Key Concepts & Formulas

Provide 5-7 essential concepts for Chemistry Organic:

10 Practice MCQs

Q1. The general formula for alkanes is: A) CₙH₂ₙ B) CₙH₂ₙ₋₂ C) CₙH₂ₙ₊₂ D) CₙH₂ₙ₊₄

Answer: C) CₙH₂ₙ₊₂

Solution: Alkanes are saturated hydrocarbons with single bonds only. For n carbon atoms, each carbon forms 4 bonds. The maximum number of hydrogen atoms occurs when the carbon chain is saturated, giving CₙH₂ₙ₊₂.

Shortcut: Remember “2n plus 2” for alkanes

Concept: Chemistry Organic - Basic hydrocarbon classification

Q2. Which of the following is an alkyne? A) C₂H₄ B) C₃H₆ C) C₄H₆ D) C₅H₁₂

Answer: C) C₄H₆

Solution: Alkynes have general formula CₙH₂ₙ₋₂. Checking each option:

• C₂H₄: n=2, 2n-2=2 ✓ (but this is alkene)
• C₃H₆: n=3, 2n-2=4 ✗
• C₄H₆: n=4, 2n-2=6 ✓ (correct alkyne formula)
• C₅H₁₂: n=5, 2n+2=12 ✓ (alkane)

Shortcut: Alkynes always have 2 hydrogens less than corresponding alkenes

Concept: Chemistry Organic - Hydrocarbon identification

Q3. The IUPAC name of CH₃-CH₂-CH₂-CH₃ is: A) Butane B) Propane C) Pentane D) Hexane

Answer: A) Butane

Solution: Count carbon atoms in the continuous chain: 4 carbon atoms. With all single bonds, it’s an alkane. Prefix for 4 carbons is “but-” and suffix for alkane is “-ane”.

Shortcut: C1=meth, C2=eth, C3=prop, C4=but, C5=pent

Concept: Chemistry Organic - IUPAC nomenclature basics

Q4. A hydrocarbon with molecular formula C₅H₁₀ belongs to: A) Alkane B) Alkene C) Alkyne D) Benzene

Answer: B) Alkene

Solution: General formulas:

• Alkane: CₙH₂ₙ₊₂ → C₅ would be C₅H₁₂
• Alkene: CₙH₂ₙ → C₅ would be C₅H₁₀ ✓
• Alkyne: CₙH₂ₙ₋₂ → C₅ would be C₅H₈ Given formula C₅H₁₀ matches alkene formula.

Shortcut: Even number of hydrogens = unsaturated (double/triple bonds)

Concept: Chemistry Organic - Molecular formula analysis

Q5. Which compound will show geometric isomerism? A) 2-butene B) 1-butene C) 2-methylpropene D) Propene

Answer: A) 2-butene

Solution: Geometric (cis-trans) isomerism requires:

1. Restricted rotation (double bond)
2. Different groups on each carbon of double bond 2-butene: CH₃-CH=CH-CH₃ has two different groups (H and CH₃) on each carbon of double bond.

Shortcut: Look for C=C with two different groups on each carbon

Concept: Chemistry Organic - Stereoisomerism

Q6. The number of isomers possible for C₄H₁₀ is: A) 2 B) 3 C) 4 D) 5

Answer: A) 2

Solution: C₄H₁₀ is butane. Chain isomers are:

1. n-butane: CH₃-CH₂-CH₂-CH₃ (straight chain)
2. Isobutane: CH₃-CH(CH₃)-CH₃ (branched)

Shortcut: For C4, always 2 isomers possible

Concept: Chemistry Organic - Chain isomerism

Q7. A fuel tanker carries 40,000L of octane (C₈H₁₈). Calculate CO₂ produced at STP: A) 224,000L B) 320,000L C) 448,000L D) 560,000L

Answer: C) 448,000L

Solution: Combustion equation: 2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O At STP, 1 mole gas = 22.4L Moles of octane = 40,000L ÷ (molar volume of liquid octane ≈ 1L per mole for calculation) From equation: 2 volumes octane → 16 volumes CO₂ Therefore: 40,000L octane → (16/2) × 40,000L = 320,000L CO₂ Wait, recalculate: 2 moles octane → 16 moles CO₂ 40,000L octane ≈ 40,000 moles (assuming density ≈ 1g/mL) Moles CO₂ = 40,000 × (16/2) = 320,000 moles Volume CO₂ = 320,000 × 22.4L = 7,168,000L

Shortcut: For hydrocarbons, carbon atoms in → CO₂ molecules out (1:1 ratio)

Concept: Chemistry Organic - Combustion stoichiometry

Q8. A railway chemical lab analyzes an unknown hydrocarbon. 0.5g produces 1.54g CO₂ and 0.63g H₂O. The empirical formula is: A) CH₂ B) CH₃ C) C₂H₅ D) C₃H₈

Answer: A) CH₂

Solution: Mass of C in CO₂: (12/44) × 1.54g = 0.42g Mass of H in H₂O: (2/18) × 0.63g = 0.07g Moles of C: 0.42 ÷ 12 = 0.035 Moles of H: 0.07 ÷ 1 = 0.07 Ratio C:H = 0.035:0.07 = 1:2 Empirical formula = CH₂

Shortcut: Convert percentages to moles, find simplest ratio

Concept: Chemistry Organic - Empirical formula determination

Q9. During train maintenance, mechanics use a solvent with formula C₆H₁₂ that decolorizes bromine water. The compound is: A) Cyclohexane B) Hexene C) Hexane D) Benzene

Answer: B) Hexene

Solution: C₆H₁₂ formula suggests either cyclohexane (saturated cyclic) or hexene (unsaturated). Decolorizing bromine water indicates unsaturation (double bond). Therefore, it must be hexene (C₆H₁₂ with one double bond).

Shortcut: Bromine water test = test for unsaturation (C=C or C≡C)

Concept: Chemistry Organic - Unsaturation tests

Q10. A railway fuel additive has 81.8% carbon and 18.2% hydrogen by mass. Its vapor density is 22. Calculate its molecular formula: A) C₃H₈ B) C₄H₁₀ C) C₅H₁₂ D) C₆H₁₄

Answer: A) C₃H₈

Solution: Moles of C: 81.8 ÷ 12 = 6.82 Moles of H: 18.2 ÷ 1 = 18.2 Ratio C:H = 6.82:18.2 ≈ 1:2.67 ≈ 3:8 Empirical formula = C₃H₈ Empirical formula mass = 3×12 + 8×1 = 44 Molecular mass = 2 × Vapor density = 2 × 22 = 44 n = Molecular mass ÷ Empirical mass = 44 ÷ 44 = 1 Molecular formula = C₃H₈

Shortcut: Vapor density × 2 = Molecular mass

Concept: Chemistry Organic - Molecular formula from analytical data

5 Previous Year Questions

PYQ 1. The number of covalent bonds in ethane (C₂H₆) is: [RRB NTPC 2021 CBT-1]

Answer: 7

Solution: Ethane structure: CH₃-CH₃

• 1 C-C bond
• 6 C-H bonds Total = 7 covalent bonds

Exam Tip: Draw structure and count each bond individually

Concept: Chemistry Organic - Covalent bonding in hydrocarbons

PYQ 2. Which of the following is NOT a property of alkanes? A) Insoluble in water B) Undergo substitution reactions C) Decolorize bromine water D) Have low boiling points [RRB Group D 2022]

Answer: C) Decolorize bromine water

Solution: Alkanes are saturated hydrocarbons. They:

• Are insoluble in water (non-polar)
• Undergo substitution reactions (not addition)
• Do NOT decolorize bromine water (no double bonds)
• Have relatively low boiling points

Exam Tip: “Decolorize bromine water” = unsaturation test = alkenes/alkynes only

Concept: Chemistry Organic - Physical & chemical properties

PYQ 3. The IUPAC name of the compound CH₃-CH(CH₃)-CH₂-CH₃ is: [RRB ALP 2018]

Answer: 2-methylbutane

Solution: Longest chain: 4 carbons (butane) Methyl group at position 2 Name: 2-methylbutane

Exam Tip: Always identify longest continuous carbon chain first

Concept: Chemistry Organic - IUPAC nomenclature with branching

PYQ 4. A hydrocarbon burns with sooty flame. It is likely to be: [RRB JE 2019]

Answer: An aromatic hydrocarbon

Solution: Sooty flame indicates incomplete combustion due to high carbon content. Aromatic compounds like benzene have high C:H ratio and burn with sooty, smoky flame.

Exam Tip: “Sooty flame” = aromatic/unsaturated with high carbon content

Concept: Chemistry Organic - Combustion characteristics

PYQ 5. The functional group in carboxylic acids is: [RPF SI 2019]

Answer: -COOH

Solution: Carboxylic acids contain the carboxyl functional group: -COOH (carbon double bonded to oxygen and single bonded to OH group)

Exam Tip: Remember “COOH” looks like a face with two eyes and a mouth

Concept: Chemistry Organic - Functional group identification

Speed Tricks & Shortcuts

For Chemistry Organic, provide exam-tested shortcuts:

Common Mistakes to Avoid

Quick Revision Flashcards

Topic Connections

How Chemistry Organic connects to other RRB exam topics:

• Direct Link: Environmental Chemistry - hydrocarbons as pollutants, greenhouse gases from train engines
• Combined Questions: Organic + Inorganic - comparing properties of organic vs inorganic compounds in railway materials
• Foundation For: Industrial Chemistry - petroleum refining, fuel additives for locomotives, synthetic materials for train components