GSEB Class 10 Science Solutions Chapter 4 Carbon and Its Compounds

Gujarat Board Class 10 Science Carbon And Its Compounds InText Questions And Answers

 

Question 1. What would be the electron dot structure of carbon dioxide which has the formula \( CO_2 \)?
Answer: The electron dot structure of \( CO_2 \) is:

In simple words: Carbon dioxide has a central carbon atom connected to two oxygen atoms. They share electrons to form two double bonds, and each oxygen atom also has two sets of unshared electrons.

Exam Tip: When drawing electron dot structures for molecules with multiple bonds, remember to distribute valence electrons to satisfy the octet rule for all atoms, especially the central atom.

 

Question 2. What would be the electron dot structure of a molecule of sulphur which is made up of eight atoms of sulphur?
Answer: The electron dot structure of a molecule of sulfur \( S_8 \) is a crown shape, where each sulfur atom forms two single bonds with adjacent sulfur atoms and has two lone pairs of electrons.

There are three common structural isomers for pentane.

n-Pentane:

Iso-Pentane:

Neo-Pentane:

In simple words: Pentane is a chemical with 5 carbon atoms. It can be arranged in three different shapes called structural isomers: a straight chain (n-Pentane), a chain with one branch (Iso-Pentane), and a central carbon with four branches (Neo-Pentane).

Exam Tip: Always remember that isomers have the same molecular formula but different structural arrangements of atoms. When drawing, count the carbons and hydrogens carefully to ensure correctness.

 

Question 3. How many structural isomers can you draw for pentane?
Answer: Three structural isomers can be drawn from pentane. Pentane: \( C_5H_{12} \)
In simple words: Pentane, which has the chemical formula \( C_5H_{12} \), can exist in three different structural arrangements. These different arrangements are called structural isomers.

Exam Tip: Be ready to draw and name the three structural isomers of pentane: n-pentane (straight chain), isopentane (2-methylbutane), and neopentane (2,2-dimethylpropane).

 

Question 4. What are the two properties of carbon which lead to the huge number of carbon compounds we see around us?
Answer: Carbon forms many compounds because of these properties:
(a) Catenation: Carbon shows catenation, which is the unique ability to form bonds with other carbon atoms. This property helps it create long chains, both branched and unbranched, and even cyclic rings.
(b) Tetravalency: Carbon has a valency of 4, meaning it can form bonds with four other carbon atoms or other monovalent elements. This gives rise to compounds with unique properties depending on the elements present.
(c) Isomerism: Carbon compounds also show isomerism, where compounds have the same molecular formula but different structural arrangements.
In simple words: Carbon can form many different chemicals because it can join up with other carbon atoms in long chains or rings (catenation), and it can make four bonds with other atoms (tetravalency). Also, it can make different shapes with the same number of atoms (isomerism).

Exam Tip: Focus on explaining catenation and tetravalency as the primary reasons for carbon's diverse compound formation, with isomerism as an important contributing factor.

 

Question 5. What would be the formula and electron dot structure of cyclopentane?
Answer: The formula of cyclopentane is \( C_5H_{10} \). The electron dot structure is:

In simple words: Cyclopentane is a molecule with 5 carbon atoms arranged in a ring, and each carbon atom is attached to two hydrogen atoms. This means it has a formula of \( C_5H_{10} \).

Exam Tip: For cyclic compounds, ensure the ring structure is clear and that each carbon atom maintains its valency of four by bonding with the correct number of hydrogen atoms.

 

Question 6. Draw the structures for the following compounds.
(i) Ethanoic acid
(ii) Bromopentane
(iii) Butanone
(iv) Hexanal
Answer:
(i) Ethanoic acid \( CH_3COOH \)

(ii) Bromopentane \( C_5H_{11}Br \)

(iii) Butanone \( C_2H_5COCH_3 \)

(iv) Hexanal \( C_5H_{11}CHO \)

Isomers of bromopentane:

1-Bromopentane:

2-Bromopentane:

3-Bromopentane:

In simple words: The structures show how atoms are connected in each compound. Ethanoic acid has a -COOH group, butanone has a C=O in the middle, and hexanal has a -CHO group at the end of a six-carbon chain. Bromopentane has a bromine atom attached to a five-carbon chain, and its position can change to create different isomers.

Exam Tip: Pay close attention to functional groups (e.g., -COOH for acid, C=O for ketone, -CHO for aldehyde, -Br for bromo) and their positions when drawing structures and their isomers.

 

Question 7. How would you name the following compounds:
(i) \( CH_3-CH_2-Br \)
(ii) \( H-C=O \)
       |\)
       H
(iii) \( H-C-C-C-C-C\equiv C-H \)
       |\  |\  |\  |\)
       H  H  H  H
Answer:
(i) Bromoethane
(ii) Methanal
(iii) Hex-1-yne
In simple words: The first compound is Bromoethane, which has two carbons and a bromine atom. The second is Methanal, a simple aldehyde. The third is Hex-1-yne, a six-carbon chain with a triple bond at the first carbon.

Exam Tip: Remember to identify the longest carbon chain, the functional group, and the position of substituents (like bromine) or multiple bonds (double or triple) to correctly name organic compounds according to IUPAC rules.

 

Question 8. Why is the conversion of ethanol to ethanoic acid an oxidation reaction?
Answer: The conversion of ethanol to ethanoic acid is an oxidation reaction because oxygen atoms are added to ethanol and hydrogen atoms are removed to change it into ethanoic acid.
\( CH_3-CH_2OH \xrightarrow{\text{Alk. } KMnO_4 + heat \text{ (or) Acidified } K_2Cr_2O_7 + heat} CH_3COOH \)
In the reaction above, alkaline \( KMnO_4 \) or acidified \( K_2Cr_2O_7 \) adds oxygen to ethanol, which is why they are known as oxidising agents.
In simple words: Turning ethanol into ethanoic acid is called oxidation because the ethanol gains oxygen and loses hydrogen. Special chemicals like \( KMnO_4 \) help this happen.

Exam Tip: Remember that oxidation in organic chemistry can mean either the gain of oxygen or the loss of hydrogen. Strong oxidising agents like acidified potassium dichromate or alkaline potassium permanganate are key reagents for this conversion.

 

Question 9. A mixture of oxygen and ethyne is burnt for welding. Can you tell why air is not used?
Answer: If air were used, incomplete combustion would take place, resulting in a sooty flame and less heat generation. When pure oxygen is used, ethyne burns completely, producing a large amount of heat and a blue flame. This heat is enough to melt metal for welding purposes.
In simple words: Air contains less oxygen, so burning ethyne in air creates a smoky flame with less heat because it doesn't burn completely. Using pure oxygen makes ethyne burn fully, giving a very hot, clean flame suitable for welding.

Exam Tip: Emphasise that complete combustion requires a high concentration of oxygen, which is not available in air (only about 21% oxygen). Pure oxygen ensures maximum heat release and a clean flame.

 

Question 10. How would you distinguish experimentally between an alcohol and a carboxylic acid?
Answer:
(a) Acid test: Reaction with carbonates/hydrogen carbonates. Take samples of alcohol and carboxylic acid in two test tubes. Add sodium carbonate or sodium bicarbonate solution to each. The compound that produces brisk effervescence of \( CO_2 \) gas will be the acid.
(b) Alcohol test: Take a small amount of ethanol and ethanoic acid in test tubes A and B. Add a 5% solution of alkaline potassium permanganate drop by drop to this solution and warm the test tube. The purple colour of potassium permanganate will disappear in the test tube containing alcohol.
In simple words: To tell an alcohol from a carboxylic acid, you can use two simple tests. First, an acid will fizz when you add baking soda, but an alcohol won't. Second, an alcohol will turn purple potassium permanganate clear, but an acid won't.

Exam Tip: Remember that carboxylic acids react with carbonates and bicarbonates to produce carbon dioxide (effervescence), a reaction not shown by alcohols. Alcohols, on the other hand, can be oxidised by alkaline potassium permanganate, leading to decolourisation, which ethanoic acid (already oxidised) would not typically show.

 

Question 12. Would you be able to check if water is hard by using detergent?
Answer: No, because detergent forms lather in both hard and soft water. Therefore, it is not useful to check for water hardness.
In simple words: No, you cannot test if water is hard using detergent. Detergents create suds in both hard and soft water, so they don't help to tell the difference.

Exam Tip: Remember that soaps form scum in hard water, but detergents do not. This difference is key for testing water hardness.

 

Question 13. People use a variety of methods to wash clothes. Usually after adding the soap, they 'beat' the clothes on a stone, or beat it with a paddle, scrub with a brush or the mixture is agitated in a washing machine. Why is agitation necessary to get clean clothes?
Answer: Soap reduces the surface tension of water. The long chain non-ionic hydrocarbon group in soap molecules attaches to oil or grease droplets and loosens them from the cloth fibres. However, this loosening alone is not enough to completely remove the grease and dirt. Therefore, clothes are agitated to fully remove these grease droplets.
In simple words: Agitation, like beating or scrubbing, is important for cleaning clothes because it helps soap molecules fully remove all the oil and dirt that has been loosened from the fabric. Without it, the dirt might not come off completely.

Exam Tip: Explain that soap loosens dirt, but physical action (agitation) provides the necessary mechanical force to dislodge and suspend the dirt particles effectively, allowing them to be rinsed away.

In-Text Activities Solved

Activity 4.1

List of ten things used or consumed daily

(C) → indicates carbon. Most substances contain carbon in it.

If we look at methanol (\( CH_3OH \)) and ethanol (\( C_2H_5OH \)):
These two compounds differ by \( CH_2 \).
Mass of \( CH_3OH = 12 + (1 \times 3) + 16 + 1 = 32 \) u
Mass of \( C_2H_5OH = (12 \times 2) + (1 \times 5) + 16 + 1 = 46 \) u
Difference = \( 46 - 32 = 14 \) u

If we look at ethanol (\( C_2H_5OH \)) and propanol (\( C_3H_7OH \)):
These two compounds differ by \( CH_2 \).
Mass of \( C_2H_5OH = (12 \times 2) + (1 \times 5) + 16 + 1 = 46 \) u
Mass of \( C_3H_7OH = (12 \times 3) + (1 \times 7) + 16 + 1 = 60 \) u
Difference = \( 60 - 46 = 14 \) u

If we look at propanol (\( C_3H_7OH \)) and butanol (\( C_4H_9OH \)):
These two compounds differ by \( CH_2 \).
Mass of \( C_3H_7OH = (12 \times 3) + (1 \times 7) + 16 + 1 = 60 \) u
Mass of \( C_4H_9OH = (12 \times 4) + (1 \times 9) + 16 + 1 = 74 \) u
Difference = \( 74 - 60 = 14 \) u

This shows a similarity: Two consecutive members of a homologous series differ by a \( CH_2 \) group and a mass of 14 u. \( CH_3OH \), \( C_2H_5OH \), \( C_3H_7OH \), \( C_4H_9OH \) form a homologous series.

Activity 4.2

Homologous Series:

Activity 4.3

Heating of different carbon compounds, observing the flame and smoke.

Alcohol and acetone burn with a non-sooty flame, showing complete combustion. Camphor and naphthalene burn with a sooty flame, which means incomplete combustion for these unsaturated hydrocarbons.

Activity 4.4

A Bunsen burner is used to observe different flame types by adjusting the holes at the base of the burner.

• When the hole is closed – A yellow, sooty flame is formed, and black deposits of carbon are seen on a spoon placed above the flame.
• When the hole is open – A blue flame is formed, and no black deposits of carbon are seen on a spoon placed above the flame.

Activity 4.5

Take 3 mL of ethanol in a test tube and warm it gently in a water bath. Add a 5% solution of alkaline potassium permanganate drop by drop to the warmed ethanol solution. The purple colour of \( KMnO_4 \) slowly fades and then completely disappears. When an excess of \( KMnO_4 \) is added, the colour will not disappear as alcohol gets oxidised to form carboxylic acid.

In this case, ethanol gets oxidised to form ethanoic acid due to \( KMnO_4 \), but the excess \( KMnO_4 \) will not decolourise.

\( CH_3CH_2OH \xrightarrow{\text{Alkaline } KMnO_4 + heat} CH_3COOH \)
(Alcohol)                                                                    (Ethanoic acid)

Activity 4.6

Take a test tube with ethanol in it and drop a small piece of sodium metal into it. A reaction occurs, and hydrogen gas is released. To test for the presence of hydrogen gas, bring a burning matchstick near the mouth of the test tube, and it will burn with a pop sound.

\( 2Na + 2CH_3CH_2OH \rightarrow 2CH_3CH_2ONa + H_2 \)
Sodium    Ethanol                               Sodium ethoxide     Hydrogen

Activity 4.7

Activity 4.8

Take 1 mL ethanol and 1 mL glacial acetic acid, along with a few drops of concentrated sulfuric acid, in a test tube. Warm it in a water-bath for at least five minutes. Pour the mixture into a beaker containing 20-50 mL of water.

1. A pleasant fruity-smelling compound, called an ester, is obtained.

2. Such a chemical reaction is called esterification.

\( CH_3-COOH + CH_3-CH_2OH \xrightarrow{\text{Conc. } H_2SO_4 \text{ dehydrating agent}} CH_3COOC_2H_5 + H_2O \)
Ethanoic acid   Ethanol                                                 Ester        Water

Activity 4.9

Take a spatula full of sodium carbonate in a test tube and add 2 mL of dilute ethanoic acid. Pass the gas produced through freshly prepared lime water. Repeat the process with sodium hydrogen carbonate.

Following reactions take place
\( 2CH_3COOH + Na_2CO_3 \rightarrow 2CH_3COONa + H_2O + CO_2 \)
\( CH_3COOH + NaHCO_3 \rightarrow CH_3COONa + H_2O + CO_2 \)

The brisk effervescence of \( CO_2 \) gas is observed, which turns lime water milky when passed through it.

Activity 4.10

Take about 10 mL of water in two test tubes. Add a drop of oil (cooking oil) to both test tubes and label them as A and B. To test tube B, add a few drops of soap solution. Now, shake both test tubes vigorously for the same period and then leave them undisturbed for some time. The separated layers of oil and water are immediately visible in test tube A, but it takes some time to separate the layers in test tube B.

Activity 4.11

Take about 10 mL of distilled water and 10 mL of hard water in separate test tubes. Add a couple of drops of soap solution to both. Shake the test tubes vigorously for an equal period. The test tube containing distilled water produces foam, while the test tube with hard water forms a curdy white precipitate.

Activity 4.12

Take two test tubes, each with 10 mL of hard water. Add five drops of soap solution to one and five drops of detergent solution to the other. Shake both test tubes for the same period.

• Solution of hard water and soap forms curdy white precipitate.
• The hard water and detergent forms foam.

Gujarat Board Class 10 Science Carbon And Its Compounds Textbook Questions And Answers

 

Question 1. Ethane, with the molecular formula \( C_2H_6 \) has –
(a) 6 covalent bonds.
(b) 7 covalent bonds.
(c) 8 covalent bonds.
(d) 9 covalent bonds
Answer: (b) 7 covalent bonds.
In simple words: Ethane, a molecule with two carbon atoms and six hydrogen atoms, forms a total of seven shared electron pairs. One bond is between the two carbon atoms, and six bonds are between carbon and hydrogen atoms.

Exam Tip: To find the number of covalent bonds, draw the Lewis structure of the molecule. For \( C_2H_6 \), there is one C-C bond and six C-H bonds, totaling 7 covalent bonds.

 

Question 2. Butanone is a four-carbon compound with the functional group –
(a) carboxylic acid.
(b) aldehyde.
(c) ketone.
(d) alcohol.
Answer: (c) ketone.
In simple words: Butanone is an organic chemical that has a main functional group called a ketone. This means it has a carbon atom double-bonded to an oxygen atom within the carbon chain.

Exam Tip: Recognize the '–one' suffix in butanone indicates a ketone functional group. Carboxylic acids end in '–oic acid', aldehydes in '–al', and alcohols in '–ol'.

 

Question 3. While cooking, if the bottom of the vessel is getting blackened on the outside, it means that –
(a) the food is not cooked completely.
(b) the fuel is not burning completely.
(c) the fuel is wet.
(d) the fuel is burning completely.
Answer: (b) The fuel is not burning completely.
In simple words: If the bottom of your cooking pot turns black, it means the fuel isn't burning properly. This incomplete burning releases soot, which is unburnt carbon particles that make the vessel black.

Exam Tip: Blackening of utensils is a classic sign of incomplete combustion, often due to an insufficient supply of oxygen to the flame, leading to the formation of soot (unburnt carbon).

 

Question 4. Explain the nature of the covalent bond using the bond formation in \( CH_3Cl \).
Answer: Bond formation in \( CH_3Cl \):
Carbon forms four covalent single bonds by sharing four electron pairs with three hydrogen atoms and one chlorine atom. Since chlorine is more electronegative, it gives a polar nature to the C-Cl bond.

In simple words: In \( CH_3Cl \), carbon shares electrons to form single bonds with three hydrogen atoms and one chlorine atom. Because chlorine pulls electrons more strongly than carbon, the carbon-chlorine bond becomes slightly uneven, or polar.

Exam Tip: When describing covalent bonds, explain how electrons are shared between atoms. For polar bonds, mention the difference in electronegativity and how it leads to partial charges.

 

Question 5. Draw the electron dot structure for
(a) ethanoic acid.
(b) \( H_2S \)
(c) Propanone
(d) \( F_2 \)
Answer: The electron dot structures are as follows:
(a) Ethanoic acid – \( CH_3COOH \)

(b) \( H_2S \)

(c) Propanone

(d) \( F_2 \)

In simple words: These diagrams show how electrons are shared and arranged around atoms in different molecules. For example, ethanoic acid has a central carbon double-bonded to one oxygen and single-bonded to another oxygen and a methyl group. \( H_2S \) is bent with sulfur having two lone pairs. Propanone has a central carbon double-bonded to an oxygen and single-bonded to two methyl groups. Fluorine gas (\( F_2 \)) has two fluorine atoms single-bonded to each other, each with three lone pairs.

Exam Tip: When drawing electron dot structures, ensure that each atom (except hydrogen) satisfies the octet rule (8 electrons) and hydrogen satisfies the duet rule (2 electrons). Represent shared electrons in bonds and unshared electrons as lone pairs.

 

Question 6. What is an homologous series? Explain with an example.
Answer: An homologous series is a group of organic compounds belonging to the same class that have similar chemical properties and share the same general formula. They also possess the same functional group. When arranged in increasing order of molecular mass, consecutive members differ by 14 atomic mass units (a.m.u.) or a \( -CH_2 \) group.
Example: Alkane – \( C_nH_{2n+2} \)
Methane \( CH_4 \)
Ethane \( C_2H_6 \)
Propane \( C_3H_8 \)
Butane \( C_4H_{10} \)
Properties:

• The difference between two consecutive members of a homologous series is a \( CH_2 \) group and a mass of 14 a.m.u.
• They all show similar chemical properties and a slight gradual change in their physical properties.

In simple words: An homologous series is a family of chemical compounds that are alike in many ways. They all have the same basic chemical formula and functional group, act similarly, and each member is just a little bigger than the last by adding a \( CH_2 \) unit. Alkanes, like methane, ethane, and propane, are a good example of this kind of series.

Exam Tip: Define homologous series by its key characteristics: same general formula, similar chemical properties, same functional group, and a \( CH_2 \) difference between successive members. Provide a clear example like the alkanes or alcohols.

 

Question 7. How can ethanol and ethanoic acid be differentiated on the basis of their physical and chemical properties?
Answer:
Physical Properties:

Chemical Properties:

In simple words: You can tell ethanol and ethanoic acid apart by how they smell (ethanoic acid is strong, ethanol is pleasant), and their melting and boiling points are different. Chemically, ethanoic acid fizzes with baking soda and doesn't change purple \( KMnO_4 \), but ethanol does not fizz and turns purple \( KMnO_4 \) clear.

Exam Tip: For differentiation questions, always provide both physical and chemical tests. The \( NaHCO_3 \) test is a reliable chemical distinction, as only carboxylic acids react to produce \( CO_2 \).

 

Question 8. Why does micelle formation take place when soap is added to water? Will a micelle be formed in other solvents such as ethanol also?
Answer: Soap molecules have two ends with different properties. One end is hydrophilic, meaning it dissolves in water, while the other end is hydrophobic, meaning it dissolves in hydrocarbons (like oil and grease). When soap is added to water, the ionic (hydrophilic) end of the soap aligns itself, keeping the hydrocarbon (hydrophobic) tail away from the water. This leads to the formation of clusters of molecules where the hydrophobic tails are in the interior of the cluster, and the ionic ends are on the surface, facing the water. This process is called micelle formation.
Soap is soluble in ethanol, so micelle formation will not occur in ethanol. Instead, the soap molecules will simply dissolve uniformly throughout the ethanol.
In simple words: Soap forms micelles in water because its molecules have a water-loving end and a water-hating end. The water-loving ends face out into the water, while the water-hating ends gather inside, forming a ball. But in solvents like ethanol, soap just dissolves evenly and doesn't form these balls because there's no need to hide the water-hating parts.

Exam Tip: Clearly explain the dual nature of soap molecules (hydrophilic head, hydrophobic tail) and how this leads to micelle formation in water. Contrast this with dissolution in organic solvents like ethanol, where micelles are not formed.

 

Question 9. Why are carbon and its compounds used as fuels for most applications?
Answer: Carbon and its compounds go through burning to make heat, and the amount of heat given off can be managed, making it suitable for many uses.
In simple words: Carbon compounds are good fuels because they produce controllable heat when burned.

Exam Tip: Remember that "combustion" is the key process for fuel usage, and "controllable heat" is why carbon compounds are practical for various applications.

 

Question 10. Explain the formation of scum when hard water is treated with soap.
Answer: Hard water has salts of calcium and magnesium. When a soap molecule meets these salts, it creates a lumpy white solid (a compound that won't dissolve in water) known as scum. Soap \( + \) Hard water \( \rightarrow \) Scum.
In simple words: Soap in hard water forms a white, insoluble solid called scum because of calcium and magnesium salts.

Exam Tip: Focus on the interaction between soap molecules and calcium/magnesium ions in hard water, leading to the insoluble precipitate (scum).

 

Question 11. What change will you observe if you test soap with litmus paper (red and blue)?
Answer: Soap is basic, so it will change red litmus paper to blue, while blue litmus paper will stay blue.
In simple words: Soap turns red litmus blue and leaves blue litmus unchanged because it is alkaline.

Exam Tip: Always remember that bases (alkaline substances) turn red litmus paper blue, while acids turn blue litmus paper red.

 

Question 12. What is hydrogenation? What is its industrial application?
Answer: When unsaturated hydrocarbons (those with double or triple bonds) are mixed with hydrogen in the presence of a catalyst like nickel, the hydrogen atoms join across the double or triple bond, transforming them into saturated hydrocarbons. This process is known as an addition reaction or hydrogenation.
Example:
\( H_2C=CH_2 + H_2 \xrightarrow{\text{Ni catalyst}} H_3C-CH_3 \)
Ethene \( \xrightarrow{\text{Hydrogenation}} \) Ethane
Industrial use: It helps to convert vegetable oil into vanaspati ghee.
Vegetable oil \( + H_2 \xrightarrow{\text{Ni}} \) Vanaspati ghee
In simple words: Hydrogenation is when hydrogen adds to unsaturated hydrocarbons (like oils) to make them saturated (like ghee), using a catalyst. Its main use is converting vegetable oils into solid fats.

Exam Tip: Key points for hydrogenation are "unsaturated to saturated hydrocarbon," "addition of hydrogen," and "catalyst (Ni or Pt)". The industrial application of converting vegetable oil to vanaspati ghee is crucial.

 

Question 13. Which of the following hydrocarbons undergo addition reactions? \( C_2H_6, C_3H_8, C_3H_6, C_2H_2 \) and \( CH_4 \).
Answer: Addition reactions occur in unsaturated hydrocarbons. Therefore, \( C_3H_6 \) and \( C_2H_2 \) are unsaturated hydrocarbons and will demonstrate an addition reaction.
In simple words: Only unsaturated hydrocarbons (those with double or triple bonds) have addition reactions, so \( C_3H_6 \) and \( C_2H_2 \) will react.

Exam Tip: Remember that addition reactions are characteristic of unsaturated compounds (alkenes and alkynes) because they have double or triple bonds that can open up to add more atoms.

 

Question 14. Give a test that can be used to differentiate chemically between saturated and unsaturated hydrocarbon.
Answer: Test: Alkaline Potassium permanganate \( + \) Unsaturated hydrocarbon (Pink colour) \( \rightarrow \) Pink colour disappears. So, if we add oil to a test tube with alkaline potassium permanganate solution, the pink color of the solution will go away. However, the pink color of alkaline potassium permanganate will not disappear if we add butter to the test tube.
In simple words: You can tell the difference using alkaline potassium permanganate solution; it turns colorless in unsaturated hydrocarbons (like oil) but stays pink in saturated ones (like butter).

Exam Tip: This is a common test called the Baeyer's test. The key observation is the decolorization of the pink potassium permanganate solution by unsaturated compounds.

 

Question 15. Explain the mechanism of cleaning action of soaps.
Answer: A soap molecule has two parts; the charged part, which loves water, is called hydrophilic, and the long carbon chain, which dislikes water, is called the hydrophobic tail. When soap mixes in water, the hydrophobic carbon chain attaches itself to oil, dirt, and grease. The hydrophilic part remains outside. This process then causes micelle formation to occur. The tail wraps around dirt, oil, or grease, and agitation helps remove it. A lot of rinsing is done with water so that water molecules attract the charged (\( Na^+ \)) end and carry the soap molecules with dirt attached to it and clean the clothes, utensils, etc.
In simple words: Soap cleans by forming micelles where the water-hating parts trap dirt, and the water-loving parts stay outside, allowing water to wash away the dirt-filled micelles.

Exam Tip: When explaining cleaning action, clearly describe the hydrophilic and hydrophobic parts of soap, micelle formation around dirt/oil, and how rinsing removes these micelles.

 

Question 1. What is a hydrocarbon?
Answer: It is a substance made up of hydrogen and carbon atoms.
In simple words: A hydrocarbon is a chemical made only of hydrogen and carbon.

Exam Tip: Simply stating the constituent elements – hydrogen and carbon – is enough for the definition of a hydrocarbon.

 

Question 2. Give different forms in which carbon occurs in nature.
Answer: Carbon can be found as graphite and diamond in its pure state, and also in combined forms like carbon dioxide and carbonates.
In simple words: Carbon naturally appears as pure forms like diamond and graphite, and in mixtures like carbon dioxide.

Exam Tip: Distinguish between elemental forms (allotropes) like diamond and graphite, and combined forms like carbon dioxide and carbonate minerals.

 

Question 3. Name two types of hydrocarbon.
Answer: The two main types of hydrocarbons are saturated and unsaturated.
In simple words: Hydrocarbons are either saturated or unsaturated.

Exam Tip: The primary classification of hydrocarbons is based on the type of bonds present: single bonds (saturated) or double/triple bonds (unsaturated).

 

Question 4. What are covalent bonds?
Answer: A covalent bond is a chemical connection created when two atoms share a pair of electrons.
In simple words: Covalent bonds form when atoms share electrons.

Exam Tip: The key idea in covalent bonding is the "sharing" of electrons between atoms, leading to a stable electron configuration for both.

 

Question 5. What is catenation?
Answer: Carbon possesses the special capacity to connect with other carbon atoms, creating large molecules. This self-joining characteristic is known as catenation.
In simple words: Catenation is carbon's special ability to link with other carbon atoms, forming long chains and large molecules.

Exam Tip: Define catenation as the self-linking property of an element, specifically carbon, to form long chains, branches, and rings.

 

Question 6. Name two allotropes of carbon.
Answer: Diamond and graphite are two examples of carbon's allotropes.
In simple words: Two different forms of carbon are diamond and graphite.

Exam Tip: Remember that allotropes are different structural forms of the same element, and diamond and graphite are the most common allotropes of carbon.

 

Question 7. Why covalent compounds have low melting and boiling points?
Answer: Since these bonds form from shared electrons between two atoms, the forces between covalent compounds are weak. These weak forces mean the bonds are easy to break, resulting in low melting and boiling points.
In simple words: Covalent compounds have low melting and boiling points because the forces holding individual molecules together are weak.

Exam Tip: Highlight that the low melting/boiling points are due to weak intermolecular forces, not weak covalent bonds themselves.

 

Question 8. Define oxidising agents.
Answer: Certain substances have the ability to give oxygen to other materials. These materials are called oxidising agents. For instance, alkaline \( KMnO_4 \) and acidified \( K_2Cr_2O_7 \) are common examples.
In simple words: Oxidizing agents are chemicals that provide oxygen to other substances, like alkaline potassium permanganate.

Exam Tip: An oxidising agent is a substance that causes oxidation by either adding oxygen or removing hydrogen/electrons from another substance.

 

Question 9. Give the reaction to show how alcohol is converted into carboxylic acid.
Answer: To convert alcohol into carboxylic acid, we use an oxidizing agent. The reaction below shows how ethanol changes into ethanoic acid using alkaline potassium permanganate or acidified potassium dichromate.
\( CH_3-CH_2OH \xrightarrow{\text{Alkaline } KMnO_4 \text{ + heat or Acidified } K_2Cr_2O_7 \text{ + heat}} CH_3COOH \)
Ethanol \( \xrightarrow{\text{Oxidation}} \) Carboxylic acid
In simple words: Alcohol becomes carboxylic acid through an oxidation reaction, usually with agents like alkaline potassium permanganate or acidified potassium dichromate.

Exam Tip: Remember that the conversion of alcohol to carboxylic acid is an oxidation reaction, requiring strong oxidizing agents.

 

Question 10. Identify the compound.
\( H-C\equiv C-CH_3 \)
Answer: The compound shown with the structure \( H-C\equiv C-CH_3 \) is propyne.
In simple words: The compound with a triple bond and three carbons is propyne.

Exam Tip: Recognize that "yne" in the name indicates a triple bond, and "prop-" signifies three carbon atoms.

 

Question 11. Name the compound.

Answer: The compound with the given hexagonal ring structure is benzene, which has the chemical formula \( C_6H_6 \).
In simple words: The compound shown, which is a six-carbon ring with alternating double bonds, is benzene, or \( C_6H_6 \).

Exam Tip: Recognize the specific hexagonal ring structure with alternating double bonds (or a circle representing resonance) as characteristic of benzene.

 

Question 12. Give two properties of ethanol.
Answer: Ethanol is a liquid at normal room temperature. It also dissolves completely in water, mixing in any amount.
In simple words: Ethanol is a liquid at room temperature and mixes completely with water.

Exam Tip: When asked for properties, consider physical states (liquid, solid, gas) and solubility in common solvents like water.

 

Question 13. Give the formula for the functional group of aldehyde.
Answer: The functional group for an aldehyde is represented by \( -CHO \).
In simple words: The aldehyde functional group is \( -CHO \).

Exam Tip: Remember the specific structure of the aldehyde group: a carbon double-bonded to an oxygen and single-bonded to a hydrogen atom, at the end of a carbon chain.

 

Question 14. What are heteroatoms? Explain with an example. Complete the following reaction: \( CH_4 \rightarrow CH_3-OH \)
Answer: Heteroatoms are elements that replace one or more hydrogen (\( H \)) atoms in a hydrocarbon, making sure that carbon's bonding capacity is still met. For example, in the reaction \( CH_4 \rightarrow CH_3-OH \), oxygen acts as a heteroatom.
In simple words: Heteroatoms are non-carbon atoms, like oxygen, that replace hydrogen in hydrocarbons, forming new compounds while keeping carbon's bonds complete.

Exam Tip: Understand that heteroatoms (like O, N, S, halogens) are atoms other than carbon and hydrogen that become part of an organic molecule's main structure.

 

Question 15. Complete the following reaction:
\( CH_3-C\equiv C-CH_3 + H_2 \xrightarrow{\text{Ni/Pt}} CH_3-CH_2-CH_2-CH_3 \)
Answer: The reaction represents the hydrogenation of an alkyne to an alkane. When hydrogen (\( H_2 \)) is added to the starting alkyne in the presence of a nickel/platinum catalyst, it becomes a saturated alkane.
In simple words: This reaction shows how a triple-bonded carbon chain (alkyne) turns into a single-bonded chain (alkane) by adding hydrogen with a metal catalyst.

Exam Tip: For hydrogenation reactions, remember the addition of hydrogen across multiple bonds and the role of a metal catalyst (Ni or Pt) to form a saturated product.

 

Question 16. Give the full form of IUPAC.
Answer: IUPAC stands for the International Union of Pure and Applied Chemistry.
In simple words: IUPAC means International Union of Pure and Applied Chemistry.

Exam Tip: Always remember the full form of IUPAC as it is fundamental for naming chemical compounds internationally.

 

Question 17. How can esters be converted into soap?
Answer: Esters can be turned into soap through a saponification reaction, which involves adding or reacting the ester with sodium hydroxide (\( NaOH \)).
In simple words: Esters become soap through a process called saponification, by reacting them with sodium hydroxide.

Exam Tip: Recall that saponification is the alkaline hydrolysis of esters (typically fats or oils) to produce soap and an alcohol (glycerol).

 

Question 18. How can we convert \( CH_3CH_2OH \) into \( C_2H_4 \)?
Answer: We can convert ethanol into ethene by adding concentrated sulfuric acid. This acid works as a dehydrating agent, removing water from the ethanol molecule to produce ethene.
\( CH_3-CH_2OH \xrightarrow{\text{Conc. } H_2SO_4} CH_2=CH_2 + H_2O \)
Ethanol \( \xrightarrow{\text{Dehydration}} \) Ethene
In simple words: To change ethanol into ethene, you use concentrated sulfuric acid to remove water from it.

Exam Tip: Remember that concentrated sulfuric acid is a powerful dehydrating agent, useful for converting alcohols into alkenes by removing a water molecule.

 

Question 19. What is the melting point of acetic acid?
Answer: The melting point of acetic acid is \( 290 \text{ K} \).
In simple words: Acetic acid melts at \( 290 \text{ Kelvin} \).

Exam Tip: Specific physical constants like melting points are useful to remember for common compounds like acetic acid.

 

Question 20. Name the given compound.

\( \begin{array}{cc} H & O \\ | & || \\ H-C-C-C-H \\ | & \\ H & H \end{array} \)

Answer: The given compound, with a ketone functional group on the second carbon, is named 2-Butanone.
In simple words: The compound with four carbons and a ketone on the second carbon is 2-Butanone.

Exam Tip: Identify the longest carbon chain and the functional group (ketone in this case) and its position to correctly name the compound according to IUPAC rules.

 

Question 21. How can you convert ethene into ethane?
Answer: You can change ethene into ethane by adding hydrogen gas to ethene, with a catalyst like nickel present. This process is called hydrogenation.
\( H_2C=CH_2 + H_2 \xrightarrow{\text{Ni Catalyst}} H_3C-CH_3 \)
Ethene \( \xrightarrow{\text{Hydrogenation}} \) Ethane
In simple words: Ethene becomes ethane by adding hydrogen in the presence of a nickel catalyst.

Exam Tip: This is a classic example of an addition reaction where an unsaturated compound (alkene) is converted to a saturated one (alkane) via hydrogenation.

 

Question 22. What is addition reaction? Give one example.
Answer: An addition reaction is a chemical process where hydrogen is added across the double bonds of unsaturated hydrocarbons. For example, when ethene reacts with hydrogen in the presence of a nickel catalyst, it forms ethane.
\( H_2C=CH_2 + H_2 \xrightarrow{\text{Ni Catalyst}} H_3C-CH_3 \)
Ethene \( \xrightarrow{\text{Addition Reaction}} \) Ethane
In simple words: An addition reaction is when atoms are added to a double or triple bond, like hydrogen adding to ethene to form ethane.

Exam Tip: Clearly define addition reaction by highlighting the breaking of multiple bonds (double/triple) and the joining of new atoms across them.

 

Question 23. What is esterification reaction?
Answer: Esterification is a reaction where an alcohol combines with a carboxylic acid to create a new compound known as an ester.
In simple words: Esterification is the reaction that makes an ester from an alcohol and a carboxylic acid.

Exam Tip: Emphasize that esterification is the process of forming an ester, usually by reacting an alcohol with a carboxylic acid, often with an acid catalyst.

 

Question 24. Give two uses of methane gas.
Answer: Methane gas is primarily used as a fuel source. Additionally, it forms a significant part of both biogas and compressed natural gas (CNG).
In simple words: Methane is used as fuel and is a main part of biogas and CNG.

Exam Tip: For uses of methane, focus on its role as a fuel and its presence in natural gas mixtures.

 

Question 25. What is isomerism?
Answer: Isomerism is a characteristic where a compound can have various structural arrangements while still possessing the same molecular formula. These distinct forms are called isomers, and they often show different physical and chemical attributes.
In simple words: Isomerism is when compounds have the same formula but different structures and properties.

Exam Tip: The core of isomerism is "same molecular formula, different structural formula," leading to different properties.

 

Question 26. Why can't we test hard water with detergents?
Answer: We cannot test hard water with detergents because detergents create foam equally well in both hard and soft water, making it impossible to tell the difference.
In simple words: Detergents foam in both hard and soft water, so they can't be used to test for water hardness.

Exam Tip: Contrast detergents with soaps: soaps form scum in hard water, allowing for differentiation, while detergents do not.

 

Question 27. What is hydrophilic?
Answer: A hydrophilic substance is one that displays a strong attraction to water.
In simple words: Hydrophilic means "water-loving," describing substances that attract and mix with water.

Exam Tip: Remember the etymology: "hydro" for water, "philic" for loving. This term is vital for understanding solubility and interactions in biological systems.

 

Question 28. Name the second member of alkyne series.
Answer: The second member in the alkyne series is propyne.
In simple words: Propyne is the second compound in the alkyne family.

Exam Tip: The alkyne series starts with ethyne (2 carbons), so the second member will have three carbons, making it propyne.

 

Question 29. Give the names of the functional group.
(i) \( -CHO \)
(ii) \( -C=O \)
Answer:
(i) The functional group \( -CHO \) is an aldehyde.
(ii) The functional group \( -C=O \) is a ketone.
In simple words: \( -CHO \) is an aldehyde, and \( -C=O \) is a ketone.

Exam Tip: Distinguish aldehydes from ketones by the position of the carbonyl group: aldehydes have it at the end of a chain (bonded to H), while ketones have it within the chain (bonded to two carbons).

 

Question 30. The structural formula of an ester is:

\( \begin{array}{ccccccc} H & H & & O & H & H \\ | & | & & || & | & | \\ H-C-C-C & - & O & - & C-C-H \\ | & | & & & | & | \\ H & H & & & H & H \end{array} \)

Name the alcohol and the acid from which it would have been formed.
Answer: The alcohol component would be ethanol (\( C_2H_5OH \)), and the acid component would be propanoic acid (\( CH_3CH_2COOH \)).
In simple words: This ester is made from ethanol and propanoic acid.

Exam Tip: To identify the parent alcohol and acid of an ester, break the ester linkage (\( -COO- \)). The part with the single-bonded oxygen comes from the alcohol, and the part with the carbonyl group comes from the acid.

 

Question 31. Give the IUPAC name of acetic acid and propyl alcohol.
Answer: The IUPAC name for acetic acid is ethanoic acid, and for propyl alcohol, it is propanol.
In simple words: Acetic acid's IUPAC name is ethanoic acid, and propyl alcohol's is propanol.

Exam Tip: Remember common trivial names and their corresponding IUPAC names, especially for widely used organic compounds like acetic acid and propyl alcohol.

 

Question 32. What will happen to the litmus solution in carboxylic acid?
Answer: When litmus solution is added to carboxylic acid, red litmus paper will stay red, but blue litmus paper will turn red, showing its acidic nature.
In simple words: Carboxylic acid turns blue litmus red and leaves red litmus unchanged.

Exam Tip: Acids turn blue litmus red. If the solution is already acidic (red litmus), no change will be observed.

 

Question 33. Give the electron dot structure of \( CH_3Cl \) and \( C_2H_2 \).
Answer: The electron dot structures for \( CH_3Cl \) and \( C_2H_2 \) show how electrons are shared between the atoms in each molecule.

Electron dot structure of \( CH_3Cl \)

\( \begin{array}{c} H \\ \cdot\cdot \\ H : \underset{\cdot\cdot}{\ddot{C}} : Cl : \\ \cdot\cdot \\ H \end{array} \)

Electron dot structure of \( C_2H_2 \)

\( H : C ::: C : H \)

In simple words: The electron dot structures illustrate the shared electrons in methyl chloride (\( CH_3Cl \)) and acetylene (\( C_2H_2 \)).

Exam Tip: For electron dot structures, ensure all valence electrons are shown, shared electrons form bonds, and all atoms achieve a stable octet (or duet for hydrogen).

 

Question 34. Draw the electron dot structure of \( N_2 \) and \( NH_3 \).
Answer: The electron dot structures for \( N_2 \) and \( NH_3 \) illustrate the arrangement of valence electrons and covalent bonds within these molecules.

Electron dot structure of \( N_2 \)

\( : \ddot{N} ::: \ddot{N} : \)

Electron dot structure of \( NH_3 \)

\( \begin{array}{c} H \\ \cdot\cdot \\ H : \underset{\cdot\cdot}{\ddot{N}} : H \\ \cdot\cdot \end{array} \)

In simple words: The electron dot structures show a triple bond in \( N_2 \) and single bonds with a lone pair in \( NH_3 \).

Exam Tip: Pay close attention to lone pairs and multiple bonds (triple bond in \( N_2 \)) when drawing electron dot structures for nitrogen-containing compounds.

 

Question 35. What happens when ethanol burns in air?
Answer: When ethanol burns in the air, it combines with oxygen to create carbon dioxide and water.
In simple words: Burning ethanol in air produces carbon dioxide and water.

Exam Tip: Remember that the complete combustion of any alcohol in excess air or oxygen produces carbon dioxide and water.

 

Question 36. Give the IUPAC name and write the functional group present in vinegar.
Answer: The IUPAC name for the acid in vinegar is ethanoic acid (\( CH_3COOH \)). The functional group present is the carboxylic acid group (\( -COOH \)).
In simple words: Vinegar's acid is ethanoic acid, and its functional group is carboxylic acid.

Exam Tip: Vinegar is an aqueous solution of acetic acid; knowing its common and IUPAC name, along with its functional group, is essential.

 

Question 37. A compound has a molecular formula \( C_2H_6O \). It is used as a fuel. Name the compound and name its functional group.
Answer: The compound with molecular formula \( C_2H_6O \) is an alcohol, specifically ethanol (\( C_2H_5OH \)). Its functional group is the hydroxyl group, \( -OH \).
In simple words: The fuel compound \( C_2H_6O \) is ethanol, and its active part is the \( -OH \) (hydroxyl) group.

Exam Tip: Remember that a molecular formula can represent different compounds (isomers), but context (like "used as a fuel") often helps identify the specific one.

 

Short Answer Type Questions

 

Question 1. What is the reactive site in the given hydrocarbon? Write its name. \( H_3C-CH_2-CH=CH-CH_3 \)
Answer: The active site for reactions in this hydrocarbon is the location of the double bond. The compound's name is pent-2-ene.
In simple words: The double bond is the reactive part of the hydrocarbon, which is named pent-2-ene.

Exam Tip: In unsaturated hydrocarbons, the double or triple bond is typically the most reactive site, undergoing addition reactions readily.

 

Question 2. What is the difference in the number of carbon and hydrogen atoms between two successive members of a homologous series? Also give the difference in their atomic masses.
Answer: Between two consecutive members of a homologous series, the difference is one carbon atom and two hydrogen atoms, meaning a \( -CH_2 \) group. This results in a mass difference of \( 14 \text{ a.m.u.} \).
In simple words: Successive members of a homologous series differ by a \( -CH_2 \) group, which means a difference of 14 atomic mass units.

Exam Tip: The defining characteristic of a homologous series is the \( -CH_2 \) difference between successive members and a gradual change in physical properties.

 

Question 3. Name the peculiar/specific chemical property exclusive in case of saturated hydrocarbons and unsaturated hydrocarbons.
Answer: Saturated hydrocarbons are uniquely known for substitution reactions, where hydrogen atoms are replaced by other elements or atoms. In contrast, unsaturated hydrocarbons are characterized by addition reactions, where hydrogen atoms join across their double or triple bonds.
In simple words: Saturated hydrocarbons do substitution reactions, while unsaturated hydrocarbons perform addition reactions.

Exam Tip: Clearly differentiate between substitution (saturated) and addition (unsaturated) reactions, as these are hallmark chemical properties for each type of hydrocarbon.

 

Question 4. Why acetic acid is called glacial acetic acid?
Answer: Acetic acid is named glacial acetic acid because it has a very low melting point of \( 290 \text{ K} \), causing it to freeze and form ice-like crystals in colder climates during winter.
In simple words: Acetic acid is called glacial acetic acid because it freezes into ice-like crystals at a relatively high temperature (below 17 °C).

Exam Tip: Emphasize the freezing property and the ice-like appearance at temperatures just below room temperature as the reason for the name "glacial."

 

Question 5. Why does carbon form large number of compounds?
Answer: Carbon forms a vast number of compounds due to its unique properties of tetravalency and catenation. Tetravalency means carbon has a valency of 4, letting it share its outer electrons with four other elements, such as hydrogen or chlorine, to achieve a stable electron arrangement. Catenation is carbon's ability to self-link, creating long, branched, or cyclic chains, which allows for the formation of countless compounds.
In simple words: Carbon forms many compounds because it can make four bonds (tetravalency) and link with other carbon atoms in long chains (catenation).

Exam Tip: Always cite catenation and tetravalency as the primary reasons for carbon's diverse compound formation, explaining what each property entails.

 

Question 6. Write the structural formula for bromopentane and ethanoic acid.
Answer: The structural formulas for bromopentane and ethanoic acid illustrate their atom arrangements and bonding.
Structural formula for Ethanoic acid (\( CH_3COOH \)):

\( \begin{array}{cc} H & O \\ | & || \\ H-C & -C-OH \\ | \\ H \end{array} \)

Structural formula for Bromopentane (\( C_5H_{11}Br \)):

\( \begin{array}{cccccc} H & H & H & H & H \\ | & | & | & | & | \\ H-C & -C & -C & -C & -C-Br \\ | & | & | & | & | \\ H & H & H & H & H \end{array} \)

In simple words: The structures show a pentane chain with a bromine atom and a two-carbon chain with a carboxylic acid group.

Exam Tip: When drawing structural formulas, ensure that carbon forms four bonds, hydrogen forms one bond, oxygen forms two bonds, and halogens (like bromine) form one bond.

 

Question 7. How does ethanoic acid react with carbonates and hydrogen carbonates? Show it with the equation.
Answer: Ethanoic acid reacts with carbonates and hydrogen carbonates to produce a salt, carbon dioxide (\( CO_2 \)), and water (\( H_2O \)). The salt formed in these reactions is sodium acetate.
\( 2CH_3COOH + Na_2CO_3 \rightarrow 2CH_3COONa + H_2O + CO_2 \)
\( CH_3COOH + NaHCO_3 \rightarrow CH_3COONa + H_2O + CO_2 \)
In simple words: Ethanoic acid reacts with carbonates and hydrogen carbonates to make sodium acetate, water, and carbon dioxide.

Exam Tip: Remember that acids react with carbonates and hydrogen carbonates to produce carbon dioxide gas, which can be identified by its effervescence.

 

Question 8. Draw the structural isomers of butane and name them.
Answer: Butane (\( C_4H_{10} \)) has two structural isomers: n-butane (a straight chain) and iso-butane (a branched chain). Their structures are shown below.
(i) n-Butane:

\( \begin{array}{cccc} H & H & H & H \\ | & | & | & | \\ H-C & -C & -C & -C-H \\ | & | & | & | \\ H & H & H & H \end{array} \)

(ii) Iso-Butane (2-Methylpropane):

\( \begin{array}{ccc} & H & \\ & | & \\ H & C & H \\ | & | & | \\ H-C & -C & -C-H \\ | & | & \\ H & H & H \end{array} \)

In simple words: Butane has two forms: a straight chain called n-butane, and a branched chain called iso-butane.

Exam Tip: For drawing isomers, ensure the molecular formula remains the same while the connectivity of atoms changes. Butane is a common example for illustrating chain isomerism.

 

Question 9. A student burns a hydrocarbon in air and obtains sooty flame. Give two reasons for this observation.
Answer: A sooty flame can result from two main reasons: incomplete combustion of saturated hydrocarbons or the burning of unsaturated hydrocarbons.
In simple words: Sooty flames happen when hydrocarbons burn incompletely, or when unsaturated hydrocarbons are burned.

Exam Tip: Sooty flames indicate incomplete combustion, which often occurs with unsaturated hydrocarbons due to their higher carbon-to-hydrogen ratio, or with saturated hydrocarbons when there isn't enough oxygen.

 

Question 10. Differentiate between saturated and unsaturated hydrocarbons. Give one example for each.
Answer:
Saturated hydrocarbon:

• It consists of single bonds between carbon atoms.
• It burns with a blue flame.
• Shows substitution reactions.
• Less reactive.
• Example: Methane (\( CH_4 \)), Ethane (\( C_2H_6 \)).

Unsaturated hydrocarbons:

• It consists of double or triple bonds between carbon atoms.
• It burns with a sooty flame.
• Shows addition reactions.
• More reactive.
• Example: Ethene (\( H_2C=CH_2 \)), Ethyne (\( HC\equiv CH \)).

In simple words: Saturated hydrocarbons have single bonds, burn with a blue flame, and are less reactive; unsaturated hydrocarbons have double/triple bonds, burn with a sooty flame, and are more reactive.

Exam Tip: Create a table in your mind to compare these points directly. The key distinctions are bond type, flame color, and type of reaction.

 

Question 11. Write the general formula for each of the following hydrocarbons and give one example for each.
(i) Alkene
(ii) Alkyne
Answer:
(i) Alkene: The general formula is \( C_nH_{2n} \). An example is ethene (\( C_2H_4 \)).
(ii) Alkyne: The general formula is \( C_nH_{2n-2} \). An example is ethyne (\( C_2H_2 \)).
In simple words: Alkenes follow \( C_nH_{2n} \) (like ethene), and alkynes follow \( C_nH_{2n-2} \) (like ethyne).

Exam Tip: Memorize the general formulas for alkanes (\( C_nH_{2n+2} \)), alkenes (\( C_nH_{2n} \)), and alkynes (\( C_nH_{2n-2} \)) as they are fundamental for classifying hydrocarbons.

 

Question 12. Name the functional groups of the following:
(a) \( CH_3-Cl \)
(b) \( CH_3-COOH \)
(c) \( CH_3-CO-CH_3 \)
(d) \( C_2H_5OH \)
Answer:
(a) The functional group in \( CH_3-Cl \) is Chloro (a halogen group).
(b) The functional group in \( CH_3-COOH \) is Carboxylic acid.
(c) The functional group in \( CH_3-CO-CH_3 \) is Ketone.
(d) The functional group in \( C_2H_5OH \) is Alcohol.
In simple words: The groups are Chloro (a halogen), Carboxylic acid, Ketone, and Alcohol, respectively.

Exam Tip: Be able to recognize and name common functional groups like halogens, carboxylic acids, ketones, and alcohols from their structural representations.

 

Question 13. Explain substitution reaction with example.
Answer: A substitution reaction occurs when a hydrogen atom in a saturated hydrocarbon is replaced by another atom, such as a chlorine atom. For instance, methane (\( CH_4 \)) reacting with chlorine (\( Cl_2 \)) forms methyl chloride (\( CH_3Cl \)) and hydrochloric acid (\( HCl \)).
\( CH_4 + Cl_2 \rightarrow CH_3Cl + HCl \)
Methane \( + \) Chlorine \( \rightarrow \) Methyl chloride \( + \) Hydrochloric acid
In simple words: In a substitution reaction, one atom (like hydrogen) in a saturated hydrocarbon is swapped with another atom (like chlorine), as seen when methane reacts with chlorine.

Exam Tip: Key aspects of substitution reactions are that they occur in saturated hydrocarbons and involve the replacement of one atom or group by another, often initiated by UV light in the case of halogens.

 

Question 14. Diamond and graphite show different physical properties although they are made up of carbon and shows same chemical properties. What is this property called?
Answer: This property is known as allotropy. The physical differences arise from the varied carbon-carbon bonding structures. In diamond, each carbon atom links to four others via strong covalent bonds, making it very hard. In contrast, graphite has each carbon atom forming three strong bonds with other carbons, creating hexagonal layers that can easily slide past each other, making it soft.
In simple words: This property is called allotropy, where different forms of carbon like diamond and graphite have varying physical properties due to how their carbon atoms are bonded.

Exam Tip: Clearly define allotropy as the existence of an element in two or more different forms in the same physical state, and explain how the structural difference (bonding pattern) leads to distinct physical properties.

 

Question 15. What is denatured alcohol?
Answer: Denatured alcohol is ethanol that has been mixed with substances like methanol, copper sulfate, or pyridine. These additions make it poisonous and unsuitable for consumption, thus rendering it unfit for drinking.
In simple words: Denatured alcohol is ethanol made undrinkable by adding toxic chemicals like methanol.

Exam Tip: Understand that denaturing alcohol is done to prevent its consumption and reduce excise duty, making it accessible for industrial and scientific uses.

 

Question 16. What is esterification and give its uses?
Answer: Esterification is a chemical reaction where a carboxylic acid combines with an alcohol to produce an ester, typically in the presence of concentrated sulfuric acid. These reactions are used to create esters, which are widely used as flavorings in food and as components in perfumes due to their sweet smell.
\( CH_3COOH + C_2H_5OH \xrightarrow{\text{Conc. } H_2SO_4} CH_3COOC_2H_5 + H_2O \)
Ethanoic acid \( + \) Ethanol \( \xrightarrow{\text{Esterification}} \) Ester \( + \) Water
In simple words: Esterification is the making of esters from carboxylic acids and alcohols, used for making perfumes and food flavorings.

Exam Tip: For esterification, key points include the reactants (acid and alcohol), product (ester and water), catalyst (concentrated \( H_2SO_4 \)), and its applications in fragrances and food.

 

Question 17. Give difference between soap and detergent.
Answer:
Soap:
• Soaps are the sodium salts of long-chain carboxylic acids.
• Soaps are not suitable for washing with hard water because they form insoluble scum.
• Soaps are prepared from fats or vegetable oils.
• Soaps are biodegradable.
Detergent:
• Detergents are generally sodium salts of long-chain sulphonic acids or long-chain ammonium salts of chloride or bromide ions.
• Detergents do not form insoluble scum with hard water.
• Detergents are not prepared from fats or vegetable oils.
• Detergents are non-biodegradable.
In simple words: Soaps come from natural fats and don't work well in hard water, while detergents are made synthetically and work in all water types without forming scum.

Exam Tip: Remember to list at least three key differences for both soap and detergent, focusing on their chemical structure, reaction with hard water, and environmental impact.

 

Question 18. Differentiate between ethanol and ethanoic acid on basis of the following test:
(i) Blue litmus test
(ii) Reaction with sodium bicarbonate
(iii) Sodium metal test
Answer:

In simple words: These tests help us tell ethanol and ethanoic acid apart. Ethanoic acid changes blue litmus to red and fizzes with baking soda, while ethanol doesn't. Both react with sodium, but ethanoic acid has distinctive acidic reactions.

Exam Tip: When differentiating compounds, always remember the key functional groups and their characteristic reactions, especially for acids (litmus, carbonates) and alcohols (sodium metal).

 

Question 19. Giving chemical equations of the reactions write what happens when
(i) Ethanol is heated with excess of concentrated sulphuric acid at 443 K.
(ii) Ethanoic acid reacts with ethanol in presence of an acid.
(iii) Ester with molecular formula \( \text{CH}_3\text{COOC}_2\text{H}_5 \) reacts with sodium hydroxide.
Answer:
(i) \( \text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{Conc. H}_2\text{SO}_4, \text{heat}} \text{H}_2\text{C}=\text{CH}_2 + \text{H}_2\text{O} \)
Ethanol forms ethene with water removal.
(ii) \( \text{C}_2\text{H}_5\text{OH} + \text{CH}_3\text{COOH} \xrightarrow{\text{acid}} \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O} \)
Ethanol and ethanoic acid create an ester and water.
(iii) \( \text{CH}_3\text{COOC}_2\text{H}_5 + \text{NaOH} \rightarrow \text{C}_2\text{H}_5\text{OH} + \text{CH}_3\text{COONa} \)
The ester reacts with sodium hydroxide to produce ethanol and sodium acetate.
In simple words: Ethanol loses water to become ethene when heated with strong acid. Ethanol and ethanoic acid join to make an ester and water. An ester reacts with sodium hydroxide to form alcohol and a sodium salt.

Exam Tip: For reactions, always balance the equations and correctly identify the products, especially for dehydration, esterification, and saponification.

 

Question 20. How can you obtain the following from pure ethanol:
(i) Ethene
(ii) Ethanoic acid
(iii) Ester?
Answer:
(i) Ethene: Ethanol, when heated with excess concentrated sulphuric acid, will form ethene.
\( \text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{Conc. H}_2\text{SO}_4, \text{heat}} \text{H}_2\text{C}=\text{CH}_2 + \text{H}_2\text{O} \)
(ii) Ethanoic acid: On oxidation of ethanol with an oxidising agent like alkaline \( \text{KMnO}_4 \) or acidified \( \text{K}_2\text{Cr}_2\text{O}_7 \), ethanoic acid is formed.
\( \text{CH}_3-\text{CH}_2\text{OH} \xrightarrow{\text{alk. KMnO}_4/\text{acidified K}_2\text{Cr}_2\text{O}_7} \text{CH}_3\text{COOH} \)
(iii) Ester: To get esters, ethanol is reacted with any carboxylic acid in the presence of acid.
Example: \( \text{C}_2\text{H}_5\text{OH} + \text{CH}_3\text{COOH} \xrightarrow{\text{acid}} \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O} \)
In simple words: To get ethene, heat ethanol with strong sulfuric acid. To get ethanoic acid, oxidize ethanol with a strong oxidizing agent. To get an ester, react ethanol with a carboxylic acid using an acid catalyst.

Exam Tip: Recall the specific reagents and conditions required for dehydration (ethene), oxidation (ethanoic acid), and esterification (ester) of ethanol.

 

Question 21. Write the chemical equations for the following reactions:
(i) Conversion of oils into fats
(ii) Oxidation of ethanol
Answer:
(i) Conversion of oils into fats (Hydrogenation):
\( \text{R}-\text{C}=\text{C}-\text{R} + \text{H}_2 \xrightarrow{\text{Ni catalyst}} \text{R}-\text{C}-\text{C}-\text{R} \)
\( \text{Oil (unsaturated)} \rightarrow \text{Fats (saturated)} \)
(ii) Oxidation of ethanol:
\( \text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{alk. KMnO}_4} \text{CH}_3\text{COOH} \)
Ethanol converts to ethanoic acid.
In simple words: Oils turn into fats by adding hydrogen with a nickel catalyst. Ethanol turns into ethanoic acid when an alkaline potassium permanganate solution is added.

Exam Tip: Remember that hydrogenation is an addition reaction converting unsaturated compounds to saturated ones, typically used for vegetable oils, and oxidation of ethanol yields ethanoic acid.

 

Question 22. An organic compound ‘X' which is also called antifreeze mixture when mixed with water has the molecular formula \( \text{C}_2\text{H}_6\text{O} \). 'X' on oxidation gives a compound Y which gives effervescence with a baking soda solution. What can X and Y be? Write their structural formula.
Answer:
X is ethanol, \( \text{C}_2\text{H}_5\text{OH} \)
Y is ethanoic acid, \( \text{CH}_3\text{COOH} \)
Structural formula:
Ethanol:
\( \text{H}-\text{C}-\text{C}-\text{OH} \)
           \(\text{H} \text{H}\)
           \(\text{H} \text{H}\)
Ethanoic acid:
\( \text{H}-\text{C}-\text{C}-\text{OH} \)
           \(\text{H} \text{O}\)
           \(\text{H}\)
In simple words: Compound X is ethanol, known as an antifreeze. When ethanol is oxidized, it becomes ethanoic acid (Compound Y), which fizzes with baking soda because it is an acid.

Exam Tip: Pay attention to keywords like "antifreeze mixture" (ethanol) and "effervescence with baking soda" (acidic nature, ethanoic acid) to identify organic compounds.

 

Question 23. Write the structures of isomers of hexane.
Answer:
The isomers of hexane (\( \text{C}_6\text{H}_{14} \)) are:
(i) n-Hexane:
\( \text{H}-\text{C}-\text{C}-\text{C}-\text{C}-\text{C}-\text{C}-\text{H} \)
    \(\text{H H H H H H}\)
    \(\text{H H H H H H}\)
(ii) 2-Methylpentane:
\( \text{H}-\text{C}-\text{C}-\text{C}-\text{C}-\text{C}-\text{H} \)
    \(\text{H H H H H}\)
    \(\text{H H H H}\)
        \(\text{H}-\text{C}-\text{H}\)
           \(\text{H}\)
(iii) 3-Methylpentane:
\( \text{H}-\text{C}-\text{C}-\text{C}-\text{C}-\text{C}-\text{H} \)
    \(\text{H H H H H}\)
        \(\text{H}-\text{C}-\text{H}\)
           \(\text{H}\)
    \(\text{H H H H}\)
(iv) 2,3-Dimethylbutane:
\( \text{H}-\text{C}-\text{C}-\text{C}-\text{C}-\text{H} \)
       \(\text{H} \text{H}\)
    \(\text{H} \text{CH}_3 \text{CH}_3 \text{H}\)
       \(\text{H}\)
(v) 2,2-Dimethylbutane:
\( \text{H}-\text{C}-\text{C}-\text{C}-\text{C}-\text{H} \)
       \(\text{H}\)
    \(\text{H} \text{CH}_3 \text{CH}_3 \text{H}\)
       \(\text{H} \text{CH}_3\)
In simple words: Hexane, which has six carbon atoms, can be arranged in five different ways. These different arrangements are called isomers, each with its own specific structure and name.

Exam Tip: When drawing isomers, start with the straight chain, then gradually shorten the main chain and add methyl branches in different positions, ensuring no two structures are identical by rotating them.

 

Question 24. Complete and balance the following equations:
(a) \( \text{CH}_3\text{CH}_2\text{OH} + \text{O}_2 \rightarrow \)
(b) \( \text{Na} + \text{CH}_3\text{CH}_2\text{OH} \rightarrow \)
(C) \( \text{CH}_3-\text{CH}_2\text{OH} \xrightarrow{\text{conc. H}_2\text{SO}_4} \)
Answer:
(a) \( \text{CH}_3\text{CH}_2\text{OH} + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O} + \text{heat} + \text{light} \)
(b) \( 2\text{Na} + 2\text{CH}_3\text{CH}_2\text{OH} \rightarrow 2\text{CH}_3\text{CH}_2\text{ONa} + \text{H}_2 \)
(C) \( \text{CH}_3-\text{CH}_2\text{OH} \xrightarrow{\text{conc. H}_2\text{SO}_4} \text{CH}_2=\text{CH}_2 + \text{H}_2\text{O} \)
In simple words: Ethanol burns in oxygen to produce carbon dioxide, water, heat, and light. Sodium reacts with ethanol to form sodium ethoxide and hydrogen gas. Concentrated sulfuric acid dehydrates ethanol to create ethene and water.

Exam Tip: For combustion reactions, always balance with \( \text{CO}_2 \) and \( \text{H}_2\text{O} \) as products. For reactions with active metals, hydrogen gas is typically released. Dehydration reactions usually involve a strong dehydrating agent and heat.

 

Question 25. Give two uses of ethanol and one harmful effect of it.
Answer:
Ethanol is a good solvent, so it is used in making medicines such as tincture iodine, cough syrups, and many tonics. Ethanol is also used in making alcoholic drinks.
Harmful effects: Consuming even a small amount of ethanol can be lethal. Long-term use or consumption can lead to severe health problems.
In simple words: Ethanol is used in medicines and alcoholic drinks. However, it can be very dangerous even in small amounts and cause serious health issues with long-term use.

Exam Tip: When discussing uses and effects, ensure you provide at least two distinct uses and one clear harmful effect, as requested by the question.

 

Question 26. (a) Why are covalent compounds generally poor conductors of electricity?
(b) Name the following compound:
\( \text{H}-\text{C}-\text{C}-\text{C}-\text{H} \)
    \(\text{H} \text{H} \text{H}\)
    \(\text{H} \text{O} \text{H}\)
(c) Name the gases evolved when ethanoic acid is added to sodium carbonate. How would you prove the presence of this gas?
Answer:
(a) Covalent compounds do not form ions, which are needed to conduct electricity.
(b) Propanone
(c) Ethanoic acid reacts with sodium carbonate to produce carbon dioxide gas. To prove the presence of this gas, pass it through freshly prepared lime water. It turns lime water milky, showing the presence of carbon dioxide.
Equation:
\( 2\text{CH}_3\text{COOH} + \text{Na}_2\text{CO}_3 \rightarrow 2\text{CH}_3\text{COONa} + \text{H}_2\text{O} + \text{CO}_2 \)
Test for the gas evolved \( \rightarrow \text{Ca(OH)}_2 + \text{CO}_2 \rightarrow \text{CaCO}_3 + \text{H}_2\text{O} \)
Lime water turns milky precipitate.
In simple words: Covalent compounds don't conduct electricity because they don't make ions. The compound is propanone. When ethanoic acid mixes with sodium carbonate, carbon dioxide gas comes out. You can check for it by bubbling the gas through limewater, which will turn cloudy.

Exam Tip: For (a), emphasize the absence of free ions. For (b), correctly identify the functional group (ketone) and count carbons. For (c), remember the characteristic test for \( \text{CO}_2 \) (limewater turns milky) and provide the chemical equation.

 

Question 1. An organic compound 'A' is widely used as a preservative in pickles and has a, molecular formula \( \text{C}_2\text{H}_4\text{O}_2 \). This compound reacts with ethanol to form a sweet, smelling compound 'B'.
(a) Identify the compound A'.
(b) Write the chemical equation for its reaction with ethanol to form compound 'B'.
(c) How can we get compound A' and 'B'?
(d) Which gas is obtained when compound A reacts with washing soda? Give the reaction.
(e) How can 'A' be obtained back from 'B'.
Answer:
(a) 'A' is \( \text{CH}_3\text{COOH} \), acetic acid.
(b) \( \text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{conc. H}_2\text{SO}_4} \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O} \)
(Ester) 'B'
(c) By oxidation of ethanol, we get A (acetic acid)
\( \text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{Alkaline KMnO}_4 + \text{heat}} \text{CH}_3\text{COOH} \)
A, Acetic acid
(d) A + washing soda \( \rightarrow \text{CO}_2 \) gas is produced
\( 2\text{CH}_3\text{COOH} + \text{Na}_2\text{CO}_3 \rightarrow 2\text{CH}_3\text{COONa} + \text{H}_2\text{O} + \text{CO}_2 \)
(e) Saponification
\( \text{CH}_3\text{COOC}_2\text{H}_5 + \text{NaOH} \rightarrow \text{C}_2\text{H}_5\text{OH} + \text{CH}_3\text{COONa} \)
B, Ester gives sodium acetate, which can be converted back to acetic acid:
\( \text{CH}_3\text{COONa} + \text{H}^+ \rightarrow \text{CH}_3\text{COOH} + \text{Na}^+ \)
A, Acetic acid
In simple words: Compound A is acetic acid, used as a preservative. It reacts with ethanol to form a sweet-smelling ester, Compound B. Acetic acid can be made by oxidizing ethanol. If acetic acid reacts with washing soda, carbon dioxide gas is released. To get acetic acid back from the ester, saponification is performed, followed by acidification.

Exam Tip: This question covers multiple fundamental organic reactions. Be sure to know the molecular formula and common name for acetic acid, the esterification reaction, the oxidation of alcohol to acid, the reaction of acids with carbonates, and saponification (hydrolysis of ester).

 

Question 2. Identify the compound A, B, C, D, and E in the following reaction:
(a) \( \text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{A}} \text{CH}_3\text{COOH} \)
(b) \( \text{CH}_3\text{CH}_2\text{OH} + \text{CH}_3\text{COOH} \xrightarrow{\text{B}} \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O} \)
(c) \( \text{B} + \text{NaOH} \rightarrow \text{C}_2\text{H}_5\text{OH} + \text{C} \)
(d) \( \text{D} + \text{Na}_2\text{CO}_3 \rightarrow \text{CH}_3\text{COONa} + \text{E} + \text{H}_2\text{O} \)
(e) \( \text{E} + \text{Ca(OH)}_2 \rightarrow \text{F} + \text{H}_2\text{O} \)
Answer:
(a) A = Alkaline \( \text{KMnO}_4 \) or acidified \( \text{K}_2\text{Cr}_2\text{O}_7 \)
(b) B = \( \text{conc. H}_2\text{SO}_4 \)
(c) C = \( \text{CH}_3\text{COONa} \)
(d) D = \( \text{CH}_3\text{COOH} \); E = \( \text{CO}_2 \)
(e) E = \( \text{CO}_2 \); F = \( \text{CaCO}_3 \)
In simple words: A is an oxidizing agent like alkaline potassium permanganate. B is concentrated sulfuric acid, used as a catalyst. C is sodium acetate. D is acetic acid and E is carbon dioxide. F is calcium carbonate.

Exam Tip: Carefully analyze each reaction step. Remember that A is an oxidizing agent (for alcohol to acid), B is an acid catalyst (for esterification), C is the salt product of saponification, D is the acid reacting with carbonate, and E is the gas produced, which then reacts with limewater to form F.

 

Question 3. What are soaps? Explain the mechanism of the cleansing action of soaps? Soaps form scum with hard water. Explain why? How this problem is overcome by use of detergents?
Answer:
Soap molecules have two ends: a charged end, called hydrophilic, that gets attracted to water, and a long carbon chain, called hydrophobic tail, that repels water. When soap is dissolved in water, the carbon chain (hydrophobic end) gets attracted towards oil, dirt, and grease. The hydrophilic end stays away from this. This leads to micelle formation.
The tail entangles dirt, oil, or grease; if needed, agitation is done. Lots of rinsing is done with water so that water molecules attract the charged \( \text{(Na}^+) \) end and carry the soap molecules with attached dirt to clean clothes, utensils, etc.
Scum formation:
Hard water contains salts of calcium and magnesium. When a soap molecule comes into contact with these salts, it forms a curdy white precipitate (a compound insoluble in water) called scum.
\( \text{Soap + Hard water} \rightarrow \text{scum} \)
In the case of detergents, the salts present in hard water do not react with the molecules of detergent to form insoluble scum. Instead, the detergent molecules remain dissolved and help in the cleansing action.
In simple words: Soaps are substances that clean things. They have one part that likes water and another part that likes oil. This helps them grab dirt and grease, which then washes away. Hard water stops soap from cleaning properly by making a sticky mess called scum. Detergents fix this problem because they don't form scum in hard water, so they clean better everywhere.

Exam Tip: Clearly define soap, then explain the micelle formation with hydrophilic/hydrophobic parts. Detail how scum forms with hard water and how detergents, unlike soaps, avoid this issue due to their different chemical structure.

 

Question 4. (a) What do you mean by allotropy?
(b) What is isomerism?
(c) Give one example of homologous series, give two properties of it.
(d) What is the full form of IUPAC?
Answer:
(a) Allotropy: It is the property of an element where it shows the same chemical properties but different physical properties, due to differences in the bonding of atoms.
Example: Diamond and graphite have the same chemical properties but appear physically different because their atomic bonding differs.
(b) Isomerism: It is the property of hydrocarbons that show the same molecular formula but exhibit different structural formulae.
Example: Butane can exist as n-butane and iso-butane, both having the formula \( \text{C}_4\text{H}_{10} \). Both of them show different properties.
(c) Homologous series: When the members of a hydrocarbon family obey the same general formula, they are said to be in a homologous series. In this series, members are arranged in increasing order of their molecular masses.
Example: Alkane – \( \text{C}_n\text{H}_{2n+2} \)
\( \text{CH}_4 \) – Methane
\( \text{C}_2\text{H}_6 \) – Ethane
\( \text{C}_3\text{H}_8 \) – Propane
\( \text{C}_4\text{H}_{10} \) – Butane
Properties:
• The difference between two consecutive members of a homologous series is \( -\text{CH}_2 \) and a mass of 14 a.m.u.
• They all show the same chemical properties and a slight gradation in their physical properties.
(d) IUPAC: International Union of Pure and Applied Chemistry.
In simple words: Allotropy is when an element exists in different forms with different physical looks, like diamond and graphite. Isomerism is when compounds have the same chemical formula but different arrangements of atoms. A homologous series is a group of compounds with similar formulas, properties, and increasing molecular weights. IUPAC stands for International Union of Pure and Applied Chemistry.

Exam Tip: Define each term clearly and concisely, providing a relevant example for allotropy, isomerism, and homologous series. For homologous series, remember the common difference of \( \text{-CH}_2 \) and 14 a.m.u. in successive members.

 

Question 5. (c) Why does carbon form large number of compounds?
Answer:
Carbon forms a large number of compounds due to:
• Catenation – This self-linking property leads to long straight chains, branched chains, and cyclic chains.
• Isomerism – Carbon compounds can exist in more than one structural formula but have the same molecular formula.
• Tetravalency – To acquire a noble gas configuration, carbon shares its outer electrons with other elements, thus forming covalent bonds with other elements.
In simple words: Carbon makes many compounds because it can link with itself in long chains (catenation), form different arrangements with the same atoms (isomerism), and bond with four other atoms (tetravalency).

Exam Tip: The main reasons for carbon's vast compound formation are catenation and tetravalency; isomerism also contributes to the diversity. Explain each property briefly.

 

Question 1. A, B, C are members of homologous series and their melting points are -183°C, -138°C, 130°C respectively. Among these
1. Which member will have least number of carbon atoms?
2. Which member will have maximum number of carbon atoms?
Answer:
1. A will have the least number of carbon atoms.
2. C will have the maximum number of carbon atoms.
In simple words: In a homologous series, as you add more carbon atoms, the melting point generally increases. So, the compound with the lowest melting point (A) will have the fewest carbons, and the one with the highest melting point (C) will have the most carbons.

Exam Tip: Remember that within a homologous series, melting and boiling points generally increase with an increasing number of carbon atoms due to stronger intermolecular forces.

 

Question 2. A hydrocarbon compound A is active ingredient of wine and cough syrup. A on oxidation with acidified \( \text{K}_2\text{Cr}_2\text{O}_7 \) forms compound B. Identify the compound A and B and write the chemical equations involved.
Answer:
A is ethanol, \( \text{C}_2\text{H}_5\text{OH} \)
B is ethanoic acid, \( \text{CH}_3\text{COOH} \)
Equation:
\( \text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{acidified K}_2\text{Cr}_2\text{O}_7} \text{CH}_3\text{COOH} \)
In simple words: Compound A is ethanol, found in wine and cough syrup. When ethanol is oxidized by acidified potassium dichromate, it becomes compound B, which is ethanoic acid.

Exam Tip: Recognize ethanol's common uses and its oxidation product, ethanoic acid. The oxidizing agent acidified \( \text{K}_2\text{Cr}_2\text{O}_7 \) is a key indicator for this conversion.

 

Question 3. Write an activity to show the acidic nature of ethanol. Give the chemical equation of the reaction taking place.
Answer:
Take ethanol in a test tube and drop a small piece of sodium, about the size of a grain of rice, into it. The reaction evolves a colorless gas, which is hydrogen. Hydrogen gas can be tested by bringing a burning splinter or matchstick near the mouth of the test tube; it will burn with a popping sound. This activity proves that ethanol, like other acids, releases \( \text{H}_2 \) gas.
\( 2\text{Na} + 2\text{CH}_3\text{CH}_2\text{OH} \rightarrow 2\text{CH}_3\text{CH}_2\text{ONa} + \text{H}_2 \)
In simple words: To show ethanol is acidic, put a tiny bit of sodium metal in it. A gas will be made. If you hold a lit match near the gas, you'll hear a "pop" sound, which proves it's hydrogen gas. This shows that ethanol acts like an acid by releasing hydrogen.

Exam Tip: To demonstrate acidic nature, the reaction with an active metal (like sodium) producing hydrogen gas is a classic test. Remember the "pop" sound for hydrogen detection.

 

Question 4. A compound 'X' has molecular formula \( \text{C}_2\text{H}_6\text{O} \) is saturated hydrocarbons and is a very good solvent. How can you convert it into unsaturated hydrocarbon? Identify X and show its conversion with the help of equation.
Answer:
'X' is \( \text{CH}_3-\text{CH}_2\text{OH} \) ethanol. It can be made unsaturated by heating it with conc. \( \text{H}_2\text{SO}_4 \), which acts as a dehydrating agent and removes water from it, thereby forming ethene.
\( \text{CH}_3-\text{CH}_2\text{OH} \xrightarrow{\text{Hot conc. H}_2\text{SO}_4} \text{H}_2\text{C}=\text{CH}_2 + \text{H}_2\text{O} \)
In simple words: Compound X is ethanol. You can change it into an unsaturated hydrocarbon (ethene) by heating it with strong sulfuric acid. The acid removes water from the ethanol.

Exam Tip: Identify ethanol (\( \text{C}_2\text{H}_6\text{O} \)) as the saturated hydrocarbon that can be dehydrated to an unsaturated alkene (ethene) using concentrated sulfuric acid as the dehydrating agent.

 

Question 5. Take about 20 mL of castor oil in a beaker. Add 30 mL of 20% sodium hydroxide solution. Heat the mixture with continuous stirring for a few minutes till the mixture thickens. Add 5-10 g of common salt to this. Stir the mixture well, allow it to cool, soaps is obtained.
Answer:
This activity describes the process of saponification, which produces soap from oil and sodium hydroxide. The addition of common salt helps in the precipitation of soap from the solution.
In simple words: This activity shows how to make soap by mixing castor oil with sodium hydroxide, heating it, and then adding salt to help the soap separate.

Exam Tip: This experiment describes saponification. The key ingredients are oil (or fat) and a strong base (like NaOH), with salt added to precipitate the soap.

 

Question 1. Give one commercial use and one domestic use of acetic acid.
Answer:
Commercial use: it is used to make ester.
Domestic use: it is used to make vinegar, which is used as an ingredient in the kitchen for many cooking dishes.
In simple words: Acetic acid is used commercially to make esters. At home, it's used as vinegar in cooking.

Exam Tip: Remember acetic acid's role in esterification (commercial) and as vinegar (domestic) for a complete answer.

 

Question 2. A student wants to find the freezing temperature of acetic acid. List the materials required to do this test in the lab.
Answer:
The materials required are a test tube, acetic acid, a thermometer, and a beaker containing ice.
In simple words: To find out when acetic acid freezes, you'll need a test tube with the acid, a thermometer to read the temperature, and a beaker filled with ice to make it cold.

Exam Tip: Basic lab setup for freezing point determination always includes the substance, a thermometer, and a cooling bath (usually ice).

 

Question 3. Draw the observation table for the above experiment and write the prediction for the same.
Answer:
The observation table is:

Prediction: The acetic acid should freeze at a temperature below 17 degrees, as its freezing temperature is 16.6 degrees C.
In simple words: An observation table would record the temperature and how the acid looks. The prediction is that acetic acid will freeze below 17 degrees Celsius because its actual freezing point is 16.6 degrees C.

Exam Tip: For an observation table, always include columns for the measured variable (temperature) and observations (e.g., state of matter). The prediction should state the expected freezing point for acetic acid.

 

Question 4. On adding sodium bicarbonate, the acetic acid releases carbon dioxide gas, suggest two ways of testing this gas in the lab.
Answer:
• Allow the gas obtained to pass through freshly prepared limewater. If it turns milky, the gas is carbon dioxide.
• Keep a burning candle near the mouth of the delivery tube through which the gas is coming out. The burning candle will extinguish.
In simple words: To check if the gas is carbon dioxide, you can bubble it through limewater; if it turns milky, it's \( \text{CO}_2 \). Another way is to hold a burning candle near the gas; if the flame goes out, it's also carbon dioxide.

Exam Tip: The two classic tests for carbon dioxide are the limewater test (turns milky) and the burning splint test (extinguishes the flame).

 

Question 5. Why is acetic acid called a weak acid?
Answer:
The pH of acetic acid is slightly higher because it dissociates partially to form \( \text{H}^+ \) and \( \text{CH}_3\text{COO}^- \) ions.
In simple words: Acetic acid is called a weak acid because it doesn't fully break apart into ions in water, making its solution less acidic than strong acids.

Exam Tip: A weak acid is one that only partially ionizes in water, releasing fewer \( \text{H}^+ \) ions compared to a strong acid, resulting in a higher pH.

 

Question 6. Soap is prepared in the lab by the reaction called saponification. Give one example of this reaction.
Answer:
The reaction of (organic acid) fatty acids with sodium hydroxide gives soap and glycerol. Soap is the sodium salt of fatty acids.
Example: \( \text{Fatty acid} + \text{NaOH} \rightarrow \text{Soap} + \text{Glycerol} \)
In simple words: Saponification is how we make soap. It involves mixing a fatty acid with sodium hydroxide to produce soap and a byproduct called glycerol.

Exam Tip: Saponification is the hydrolysis of an ester (like a fat or oil) with a strong base to produce a carboxylic acid salt (soap) and an alcohol (glycerol).

 

Question 7. How is soap helpful in cleaning action?
Answer:
Soap molecules form micelles when added to a bucket containing dirty/oily cloth soaked in water. The hydrophobic tail attracts the oil and the hydrophilic part of the soap molecule attracts the water, and hence on rinsing, it removes the dirt.
In simple words: Soap cleans by forming tiny balls called micelles around dirt and oil. The oil-loving part of the soap grabs the dirt, and the water-loving part then lets water carry it away during rinsing.

Exam Tip: Focus on the micelle formation, explaining how the hydrophobic tails encapsulate grease/dirt and the hydrophilic heads interact with water, allowing the dirt to be washed away.

 

Question 8. State the conditions for obtaining the best cleansing action of soap.
Answer:
The water used should be soft water and not hard water. The soap should have more soap molecules and fewer impurities or adulterants in it.
In simple words: For soap to clean best, you should use soft water and the soap itself should be pure with many active cleaning molecules.

Exam Tip: Hard water interferes with soap's action by forming scum; thus, using soft water is crucial for effective cleansing. Also, the purity of the soap affects its performance.

 

Question 9. What is the environmental impact of excess use of detergents?
Answer:
Detergents are non-biodegradable and pollute water, thereby killing many aquatic life creatures.
In simple words: Too much detergent pollutes water because it doesn't break down easily, harming and killing many animals and plants living in the water.

Exam Tip: Remember that the primary environmental concern with detergents is their non-biodegradable nature, leading to water pollution and harm to aquatic ecosystems.

 

Question 10. How can you prepare a hard water in the lab and show the ineffective action of soap in it.
Answer:
To make hard water in the lab, take normal water and dissolve some calcium and magnesium salts of sulphates and phosphates in it. To show the ineffective action of soap with this water, just add a little sample of soap solution to it. The sample will not form any froth, and a curdy white precipitate will be formed.
In simple words: You can make hard water by adding calcium and magnesium salts to regular water. To show soap doesn't work well in it, add soap to this hard water; it won't foam, and you'll see a white, curdy mess instead.

Exam Tip: The key to preparing hard water is dissolving calcium and magnesium salts. The indicator of soap's ineffective action is the lack of lather and the formation of a curdy precipitate (scum).