microorganisms or stop their growth but are harmful to human tissues.
  • Disinfectants are applied to inanimate objects such as floor, drainage system, instruments, etc.
  • Chlorine in the concentration of 0.2 to 0.4 ppm in aqueous solution and SO 2 in very low concentration are disinfectants.

Question 11.
Why are cimetidine and ranitidine better antacids than sodium hydrogen carbonate or magnesium or aluminium hydroxide?
Answer:
Excessive use of sodium hydrogen carbonate or a mixture of aluminium and magnesium hydroxide can make the stomach alkaline and trigger the production of even more acid. On the other hand, ranitidine and cimetidine prevent the interaction of histamine with the receptors present in the stomach wall. This results in release of lesser amount of acid. Thus, these are better antacids.

PSEB 12th Class Chemistry Solutions Chapter 16 Chemistry in Everyday Life

Question 12.
Name a substance which can be used as an antiseptic as well as disinfectant.
Answer:
0.2% solution of phenol acts as an antiseptic while 1% of the solution acts as a disinfectant.

Question 13.
What are the main constituents of Dettol?
Answer:
Chloroxylenol and a-terpineol in a suitable solvent.

Question 14.
What is tincture of iodine? What is its use?
Answer:
A 2-3 per cent solution of iodine in alcohol-water mixture is known as tincture of iodine. It is used 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 food preservatives. For example, sodium metabisulphite, sodium benzoate, etc.

Question 16.
Why is the use of aspartame limited to cold foods and drinks?
Answer:
Aspartame decomposes on heating and may not work well. So, its use as an artificial sweetener is limited to foods and drinks at low temperatures.

Question 17.
What are artificial sweetening agents? Give two examples.
Answer:
Artificial sweetening agents are the chemical substances which provide sweetness to the food without increasing the calories to the body. For example, saccharin, aspartame, sucralose etc.

Question 18.
Name the sweetening agent used in the preparation of sweets for a diabetic patient.
Answer:
Saccharin.

Question 19.
What problem arises in using alitame as artificial sweetener?
Answer:
Alitame is a high potency artificial sweetener. Therefore, it becomes difficult to control die level of sweetness while using it.

Question 20.
How are synthetic detergents better than soaps?
Answer:
Advantages of synthetic detergents over soaps :

  • Detergents can work with hard water too while soaps cannot.
  • They can work even in an acidic medium while soaps cannot.
  • Synthetic detergents are stronger cleansing agents than soaps.
  • Their solubility is higher than that of soaps.
  • They are prepared from hydrocarbons (petroleum) so their use is to save vegetable oils which are used during the preparation of soaps.

Question 21.
Explain the following terms with suitable examples :
(i) Cationic detergents
(ii) Anionic detergents and
(iii) Non-ionic detergents
Answer:
(i) Cationic detergents: These are quaternary ammonium salts of amines with acetates, chlorides or bromides.
Example: Cetyl trimethyl ammonium bromide
(ii) Anionic detergents: These detergents have large anionic part in their molecules. These are of two types :
(a) Sodium alkyl sulphates: For example, Sodium lauryl sulphate CH3 (CH2)10 CH2OSO3–Na+.
(b) Sodium alkyl benzene sulphonates: For example, sodium dodecylbenzene sulphonate
PSEB 12th Class Chemistry Solutions Chapter 16 Chemistry in Everyday Life 1
(iii) Non-ionic detergents: These are esters of high molecular mass alcohols with fatty acids. For example, Polyethylene glycol stearate.
CH3 (CH2)16 COO(CH2CH2O)nCH2CH2OH.

PSEB 12th Class Chemistry Solutions Chapter 16 Chemistry in Everyday Life

Question 22.
What are biodegradable and non-biodegradable detergents? Give one example of each.
Answer:
Detergents having straight hydrocarbon chains are easily degraded by microorganisms and hence are called biodegradable detergents, whereas detergents containing branched hydrocarbon chains are not easily degraded by the microorganisms and hence are called non-biodegradable detergents. As a result, non-biodegradable detergents accumulate in rivers and waterways thereby causing severe water pollution.

Examples of biodegradable detergents are : sodium lauryl sulphate, sodium 4-(l-dodecyl) benzene-sulphonate and sodium 4-(2-dodecyl) benzenesulphonate. An example of non-biodegradable detergent is sodium 4-(l, 3, 5, 7-tetramethyloctyl) benzenesulphonate.

Question 23.
Why do soaps not work with hard water?
Answer:
Calcium and magnesium salts present in hard water react with soaps to form insoluble compounds, which form curdy white precipitates and are difficult to remove from the clothes.

PSEB 12th Class Chemistry Solutions Chapter 16 Chemistry in Everyday Life 2

Question 24.
Can you use soaps and synthetic detergents to check the hardness of water?
Answer:
We can use soaps to check the hardness of water because with hard water, soaps give a gummy mass (sticky precipitate) but we cannot use detergents for this purpose because they give foam with both hard and soft water.

Question 25.
Explain the cleansing action of soaps.
Answer:
The cleansing action of soap is due to the fact that soap molecules, such as sodium stearate form micelle around the oil droplet in such a way that hydrophobic part of the stearate ions is in the oil droplet and hydrophilic part projects out of the oil droplet like the bristles.

Since the polar groups can interact with water, the oil droplet surrounded by stearate ions is now pulled in water and removed from the dirty surface. Thus, soap helps in emulsification and washing away of oils and fats. The negatively charged sheath around the globules prevents them from coming together and forming aggregates.
PSEB 12th Class Chemistry Solutions Chapter 16 Chemistry in Everyday Life 3

Question 26.
If the water contains dissolved calcium hydrogen carbonate, out of soaps and synthetic detergents which one will you use for cleaning clothes?
Answer:
Synthetic detergents.

PSEB 12th Class Chemistry Solutions Chapter 16 Chemistry in Everyday Life

Question 27.
Label the hydrophilic and hydrophobic parts in the following compounds.
(i) CH3(CH2)10CH2OSO3–Na+
(ii) CH3(CH2)15N+(CH3)3 Br–
(iii) CH3(CH2)16COO(CH2CH2O)n CH2CH2OH
Answer:

Chemistry Guide for Class 12 PSEB Chemistry in Everyday Life Textbook Questions and Answers

Question 1.
Sleeping pills are recommended by doctors to the patients suffering from sleeplessness but It is not advisable to take Its doses without consultation with the doctor. Why?
Answer:
Sleeping pills contain drugs that may be tranquillizers or antidepressants. They affect the nervous system, relieve anxiety, stress, irritability or excitement. But they should strictly be used under the supervision of a doctor. If not, the uncontrolled and overdosage can cause harm to the body and mind because in higher doses these drugs act as poisons.

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

Question 3.
Why do we require artifical sweetening agents?
Answer:
Natural sweeteners (sucrose etc.) provide calories to the body. Taking extra calories is harmful for diabetic patients. So, artificial sweeteners are used (i) to control intake of calories and (ii) as a substitute of sugar for diabetics.

Question 4.
Write the chemical equation for preparing sodium soap from glyceryl oleate and glyceryl palmitate. Structural formulae of these compounds are given below.
(i) (C15H31COO)3C3H5 – Glyceryl pfi]mitate
(ii) (C17H32COO)3C3H5 – Glyceryl oleate
Answer:

PSEB 12th Class Chemistry Solutions Chapter 16 Chemistry in Everyday Life 5

Question 5.
Following types of non-ionic detergents are present in liquid detergents, emulsifying agents and wetting agents. Label the hydrophilic and hydrophobic parts In the molecule. Identify the functional group(s) present In the molecule.
PSEB 12th Class Chemistry Solutions Chapter 16 Chemistry in Everyday Life 6
Answer:
PSEB 12th Class Chemistry Solutions Chapter 16 Chemistry in Everyday Life 7
(b) Functional groups: Ether and alcohol.

PSEB 12th Class Chemistry Important Questions Chapter 2 Solutions

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Punjab State Board PSEB 12th Class Chemistry Important Questions Chapter 2 Solutions Important Questions and Answers.

PSEB 12th Class Chemistry Important Questions Chapter 2 Solutions

Very short answer type questions

Question 1.
Define mole fraction.
Answer:
Mole fraction of a component in a solution may be defined as the ratio of moles of that compdnent to the total number of moles of all the components present in the solution.

Question 2.
What is the similarity between Raoult’s law and Henry’s law?
Answer:
The similarity between Raoult’s law and Henry’s law is that both state that the partial vapour pressure of the volatile component or gas is directly proportional to its mole fraction in the solution.

Question 3.
Why is the vapour pressure of a solution of glucose in water lower than that of water? (NCERT Exemplar)
Answer:
This is due to decrease in the escaping tendency of the water molecules from the surface of solution as some of the surface area is occupied by non-volatile solute, glucose particles.

PSEB 12th Class Chemistry Important Questions Chapter 2 Solutions

Question 4.
What type of deviation is shown by a mixture of ethanol and acetone? What type of azeotrope is formed by mixing ethanol and acetone?
Answer:
Positive deviation.
Minimum boiling azeotropes.

Question 5.
State how does osmotic pressure vary with temperature.
Answer:
Osmotic pressure increases with increase in temperature.

Question 6.
What are isotonic solutions?
Answer:
The solutions of the same osmotic pressure at a given temperature are called isotonic solutions.

Question 7.
Define van’t Hoff factor.
Answer:
van’t Hoff factor may be defined as the ratio of normal molecular mass to observed molecular mass or the ratio of observed colligative property to normal colligative property.

Question 8.
Why are aquatic species more comfortable in cold water in comparison to warm water? (NCERT Exemplar)
Answer:
Solubility of oxygen in water increases with decrease in temperature. Presence of more oxygen at lower temperature makes the aquatic species more comfortable in cold water.

PSEB 12th Class Chemistry Important Questions Chapter 2 Solutions

Question 9.
What is semipermeable membrane? (NCERT Exemplar)
Answer:
Continuous sheets or films (natural or synthetic) which contain a network of submicroscopic holes or pores through which small solvent molecules (water etc.) can pass, but solute molecules of bigger size cannot pass are called semipermeable membrane.

Question 10.
Give an example of a material used for making semipermeable membrane for carrying out reverse osmosis. (NCERT Exemplar)
Answer:
Cellulose acetate.

Short answer type questions

Question 1.
State Henry’s law. Write its one application. What is the effect of temperature on solubility of gases in liquid?
Answer:
It states that the partial pressure of a gas in vapour phase (p) is proportional to the mole fraction of the gas (χ) in the solution.
p ∝ χ or p = KHχ where KH is the
Henry’s constant.
Application of Henry’s law:
To increase the solubility of CO2 in soft drinks and soda water, the bottle is sealed under high pressure.

Effect of temperature on solubility:
As dissolution is an exothermic process, therefore, according to Le Chatelier’s principle solubility should decrease with rise in temperature.

Question 2.
Why does a solution containing non-volatile solute have higher boiling point than the pure solvent? Why is elevation of boiling point a colligative property?
Answer:
The addition of a non-volatile solute to a volatile solvent lowers its vapour pressure. In order to boil the solution, i.e., to make its vapour pressure equal to atmospheric pressure, the solution has to be heated at a higher temperature. In other words, the boiling point of solution becomes higher than solvent.

As elevation in boiling point depends on the number of moles of solute particles and independent of their nature, therefore, it is a colligative property.

PSEB 12th Class Chemistry Important Questions Chapter 2 Solutions

Question 3.
Will the elevation in boiling point be same if 0.1 mol of sodium chloride or 0.1 mol of sugar is dissolved in 1L of water? Explain.
Answer:
No, the elevation in boiling point is not the same. NaCl, being an electrolyte, dissociates almost completely to give Na+ and Cl– ions whereas glucose, being non-electrolyte does not dissociate. Hence, the number of particles in 0.1 M NaCl solution is nearly double than 0.1 M glucose solution. Elevation in boiling point being a colligative property, is therefore, nearly twice for 0.1 M NaCl solution than for 0.1 M glucose solution.

Question 4.
Explain the solubility rule ‘like dissolves like’ in terms of intermolecular forces that exist in solutions. (NCERT Exemplar)
Answer:
If the intermolecular interactions are similar in both constituents, i.e., solute and solvent then solute dissolves in the solvent. e.g., polar solutes dissolve in polar solvents and non-polar solutes in non-polar solvents. Thus, the statement ‘like dissolved like’ proves to be true.

Question 5.
Concentration terms such as mass percentage, ppm, mole fraction and molality are independent of temperature. However, molarity is a function of temperature. Explain.
(NCERT Exemplar)
Answer:
Molarity is defined as the number of moles of solute dissolved per litre of a solution. Since, volume depends on temperature and changes with change in temperature, the molarity will also change with change in temperature. On the other hand, mass does not change with change in temperature, so other concentration terms given in the question also do not do so. According to the definition of all these terms, mass of solvent used for making the solution is related to the mass of solute.

PSEB 12th Class Chemistry Important Questions Chapter 2 Solutions

Question 6.
Which of the following solutions has higher freezing point?
0.05 M Al2(SO4)3, 0.1 M K3 [Fe(CN)6] Justify.
Answer:
0.05 M Al2(SO4)3 has higher freezing point.

Justification:
For 0.05 MAl2(SO4) 3,i = 5
Number of particles = i × concentration
= 5 × 0.05
= 0.25 moles of ions
For 0.1MK3 [Fe(CN)6], i = 4
Number of particles = i × concentration
= 4 × 0.1
= 0.4 moles of ions
We know that, ΔTf. Number of particles
Hence, 0.05 M Al2(SO4)3 has higher freezing point because it has lower number of particles than 0.1 M K3 [Fe (CN)6].

Question 7.
The freezing point of benzene decreases by 2.12 K when 2.5 g of benzoic acid (C6H5COOH) is dissolved in 25 g of benzene. If benzoic acid forms a dimer in benzene, calculate the van’t Hoff factor and the percentage association of benzoic acid. (Kf for benzene = 5.12 K kg mol-1)
Answer:
Given, ΔTf = 2.12 K, Kf = 5.12 K kg mol-1
We know that, ΔTf = i Kf m
2.12 = \(\frac{i \times 5.12 \times 2.5 \times 1000}{122 \times 25}\)
or i = 0.505
For association,
i = 1 – \(\frac{\alpha}{2}\)
0.505 = 1 – \(\frac{\alpha}{2}\)
or α =0.99
Hence, percentage association of benzoic acid is 99%.

Long answer type questions

Question 1.
(a) What type of deviation is shown by a mixture of ethanol and acetone? Give reason.
(b) A solution of glucose (molar mass = 180 g mol-1) in water is labelled as 10% (by mass). What would be the molality and molarity of the solution?
(Density of solution = 1.2 g mL-1)
Answer
(a) A mixture of ethanol and acetone shows positive deviation from Raoult’s law.
In pure ethanol hydrogen bond exist between the molecules. On adding acetone to ethanol, acetone molecules get in between the molecules of . ethanol thus breaking some of the hydrogen bonds and weakening molecular interactions considerably. Weakening of molecular interactions leads to increase in vapour pressure resulting in positive deviation from Raoult’s law.

(b) Let the mass of solution = 100 g
∴ Mass of glucose = 10 g
Number of moles of glucose \(=\frac{\text { Mass of glucose }}{\text { Molar mass }}\)
= \(\frac{10 \mathrm{~g}}{180 \mathrm{~g} \mathrm{~mol}^{-1}}\) = 0.056 mol
PSEB 12th Class Chemistry Important Questions Chapter 2 Solutions 1

PSEB 12th Class Chemistry Important Questions Chapter 2 Solutions

Question 2.
Discuss biological arid industrial importance of osmosis.
(NCERT Exemplar)
Answer:
The process of osmosis is of great biological and industrial importance as is evident from the following examples:

  • Movement of water from soil into plant roots and subsequently into upper portion of the plant occurs partly due to osmosis.
  • Preservation of meat against bacterial action by adding salt.
  • Preservation of fruits against bacterial action by adding sugar.
    Bacterium in canned fruit loses water through the process of osmosis, shrivels and dies.
  • Reverse osmosis is used for desalination of water.

PSEB 11th Class Political Science Book Solutions Guide in Punjabi English Medium

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Punjab State Board Syllabus PSEB 11th Class Political Science Book Solutions Guide Pdf in English Medium and Punjabi Medium are part of PSEB Solutions for Class 11.

PSEB 11th Class Political Science Guide | Political Science Guide for Class 11 PSEB

Political Science Guide for Class 11 PSEB | PSEB 11th Class Political Science Book Solutions

PSEB 11th Class Political Science Book Solutions in English Medium

Part A Foundations of Political Science

Part B Indian Constitution and Government

PSEB 11th Class Political Science Book Solutions in Hindi Medium

PSEB 11th Class Political Science Syllabus

Part – A Foundations of Political Science

Unit I: Meaning, Scope, and Significance of Political Science
(a) Meaning of Political Science.
(b) Scope and Significance of Political Science.
(c) Relationship of Political Science with History, Economics, Sociology.
(d) Citizen and his Rights and Duties.
(e) Citizen and Citizenship.

Unit II: (a) Meaning of Rights and Duties.
(b) Relation between Rights and Duties.
Basic Concepts
(a) Law – Meaning and its Kinds.
(b) Liberty – Meaning. Kinds and Safeguards.
(c) Equality – Meaning, Kinds, Liberty, and Equality.
(d) Justice.

Unit III: State, Forms of Governments
(a) State and its attributes.
(b) State and Government differences.
(c) Forms of Governments.
a. Democratic and Authoritarian (Dictatorial)
b. Parliamentary and Presidential.
c. Unitary and Federal.

Unit IV: Organs of Government
(a) Executive – Types of Executive, Functions.
(b) Legislature, Types of Legislature – Unicameral and Bicameral, Functions.
(c) Judiciary, Importance, and Functions, Independence of the Judiciary.

Part – B Indian Constitution and Government

Unit V: (a) Preamble
(b) Salient features of the Indian Constitution.
(c) Fundamental Rights and Directive Principles of State Policy.
(a) Fundamental Rights: Nature and Kinds.
(b) Fundamental Duties.
(c) Directive Principles of the State Policy – Importance and Sanctions behind them.
(d) Distinction and relationship between Fundamental Rights and Directive Principles.

Unit VI: Indian Federal System
(a) Nature of Indian Federation.
(b) Union-State Relations: Legislative, Administrative, and Financial.
Union Government.
(c) The Union Executive – President, Prime Minister, and Council of Ministers.

Unit VII: The Union Legislature
(Lok Sabha, Rajya Sabha)
Law Making procedure (ordinary bill and money bill)
The State Government.
(a) State Executive – Governor, Chief Minister, and Council of Ministers.
(b) State Legislature.

Unit VIII: District Administration.
Indian Judicial System.
(a) The Supreme Court.
(b) State High Court.

PSEB 12th Class Chemistry Important Questions Chapter 14 Biomolecules

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Punjab State Board PSEB 12th Class Chemistry Important Questions Chapter 14 Biomolecules Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 14 Biomolecules

Very short answer type questions

Question 1.
What are oligosaccharides?
Answer:
Carbohydrates which on hydrolysis give two to ten molecules of monosaccharides are called oligosaccharides e. g., sucrose.

Question 2.
How do you explain the presence of all the six carbon atoms in glucose in a straight chain? (NCERT Exemplar)
Answer:
On prolonged heating with HI, glucose gives n-hexane.
PSEB 12th Class Chemistry Important Questions Chapter 14 1

Question 3.
Write the product obtained when D-glucose reacts with H2N—OH.
Answer:
PSEB 12th Class Chemistry Important Questions Chapter 14 2

Question 4.
What are anomers?
Answer:
A pair of stereoisomers such as α-D-(+) glucose and β-D-(+) glucose which differ in configuration only around C1 are called anomers.

PSEB 12th Class Chemistry Important Questions Chapter 14 Biomolecules

Question 5.
What type of linkage is present in proteins?
Answer:
PSEB 12th Class Chemistry Important Questions Chapter 14 3

Question 6.
What are Vitamins?
Answer:
Vitamins are generally regarded as organic compounds required in the diet in small amounts to perform specific biological functions for normal maintenance of optimum growth and health of the organism.

Question 7.
Why must vitamin C be supplied regularly in diet? [NCERT Exemplar]
Answer:
Vitamin C is water-soluble hence, they are regularly excreted in urine and can not be stored in our body, so, they are supplied regularly in diet.

Question 8.
Of the two bases named below, which one is present in RNA and which one is present in DNA? Thymine, Uracil.
Answer:

  1. Thymine is present in DNA.
  2. Uracil is present in RNA.

Question 9.
The activation energy for the acid-catalyzed hydrolysis of sucrose is 6.22 kJ mol-1, while the activation energy is only 2.15 kJ mol-1 when hydrolysis is catalyzed by the enzyme sucrase. Explain. [NCERT Exemplar]
Answer:
Enzymes, the biocatalysts reduce the magnitude of activation energy by providing alternative paths. In the hydrolysis of sucrose, the enzyme sucrase reduces the activation energy from 6.22 kJ mol-1 to 2.15 kJ mol-1.

Question 10.
Name the bases present in RNA. Which one of these is not present in DNA?
Answer:
The bases present in RNA are Adenine (A), Guanine (G), Cytosine (C) and Uracil (U). Uracil is not present in DNA.

PSEB 12th Class Chemistry Important Questions Chapter 14 Biomolecules

Short answer type questions

Question 1.
What is essentially the difference between α-glucose and β-glucose? What is meany by pyranose structure of glucose?
Answer:
α -Glucose and β-Glucose differ only in the configuration of hydroxy group at C1 and are called anomers and the C1 carbon is called anomeric carbon. The six-membered cyclic structure of glucose is called pyranose (α-or β), in analogy with pyran. The cyclic structure of glucose is more correctly represented by Haworth structure as given below:
PSEB 12th Class Chemistry Important Questions Chapter 14 4

Question 2.
Describe the term D- and L-configuration used for amino acids with examples. [NCERT Exemplar]
Answer:
All naturally occurring a-amino acids (except glycine) are optically active due to the presence of chiral carbon atom. These have either D- or  L-configuration. D-form means that the amino (-NH2) group is present towards the right-hand side. L-form shows the presence of -NH2 group on the left-hand side.
PSEB 12th Class Chemistry Important Questions Chapter 14 5

Question 3.
Aldopentoses named as ribose and 2-deoxyribose are found in nucleic acids. What is their relative configuration? [NCERT Exemplar]
Answer:
In case of cyclic structure of saccharide, if -OH group present at second last carbon is present at bottom side, then it is considered as D configuration (as shown above).
PSEB 12th Class Chemistry Important Questions Chapter 14 6

Question 4.
How do enzymes help a substrate to be attacked by the reagent effectively? [NCERT Exemplar]
Answer:
At the surface of enzyme, active sites are present. These active sites of enzymes hold the substrate molecule in a suitable position so that it can be attacked by the reagent effectively. This reduces the magnitude of activation energy.

Enzymes contains cavities of characteristics shape and possessing active groups known as active centre on the surface. The molecules of the reactant (substrate) having complementary shape, fit into these cavities. On account of these active groups, an activated complex is formed which then decomposes to yield the products.

PSEB 12th Class Chemistry Important Questions Chapter 14 Biomolecules

Question 5.
Amino acids can be classified as α-, β-, γ-, δ- and so on depending upon the relative position of amino group with respect to carboxyl group. Which type of amino acids form polypeptide chain in proteins? [NCERT Exemplar]
Answer:
α-amino acid forms a polypeptide chain by elimination of water molecules.
PSEB 12th Class Chemistry Important Questions Chapter 14 7

Question 6.
(i) Winch vitamin deficiency causes rickets?
(ii) Name the base that is found in nucleotide of RNA only.
(iii) Glucose on reaction with acetic acid gives glucose Penta acetate. What does it suggest about the structure of glucose?
Answer:
(i) Vitamin D
(ii) Uracil
(iii) 5-OH groups are present in glucose.

Long answer type questions

Question 1.
Which moieties of nucleosides are involved in the formation of phosphodiester linkages present in dinucleotides? What does the word diester in the name of linkage indicate? Which acid is involved in the formation of this linkage?
Answer:
(i) 5′ and 3′ carbon atoms of pentose sugar.
(ii) Most probably the resemblance of with 2 esters (-COO)2 groups joined together.
PSEB 12th Class Chemistry Important Questions Chapter 14 8
(iii) Phosphoric acid (H3PO4).
Nucleosides are joined together by phosphodiester linkage between 5′ and 3′ carbon atoms of pentose sugar and a dinucleotide with phosphoric acid (CH3PO4) is formed.
PSEB 12th Class Chemistry Important Questions Chapter 14 9

Question 2.
Write the structures of fragments produced on complete hydrolysis of DNA. How are they linked in DNA molecules? Draw a diagram to show pairing of nucleotide bases in double helix of DNA. [NCERT Exemplar]
Answer:
On complete hydrolysis of DNA, following fragments are formed-pentose sugar (β-D-2-deoxyribose), phosphoric acid (H3PO4) and bases (nitrogen-containing heterocyclic compounds).
Structures
(i) Sugar
PSEB 12th Class Chemistry Important Questions Chapter 14 10
(ii) Phosphoric acid
PSEB 12th Class Chemistry Important Questions Chapter 14 11
(iii) Nitrogen bases: DNA contains four bases.
Adenine (A), Guanine (G), Cytosine (C), and Thymine (T).
PSEB 12th Class Chemistry Important Questions Chapter 14 12
A unit formed by the attachment of a base to 1′-position of sugar is called nucleoside. When nucleoside links to phosphoric acid at 5′-position of sugar moiety, a nucleotide is formed. Nucleotides are joined together by phosphodiester linkage between 5′ and 3′ carbon atoms of the pentose sugar. In DNA two nucleic acid chains are wound about each other and held together by hydrogen bonds between pairs of bases.

PSEB 12th Class Chemistry Important Questions Chapter 14 Biomolecules

The two strands are complementary to each other because hydrogen bonds are formed between specific pairs of base adenine form hydrogen bonds with thymine whereas cytosine forms hydrogen bonds with guanine.
PSEB 12th Class Chemistry Important Questions Chapter 14 13

PSEB 12th Class Chemistry Solutions Chapter 14 Biomolecules

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Punjab State Board PSEB 12th Class Chemistry Book Solutions Chapter 14 Biomolecules Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Chemistry Chapter 14 Biomolecules

PSEB 12th Class Chemistry Guide Biomolecules InText Questions and Answers

Question 1.
What are monosaccharides?
Answer:
Monosaccharides are carbohydrates which cannot be further hydrolysed to simpler molecules. The general formula of monosaccharides is (CH2O)n where n = 3-7.

Question 2.
What are reducing sugars?
Answer:
Carbohydrates which reduce Fehling’s solution to red precipitate of Cu2O or Tollen’s reagent to metallic Ag are called reducing sugars. All monosaccharides (both aldoses and ketoses) and disaccharides except sucrose are reducing sugars.

Question 3.
Write two main functions of carbohydrates in plants.
Answer:
Two main functions of carbohydrates in plants are:

  1. Polysaccharides such as starch serve as storage molecules.
  2. Cellulose, a polysaccharide, is used to build the cell wall.

Question 4.
Classify the following into monosaccharides and disaccharides. Ribose, 2-deoxyribose, maltose, galactose, fructose and lactose
Answer:
Monosaccharides: Ribose, 2-deoxyribose, galactose, fructose
Disaccharides: Maltose, lactose.

PSEB 12th Class Chemistry Solutions Chapter 14 Biomolecules

Question 5.
What do you understand by the term glycosidic linkage?
Answer:
Glycosidic linkage refers to the linkage formed between two monosaccharide units through an oxygen atom by the loss of a water molecule. For example, in a sucrose molecule, two monosaccharide units, α-D-glucose and β-D-fructose, are joined together by a glycosidic linkage.
PSEB 12th Class Chemistry Solutions Chapter 14 Biomolecules 1

Question 6.
What is glycogen? How is it different from starch?
Answer:
Glycogen is a carbohydrate (polysaccharide). In animals, carbohydrates are stored as glycogen. Starch is a carbohydrate consisting of two components-amylose (15 – 20%) and amylopectin (80 – 85%). However, glycogen consists of only one component whose structure is similar to amylopectin. Also, glycogen is more branched than amylopectin. Glycogen chain consists of 10-14 glucose units whereas amylopectin chains consist of 20-25 glucose units.

Question 7.
What are the hydrolysis products of (i) sucrose and (ii) lactose?
Answer:
(i) On hydrolysis, sucrose gives one molecule of α-D glucose and one molecule of β-D-fructose.
PSEB 12th Class Chemistry Solutions Chapter 14 Biomolecules 2
(ii) On hydrolysis, lactose gives β-D-galactose and β-D-glucose.
PSEB 12th Class Chemistry Solutions Chapter 14 Biomolecules 3

Question 8.
What is the basic structural difference between starch and cellulose?
Answer:
Starch consists of two components-amylose and amylopectin. Amylose is a long linear chain of ≅α-D-(+)-glucose units joined by C1 -C4 glycosidic linkage (α-link).
PSEB 12th Class Chemistry Solutions Chapter 14 Biomolecules 4
Amylopectin is a branched-chain polymer of []α-D-glucose units, in which the chain is formed by C1 -C4 glycosidic linkage and the branching occurs by C1 – C6 glycosidic linkage.
PSEB 12th Class Chemistry Solutions Chapter 14 Biomolecules 5
On the other hand, cellulose is a straight-chain polysaccharide of β-D-glucose units joined by C1– C4 glycosidic linkage (β-link).
PSEB 12th Class Chemistry Solutions Chapter 14 Biomolecules 6

PSEB 12th Class Chemistry Solutions Chapter 14 Biomolecules

Question 9.
What happens when D-glucose is treated with the following reagents?
(i) HI
(ii) Bromine water
(iii) HNO3
Answer:
(i) HI s On prolonged heating with HI, glucose forms n-hexane.
PSEB 12th Class Chemistry Solutions Chapter 14 Biomolecules 7
(ii) Br2 water: Glucose gets oxidised (gluconic acid) on reaction with a mild oxidising agent like bromine water.
PSEB 12th Class Chemistry Solutions Chapter 14 Biomolecules 8
(iii) HNO3: On oxidation with nitric acid, glucose yields a dicarboxylic acid, that is saccharic acid.
PSEB 12th Class Chemistry Solutions Chapter 14 Biomolecules 9

Question 10.
Enumerate the reactions of D-glucose which cannot be explained by its open-chain structure.
Answer:

  • Aldehydes give 2, 4-DNP test, Schiff’s test, and react with NaHSO4 to form the hydrogen sulphite addition product. However, glucose does not undergo these reactions.
  • The pentaacetate of glucose does not react with hydroxylamine. This indicates that a free -CHO group is absent from glucose.
  • Glucose exists in two crystalline forms-α and β. The α-form (m.p. = 419 K) crystallises from a concentrated solution of glucose at 303 K and the β-form (m.p = 423 K) crystallises from a hot and saturated aqueous solution at 371 K. This behaviour cannot be explained by the open chain structure of glucose.

Question 11.
What are essential and non-essential amino acids? Give two examples of each type.
Answer:
Essential amino acids are required by the human body, but they cannot be synthesised in the body. They must be taken through food, e.g., valine and leucine. Non-essential amino acids are also required by the human body, but they can be synthesised in the body e.g., glycine and alanine.

Question 12.
Define the following as related to proteins
(i) Peptide linkage
(ii) Primary structure
(iii) Denaturation.
Answer:
(i) Peptide linkage: The amide formed between —COOH group of one molecule of an amino acid and -NH2 group of another molecule of the amino acid by the elimination of a water molecule is called a peptide linkage.
PSEB 12th Class Chemistry Solutions Chapter 14 Biomolecules 10
PSEB 12th Class Chemistry Solutions Chapter 14 Biomolecules 11
(ii) Primary structure: The primary structure of protein refers to the specific sequence in which various amino acids are present in it, i.e., the sequence of linkages between amino acids in a polypeptide chain. The sequence in which amino acids are arranged is different in each protein. A change in the sequence creates a different protein.

(iii) Denaturation: In a biological system, a protein is found to have a unique 3-dimensional structure and a unique biological activity. In such a situation, the protein is called native protein. However, when the native protein is subjected to physical changes such as change in temperature or chemical changes such as change in pH, its H-bonds are disturbed.

This disturbance unfolds the globules and uncoils the helix. As a result, the protein loses its biological activity. This loss of biological activity by the protein is called denaturation. During denaturation, the secondary and the tertiary structures of the protein get destroyed, but the primary structure remains unaltered. One of the examples of denaturation of proteins is the coagulation of egg white when an egg is boiled.

PSEB 12th Class Chemistry Solutions Chapter 14 Biomolecules

Question 13.
What are the common types of secondary structures of proteins?
Answer:
There are two common types of secondary structure of proteins:
(i) α- helix structure: In this structure, the -NH group of an amino acid residue forms H-bond with the img group of the adjacent turn of the right-handed screw (α-helix).
PSEB 12th Class Chemistry Solutions Chapter 14 Biomolecules 12
(ii) β-pleated sheet structure: This structure is called so because it looks like the pleated folds of drapery. In this structure, all the peptide chains are stretched out to nearly the maximum extension and then laid side by side. These peptide chains are held together by intermolecular hydrogen bonds.
PSEB 12th Class Chemistry Solutions Chapter 14 Biomolecules 13

Question 14.
What type of bonding helps in stabilising the α-helix structure of proteins?
Answer:
The α-helix structure of proteins is stabilised by intramolecular H-bonding between C-O of one amino acid residue and the N-H of the fourth amino acid residue in the chain.

Question 15.
Differentiate between globular and fibrous proteins.
Answer:

Globular proteins Fibrous proteins
1. These are water-soluble proteins. These are water-insoluble proteins.
2. These are spherical in shape. These are linear in shape.
3. Globular proteins are highly unstable. Fibrous proteins are stable to moderate changes in temperature and pH.

Question 16.
How do you explain the amphoteric behaviour of amino acids?
Answer:
In aqueous solution, the carboxyl group of an amino acid can lose a proton and the amino group can accept a proton to give a dipolar ion known as zwitterion.
PSEB 12th Class Chemistry Solutions Chapter 14 Biomolecules 14
Therefore, in zwitterionic form, the amino acid can act both as an acid and as a base.
PSEB 12th Class Chemistry Solutions Chapter 14 Biomolecules 15
Thus, amino acids show amphoteric behaviour.

Question 17.
What are enzymes?
Answer:
Enzymes are proteins that catalyse biological reactions. They are very specific in nature and catalyse only a particular reaction for a particular substrate. Enzymes are usually named after the particular substrate or class of substrate and sometimes after the particular reaction. For example, the enzyme used to catalyse the hydrolysis of maltose into glucose is named as maltase.
PSEB 12th Class Chemistry Solutions Chapter 14 Biomolecules 16
Again, the enzymes used to catalyse the oxidation of one substrate with the simultaneous reduction of another substrate are named as oxidoreductase enzymes. The name of an enzyme ends with ‘ – ase’.

PSEB 12th Class Chemistry Solutions Chapter 14 Biomolecules

Question 18.
What is the effect of denaturation on the structure of proteins?
Answer:
During denaturation, 2° and 3° structures of proteins are destroyed but 1° structure remains intact. Due to denaturation, the globular proteins (soluble in H20) are converted into fibrous proteins (insoluble in H2O) and their biological activity is lost. For example, boiled egg which contains coagulated proteins cannot be hatched.

Question 19.
How are vitamins classified? Name the vitamin responsible for the coagulation of blood.
Answer:
Vitamins are classified into two groups depending upon their solubility in water or fat.

  1. Water-soluble vitamins: These include vitamin B-complex (B1, B2, B5, i.e., nicotinic acid, B6, B12, pantothenic acid, biotin, i.e., vitamin H and folic acid) and vitamin C.
  2. Fat-soluble vitamins: These include vitamins A, D, E and K. These are stored in liver and adipose tissues (fat-storing tissues). Vitamin K is responsible for coagulation of blood.

Question 20.
Why are vitamin A and vitamin C essential to us? Give their important sources.
Answer:
Vitamin A: Vitamin A is essential for us because its deficiency can cause xerophthalmia (hardening of cornea of eye) and night blindness. Sources: Carrots, fish liver oil, butter and milk.
Vitamin C: Vitamin C is essential for us because its deficiency causes scurvy (bleeding gums) and pyorrhea (loosening and bleeding of teeth).
Sources: Amla, citrus fruits and green leafy vegetables.

Question 21.
What are nucleic acids? Mention their two important functions.
Answer:
Nucleic acids are biomolecules which are found in the nuclei of all living cells in the form of nucleoproteins or chromosomes (proteins containing nuclei acids as the prosthetic group).

These are of two types: deoxyribonucleic acid (DNA) and ribonucleic acid (RNA).
The two main functions of nucleic acids are :

  1. DNA is responsible for transmission of hereditary effects from one generation to another. This is because of the unique property of replication during cell division and the transfer of two identical DNA strands to the daughter cells.
  2. DNA and RNA are responsible for synthesis of all proteins essential for the growth and maintenance of our body. Actually, the proteins are synthesised by various RNA molecules (rRNA, mRNA and tRNA) in the cell but the message for the synthesis of a particular protein is present in DNA.

Question 22.
What is the difference between a nucleoside and a nucleotide?
Answer:
A nucleoside is formed when 1-position of pyrimidine (cytosine, thymine or uracil) or 9-position of purine (guanine or adenine) base is connected to C-1 of sugar (ribose or deoxyribose) by a β-linkage. Hence, in general, nucleosides may be represented as: Sugar-Base.

A nucleotide contains all the three basic compounds of nucleic acids, i.e., a phosphoric acid group, a pentose sugar and a nitrogenous base. These are obtained by esterification of C’5– OH group of the pentose sugar by phosphoric acid.

Question 23.
The two strands in DNA are not identical but are complementary. Explain.
Answer:
In the helical structure of DNA, the two strands are held together by hydrogen bonds between specific pairs of bases. Cytosine forms hydrogen bond with guanine, while adenine forms hydrogen bond with thymine. As a result, the two strands are complementary to each other.

Question 24.
Write the important structural and functional differences between DNA and RNA.
Answer:
Structural differences

DNA RNA
(i) The sugar present in DNA is 2-deoxy-D-(-)-ribose. The sugar present in RNA is D-(-)-ribose.
(ii) DNA contains cytosine and thymine as pyrimidine bases. RNA contains cytosine and uracil as pyrimidine bases.
(iii) DNA has a double-stranded a-helix structure. RNA has a single-stranded a-helix structure.
(iv) DNA molecules are very large; their molecular mass may vary from 6 x 106 -16 x 106 u. RNA molecules are much smaller with molecular mass ranging from 20,000 to 40,000 u.

Functional differences:

(i) DNA has unique property of replication. RNA usually does not replicate.
(ii) DNA controls the transmission of hereditary effects. RNA controls the synthesis of proteins.

Question 25.
What are the different types of RNA found in the cell?
Answer:
Three types of RNA present in the cell are :

  1. Messenger RNA (m-RNA)
  2. Ribosomal RNA (r-RNA)
  3. Transfer RNA (t-RNA)

PSEB 12th Class Chemistry Solutions Chapter 14 Biomolecules

Chemistry Guide for Class 12 PSEB Biomolecules Textbook Questions and Answers

Question 1.
Glucose or sucrose are soluble in water but cyclohexane or benzene (simple six-membered ring compounds) are insoluble in water. Explain.
Answer:
Glucose and sucrose have hydroxyl (-OH) groups (5 in glucose and 8 in sucrose) which form strong hydrogen bonds with water molecules and hence these are soluble in water. Cyclohexane and benzene are non-polar compounds and do not have hydroxyl groups and hence they are not able to form hydrogen bonding with water molecules and, therefore, these are not soluble in water.

Question 2.
What are the expected products of hydrolysis of lactose?
Answer:
Lactose is composed of β-D galactose and β-D glucose. Thus, on hydrolysis, it gives D-(+)- galactose and D-(+)- glucose.
PSEB 12th Class Chemistry Solutions Chapter 14 Biomolecules 17

Question 3.
How do you explain the absence of aldehyde group in the pentaacetate of D-glucose?
Answer:
D-glucose reacts with hydroxylamine (NH2OH) to form an oxime because of the presence of aldehydic (-CHO) group or carbonyl carbon. This happens as the cyclic structure of glucose forms an open-chain structure in an aqueous medium, which then reacts with NH2OH to give an oxime.
PSEB 12th Class Chemistry Solutions Chapter 14 Biomolecules 18
But pentaacetate of D-glucose does not react with NH2OH. This is because pentaacetate does not form an open-chain structure.
PSEB 12th Class Chemistry Solutions Chapter 14 Biomolecules 19

PSEB 12th Class Chemistry Solutions Chapter 14 Biomolecules

Question 4.
The melting points and solubility in water of amino acids are generally higher than that of the corresponding halo acids. Explain.
Answer:
Both acidic (carboxyl) as well as basic (amino) groups are present in the same molecule of amino acids. In aqueous solutions, the carboxyl group can lose a proton and the amino group can accept a proton, thus giving rise to a dipolar ion known as a zwitterion.
PSEB 12th Class Chemistry Solutions Chapter 14 Biomolecules 20
Due to this dipolar behaviour, they have strong electrostatic interactions within them and with water. But halo-acids do not exhibit such dipolar behaviour. For this reason, the melting points and the solubility of amino acids in water is higher than those of the corresponding halo-acids.

Question 5.
Where does the water present in the egg go after boiling the egg?
Answer:
When an egg is boiled, the proteins present inside the egg get denatured and coagulated. After boiling the egg, the water present in it is absorbed by the coagulated protein through H-bonding.

Question 6.
Why cannot vitamin C be stored in our body?
Answer:
Because vitamin C are water-soluble vitamins and so these are readily excreted through urine and hence cannot be stored in our body.

Question 7.
What products would be formed when a nucleotide from DNA containing thymine is hydrolysed?
Answer:
When a nucleotide from the DNA containing thymine is hydrolysed, thymine β-D-2-deoxyribose and phosphoric acid are obtained as products.

Question 8.
When RNA is hydrolysed, there is no relationship among the quantities of different bases obtained. What does this fact suggest about the structure of RNA?
Answer:
On complete hydrolysis of RNA six compounds are isolated. These compounds are adenine, guanine, cytosine, uracil, D-ribose and phosphoric acid. There is, no relationship among the quantities of different bases obtained on hydrolysis of RNA suggests that RNA is single-stranded. If it would have been double-stranded like DNA the complementary pairing of bases would have given equal proportion of complementary bases.

PSEB 12th Class History Book Solutions Guide in Punjabi English Medium

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Punjab State Board Syllabus PSEB 12th Class History Book Solutions Guide Pdf in English Medium and Punjabi Medium are part of PSEB Solutions for Class 12.

PSEB 12th Class History Guide | History Guide for Class 12 PSEB

History Guide for Class 12 PSEB | PSEB 12th Class History Book Solutions

PSEB 12th Class History Book Solutions in English Medium

PSEB 12th Class History Book Solutions in Hindi Medium

PSEB 12th Class History Book Solutions in Punjabi Medium

Map Question Topics
In this part there will be one question on the map. The question on the map will be compulsory. The question of Map will be set out of the following Map of Punjab.
1. Battles of Guru Gobind Singh Ji.
2. Important Battles of Banda Singh Bahadur.
3. Ranjit Singh’s Kingdom.
4. First Anglo-Sikh War.
5. Second Anglo-Sikh War.

PSEB 12th Class Chemistry Important Questions Chapter 15 Polymers

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Punjab State Board PSEB 12th Class Chemistry Important Questions Chapter 15 Polymers Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 15 Polymers

Very short answer type questions

Question 1.
Which of the following is a natural polymer? Buna-S, Proteins, PVC
Answer:
Proteins.

Question 2.
Can enzyme be called a polymer? [NCERT Exemplar]
Answer:
Enzymes are biocatalysts which are proteins and are thus polymers.

Question 3.
PSEB 12th Class Chemistry Important Questions Chapter 15 Polymers 1
Answer:
Homopolymer.

Question 4.
Identify the type of polymer -A-B-B-A-A-A-B-A- [NCERT Exemplar]
Answer:
Copolymer.

Question 5.
Out of chain growth polymerisation and step-growth polymerisation, in which type will you place the following: [NCERT Exemplar]
PSEB 12th Class Chemistry Important Questions Chapter 15 Polymers 2
Answer:
Chain growth polymerisation, as there is no loss of small molecules like water; methanol, etc.
PSEB 12th Class Chemistry Important Questions Chapter 15 Polymers 3

Question 6.
What is the role off-butyl peroxide in the polymerisation of ethene?
Answer:
It acts as a free radical generating initiator in the chain initiation step of polymerisation of ethene.

PSEB 12th Class Chemistry Important Questions Chapter 15 Polymers

Question 7.
Can nucleic acids, proteins and starch be considered as step growth polymers? [NCERT Exemplar]
Answer:
Yes, step-growth polymers are condensation polymers and they are formed by the loss of simple molecules like water leading to the formation of high molecular mass polymers.

Question 8.
Write the structure of the monomer used for getting the melamine-formaldehyde polymer.
Answer:
Melamine and formaldehyde
PSEB 12th Class Chemistry Important Questions Chapter 15 Polymers 4

Question 9.
How is the following resin intermediate prepared and which polymer is formed by this monomer unit?
PSEB 12th Class Chemistry Important Questions Chapter 15 Polymers 5
Answer:
Melamine and formaldehyde are starting materials for this intermediate. Its polymerisation gives melamine polymer.

Question 10.
Why does cis-polyisoprene possess elastic property? [NCERT Exemplar]
Answer:
cis-polyisoprene is also known’ as natural rubber. Its elastic property is due to the existence of weak van der Waals’ interactions between their various polymer chains.

Short answer type questions

Question 1.
Which of the following polymers soften on heating and harden on cooling? What are the polymers with this property collectively called? What are the structural similarities between such polymers? Bakelite, urea-formaldehyde resin, polythene, polyvinyls, polystyrene. [NCERT Exemplar]
Answer:
Polythene, polyvinyls and polystyrene soften on heating and harden on cooling. Such polymers are called thermoplastic polymers. These polymers are linear or slightly branched long-chain molecules. These possess intermolecular forces whose strength lies between strength of intermolecular forces of elastomers and fibres.

Question 2.
What is the role of benzoyl peroxide in addition polymerisation of alkenes? Explain its mode of action with the help of an example. [NCERT Exemplar]
Answer:
Role of benzoyl peroxide is to initiate the free radical polymerisation reaction which can be easily understood by taking an example of polymerisation of ethene of polythene.
(i) Chain initiation
PSEB 12th Class Chemistry Important Questions Chapter 15 Polymers 6

(ii) Chain propagation
PSEB 12th Class Chemistry Important Questions Chapter 15 Polymers 7

(iii) Chain terminator step
PSEB 12th Class Chemistry Important Questions Chapter 15 Polymers 8

Question 3.
Low-density polythene and high-density polythene, both are polymers of ethene but there is marked difference in their properties. Explain. [NCERT Exemplar]
Answer:
Low density and high-density polythenes are obtained under different conditions. These differ in their structural features. Low-density polythenes are highly branched structures while high-density polythene consists of closely packed linear molecules. Close packing increases the density.

PSEB 12th Class Chemistry Important Questions Chapter 15 Polymers

Question 4.
Differentiate between rubbers and plastics on the basis of intermolecular forces. [NCERT Exemplar]
Answer:
Rubber is a natural polymer which possess elastic properties. Natural polymer is a linear polymer of isoprene (2-methyl-1, 3-butadiene).
In natural rubber cis-polyisoprene molecules consists of various chains held together by weak van der Waals’ interaction and has coiling structure. So, it can be stretched like a spring.
PSEB 12th Class Chemistry Important Questions Chapter 15 Polymers 9
Plastics are generally polymers of ethene known as polythene. Polythene is thermoplastic polymer which may be linear (HDP) or branched (LDP) these type of polymers. Possesses intermediate intermolecular forces of attraction. It has linear, structure that can be moulded but can’t be regained on its original shape after stretching.

Question 5.
A natural linear polymer of 2-methyl-l, 3-butadiene becomes hard on treatment with sulphur between 373 to 415 K and -S-S- bonds are formed between chains. Write the structure of the product of this treatment? [NCERT Exemplar]
Answer:
The product is called vulcanised rubber. Its structure is as follows :
PSEB 12th Class Chemistry Important Questions Chapter 15 Polymers 10

Question 6.
Name the type of reaction involved in the formation of the following polymers from their respective monomers
(i) PVC.
(ii) Nylon6.
(iii) PHBV.
Answer:
(i) Addition
(ii) Condensation/Hydrolysis
(iii) Condensation.

Long answer type questions

Question 1.
Explain the following terms giving a suitable example for each:
(i) Elastomers
(ii) Condensation polymers
(iii) Addition polymers
Answer:
(i) Elastomers: These are the polymers having the weakest intermolecular forces of attraction between the polymer chains. The weak forces permit the polymer to be stretched. A few ‘cross links’ are introduced between the chains, which help the polymer to retract to its original position after the force is released as in vulcanised rubber. Elastomers thus possess an elastic character. For example, buna-S, buna-N, neoprene, etc.

(ii) Condensation polymers: The condensation polymers are formed by the repeated condensation reaction between different bi-functional or tri-functional monomer units usually with elimination of small molecules such as water, alcohol, hydrogen chloride, etc. For example, Nylon-6,6, nylon 6, terylene, etc.

(iii) Addition polymers: Addition polymers are formed by repeated addition of same or different monomer molecules. The monomers used are unsaturated compounds. For example, alkenes alkadienes and their derivatives. Polythene is an example of addition polymer.
PSEB 12th Class Chemistry Important Questions Chapter 15 Polymers 11

PSEB 12th Class Chemistry Important Questions Chapter 15 Polymers

Question 2.
Phenol and formaldehyde undergo condensation to give a polymer (A) which on heating with formaldehyde gives a thermosetting polymer (B). Name the polymers. Write the reactions involved in the formation of (A). What is the structural difference between two polymers? [NCERT Exemplar]
Answer:
Phenol and formaldehyde undergo condensation to give a polymer novolac (A) which on heating with formaldehyde gives bakelite (B) as a thermosetting polymer.
A sequence of the reaction can be written as follows :
PSEB 12th Class Chemistry Important Questions Chapter 15 Polymers 12
Structural difference in between these two is that novolac is a linear polymer while bakelite is a cross-linked polymer.
PSEB 12th Class Chemistry Important Questions Chapter 15 Polymers 13

PSEB 12th Class Chemistry Important Questions Chapter 1 The Solid State

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Punjab State Board PSEB 12th Class Chemistry Important Questions Chapter 1 The Solid State Important Questions and Answers.

PSEB 12th Class Chemistry Important Questions Chapter 1 The Solid State

Very short answer type questions

Question 1.
Give an example each of a molecular solid and an ionic solid.
Answer:
Molecular solids: CO2, I2, HCl
Ionic solids: NaCl, ZnS, CaF2

Question 2.
Why does the window glass of the old buildings look milky?
Answer:
It is due to heating during the day and cooling at night, i.e., due to annealing over a number of years, glass acquires crystalline character.

Question 3.
What would be the nature of solid if there is no energy gap between valence band and conduction band?
Answer:
Conductor.

PSEB 12th Class Chemistry Important Questions Chapter 1 The Solid State

Question 4.
Express the relationship between atomic radius (r) and the edge length (a) in the bcc unit cell,
Answer:
Atomic radius (r) = \(\frac{\sqrt{3}}{4}\) a (edge length of unit cell).

Question 5.
Which point defect in its crystal units increases the density of a solid?
Answer:
Interstitial defect.

Question 6.
What is meant by the term ‘forbidden zone’ in reference to band theory of solids?
Answer:
The energy gap between valence band and conduction band is known as forbidden zone.

Question 7.
‘Crystalline solids are anisotropic in nature.’ What does this statement mean?
Answer:
It means that some of their physical properties like electrical conductivity, refractive index, etc., are diferent in different directions.

Question 8.
Why does the electrical conductivity of semiconductors increase with rise in temperature? [NCERT Exemplar)
Answer:
The gap between conduction band and valence band is small in semiconductors. Therefore, electrons from the valence band can jump to the conduction band on increasing temperature. Thus, they become more conducting as the temperature increases.

Question 9.
Why does table salt NaCl sometimes appear yellow in colour?
(NCERT Exemplar)
Answer:
Yellow colour in NaCl is due to metal excess defect due to which unpaired electrons occupy anionic sites, known as F-centres. These electrons absorb energy from the visible region for the excitation which makes crystal appear yellow.

PSEB 12th Class Chemistry Important Questions Chapter 1 The Solid State

Question 10.
Why are liquids and gases categorised as fluids? (NCERT Exemplar)
Answer:
Liquids and gases have the tendency to flow, i.e., their molecules can move freely from one place to another. Therefore, they are known as fluids.

Short answer type questions

Question 1.
(i) What type of stoichiometric defect is shown by KC1 and why?
(ii) What type of semiconductor is formed when silicon is doped with As?
(iii) Which one of the following is an example of molecular solid : CO2 or SiO2?
Answer:
(i) KCl shows Schottky defect as the cation, K+ and anion, Cl– are of almost similar sizes.
(ii) n-type semiconductor.
(iii) CO2

Question 2.
Iron has a body centred cubic unit cell with a cell dimension of 286.65 pm. The density of iron is 7.874 g cm-3. Use this information to calculate Avogadro’s number. (At. mass of Fe = 55.845 u)
Solution:
Given, a =286.65 pm = 286.65 × 10-10 cm; M = 55.845 g mol-1;
d = 7.874 g cm-3
For bee unit cell, z = 2
Substituting the values in the expression, NA = \(\frac{z \times M}{a^{3} \times d}\), we get
NA = \(\frac{2 \times 55.845 \mathrm{~g} \mathrm{~mol}^{-1}}{\left(286.65 \times 10^{-10} \mathrm{~cm}\right)^{3} \times 7.874 \mathrm{~g} \mathrm{~cm}^{-3}}\)
NA = 6.022 × 1023 mol-1

Question 3.
An element crystallises in a fee lattice with cell edge of 400 pm. The density of the element is 7 g cm-3. How many atoms are present in 280 g of the element?
Solution:
Given, a = 400 pm = 400 × 10-10 cm = 4 x× 10-8 cm
Volume of the unit cell = a3
= (4 × 10-8 cm)3 = 6.4 × 10-23cm3
PSEB 12th Class Chemistry Important Questions Chapter 1 The Solid State 1
Since each fee unit cell contains 4 atoms, therefore, the total number of atoms in 280 g = 4 × 6.25 × 1023 = 2.5 × 1024 atoms

PSEB 12th Class Chemistry Important Questions Chapter 1 The Solid State

Question 4.
An element crystallises in a fee lattice with cell edge of 400 pm. Calculate the density if 200 g of this element contain 2.5 × 1024 atoms.
PSEB 12th Class Chemistry Important Questions Chapter 1 The Solid State 2
Molar mass M = 48.18 g mol-1
Here, z = 4, M = 48.18 g mol-1, NA = 6.022 × 1023 mol-1
a = 400 pm = 400 × 10-10 cm = 4 × 10-8 cm
Substituting these values in the expression,
d = \(\frac{z \times M}{a^{3} \times N_{A}}\) , we get
d = \(\frac{4 \times 48.18 \mathrm{~g} \mathrm{~mol}^{-1}}{\left(4 \times 10^{-8} \mathrm{~cm}\right)^{3} \times\left(6.022 \times 10^{23} \mathrm{~mol}^{-1}\right)}\) = 5 g cm-3

Question 5.
Explain why does conductivity of germanium crystals increase on doping with galium?
Answer:
On doping germanium with galium some of the positions of lattice of germanium are occupied be galium. Galium atom has only three valence electrons. Therefore, fourth valency of nearly germanium atom is not satisfied. The place remains vacant. This place is deficient of electrons and is therefore called electron hole or electron vacancy. Electron from neighbouring atom comes and fills the hole, thereby creating a hole in its original position. Under the influence of electric field electrons move towards positively charged plates through these and conduct electricity. The holes appear to move towards negatively charged plates.

Long answer type questions

Question 1.
A sample of ferrous oxide has actual formula Fe0.93O1.00In this
sample what fraction of metal ions are Fe2+ ions? What type of non-stoichiometric defect is present in this sample? (NCERT Exemplar)
Solution:
Let the formula of the sample be (Fe2+ )x (Fe3+ )y O
On looking at the given formula of the compound
x + y = 0.93 ………….. (i)
Total positive charge on ferrous and ferric ions should balance the two units of negative charge on oxygen.
Therefore, 2x + 3y = 2 ……….. (ii)
x + \(\frac{3}{2}\)y = 1 ……………..(iii)
On subtracting equation (i) from equation (iii) we have
\(\frac{3}{2}\) y – y = 1 – 0.93 ⇒ \(\frac{1}{2}\)y = 0.07 ⇒ y = 0.14
On putting the value of y in equation (i), we get
x + 0.14 = 0.93
⇒ x = 0.93 – 0.14 ⇒ x = 0.79
Fraction of Fe2+ ions present in the sample = \(\frac{0.79}{0.93}\) = 0.849
Metal deficiency defect is present in the sample because iron is less in amount than that required for stoichiometric composition.

PSEB 12th Class Chemistry Important Questions Chapter 1 The Solid State

Question 2.
(i) Following is the schematic alignment of magnetic moments:
PSEB 12th Class Chemistry Important Questions Chapter 1 The Solid State 3
Identify the type of magnetism. What happens when these substances are heated?
(ii) If the radius of the octahedral void is ‘r’ and radius of the atoms in close packing is ‘R’ What is the relation between ‘r’ and ‘R’?
(iii) Tungsten crystallises in body centred cubic unit cell. If the edge of the unit cell is 316.5 pm. What is the radius of tungsten atom?
Answer:
(i) The given schematic alignment of magnetic moments shows ferrimagnetism. When these substances are heated they lose ferrimagnetism and become paramagnetic.

(ii) The radius of the octahedral void = r
The radius of the atoms in close packing = R
Relation between r and R is given as :
r = 0.414 R

(iii) Given, a = 316.5 pm
Vs
We know that for body centred cubic unit cell r = \(\frac{\sqrt{3}}{4}\) a
∴ r = \(\frac{\sqrt{3}}{4}\) × 316.5 pm = 136.88 pm

PSEB 12th Class Chemistry Important Questions Chapter 1 The Solid State

Question 3.
(i) Identify the type of defect shown in the following figure:
PSEB 12th Class Chemistry Important Questions Chapter 1 The Solid State 4

What type of substances show this defect?
(ii) A metal crystallises in a body centred cubic structure. If ‘a’ is the edge length of its unit cell, ‘r’ is the radius of the sphere. What is the relationship between ‘r’ and ‘a’?
(iii) An element with molar mass 63 g/ mol forms a cubic unit cell with edge length of 360.8 pm. If its density is 8.92 g/cm 3. What is the nature of the cubic unit cell?
Answer:
(i) The given figure shows Schottky defect. This defect arises when equal number of cations and anions are missing from the lattice. It is a common defect in ionic compounds of high coordination number.
Where born cations and anion are of the same size, e.g., KCl, NaCl, KBr etc.

(ii) Edge length of the unit cell = a
Radius of the sphere = r
For body centred cubic structure ,
r = \(\frac{\sqrt{3}}{4}\)

(iii) We know that,
Density d = \(\frac{z \times M}{a^{3} \times N_{A}}\) or z = \(\frac{d \times a^{3} \times N_{A}}{M}\)
Given,
M = 63 g-mol-1 6.3 × 10-2 kg mol-1
a = 360.8 pm = 360.8 × 10-12 m = 3.608 × 10-10m
d = 8.92 g/cm2 = 0.892 kgm-3
NA = 6.022 × 1023 mol-1
On putting the given values in formula,
PSEB 12th Class Chemistry Important Questions Chapter 1 The Solid State 5
= 3.97 ≅ 4.
Since, 4 atoms of the element present per unit cell. Hence, the cubic unit cell must be face-centered.

PSEB 12th Class Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes

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Punjab State Board PSEB 12th Class Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes Important Questions and Answers.

PSEB 12th Class Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes

Very Short Answer Type Questions

Question 1.
Out of PSEB 12th Class Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 1 and PSEB 12th Class Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 2, which is an example of allylic halide?
X
Answer:
PSEB 12th Class Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 1 is an example of allylic halide.

Question 2.
Which of the following reactions is SN1 type ?
PSEB 12th Class Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 3
Answer:
Reaction (ii) is SN1 reaction.

Question 3.
Which one of the following compounds is more easily hydrolysed by KOH and why?
CH3CHClCH2CH3 or CH3CH2CH2CH2Cl
Answer:
Due to +1 effect of alkyl groups the 2° carbonium ion CH3—CH—CH2—CH3 derived from sec-butyl chloride is more stable than the 1° carbonium ion \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\stackrel{+}{\mathrm{C}} \mathrm{H}_{2}\) derived from n-propyl chloride. Therefore sec-butyl chloride gets hydrolysed more easily than n-propyl chloride under SN1 conditions.

PSEB 12th Class Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes

Question 4.
What is known as a racemic mixture ? Give an example.
Answer:
equimolar mixture of a pair of enantiomers is called racemic mixture. A racemic mixture is optically inactive due to external compensation.

Question 5.
Consider the three types of replacement of group X by group Y as shown here.
PSEB 12th Class Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 4
This can result in giving compound (A) or (B) or both. What is the process called if
(A) is the only compound obtained ?
(B) is the only compound obtained ?
(A) and (B) are formed in equal proportions ?
Answer:
(i) Retention
(ii) Inversion
(iii) Racemisation.

Question 6.
What is an asymmetric carbon?
Answer:
A carbon which is attached to four different atoms/groups is called asymmetric carbon. For example, the carbon atom in BrCHClI.

PSEB 12th Class Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes

Question 7.
What is plane polarized light?
Answer:
A beam of light which has vibration in only one plane is called plane polarized light.

Question 8.
Why iodoform has appreciable antiseptic property ?
Answer:
Iodoform liberate I2 when it comes in contact with skin. Antiseptic property of iodine is due to the liberation of I2 not because of iodoform itself.
PSEB 12th Class Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 5 (responsible for antiseptic property)

Question 9.
How does the ordinary light differ from the plane polarized light?
Answer:
Ordinary light has oscillations in all the directions perpendicular to the path of propagation whereas plane polarised light has all oscillations in the same plane.

PSEB 12th Class Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes

Question 10.
What do you .understand by the term optical activity of compounds?
Answer:
The property of certain compounds to rotate the plane of polarzed light in a characteristic way when it is passed through their solutions is called optical activity of compounds.

Short Answer Type Questions

Question 1.
Give reasons for the following:
(i) Ethyl iodide undergoes SN2 reaction faster than ethyl bromide.
(ii) (±) 2-Butanol is optically inactive.
(iii) C—X bond length in halobenzene is smaller than C—X bond length in CH3 —X.
Answer:
(i) Since I– ion is a better leaving group than Br– ion, hence, CH3I reacts faster than CH3Br in SN2 reaction with OH– ion.

(ii) (±) 2-Butanol is a racemic mixture, i.e., there are two enantiomers in equal proportions. The rotation by one enantiomer will be cancelled by the rotation due to the other isomer, making the mixture optically inactive.

(iii) In CH3—X the carbon atom is sp2-hybridised while in halobenzene the carbon atom is sp3-hybridised. The sp2-hybridised carbon is more electronegative due to greater s-character and holds the electron pair of C—X bond more tightly than sp3-hybridised carbon with less s-character. Thus, C—X bond length in CH3—X is bigger than C—X in halobenzene.

Question 2.
Give reasons for the following:
(i) Haloalkanes easily dissolve in organic solvents.
(ii) Halogen compounds used in industry as solvents are alkyl chlorides rather than bromides and iodides.
Answer:
(i) Haloalkanes dissolve in organic solvents because the new intermolecular attractions between haloalkanes and organic solvent molecules have much the same strength as ones being broken in the separate haloalkanes and solvent molecules.
(ii) Because alkyl chlorides are more stable and more volatile than bromides and iodides.

PSEB 12th Class Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes

Question 3.
Write the mechanism of the following reaction
PSEB 12th Class Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 6
Answer:
SN2 mechanism
PSEB 12th Class Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 7

Question 4.
Which would undergo SN1 reaction faster in the following pairs and why ?
PSEB 12th Class Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 8
Answer:
(i) PSEB 12th Class Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 9
Tertiary halide reacts faster than primary halide because of greater stability of 3°-carbocation.

(ii) PSEB 12th Class Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 10
Because the secondary carbocation formed in the slowest step is more stable than the primary carbocation.

Question 5.
Give reasons:
(i) n-Butyl bromide has higher boiling point than f-butyl bromide. Racemic mixture is optically inactive.
(ii) The presence of nitro group (—NO2) at o/p positions increases the reactivity of haloarenes towards nucleophilic substitution reactions.
Answer:
(i) n-butyl bromide being a straight chain alkyl halide has larger surface area than tert butyl bromide. Larger the surface area, larger the magnitude of the van der Waal’s forces and hence higher is the boiling point.

(ii) A racemic mixture contains the two enantiomers d and l in equal •proportions. As the rotation due to one enantiomer is cancelled by equal and opposite rotation of another enantiomer, therefore, it is optically inactive.

(iii) The presence of NO2 group at o/p position in haloarenes helps in the stabilisation of resulting carbanion by -R and -I effects and hence increases the reactivity of haloarenes towards nucleophilic substitution reactions.

PSEB 12th Class Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes

Question 6.
(i) Why are alkyl halides insoluble in water ?
(ii) Why is butan-l-ol optically inactive but butan-2-ol is optically active ?
(iii) Although chlorine is an electron withdrawing group, yet it is ortho, para directing in electrophilic aromatic substitution reactions. Why ?
Answer:
(i) This is due to the inability of alkyl halide molecule to form intermolecular hydrogen bonds with water molecules.
(ii) PSEB 12th Class Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 11
due to presence of a chiral carbon butan-2-ol is an optically active compound.

(iii) As the weaker resonance (+R) effect of Cl which stabilise the carbocation formed tends to oppose the stronger inductive (-I) effect of Cl which destabilise the carbocation at ortho and para positions and makes deactivation less for ortho and para position.

Question 7.
(i) Allyl chloride can be distinguished from vinyl chloride by NaOH and silver nitrate test. Comment.
(ii) Alkyl halide reacts with lithium aluminium hydride to give alkane. Name the attacking reagent which will bring out this change.
Answer:
(i) Vinyl chloride does not respond to NaOH and silver nitrate test because of partial double bond character due to resonance.
(ii) Hydride ion (H )

Question 8.
Give the ITJPAC name of the product formed when :
(i) 2-Methyl-l-bromopropane is treated with sodium in the presence of dry ether.
(ii) 1-Methyl cyclohexene is treated with HI.
(iii) Chloroethane is treated with silver nitrite.
Answer:
(i) 2, 5-dimethylhexane
(ii) 1 -Methyl-1 -iodocyclohexane
(iii) Nitroethane

PSEB 12th Class Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes

Long Answer Type Questions

Question 1.
Some alkyl halides undergo substitution whereas some undergo elimination reaction on treatment with base. Discuss the structural features of alkyl halides with the help of examples which are responsible for this difference.
Answer:
Primary alkyl halides follow SN2 mechanism in which a nucleophile attacks at 180° to the halogen atom. A transition state is formed in which carbon is bonded to two nucleophiles and finally halogen atom is pushed out. In SN2 mechanism, substitution of nucleophile takes place as follows :
PSEB 12th Class Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 12
Thus, in SN2 mechanism, substitution takes place. Tertiary alkyl halides follow? SN1 mechanism, In this case, tert-alkyl halides form 3° carbocation. Now, if the reagent used is a weak base then substitution occur while if it is a strong base then instead of substitution elimination occur.
PSEB 12th Class Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 13
As alc. KOH is a strong base, so elimination competes over substitution and alkene is formed.

PSEB 12th Class Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes

Question 2.
Why are aryl halides less reactive towards nucleophilic substitution reactions than alkyl halides ? How can we enhance the reactivity of aryl halides ?
Answer:
Aryl halides are less reactive towards nucleophilic substitution reaction due to the following reasons :
(i) In haloarenes, the lone pair of electron on halogen are in resonance with benzene ring. So, C—Cl bond acquires partial double bond character which strengthen C—Cl bond. Therefore, they are less reactive towards nucleophilic substitution reaction.
PSEB 12th Class Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 14
(ii) In haloarenes, the carbon atom attached to halogen is sp2-hybridised. The sp2-hybridised carbon is more electronegative than sp3-hybridised carbon. This sp 2-hybridised carbon in haloarenes can hold the electron pair of C—X bond more tightly and make this C—Cl bond shorter than C—Cl bond haloalkanes.

(iii) Since, it is difficult to break a shorter bond than a longer bond therefore haloarenes are less reactive than haloarenes.
In haloarenes, the phenyl cation will not be stabilised by resonance therefore SN1 mechanism ruled out.

(iv) Because of the repulsion between the nucleophile and electron rich arenes, aryl halides are less reactive than alkyl halides.

The reactivity of aryl halides can be increased by the presence of an electron withdrawing group (—NO2) at ortho and para positions. However, no effect on reactivity of haloarenes is observed by the presence of electron withdrawing group at meta-position. Mechanism of the reaction is as depicted with OH– ion.
PSEB 12th Class Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 15
PSEB 12th Class Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 16
PSEB 12th Class Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 17
From the above resonance, it is very clear that electron density is rich at ortho and para positions. So, presence of EWG will facilitate nucleophilic at ortho and para positions not on meta position.