Carbon and its Compounds

Step-by-Step Solution

Step 1: Identify number of carbon atoms
Prefix “but-” indicates: \[ \text{Number of carbons} = 4 \]

Step 2: Identify functional group from suffix
Suffix “-one” represents a ketone functional group.

Step 3: Understand ketone structure
A ketone contains a carbonyl group: \[ \ce{C=O} \] placed between two carbon atoms (not at the end).

Step 4: Structure of butanone (2-butanone)
\[ \ce{CH3 - CO - CH2 - CH3} \] Here, the carbonyl carbon is bonded to two carbon atoms.

Step 5: Compare with options
Only ketone matches this structure.

Final Answer: (c) ketone

Step-by-Step Solution

Step 1: Define complete combustion
In complete combustion, fuel burns in sufficient oxygen: \[ \text{Fuel} + \ce{O2} \rightarrow \ce{CO2} + \ce{H2O} + \text{heat} \] This produces a clean blue flame with no soot.

Step 2: Define incomplete combustion
When oxygen supply is insufficient: \[ \text{Fuel} \rightarrow \ce{CO} + \ce{C} + \text{heat} \] Here, carbon particles (\(\ce{C}\)) are produced.

Step 3: Formation of black deposit
The unburnt carbon particles (soot) stick to the bottom of the vessel, causing blackening.

Step 4: Interpretation of observation
Blackened vessel indicates: \[ \text{Incomplete combustion of fuel} \]

Final Answer: (b) the fuel is not burning completely.

Step-by-Step Solution

Step 1: Valence electrons of atoms
Carbon: \[ 4 \text{ valence electrons} \]
Hydrogen: \[ 1 \text{ valence electron} \]
Chlorine: \[ 7 \text{ valence electrons} \]

Step 2: Requirement for stability
Carbon needs 4 electrons to complete octet.
Each hydrogen needs 1 electron (duplet).
Chlorine needs 1 electron to complete octet.

Step 3: Bond formation by sharing
Carbon shares electrons as follows:
With 3 hydrogen atoms → forms 3 C–H single bonds.
With 1 chlorine atom → forms 1 C–Cl single bond.

Step 4: Total bonds formed
\[ 3 \text{ (C–H bonds)} + 1 \text{ (C–Cl bond)} = 4 \text{ covalent bonds} \]

Step 5: Representation
\[ \begin{array}{cccc} & H & \\ & | & \\ H - C - Cl \\ & | & \\ & H & \end{array} \]

Step 6: Nature of covalent bond
Covalent bonds are formed due to mutual sharing of electrons.
These bonds are strong and directional.
No ions are formed in the molecule.

Final Conclusion: \(\ce{CH3Cl}\) demonstrates that covalent bonds are formed by sharing of electron pairs to achieve stable electronic configuration.

Step-by-Step Structures

(a) Ethanoic acid \(\ce{CH3COOH}\)

(b) \(\ce{H2S}\)

(c) Propanone \(\ce{CH3COCH3}\)

(d) \(\ce{F2}\)

Each structure shows shared electron pairs (bonds) and lone pairs, satisfying octet (or duplet for hydrogen).

Final Conclusion: Electron dot structures represent covalent bonding through shared electron pairs and help visualize molecular structure.

Exam Significance

• Very frequently asked in CBSE board exams (diagram-based question).
• Tests understanding of octet rule and bonding.
• Important for identifying functional groups visually.
• Foundation for chemical bonding and organic chemistry in competitive exams.

Step-by-Step Explanation

Step 1: Definition
A homologous series is a family of organic compounds having the same functional group and similar chemical properties, where successive members differ by a \(\ce{CH2}\) unit.

Step 2: General difference between members
\[ \text{Difference between consecutive members} = \ce{CH2} \]

Step 3: Characteristics

• Same functional group → similar chemical properties.
• Gradual change in physical properties (boiling point, melting point).
• General formula exists for the series.

Step 4: Example (Alcohol series)
\[ \ce{CH3OH,\ C2H5OH,\ C3H7OH} \] Each successive compound differs by: \[ \ce{+CH2} \]

Step 5: Reason for similarity
All members contain the same functional group \(\ce{-OH}\), hence they undergo similar chemical reactions.

Final Conclusion: A homologous series is a systematic progression of compounds differing by \(\ce{CH2}\) with similar chemical behavior due to the same functional group.

Detailed Comparison

Key distinguishing test: \[ \text{Only ethanoic acid gives CO₂ with } \ce{NaHCO3} \]

Final Conclusion: Ethanoic acid shows acidic properties (reacts with carbonates and changes litmus), whereas ethanol does not, making them easily distinguishable.

Step-by-Step Explanation

Step 1: Structure of soap molecule
Soap molecules contain:

• Hydrophilic head \(\ce{-COO^-}\) (water-attracting)
• Hydrophobic tail (long hydrocarbon chain)

Step 2: Behaviour in water
Water is polar, so:

• Hydrophilic heads dissolve in water.
• Hydrophobic tails avoid water.

Step 3: Formation of micelles
To minimize repulsion with water:

• Tails aggregate inward.
• Heads remain outward in contact with water.

Step 4: Cleaning action
Oil/dirt gets trapped inside micelle: \[ \text{Oil} \rightarrow \text{inside micelle} \rightarrow \text{washed away} \]

Step 5: Behaviour in ethanol
Ethanol is less polar and can dissolve both parts of soap molecules.
Therefore:

• No separation of head and tail occurs.
• No aggregation into micelles occurs.

Final Conclusion: Micelles form in water due to the hydrophobic effect and polarity difference, but they do not form in ethanol because this driving force is absent.

Step-by-Step Explanation

Step 1: Combustion reaction
Carbon compounds burn in oxygen to produce carbon dioxide and water: \[ \ce{C_xH_y + O2 -> CO2 + H2O + heat} \]

Step 2: High calorific value
These reactions release a large amount of heat energy, making them efficient fuels.

Step 3: Controlled combustion
Carbon fuels generally burn at a steady rate, allowing controlled use in cooking, engines, and industries.

Step 4: Availability
Fuels like coal, petroleum, and natural gas are widely available in nature.

Step 5: Ease of storage and transport
Carbon fuels exist in solid, liquid, and gaseous forms, making them convenient to handle.

Step 6: Clean burning (in sufficient oxygen)
Many carbon fuels burn with little smoke when oxygen supply is sufficient.

Final Conclusion: Carbon and its compounds are used as fuels because they release large amounts of energy, burn efficiently, and are easily available and convenient to use.

Step-by-Step Explanation

Step 1: Nature of hard water
Hard water contains dissolved salts of calcium and magnesium: \[ \ce{Ca^{2+},\ Mg^{2+}} \]

Step 2: Composition of soap
Soap is sodium or potassium salt of fatty acids: \[ \ce{RCOO^- Na^+} \]

Step 3: Reaction with hard water
Calcium and magnesium ions react with soap: \[ \ce{2RCOONa + Ca^{2+} -> (RCOO)2Ca + 2Na^+} \]

Step 4: Formation of scum
The product \(\ce{(RCOO)2Ca}\) (calcium salt) is insoluble in water and appears as a greyish-white deposit called scum.

Step 5: Effect on cleaning

• Soap is wasted in forming scum.
• Lather formation is reduced.
• Cleaning efficiency decreases.

Final Conclusion: Scum forms because soap reacts with calcium and magnesium ions in hard water to produce insoluble salts, which reduce the effectiveness of soap.

Step-by-Step Explanation

Step 1: Nature of soap
Soap is a salt of fatty acids and is basic in aqueous solution: \[ \ce{RCOO^- + H2O -> RCOOH + OH^-} \]

Step 2: Effect on red litmus
Bases turn red litmus blue.
Therefore: \[ \text{Red litmus} \rightarrow \text{Blue} \]

Step 3: Effect on blue litmus
Bases do not affect blue litmus.
Therefore: \[ \text{Blue litmus} \rightarrow \text{No change} \]

Final Conclusion: Soap solution turns red litmus blue but shows no change with blue litmus, confirming its basic nature.

Step-by-Step Explanation

Step 1: Definition
Hydrogenation is the addition of hydrogen \(\ce{H2}\) to an unsaturated compound in the presence of a catalyst such as nickel (Ni).

Step 2: Chemical change
Double or triple bonds are converted into single bonds: \[ \ce{C=C + H2 -> C-C} \]

Step 3: Nature of conversion
\[ \text{Unsaturated compound} \rightarrow \text{Saturated compound} \]

Step 4: Industrial application
Hydrogenation is used in the manufacture of vanaspati ghee.

Step 5: Explanation of application

• Vegetable oils are unsaturated and liquid.
• Hydrogenation converts them into saturated fats.
• The product becomes semi-solid and more stable.

Final Conclusion: Hydrogenation converts unsaturated compounds into saturated ones and is widely used industrially to convert vegetable oils into vanaspati ghee.

Step-by-Step Solution

Step 1: Definition
Addition reactions occur in compounds having multiple bonds: \[ \ce{C=C \; or \; C#C} \]

Step 2: Analyze each compound

• \(\ce{C2H6}\) (Ethane): Saturated → only single bonds → no addition
• \(\ce{C3H8}\) (Propane): Saturated → no addition
• \(\ce{CH4}\) (Methane): Saturated → no addition
• \(\ce{C3H6}\) (Propene): Contains \(\ce{C=C}\) → undergoes addition
• \(\ce{C2H2}\) (Ethyne): Contains \(\ce{C#C}\) → undergoes addition

Step 3: Final selection
\[ \text{Hydrocarbons undergoing addition} = \ce{C3H6,\ C2H2} \]

Final Answer: \(\ce{C3H6}\) and \(\ce{C2H2}\)

Step-by-Step Solution

Step 1: Test used
Bromine water test is used to differentiate between saturated and unsaturated hydrocarbons.

Step 2: Reaction with unsaturated hydrocarbons
Unsaturated hydrocarbons undergo addition reaction with bromine: \[ \ce{C=C + Br2 -> C-CBr2} \] Observation: Reddish-brown colour of bromine disappears.

Step 3: Reaction with saturated hydrocarbons
Saturated hydrocarbons do not react with bromine under normal conditions.
Observation: Colour remains unchanged.

Step 4: Conclusion
\[ \text{Decolourisation} \Rightarrow \text{Unsaturated hydrocarbon} \] \[ \text{No change} \Rightarrow \text{Saturated hydrocarbon} \]

Final Conclusion: Bromine water test distinguishes hydrocarbons based on their ability to undergo addition reactions.

Step-by-Step Explanation

Step 1: Structure of soap molecule
Soap contains:

• Hydrophobic tail (non-polar hydrocarbon chain)
• Hydrophilic head \(\ce{-COO^-}\) (polar, water-attracting)

Step 2: Interaction with water and oil

• Tails dissolve in oil/grease.
• Heads remain in water.

Step 3: Micelle formation
Soap molecules arrange themselves into micelles:

• Tails point inward trapping oil.
• Heads remain outward in contact with water.

Step 4: Emulsification of grease
Large grease droplets break into smaller droplets surrounded by soap molecules.

Step 5: Removal during washing
\[ \text{Grease + soap} \rightarrow \text{micelles} \rightarrow \text{washed away with water} \]

Final Conclusion: Soap cleans by forming micelles that trap oily dirt and allow it to be washed away with water.