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Punjab State Board PSEB 11th Class Chemistry Book Solutions Chapter 14 Environmental Chemistry Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Chemistry Chapter 14 Environmental Chemistry

PSEB 11th Class Chemistry Guide Environmental Chemistry InText Questions and Answers

Question 1.
Define environmental chemistry.
Answer:
Environmental chemistry is the study of chemical and biochemical processes occurring in nature. It deals with the study of origin, transport, reaction, effects, and fates of various chemical species in the environment.

Question 2.
Explain tropospheric pollution in 100 words.
Answer:
Tropospheric pollution occurs due to the presence of undesirable solid or gaseous particles in the air. The major gaseous and particulate pollutants present in the troposphere are :
(i) Gaseous air pollutants : These are oxides of sulphur, nitrogen and carbon, hydrogen sulphide, hydrocarbons, ozone and other oxidants.
(ii) Particulate pollutants : These are dust, mist, fumes, smoke, smog, etc.

Gaseous Air Pollutants
(a) Oxides of sulphur : These are produced when sulphur containing fossil fuel is burnt. S02 gas is poisonous to both animals and plants.
(b) Oxides of nitrogen : These are produced by the reaction of nitrogen and oxygen at high altitudes when lightning strikes.

(c) Hydrocarbons : Incomplete combustion of fuel used in automobiles is the major source for the release of hydrocarbon. These are carcinogenic and cause cancer. They also harm plants.
(d) Oxides of carbon : Carbon monoxide is one of the most serious air pollutants. It is highly poisonous to living beings because it blocks the supplyof oxygen to the organs and tissues. It is produced due to the incomplete combustion of carbon.
Carbon dioxide is the main contributor towards green house effect and global warming. It is released into the atmosphere by respiration, burning of fossil fuels and by decomposition of limestone during cement manufacturing.

Question 3.
Carbon monoxide gas is more dangerous than carbon dioxide gas. Why?
Answer:
Carbon dioxide (CO₂) and carbon monoxide (CO) gases are emitted during the combustion of various fuels. Carbon monoxide is poisonous, whereas carbon dioxide is non-toxic in nature.
Carbon monoxide is poisonous because it is capable of forming a complex with haemoglobin (carboxyhaemoglobin), which is more stable than the
oxygen-heamoglobin complex. The concentration range (3-4% )of carboxyhaemoglobin decreases the oxygen-carrying capacity of blood. This results in headaches, weak eyesight, nervousness, and cardiovascular disorders. A more increased concentration may even lead to death.
Carbon dioxide is not poisonous. It proves harmful only at very high concentrations.

Question 4.
List gases which are responsible for greenhouse effect.
Answer:
The major greenhouse gases are:
1. Carbon dioxide (CO₂)
2. Methane (CH₄)
3. Nitrous oxide (NO)
4. Ozone (O₃)
5. Chlorofluorocarbons (CFCs)

Question 5.
Statues and monuments in India are affected by acid rain. How?
Answer:
Acid rain is a byproduct of various human activities that leads to the emission of oxides of sulphur and nitrogen in the atmosphere. These oxides undergo oxidation and then react with water vapour to form acids.
2SO₂(g) + O₂(g) + 2H₂O(l) > 2H₂SO₄(aq)
4NO₂(g) + O₂(g) + 2H₂O(l) > 4HNO₃(aq)
Acid rain causes damage to buildings and structures made of stone and metal.
In India, limestone is a major stone used in the construction of various monuments and statues, including the Taj Mahal.
Acid rain reacts with limestone as:
CaCO₃ + H₂SO₄ > CaSO₄ + H₂O + CO₂
This results in the loss of lustre and colour of monuments, leading to their disfiguration.

Question 6.
What is smog? How is classical smog different from photochemical smog?
Answer:
Smog is a kind of air pollution. It is the blend of smoke and fog. There are two kinds of smog.
(a) Classical smog
(b) Photochemical smog ^
The two smogs can be differentiated as follows :
table

Question 7.
Write down the reactions involved during the formation of photochemical smog.
Answer:
Photochemical smog is formed as a result of the reaction of sunlight with hydrocarbons and nitrogen oxides. Ozone, nitric oxide, acrolein, formaldehyde, and peroxyacetyl nitrate (PAN) are common components of photochemical smog. The formation of photochemical smog can be summarized as follows:

Burning of fossil fuels leads to the emission of hydrocarbons and nitrogen dioxide in the atmosphere. High concentrations of these pollutants in air results in their interaction with sunlight as follows:

Question 8.
What are the harmful effects of photochemical smog and how can they be controlled?
Answer:
Effects of photochemical smog : Photochemical smog is oxidizing smog owing to the presence of N02 and 03 causing corrosion of metals, stones, rubber, and painted surfaces. The other major components of photochemical smog are PAN, acrolein, and formaldehyde. Both PAN and ozone are eye irritants, while nitric oxide (formed from NO₂) causes nose and throat irritation. At higher concentrations, photochemical smog causes chest pain, headaches, throat dryness, and various respiratory ailments.
Control measures : Photochemical smog results from the burning of fossil fuels and automobile fuels that emit NO₂ and hydrocarbons, which in turn form ozone, PAN and other chemicals. The use of catalytic converters in automobiles is recommended to prevent the release of NO₂ and hydrocarbons into the atmosphere.
Plantation of plants such as Finns, Juniparus, Quercus, Pyrus, and Vitis is also advised as these plants have the capability to metabolize NO₂.

Question 9.
What are the reactions involved for ozone layer depletion in the stratosphere?
Answer:
In the stratosphere, ozone is a product of the action of UV radiations on dioxygen as:

Reaction (ii) indicates the dynamic equilibrium existing between the production and decomposition of ozone molecules. Any factor that disturbs the equilibrium may cause depletion of ozone layer by its decomposition. One such factor is the release of chlorofluorocarbon compounds (CFCs). These are non-reactive, non-flammable molecules that are used in refrigerators, air conditioners, plastics, and electronic industries.
Once released CFCs mix with atmospheric gases and reach the stratosphere, where they are decomposed by UV radiations.

The chlorine free radical produced in reaction (iii) reacts with ozone as:

The radicals further react with atomic oxygen to produce more chlorine radicals as:

The regeneration of causes a continuous breakdown of ozone present in the stratosphere damaging the ozone layer.

Question 10.
What do you mean by ozone hole? What are its consequences? Ans. In Polar regions, stratospheric clouds provide the surface for chlorine nitrate and hypochlorous acid, which react further to give molecular chlorine. Molecular chlorine and H0C1 are photolysed to give chlorine-free radicals.

Hence, a chain reaction is initiated. The chlorine-free radical is continuously regenerated, thereby depleting the ozone layer. This phenomenon is known as the ozone hole.

Effects of depletion of ozone layer : The ozone layer protects the Earth from the harmful UV radiations of the sun. With the depletion of the layer, more radiation will enter the Earth’s atmosphere. UV radiations are harmful because they lead to the ageing of skin, cataract, skin cancer, and sunburns. They cause death of many phytoplanktons, which leads to a decrease in fish productivity. Excess exposure may even causes mutation in plants.
Increase in UV radiations, decreases the moisture content of the soil and damages both plants and fibres.

Question 11.
What are the major causes of water pollution? Explain.
Answer:
Several human activities caused water population which leads to the presence of several undesirable substances in water.
Major water pollutants with their sources have been tabulated as follows:
table

Roles played by major pollutants are :
1. Pathogens : These water pollutants include bacteria and other organisms. They enter water from animal excreta and domestic sewage. Bacteria present in human excreta (for example, Escherichia coli and Streptococcus faecalis) cause gastrointestinal diseases.
2. Organic wastes : These are biodegradable wastes that pollute water as a result of run off. The presence of excess organic wastes in water decreases the amount of oxygen held by water. This decrease in the amount of dissolved oxygen inhibits aquatic life.
3. Chemical pollutants : These are water soluble chemicals like heavy metals such as cadmium, mercury, nickel, etc. The presence of these chemicals (above the tolerance limit) can damage the kidneys, central nervous system, and liver.

Question 12.
Have you ever observed any water pollution in your area? What measures would you suggest to control it?
Answer:
Water pollution arises as a result of various human activities. This includes discharges from waste water treatment plants, run-off from agricultural fields, storm water drainage, etc. Pollutants from these sources enter the water bodies, thereby contaminating the water and rendering it impure.
Industries and chemical factories discharge toxic, heavy metals such as Fe, Mn, Al, etc., along with organic wastes into water. Domestic sewage and animal excreta are also responsible for pathogenic contamination of water.
These pollutants make water unfit for drinking.
Therefore, all industrial and chemical discharges should be made free from toxic metals before allowing them to enter a water body. The concentration of these pollutants should be checked regularly. Compost should be preferred over chemical fertilizers in gardens and agricultural fields to avoid harmful chemicals from entering ground water.

Question 13.
What do you mean by Biochemical Oxygen Demand (BOD)?
Answer:
Biochemical oxygen demand is the amount of oxygen required by bacteria to decompose organic matter in a certain volume of sample of water. Clean water would have a BOD value of less than 5 ppm, whereas highly polluted water has a BOD value of 17 ppm or more.

Question 14.
Do you observe any soil pollution in your neighbourhood? What efforts will you make for controlling the soil pollution?
Answer:
Major sources of soil pollution are industrial wastes and agricultural pollutants such as pesticides, fertilizers, etc.
It is very important to maintain the quality and fertility of soil to ensure and sustain the growth of plants and food crops.
Insecticides like DDT are not soluble in water. For this reason, they remain in soil for a long time contaminating the root crops. Pesticides like Aldrin and Dieldrin are non-biodegradable and highly toxic in nature. They can enter the higher trophic levels through food chains, causing metabolic and physiological disorders. The same is true for industrial wastes that comprises of several toxic metals like Pb, As, Hg, Cd, etc.
Hence, the best way to check soil pollution is to avoid direct addition of pollutants to the soil. Also, wastes should undergo proper treatment. They should be recycled and only then, allowed to be dumped. i

Question 15.
What are pesticides and herbicides? Explain giving examples.
Answer:
Pesticides are a mixture of two or more substances. They are used for killing pests. Pests include insects, plant pathogens, weeds, molluscs, etc., that destroy the plant crop and spread diseases. Aldrin and dieldrin are the names of some common pesticides.
Herbicides are pesticides specially meant for killing weeds. For example, sodium chlorate (NaClO₃), sodium arsenite (Na₃AsO₃) etc.

Question 16.
What do you mean by green chemistry? How will it help in reducing environmental pollution?
Answer:
Green chemistry is a production process that aims at using the existing knowledge and principles of chemistry for developing and implementing chemical products and processes to reduce the use and generation of substances hazardous to the environment.
The release of different harmful chemicals (particulates, gases, organic and inorganic wastes) causes environmental pollution. In green chemistry, the reactants to be used in chemical reactions are chosen in such a way that the yield of the end products is up to 100%. This prevents or limits chemical pollutants from being introduced into the environment. Through the efforts of green chemists, H₂O₂ has replaced tetrachloroethane and chlorine gas in drying and bleaching of paper.

Question 17.
What would have happened if the greenhouse gases were totally missing in the earth’s atmosphere? Discuss.
Answer:
Earth’s most abundant greenhouse gases are CO₂, CH₄, O₃ , CFCs and water vapour. These gases are present near the Earth’s surface. They absorb solar energy that is radiated back from the surface of the Earth. The absorption of radiation results in the heating up of the atmosphere. Hence, greenhouse gases are essential for maintaining the temperature of the Earth for the sustenance of life.
In the absence of greenhouse gases, the average temperature of the Earth will decrease drastically, making it uninhabitable. As a result, life on Earth would be impossible.

Question 18.
A large number of fish are suddenly floating dead on a lake. There is no evidence of toxic dumping but you find an abundance of phytoplankton. Suggest a reason for the fish kill.
Answer:
The amount of dissolved oxygen present in water is limited. The abundance of phytoplanktons causes depletion of this dissolved oxygen. This is because phytoplanktons are degraded by bacteria present in water. For their decomposition they require a large amount of oxygen. Hence, they consume the oxygen dissolved in water. As a result, the BOD level of water drops below 6 ppm, inhibiting the growth of fish and causing excessive fish kill.

Question 19.
How can domestic waste be used as manure?
Answer:
Depending upon the nature of the waste, domestic waste can be segregated into two categories i.e., biodegradable and non-biodegradable. Biodegradable waste such as leaves, rotten food, etc. should be deposited in land fills, where they get decomposed aerobically and anaerobically into manure. Non-biodegradable waste (which cannot be degraded) such as plastic, glass, metal scraps etc. should be sent for recycling.

Question 20.
For your agricultural field or garden you have developed a compost producing pit. Discuss the process in the light of bad ’ odour, flies and recycling of wastes for a good produce.
Answer:
It is essential to take proper care of the compost producing pit in order to protect ourselves from bad odour and flies.
It should be kept covered to minimize bad odour and prevent flies from entering it.
The recyclable waste should not be dumped in the compost producing pit. It should be sent to the industries through vendors for recycling.

Punjab State Board PSEB 11th Class Chemistry Book Solutions Chapter 13 Hydrocarbons Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Chemistry Chapter 13 Hydrocarbons

PSEB 11th Class Chemistry Guide Hydrocarbons InText Questions and Answers

Question 1.
How do you account for the formation of ethane during the chlorination of methane?
Answer:
Chlorination of methane proceeds via a free radical chain mechanism. The whole reaction takes place in the given three steps :
Step 1: Initiation : The reaction begins with the homolytic cleavage of Cl—Cl bond as:

Step 2: Propagation : In the second step, chlorine free radicals attack methane molecules and break down the C—H bond to generate methyl radicals as:

These methyl radicals react with other chlorine free radicals to form methyl chloride along with the liberation of a chlorine free radical.

Hence, methyl free radicals and chlorine free radicals set up a chain reaction. While HC1 and CH3C1 are the major products formed, other higher
halogenated compounds are also formed as:

Step 3: Termination : Formation of ethane is a result of the termination
of chain reactions taking place as a result of the consumption of reactants as:

Hence, by this process, ethane is obtained as a by-product of chlorination of methane.

Question 2.
Write IUPAC names of the following compounds : ,
(a) CH₃CH = C(CH₃)₂
(b) CH₂ = CH — C = C — CH₃

Answer:

Question 3.
For the following compounds, write structural formulas and IUPAC names for all possible isomers having the number of double or triple bond as indicated:
(a) C₄H₈ (one double bond)
(b) C₅H₈ (one triple bond)
Answer:
(a) The following structural isomers are possible for C₄H₈ with one double bond:

(b) The following structural isomers are possible for C5H8 with one triple bond:

Question 4.
Write IUPAC names of the products obtained by the ozonolysis of the following compounds:
(i) Pent-2-ene
(ii) 3,4-Dimethylhept-3-ene
(iii) 2-Ethylbut-l-ene
(iv) 1-Phenylbut-l-ene
Answer:
(i) Pent-2-ene undergoes ozonolysis as:

The IUPAC name of product (I) is ethanal and product (II) is propanal.
(ii) 3, 4-Dimethylhept-3-ene undergoes ozonolysis as:

The IUPAC name of product (I) is butan-2-one and product (II) is pentan-2-one.
(iii) 2-Ethylbut-l-ene undergoes ozonolysis as:

The IUPAC name of product (I) is pentan-3-one and product (II) is methanal.
(iv) 1-Phenylbut-l-ene undergoes ozonolysis as:

The IUPAC name of product (I) is benzaldehyde and product (II) is propanal.

Question 5.
An alkene ‘A’ on ozonolysis gives a mixture of ethanal and pentan-3-one. Write structure and IUPAC name of ‘A’.
Answer:

Question 6.
An alkene ‘A’ contains three C — C, eight C — Hσ bonds and one C — C π bond. ‘A’ on ozonolysis gives two moles of an aldehyde of molar mass 44 u. Write IUPAC name of ‘A’
Answer:
As per the given information, ‘A’ on ozonolysis gives two moles of an aldehyde of molar mass 44 u. The formation of two moles of an
aldehyde indicates the esence of identical structural units on both sides of the double bond containing carbon atoms.
Hence, the structure of ‘A’ can be represented as :
XC = CX
There are eight C—Ha bonds. Hence, there are 8 hydrogen atoms in ‘A’. Also, there are three C—C bonds. Hence, there are four carbon atoms present in the structure of‘A’.

Combining the inferences, the structure of A’ can be represented as:

‘A’ has 3 C — C bonds, 8 C — H bonds, and one C— C it bond. Hence, the IUPAC name of ‘A’ is But-‘ 2-ene.
Ozonolysis of ‘A’ takes place as:

The final product is ethanal with molecular mass
= [(2 x 12) + (4 x 1) + (1 x 16)]= 44 u

Question 7.
Propanal and pentan-3-one are the ozonolysis products of an alkene? What is the structural formula of the alkene?
Answer:

Question 8.
Write chemical equations for combustion reaction of the following hydrocarbons:
(i) Butane
(ii) Pentene
(iii) Hexyne
(iv) Toluene
Answer:
Combustion can be defined as a reaction of a compound with oxygen.

Question 9.
Draw the cis and trans structures of hex-2-ene. Which isomer will have higher b.p. and why?
Answer:
Hex-2-ene is represented as:

The cis-isomer will have higher boiling point due to more polar nature leading to stronger intermolecular dipole-dipole interactions thus requiring more heat energy to separate them, whereas trans form being non-polar have weak induced dipole interactions and so have lower boiling point.

Question 10.
Why is benzene extra ordinarily stable though it contains three double bonds?
Answer:
Resonance and delocalization of electrons generally leads to the stability of benzene molecule.

The dotted circle in the hybrid structure represents the six electrons which are delocalised between the six carbon atoms of the benzene ring. Therefore, presence of delocalised Jt electrons in benzene makes it more stable than the hypothetical cyclohexatriene.

Question 11.
What are the necessary conditions for any system to be aromatic?
Answer:
A compound is said to be aromatic if it satisfies the following three conditions:
(i) It should have a planar structure.
(ii) The π-electrons of the compound are completely delocalized in the ring.
(iii) The total number of π-electrons present in the ring should be equal to (4n + 2), where n = 0,1, 2….etc. This is known as Huckel’s rule.

Question 12.
Explain why the following systems are not aromatic?

Answer:

For the given compound, the number of π-electrons is six. But only four Tt-electrons are present within the ring. Also there is no conjugation of π-electrons within the ring and the compound is not planar in shape. Hence, the given compound is not aromatic in nature.

For the given compound, the number of π-electrons is four.
By Huckel’s rule,
4n + 2 = 4,
4n = 2, n = 1/2
For a compound to be aromatic, the value of n must be an integer (n = 0,1,2 …), which is not true for the given compound. Hence, it is not aromatic in nature.

For the given compound, the number of it-electrons is eight.
By Huckel’s rule.
4n + 2 = 8
4n = 6, n = 3/2
For a compound to be aromatic, the value of n must be an integer (n = 0,1, 2…). Since, the value of n is not an integer, the given compound is not aromatic in nature.

Question 13.
How will you convert benzene into
(i) p-nitrobromobenzene
(ii) m-nitrochlorobenzene
(iii) acetophenone
(iv) p-nitrotoluene
Answer:
(i) Benzene can be converted into p-nitrobromobenzene as:

(ii) Benzene can be converted into m-nitrochlorobenzene as:

(iii) Benzene can be converted into acetophenone as:

(iv) Benzene can be converted into p-nitrotoluene as:

Question 14.
In the alkane H₃C — CH₂ — C(CH₃)₂ — CH₂ — CH(CH₃)₂ identify 1°, 2°, 3° carbon atoms and give the number of H atoms bonded to each one of these.
Answer:

1° carbon atoms are those which are bonded to only one carbon atom, i.e., they have only one carbon atom as their neighbour. The given structure has five 1° carbon atoms and fifteen hydrogen atoms are attached to it.
2° carbon atoms are those which are bonded to two carbon atoms, i.e., they have two carbon atoms as their neighbours. The given structure has two 2° carbon atoms and four hydrogen atoms are attached to it.
3° carbon atoms are those which are bonded to three carbon atoms, i.e., they have three carbon atoms as their neighbours. The given structure has one 3° carbon atom and only one hydrogen atom is attached to it.

Question 15.
What effect does branching of an alkane chain has on its boiling point?
Answer:
Alkanes experience inter-molecular Van der Waals forces. The stronger the force, the greater will be the boiling point of the alkane.
As branching increases, the surface area of the molecule decreases which results in a small area of contact. As a result, the Van der Waals force also decreases which can be overcome at a relatively lower temperature. Hence, the boiling point of an alkane chain decreases with an increase in branching.

Question 16.
Addition of HBr to propene yields 2-bromopropane, while in the presence of benzoyl peroxide, the same .reaction yields
1- bromopropane. Explain and give mechanism.
Answer:
Addition of HBr to propene is an example of an electrophilic substitution reaction.
Hydrogen bromide provides an electrophile, H⁺. This electrophile attacks the double bond to form 1° and 2° carbocations as shown:

Secondary carbocations are more stable than primary carbocations. Hence, the former predominates since it will form at a faster rate. Thus, in the next step, Br^– attacks the carbocation to form 2 – bromopropane as the major product.

This reaction follows Markovnikov’s rule where the negative part of the addendum is attached to the carbon atom having a lesser number of hydrogen atoms.

In the presence of benzoyl peroxide an addition reaction takes place against Markovnikov’s rule. The reaction follows a free radical chain mechanism as:

Secondary free radicals are more stable than primary radicals. Hence, the former predominates since it forms at a faster rate. Thus, 1-bromopropane is obtained as the major product.

In the presence of peroxide, Br free radical acts as an electrophile. Hence, two different products are obtained on addition of HBr to propene in the absence and presence of peroxide.

Question 17.
Why does benzene undergo electrophilic substitution reactions easily and nucleophilic substitutions with difficulty?
Answer:
Benzene is a planar molecule having delocalized electrons above and below the plane of ring. Hence, it is electron-rich. As a result, it is highly attractive to electron deficient species i.e., electrophiles. Therefore, it undergoes electrophilic substitution reactions very easily. Nucleophiles are electron-rich. Hence, they are repelled by benzene. Hence, benzene undergoes nucleophilic substitutions with difficulty.

Question 18.
Write down the products of ozonolysis of 1, 2-dimethylbenzene (o-xylene). How does the result support Kekule structure for benzene?
Answer:
o-xylene has two resonance structures:

All three products, i.e., methyl glyoxal, 1, 2-dimethyl glyoxal, and glyoxal are obtained from two Kekule structures. Since all three products cannot be obtained from any one of the two structures, this proves that o-xylene is a resonance hybrid of two Kekule structures (I and II).

Question 18.
Arrange benzene, n-hexane and ethyne in decreasing order of acidic behaviour. Also give reason for this behaviour.
Answer:
Acidic character of a species is defined on the basis of ease with which it can lose its H-atoms.
The hybridization state of carbon in the given compound is:

As the s-character increases, the electronegativity of carbon increases and the electrons of C—H bond pair lie closer to the carbon atom. As a result, partial positive charge of H-atom increases and H+ ions are set free.
The s-character increases in the order:
sp³ < sp² < sp Hence, the decreasing order of acidic behaviour is Ethyne > Benzene > Hexane.

Question 20.
How would you convert the following compounds into benzene?
(i) Ethyne (ii) Ethene (iii) Hexane
Answer:
(i) Benzene from Ethyne:

Question 21.
Write structures of all the alkenes which on hydrogenation give 2-methyl butane.
Answer:
The basic skeleton of 2-methyl butane is

Putting double bonds at various different positions and satisfying the tetracovalency of each carbon, the structures of various alkenes which give 2-methyl butane on hydrogenation are:

Question 22.
Arrange the following set of compounds in order of their decreasing relative reactivity with an electrophile, E+
(a) Chlorobenzene, 2, 4-dinitrochlorobenzene, p-nitrochlorobenzene
(b) Toluene, p-H₃C—C₆H₄—NO₂, p-O₂N — C₆H₄ — NO₂
Answer:
Electrophiles are reagents that participate in a reaction by accepting an electron pair in order to bond to nucleophiles.
The higher the electron density on a benzene ring, the more reactive is the compound towards an electrophile, E+ (Electrophilic reaction).
(a) The presence of an electron withdrawing group (i.e., —NO₂ and Cl) deactivates the aromatic ring by decreasing the electron density. Since, —NO₂ group is more electron withdrawing (due to resonance effect) than the —Cl group (due to inductive effect) the decreasing order of reactivity is as follows :
Chlorobenzene > p-nitrochlorobenzene > 2, 4-dinitrochlorobenzene
(b) While —CH₃ is an electron donating group, —NO₃ group is electron withdrawing. Hence, toluene will have the maximum electron density and is most easily attacked by E⁺.
—NO₂ is an electron withdrawing group. Hence, when the number of —NO₂ substituents is greater, the order is as follows:
Toluene > p-CH₃—C₆H₄—NO₂ > p-ON₂—C₆H₄—NO₂

Question 23.
Out of benzene, m-dinitrobenzene and toluene which will undergo nitration most easily and why?
Answer:
The ease of nitration depends on the presence of electron density on the compound to form nitrates. Nitration reactions are examples of electrophilic substitution reactions where an electron-rich species is attacked by a nitronium ion (\(\mathrm{NO}_{2}^{-}\)).
Now, —CH₃ group is electron donating and —NO₂ is electron withdrawing. Therefore, toluene will have the maximum electron density among the three compounds followed by benzene. On the other hand, m-Dinitrobenzene will have the least electron density. Hence, it will undergo nitration with difficulty. Hence, the increasing order of nitration is as follows:

Question 24.
Suggest the name of a Lewis acid other than anhydrous aluminium chloride which can be used during ethylation of benzene.
Answer:
The ethylation reaction of benzene involves the addition of an ethyl group on the benzene ring. Such a reaction is called a Friedel-Crafts alkylation reaction. This reaction takes place in the presence of a Lewis acid.
Any Lewis acid like anhydrous FeCl₃, SnCl₄, BF₃ etc. can be used during the ethylation of benzene.

Question 25.
Why is Wurtz reaction not preferred for the preparation of alkanes containing odd number of carbon atoms? Illustrate your answer by taking one example.
Answer:
Wurtz reaction is limited for the synthesis of symmetrical alkanes (alkanes with an even number of carbon atoms) in the reaction, two
similar alkyl halides are taken as reactants and an alkane, containing double the number of carbon atoms, are formed. Example:

Wurtz reaction cannot be used for the preparation of unsymmetrical alkanes because if two dissimilar alkyl halides are taken as the reactants, then a mixture of alkanes is obtained as the products. Since the reaction involves free radical species, a side reaction also occurs to produce an alkene. For example, the reaction of bromomethane and iodoethane gives a mixture of alkanes.

The boiling point of alkanes (obtained in the mixture) are very close. Hence, it becomes difficult to separate them.

Punjab State Board PSEB 11th Class Chemistry Book Solutions Chapter 12 Organic Chemistry: Some Basic Principles and Techniques Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Chemistry Chapter 12 Organic Chemistry: Some Basic Principles and Techniques

PSEB 11th Class Chemistry Guide Organic Chemistry: Some Basic Principles and Techniques InText Questions and Answers

Question 1.
What are hybridisation states of each carbon atom in the following compounds?
CH₂ = C = O, CH₃CH = CH₂, (CH₃)₂CO, CH₂ = CHCN, C₆H₆
Answer:

C — 1 is sp² hybridised.
C — 2 is sp hybridised.

C — 1 is sp³ hybridised
C — 2 is sp² hybridised
C — 3 is sp² hybridised

C — 1 and C — 3 are sp³ hybridised.
C — 2 is sp² hybridised.

C — 1 is sp² hybridised.
C — 2 is sp² hybridised.
C — 3 is sp hybridised.

(v) C₆H₆
All the 6 carbon atoms in benzene are sp² hybridised.

Question 2.
Indicate the o and n bonds in the following molecules:
C₆H₆, C₆H₁₂, CH₂Cl₂, CH₂ = C = CH₂, CH₃NO₂, HCONHCH₃
Answer:

Question 3.
Write bond line formulas for : isopropyl alcohol, 2,3-Dimethyl butanal, Heptan-4-one.
Answer:

Question 4.
Give the IUPAC names of the following compounds:

Answer:

Question 5.
Which of the following represents the correct IUPAC name for the compounds concerned?
(a) 2, 2 -Dimethylpentane or
2-Dimethylpentane (b) 2,4,7-Trimethyloctane or
2,5,7-Trimethyloctane (c) 2-Chloro-4-methylpentane or
4-Chloro-2-methylpentane (d) But-3-yn-1-ol or But-4-ol-1-yne.
Answer:
(a) The prefix di in the JUPAC name indicates that two identical substituent groups are present in the parent chain. Since two methyl groups are present in the C-2 of the parent chain of the given compound, the correct IUPAC name of the given compound is 2-2-dimethylpentane.
(b) Locant number 2, 4, 7 is lower than 2, 5, 7. Hence, the IUPAC name of the given compound is 2, 4, 7-trimethyloctane.
(c) If the substituents are present in the equivalent position of the parent chain, then the lower number is given to the one that comes first in the name according to the alphabetical order. Hence, the correct IUPAC name of the given compound is 2-chloro-4-methylpentane.
(d) Two functional groups—alcoholic and alkyne—are present in the given compound. The principal functional group is the alcoholic group. Hence, the parent chain will be suffixed with ol. The alkyne group is present in the C—3 of the parent chain. Hence, the correct IUPAC name of the given compound is But – 3-yn -1 – ol.

Question 6.
Draw formulas for the first five members of each homologous series beginning with the following compounds :
(a) H—COOH
(b) CH₃COCH₃
(c) H—CH = CH₂
Answer:
The first five members of each homologous series beginning with the given compounds are shown as follows:

Question 7.
Give condensed and bond line structural formulas and identify the functional group (s) present, if any, for :
(a) 2,2, 4-Trimethylpentane
(b) 2-Hydroxy-1, 2, 3 -propanetricarboxylic acid
(c) Hexanedial
Answer:

Question 8.
Identify the functional groups in the following compounds:

Answer:
The functional groups present in the given compounds are:
(a) Aldehyde (—CHO) Hydroxyl (—OH)
Methoxy (—OMe)
(b) Amino (—NH₂); primary amine (1° amine).
Ester (— O—CO—)
Triethylamine [N(C₂H₅)₂]; tertiary amine (3° amine).
(C) Nitro (—NO₂)
C = C double bond (—C = C—) .

Question 9.
Which of the two: 02NCH2CH20_ or CH3CH20 is expected to be more stable and why?
Answer:
NO₂ group is an electron-withdrawing group. Hence, it shows-I effect. By withdrawing the electrons toward it, the NO₂ group decreases the negative charge on the compound, thereby stabilising it. On the other hand, ethyl group is an electron-releasing group. Hence, the ethyl group shows +1 effect. This increases the negative charge on the compound, thereby destabilising it. Hence, O₂NCH₂CH₂O^– is expected to be more stable than CH₃CH₂O^–.

Question 10.
Explain why alkyl groups act as electron donors when attached to a π system,
Answer:
Due to hyperconjugation, alkyl groups act as electron donors when attached to a π-system as shown below :

Question 11.
Draw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation.
(a) C₆H₂OH
(b) C₆H₅NO₂
(c) CH₃CH = CH CHO
(d) C₆H₅ -CHO

Answer:

Question 12.
What are electrophiles and nucleophiles? Explain with examples?
Answer:
Electrophiles: An electrophile is a reagent that takes away an electron pair. In other words, an electron-seeking reagent is called an electrophile (E⁺). Electrophiles are electron-deficient and can receive an electron pair.
Neutral electrophiles :

Charged electrophiles :

Nucleophiles : A nulceophile is a reagent that brings an electron pair. In other words, a nucleus-seeking reagent is called a nucleophile.

Question 13.
Identify the reagents shown in bold italic in the following equations as nucleophiles or electrophiles:

Answer:
Electrophiles are electron-deficient species and can receive an electron pair. On the other hand, nucleophiles are electron-rich species and can donate their electrons.

Here, HO^– acts as a nucleophile as it is an electron-rich species, i.e., it is a nucleus-seeking species.

Here, CN^– acts as a nucleophile as it is an electron-rich species i.e., it is a nucleus-seeking species.

Here, acts as an electrophile as it is an electron-deficient species.

Question 14.
Classify the following reactions in one of the reaction type studied in this unit.
(a) CH₃CH₂Br + HS^– → CH₃CH₂SH + Br^–
(b) (CH₃)₂ C = CH₂ +HCl → (CH₃)₂ ClC – CH₃
(c) CH₃CH₂Br + HO^– → CH₂ = CH₂ +H₂O + Br^–
(d) (CH₃)₃C— CH₂OH +HBr →(CH₃)₂CBrCH₂CH₂CH₃ + H₂O
Ans. (a) Nucleophilic substitution reaction
(b) Electrophilic addition reaction
(c) β-elimination reaction
(d) Nucleophilic substitution reaction with rearrangement.

Question 15.
What is the relationship between the members of following pairs of structures? Are they structural or geometrical isomers or resonance contributors?

Answer:
(a) Compounds having the same molecular formula but with different structures are called structural isomers. The given compounds have the same molecular formula but they differ in the position of the functional group (ketone group.)

In structure I, ketone group is at the C—3 of the parent chain (hexane chain) and in structure II, ketone group is at the C—2 of the parent chain (hexane chain). Hence, the given pair represents structural isomers.
(b) Compounds having the same molecular formula, the same constitution, and the sequence of covalent bonds, but with different relative positions of their atoms in space are called geometrical isomers.

In structures I and II, the relative position of deuterium (D) and hydrogen (H) in space are different. Hence, the given pairs represent geometrical isomers.
(c) The given structures are canonical structures or contributing structures. They are hypothetical and individually do not represent any real molecule. Hence, the given pair represents resonance structures, called resonance isomers.

Question 16.
For the following bond cleavages, curved-arrows are used to show the electron flow and classify each as homolysis or heterolysis. Identify reactive intermediate produced as free radical, carbocation and carboanion.


Answer:
(a) The bond cleavage using curved-arrows to show the electron flow of the given reaction can be represented as
It is an example of homolytic cleavage as one of the shared pair in a covalent bond goes with the bonded atom. The reaction intermediate formed is a free radical.

(b) The bond cleavage using curved-arrows to show the electron flow of the given reaction can be represented as

It is an example of heterolytic cleavage as the bond breaks in such a manner that the shared pair of electrons remains with the carbon of propanone. The intermediate formed is carbanion.
(c) The bond cleavage using curved-arrows to show the electron flow of the given reaction can be represented as

it is an example of heterolytic cleavage as the bond breaks in such a manner that the shared pair of electrons remains with the bromine ion. The reaction intermediate formed is a carbocation.
(d) The bond cleavage using curved-arrows to show the electron flow of the given reaction can be represented as

It is a heterolytic cleavage as the bond breaks in such a manner that the shared pair of electrons remains with one of the fragments. The intermediate formed is a carbocation.

Question 17.
Explain the terms inductive and electromeric effects. Which electron displacement effect explains the following correct orders of acidity of the carboxylic acids?
(a) Cl₃CCOOH > Cl₂CHCOOH > ClCH₂COOH
(b) CH₃CH₂OOH > (CH₃)₂ CHCOOH > (CH₃)₃C. COOH
Answer:
Inductive effect : The permanent displacement of sigma electrons along a saturated chain, whenever an electron withdrawing or electron donating group is present, is called inductive effect.
Inductive effect could be +I effect or -I effect. When an atom or group attracts electrons towards itself more strongly than hydrogen, it is said to possess -I effect. For example,

When an atom or group attracts electrons towards itself less strongly than hydrogen, it is said to possess +I effect. For example,

Electromeric effect : It involves the complete transfer of the shared pair of n electrons to either of the two atoms linked by multiple bonds in the presence of an attacking agent. For example,

Electromeric effect could be + E effect or – E effect.
+E effect : When the electrons are transferred towards the attacking reagent.

-E effect : When the electrons are transferred away from the attacking reagent.

(a) Cl₃CCOOH > Cl₂CHCOOH > ClCH₂COOH
The order of acidity can be explained on the basis of inductive effect (-I effect). As the number of chlorine atoms increases, the -I effect increases. With the increase in -7 effect, the acid strength also increases accordingly.

(b) CH₃CH₂COOH > (CH₃)₂CHCOOH > (CH₃)₃C.COOH
The order of acidity can be explained on the basis of inductive effect (+I effect.) As the number of alkyl groups increases, the +I effect also increases. With the increase in +I effect, the acid strength also increases accordingly.

Question 18.
Give a brief description of the principles of the following techniques taking an example in each case.
(a) Crystallisation (b) Distillation (c) Chromatography
Answer:
(a) Crystallisation : Crystallisation is one of the most commonly used techniques for the purification of solid organic compounds. Principle : It is based on the difference in the solubilities of the compound and the impurities in a given solvent. The impure compound gets dissolved in the solvent in which it is sparingly soluble at room temperature, but appreciably soluble at higher temperature. The solution is concentrated to obtain a nearly saturated solution. On cooling the solution, the pure compound crystallises out and is removed by filtration.
For example, iodoform is crystallised with alcohol, benzoic acid mixed with naphthalene can be purified by hot water.
(b) Distillation : This method is used to separate volatile liquids from non-volatile impurities or a mixture of those liquids that have a sufficient difference in their boiling points.
Principle : It is based on the fact that liquids having different boiling points vapourise at different temperatures. The vapours are then cooled and the liquids so formed are collected separately.
For example, chloroform (b.pt. 334 K) and aniline (b.pt. 457 K) are easily separated by the distillation. On boiling the vapours of lower boiling component are formed first so it is collected first in the receiver.
(c) Chromatography : It is one of the most useful methods for the separation and purification of organic compounds.
Principle : It is based on the difference in movement of individual components of a mixture through the stationary phase under the influence of mobile phase.
For example, a mixture of red and blue ink can be separated by chromatography. A drop of the mixture is placed on the chromatogram. The component of the ink, which is less adsorbed on the chromatogram, moves with the mobile phase while the less adsorbed component remains almost stationary.

Question 19.
Describe the method, which can be used to separate two compounds with different solubilities in a solvent S.
Answer:
Fractional crystallisation is the method used for separating two compounds with different solubilities in a solvent S.
The process of fractional crystallisation is carried out in four steps.
(a) Preparation of the solution: The powdered mixture is taken in a flask and the solvent is added to it slowly and stirred simultaneously. The solvent is added till the solute is just dissolved in the solvent. This saturated solution is then heated.
(b) Filtration of the solution: The hot saturated solution is then filtered through a filter paper in a China dish.
(c) Fractional crystallisation: The solution in the China dish is now allowed to cool. The less soluble compound crystallises first while the more soluble compound remains in the solution. After separating these crystals from the mother liquor, the latter is concentrated once again. The hot solution is allowed to cool and consequently, the crystals of the more soluble compound are obtained.
(d) Isolation and drying: These crystals are separated from the mother liquor by filtration. Finally, dried the crystals.

Question 20.
What is the difference between distillation, distillation under reduced pressure and steam distillation?
Answer:
The difference among distillation, distillation under reduced pressure, and steam distillation are given in the following table :

Question 21.
Discuss the chemistry of Lassaigne’s test.
Answer:
Lassaigne’s test : This test is employed to detect the presence of nitrogen, sulphur and halogens in an organic compound. These elements are present in the covalent form in an organic compound. These are converted into the ionic form by fusing the compound with sodium metal.

The cyanide, sulphide, and halide of sodium formed are extracted from the fused mass by boiling it in distilled water. The extract so obtained is called Lassaigne’s extract. This Lassaigne’s extract is then tested for the presence of nitrogen, sulphur and halogens.

(a) Test for nitrogen

Chemistry of the test: In the Lassaignes’s test for nitrogen in an organic compound, the sodium fusion extract is boiled with iron (II) sulphate and then acidified with sulphuric acid. In the process, sodium cyanide first reacts with iron (II) sulphate and forms sodium hexacyanoferrate(II). Then, on heating with sulphuric acid, some iron (II) gets oxidised to form iron (III) hexacyanoferrate(II) which is Prussian blue in colour. The chemical equations involved in the reaction can be represented as

(b) Test for sulphur

Chemistry of the test: In the Lassaigne’s test for sulphur in an organic compound, the sodium fusion extract is acidified with acetic acid and then lead acetate is added to it. The precipitation of lead sulphide, which is black in colour, indicates the presence of sulphur in the compound.

(ii) Lassaigne’s extract + Sodium nitroprusside → Violet colour
Chemistry of the test: The sodium fusion extract is treated with sodium nitroprusside. Appearance of violet colour also indicates the presence of sulphur in the compound.

If in an organic compound, both nitrogen and sulphur are present, then instead of NaCN, formation of NaSCN takes place.

This NaSCN (sodium thiocyanate) gives a blood red colour. Prussian colour is not formed due to the absence of free cyanide ions.

(c) Test for halogens

Chemistry of the test: In the Lassaigne’s test for halogens in an organic compound, the sodium fusion extract is acidified with nitric acid and then treated with silver nitrate.
x^– + Ag⁺ → AgX (X = Cl, Br, I)
If nitrogen and sulphur both are present in the organic compound, then the Lassaigne’s extract is boiled to expel nitrogen and sulphur, which would otherwise interfere in the test for halogens.

Question 22.
Differentiate between the principle of estimation of nitrogen in an organic compound by (i) Dumas method and (ii) Kjeldahl’s method.
Answer:
(i) In Dumas method, when a known mass of the nitrogen containing organic compound is heated with excess of CuO in an atmosphere of CO₂, nitrogen of the organic compound is converted into N₂ gas. The volume of N₂ thus obtained is converted into STP and the percentage of nitrogen is determined.
%N = \(\frac{28}{22400} \times \frac{\text { Vol. of } \mathrm{N}_{2} \text { at STP }}{\text { Mass of the substance taken }}\) x 100

(ii) In Kjeldahl’s method, a known mass of the nitrogen containing
organic substance is digested (heated) with conc. H₂SO₄ and CuSO₄ (in little amount) in Kjeldahl’s flask. Nitrogen present in the organic compound is quantitatively converted into (NH₄ )₂SO₄. (NH₄ )₂SO₄ thus obtained is boiled with excess of NaOH solution to liberate NH3 gas which is absorbed in a known excess of a standard solution of H₂SO₄ or HCl.
The volume of acid left after absorption of NH₃ is estimated by titration against a standard alkali solution. From the volume of the acid used, the percentage of nitrogen is determined by applying the equation,

Question 23.
Discuss the principle of estimation of halogens, sulphur and phosphorus present in an organic compound.
Answer:
Estimation of halogens : Halogens are estimated by the Carius method. In this method, a known quantity of organic compound is heated with fuming nitric acid in the presence of silver nitrate, contained in a hard glass tube called the Carius tube, taken in a furnace. Carbon and hydrogen that are present in the compound are oxidized to form CO₂ and H₂O respectively and the halogen present in the compound is converted into AgX. This AgX is then filtered, washed, dried, and weighed.
Let the mass of organic compound taken = m g

Estimation of sulphur : In this method, a known quantity of organic compound is heated with either fuming nitric acid or sodium peroxide in a hard glass tube called the Carius tube. Sulphur present in the compound, is oxidized to form sulphuric acid. On addition of excess of barium chloride to it, the precipitation of barium sulphate takes place. This precipitate is then filtered, washed, dried, and weighed.
Let the mass of organic compound taken -mg
Mass of BaSO₄ formed = m₁ g
% of sulphur = \(\frac{32 \times m_{1} \times 100}{233 \times m}\)

Estimation of phosphorus : In this method, a known quantity of organic compound is heated with fuming nitric acid. Phosphorus, present in the compound, is oxidized to form phosphoric acid. By adding ammonia and ammonium molybdate to the solution, phosphorus can be precipitated as ammonium phosphomolybdate.

Phosphorus can also be estimated by precipitated it as MgNH4P04 by adding magnesia mixture, which on ignition yields Mg2P207.
Let the mass of organic compound = m g
Mass of ammonium phosphomolybdate = m₁ g
Molar mass of ammonium phosphomolybdate = 1877 g

Question 24.
Explain the principle of paper chromatography.
Answer:
In paper chromatography, chromatography paper is used. This paper contains water trapped in it, which acts as the stationary phase. On the base of this chromatography paper, the solution of the mixture is spotted. The paper strip is then suspended in a suitable solvent, which acts the mobile phase. This solvent rises up the chromatography paper by capillary action and in this procedure it flows over the spot. The components are selectively retained on the paper (according to their differing partition in these two phases). The spots of different components travel with the mobile phase to different heights. The paper so obtained (shown in the given figure) is known as a chromatogram.

Question 25.
Why is nitric acid added to sodium extract before adding silver nitrate for testing halogens?
Answer:
While testing the Lassaigne’s extract for the presence of halogens, it is first boiled with dilute nitric acid. This is done to decompose NaCN to HCN and NA₂S to H₂S and to expel these gases. That is, if any nitrogen and sulphur are present in the form of NaCN and Na₂S they are removed. The chemical equations involved in the reaction are represented as
NaCN + HNO₃ > NaNO₃ + HCN
Na₂S + 2HNO₃ > 2NaNO₃ + H₂S

Question 26.
Explain the reason for the fusion of an organic compound with metallic sodium for testing nitrogen, sulphur and halogens.
Answer:
Nitrogen, sulphur, and halogens are covalently bonded in organic compounds. For their detection, they have to be first converted to ionic form. This is done by fusing the organic compound with sodium metal. This is called Lassaigne’s test. The chemical equations involved in the test are
Na + C + N → NaCN
Na + S + C + N → NaSCN
2Na + S → Na₂S
Na + X → NaX (X = Cl, Br, I)
Carbon, nitrogen, sulphur, and halogen come from organic compounds.

Question 27.
Name a suitable technique of separation of the components from a mixture of calcium sulphate and camphor.
Answer:
Calcium sulphate (CaSO₄) which is an inorganic compound and does not sublime can be separated from camphor which is an organic compound and sublimes by the simple technique of sublimation. This technique involves the direct conversion of a solid into the gaseous state on heating without passing through the intervening liquid state and vice-versa on cooling. Camphor sublimes and gets-deposited on the walls of the cooler portion of the funnel and CaSO₄ remains as such. The two components can thus be separated.

Question 28.
Explain, why an organic liquid vaporises at a temperature below its boiling point in its steam distillation?
Answer:
In steam distillation, the organic liquid starts to boil when the sum of vapour pressure due to the organic liquid (p₁) and the vapour pressure due to water (p₂) becomes equal to atmospheric pressure (p), that is P = P₁ + P₂– Since P₁ < p₂, organic liquid will vapourise at a lower temperature than its boiling point.

Question 29.
Will CCl₄ give white precipitate of AgCl on heating it with silver nitrate? Give reason for your answer.
Answer:
CCl₄ will not give the white precipitate of AgCl on heating it with silver nitrate. This is because the chlorine atoms are covalently bonded to carbon in CCl₄. To obtain the precipitate, it should be present in ionic form and for this, it is necessary to prepare the Lassaigne’s extract of CCl₄.

Question 30.
Why does solution of potassium hydroxide is used to absorb carbon dioxide evolved during the estimation of carbon present in an organic compound?
Answer:
Carbon dioxide is acidic in nature and potassium hydroxide is a strong base. Hence, carbon dioxide reacts with potassium hydroxide to form potassium carbonate and water as
2KOH + CO₂ → K₂CO₃ + H₂O
Thus, the mass of the U-tube containing KOH increases. This increase in the mass of U-tube gives the mass of CO₂ produced. From its mass, the percentage of carbon in the organic compound is calculated as
%C = \(\frac{12}{44} \times \frac{\text { Weight of } \mathrm{CO}_{2} \text { formed }}{\text { Weight of substance taken }}\) x 100

Question 31.
Why is it necessary to use acetic acid and not sulphuric acid for acidification of sodium extract for testing sulphur by lead acetate test?
Answer:
For testing of sulphur present in the organic compound the Lassaigne’s extract is acidified with acetic acid (CH₃COOH) and not sulphuric acid because lead acetate is soluble and does not interfere with the test. If sulphuric acid (H₂SO₄) is used, lead acetate will react with H₂SO₄ itself to form white ppt. of PbSO₄ as follows :

White ppt. of PbSO₄ so formed will interfere with this test of sulphur.

Question 32.
An organic compound contains 69% carbon and 48% hydrogen, the remainder being oxygen. Calculate the masses of carbon dioxide and water produced when 0.20 g of this substance is subjected to complete combustion.
Answer:

Question 33.
A sample of 0.50 g of an organic compound was treated according to Kjeldahl’s method. The ammonia evolved was absorbed in 50 ml of 0.5 M H₂SO₄. The residual acid required 60 mL of 0.5 M solution of NaOH for neutralisation. Find the percentage composition of nitrogen in the compound.
Ans. Volume of the acid taken = 50 mL of 0.5 M H₂SO₄
= 25 mL of 1.0 M H₂SO₄
Volume of alkali used for neutralization of excess acid = 60 mL of 0.5 M NaOH = 30 mL of 1.0 M NaOH
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
1 mole of H₂SO₄ = 2 moles of NaOH
Hence, 30 mL of 1.0 M NaOH = 15 mL of 1.0 M H₂SO₄
∴ Volume of acid used by ammonia = 25 – 15 = 10 mL
%of nitrogen = \(\frac{1.4 \times \mathrm{N}_{1} \times \text { Volume of acid used }}{\mathrm{W}}\)
= \(\frac{1.4 \times 2 \times 10}{0.5}\) (N₁ = normality of acid and W = mass of organic compound)
= 56

Question 34.
0.3780 g of an organic chloro compound gave 0.5740 g of silver chloride in Carius estimation. Calculate the percentage of chlorine present in the compound.
Answer:
Given that,
Mass of organic compound = 0.3780 g
Mass of AgCl formed = 0.5740 g
1 mol of AgCl contains 1 mol of Cl
Thus, mass of chlorine in 0.5740 g of AgCl = \(\frac{35.5 \times 0.5740}{143.5}\) = 0.142g
∴ Percentage of chlorine = \(\frac{0.142}{0.3780}\) x 100 = 37.57%
Hence, the percentage of chlorine present in the given organic chloro compound is 37.57%.

Question 35.
In the estimation of sulphur by Carius method, 0.468 g of an organic sulphur compound afforded 0.668 g of barium sulphate. Find out the percentage of sulphur in the given compound.
Answer:
The mass of organic compound = 0.468 g
Mass of barium sulphate formed = 0.668 g
Percentage of sulphur = \(\frac{32}{233} \times \frac{\text { Mass of } \mathrm{BaSO}_{4} \text { formed }}{\text { Mass of substance taken }}\) x 100
= \(\frac{32}{233} \times \frac{0.668}{0.468}\) x 100 = 19.59%
Hence, the percentage of sulphur in the given compound is 19.59%.

Question 36.
In the organic compound,
CH₂ = CH – CH₂ – CH₂ – C = CH, the pair of hybridised orbitals involved in the formation of C₂ – C₃ bond is (a) sp – sp² (b) sp – sp³ (c) sp² – sp³ (d) sp³ – sp³
Answer:

In the given organic compound, the carbon atoms numbered as 1, 2, 3, 4, 5, and 6 are sp², sp², sp³, sp³, sp, and sp hybridised respectively. Thus, the pair of hybridised orbitals involved in the formation of C₂ – C₃ bond is sp² – sp².

Question 37.
In the Lassaigne’s test for nitrogen in an organic compound, the Prussian blue colour is obtained due to the formation of:
(a) Na₄[Fe(CN)₆]
(b)Fe₄[Fe(CN)₆]₃
(c) Fe₂[FeCN₆]
(d) Fe₃[Fe(CN)₆]₄
Answer:
(b) The Prussian blue colour is due to the formation of ferric ferro cyanide, Fe₄[Fe(CN)₆]₃.

Question 38.
Which of the following carbocations is most stable?

Answer:
(b) The order of stability of carbocation is 3° > 2° > 1°. 3° carbocation,

Question 39.
The best and latest technique for isolation, purification and separation of organic compounds is:
(a) Crystallisation
(b) Distillation
(c) Sublimation
(d) Chromatography
Answer:
(d) Chromatography is the best and latest technique for isolation, purification and separation of organic compounds.

Question 40.
The reaction:
CH₃CH₂I+KOH(aq) > CH₃CH₂OH +KI is classified as :
(a) electrophilic substitution
(b) nucleophilic substitution
(c) elimination
(d) addition
Answer:
(b) This is an example of nucleophilic substitution reaction. The hydroxyl group of KOH (OH^–) with a lone pair of itself acts as a nucleophile and substitutes iodide ion in CH₃CH₂I to form ethanol.