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JEE Main Previous Year Questions (2025): General Organic Chemistry (GOC)
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Page 1 I : 1 Q n Carius method for estimation of halogens, 180 mg of an organic compound produced 143.5 mg of AgCl . The percentage composition of chlorine in the compound is _________ %. (Given : molar mass in of ) JEE Main 2025 (Online) 22nd January Morning Shift Ans: 20 Solution: To find the percentage composition of chlorine in an organic compound using the Carius method, we follow these steps: Calculate Millimoles of AgCl: Since 143.5 mg of AgCl is produced, and the molar mass of AgCl is 143.5 g/mol, the millimoles of AgCl is: Determine Millimoles of Cl: In AgCl, there is a 1:1 ratio of Ag to Cl, so the millimoles of Cl are equal to that of AgCl: Calculate Mass of Cl: Using the molar mass of Cl (35.5 g/mol), the mass of Cl is: Compute Percentage by Mass of Cl: The total mass of the organic compound is 180 mg, so the percentage of chlorine is: g m o l - 1 A g : 1 0 8 , C l : 3 5 . 5 Millimoles of AgCl = 1 4 3 . 5 mg 1 4 3 . 5 mg/mmol = 1 mmol Millimoles of Cl = 1 mmol Mass of Cl = 3 5 . 5 × 1 0 - 3 g = 3 5 . 5 mg Page 2 I : 1 Q n Carius method for estimation of halogens, 180 mg of an organic compound produced 143.5 mg of AgCl . The percentage composition of chlorine in the compound is _________ %. (Given : molar mass in of ) JEE Main 2025 (Online) 22nd January Morning Shift Ans: 20 Solution: To find the percentage composition of chlorine in an organic compound using the Carius method, we follow these steps: Calculate Millimoles of AgCl: Since 143.5 mg of AgCl is produced, and the molar mass of AgCl is 143.5 g/mol, the millimoles of AgCl is: Determine Millimoles of Cl: In AgCl, there is a 1:1 ratio of Ag to Cl, so the millimoles of Cl are equal to that of AgCl: Calculate Mass of Cl: Using the molar mass of Cl (35.5 g/mol), the mass of Cl is: Compute Percentage by Mass of Cl: The total mass of the organic compound is 180 mg, so the percentage of chlorine is: g m o l - 1 A g : 1 0 8 , C l : 3 5 . 5 Millimoles of AgCl = 1 4 3 . 5 mg 1 4 3 . 5 mg/mmol = 1 mmol Millimoles of Cl = 1 mmol Mass of Cl = 3 5 . 5 × 1 0 - 3 g = 3 5 . 5 mg Round to the Nearest Integer: Rounding 19.72% to the nearest integer gives 20%. Thus, the percentage composition of chlorine in the compound is approximately 20%. T : 2 Q he possible number of stereoisomers for 5-phenylpent-4-en-2-ol is ________. JEE Main 2025 (Online) 24th January Evening Shift Ans: 4 Solution: stereogenic unit stereoisomers are possible. : 3 Q has _______ degree of unsaturation. JEE Main 2025 (Online) 24th January Evening Shift % Cl by mass = ( 3 5 . 5 mg 1 8 0 mg ) × 1 0 0 = 1 9 . 7 2 % n ( ) = 2 , 2 2 = 4 Page 3 I : 1 Q n Carius method for estimation of halogens, 180 mg of an organic compound produced 143.5 mg of AgCl . The percentage composition of chlorine in the compound is _________ %. (Given : molar mass in of ) JEE Main 2025 (Online) 22nd January Morning Shift Ans: 20 Solution: To find the percentage composition of chlorine in an organic compound using the Carius method, we follow these steps: Calculate Millimoles of AgCl: Since 143.5 mg of AgCl is produced, and the molar mass of AgCl is 143.5 g/mol, the millimoles of AgCl is: Determine Millimoles of Cl: In AgCl, there is a 1:1 ratio of Ag to Cl, so the millimoles of Cl are equal to that of AgCl: Calculate Mass of Cl: Using the molar mass of Cl (35.5 g/mol), the mass of Cl is: Compute Percentage by Mass of Cl: The total mass of the organic compound is 180 mg, so the percentage of chlorine is: g m o l - 1 A g : 1 0 8 , C l : 3 5 . 5 Millimoles of AgCl = 1 4 3 . 5 mg 1 4 3 . 5 mg/mmol = 1 mmol Millimoles of Cl = 1 mmol Mass of Cl = 3 5 . 5 × 1 0 - 3 g = 3 5 . 5 mg Round to the Nearest Integer: Rounding 19.72% to the nearest integer gives 20%. Thus, the percentage composition of chlorine in the compound is approximately 20%. T : 2 Q he possible number of stereoisomers for 5-phenylpent-4-en-2-ol is ________. JEE Main 2025 (Online) 24th January Evening Shift Ans: 4 Solution: stereogenic unit stereoisomers are possible. : 3 Q has _______ degree of unsaturation. JEE Main 2025 (Online) 24th January Evening Shift % Cl by mass = ( 3 5 . 5 mg 1 8 0 mg ) × 1 0 0 = 1 9 . 7 2 % n ( ) = 2 , 2 2 = 4 Ans: 3 Solution: Mass of carbon Number of C-atoms Mass of hydrogen Number of H -atoms So molecular formula D.U. T : 4 Q he total number of structural isomers possible for the substituted benzene derivatives with the molecular formula is________. JEE Main 2025 (Online) 3rd April Evening Shift Ans: 8 Solution: = 8 0 × 9 0 1 0 0 = 7 2 g m = 7 2 1 2 = 6 = 8 0 × 1 0 8 0 0 = 8 g m = 8 1 = 8 C 6 H 8 = 6 + 1 - 8 / 2 = 7 - 4 = 3 C 9 H 1 2 M F = C 9 H 1 2 Page 4 I : 1 Q n Carius method for estimation of halogens, 180 mg of an organic compound produced 143.5 mg of AgCl . The percentage composition of chlorine in the compound is _________ %. (Given : molar mass in of ) JEE Main 2025 (Online) 22nd January Morning Shift Ans: 20 Solution: To find the percentage composition of chlorine in an organic compound using the Carius method, we follow these steps: Calculate Millimoles of AgCl: Since 143.5 mg of AgCl is produced, and the molar mass of AgCl is 143.5 g/mol, the millimoles of AgCl is: Determine Millimoles of Cl: In AgCl, there is a 1:1 ratio of Ag to Cl, so the millimoles of Cl are equal to that of AgCl: Calculate Mass of Cl: Using the molar mass of Cl (35.5 g/mol), the mass of Cl is: Compute Percentage by Mass of Cl: The total mass of the organic compound is 180 mg, so the percentage of chlorine is: g m o l - 1 A g : 1 0 8 , C l : 3 5 . 5 Millimoles of AgCl = 1 4 3 . 5 mg 1 4 3 . 5 mg/mmol = 1 mmol Millimoles of Cl = 1 mmol Mass of Cl = 3 5 . 5 × 1 0 - 3 g = 3 5 . 5 mg Round to the Nearest Integer: Rounding 19.72% to the nearest integer gives 20%. Thus, the percentage composition of chlorine in the compound is approximately 20%. T : 2 Q he possible number of stereoisomers for 5-phenylpent-4-en-2-ol is ________. JEE Main 2025 (Online) 24th January Evening Shift Ans: 4 Solution: stereogenic unit stereoisomers are possible. : 3 Q has _______ degree of unsaturation. JEE Main 2025 (Online) 24th January Evening Shift % Cl by mass = ( 3 5 . 5 mg 1 8 0 mg ) × 1 0 0 = 1 9 . 7 2 % n ( ) = 2 , 2 2 = 4 Ans: 3 Solution: Mass of carbon Number of C-atoms Mass of hydrogen Number of H -atoms So molecular formula D.U. T : 4 Q he total number of structural isomers possible for the substituted benzene derivatives with the molecular formula is________. JEE Main 2025 (Online) 3rd April Evening Shift Ans: 8 Solution: = 8 0 × 9 0 1 0 0 = 7 2 g m = 7 2 1 2 = 6 = 8 0 × 1 0 8 0 0 = 8 g m = 8 1 = 8 C 6 H 8 = 6 + 1 - 8 / 2 = 7 - 4 = 3 C 9 H 1 2 M F = C 9 H 1 2 T : 5 Q he IUPAC name of the following compound is : JEE Main 2025 (Online) 22nd January Morning Shift Options: A. 5-Methoxycarbonly-2-methylhexanoic acid B. Methyl-5-carboxy-2-methylhexanoate. C. 2-Carboxy-5-methoxycarbonylhexane. D. Methyl-6-carboxy-2,5-dimethylhexanoate. Ans: A Solution: Page 5 I : 1 Q n Carius method for estimation of halogens, 180 mg of an organic compound produced 143.5 mg of AgCl . The percentage composition of chlorine in the compound is _________ %. (Given : molar mass in of ) JEE Main 2025 (Online) 22nd January Morning Shift Ans: 20 Solution: To find the percentage composition of chlorine in an organic compound using the Carius method, we follow these steps: Calculate Millimoles of AgCl: Since 143.5 mg of AgCl is produced, and the molar mass of AgCl is 143.5 g/mol, the millimoles of AgCl is: Determine Millimoles of Cl: In AgCl, there is a 1:1 ratio of Ag to Cl, so the millimoles of Cl are equal to that of AgCl: Calculate Mass of Cl: Using the molar mass of Cl (35.5 g/mol), the mass of Cl is: Compute Percentage by Mass of Cl: The total mass of the organic compound is 180 mg, so the percentage of chlorine is: g m o l - 1 A g : 1 0 8 , C l : 3 5 . 5 Millimoles of AgCl = 1 4 3 . 5 mg 1 4 3 . 5 mg/mmol = 1 mmol Millimoles of Cl = 1 mmol Mass of Cl = 3 5 . 5 × 1 0 - 3 g = 3 5 . 5 mg Round to the Nearest Integer: Rounding 19.72% to the nearest integer gives 20%. Thus, the percentage composition of chlorine in the compound is approximately 20%. T : 2 Q he possible number of stereoisomers for 5-phenylpent-4-en-2-ol is ________. JEE Main 2025 (Online) 24th January Evening Shift Ans: 4 Solution: stereogenic unit stereoisomers are possible. : 3 Q has _______ degree of unsaturation. JEE Main 2025 (Online) 24th January Evening Shift % Cl by mass = ( 3 5 . 5 mg 1 8 0 mg ) × 1 0 0 = 1 9 . 7 2 % n ( ) = 2 , 2 2 = 4 Ans: 3 Solution: Mass of carbon Number of C-atoms Mass of hydrogen Number of H -atoms So molecular formula D.U. T : 4 Q he total number of structural isomers possible for the substituted benzene derivatives with the molecular formula is________. JEE Main 2025 (Online) 3rd April Evening Shift Ans: 8 Solution: = 8 0 × 9 0 1 0 0 = 7 2 g m = 7 2 1 2 = 6 = 8 0 × 1 0 8 0 0 = 8 g m = 8 1 = 8 C 6 H 8 = 6 + 1 - 8 / 2 = 7 - 4 = 3 C 9 H 1 2 M F = C 9 H 1 2 T : 5 Q he IUPAC name of the following compound is : JEE Main 2025 (Online) 22nd January Morning Shift Options: A. 5-Methoxycarbonly-2-methylhexanoic acid B. Methyl-5-carboxy-2-methylhexanoate. C. 2-Carboxy-5-methoxycarbonylhexane. D. Methyl-6-carboxy-2,5-dimethylhexanoate. Ans: A Solution: T : 6 Q he incorrect statements regarding geometrical isomerism are : (A) Propene shows geometrical isomerism. (B) Trans isomer has identical atoms/groups on the opposite sides of the double bond. (C) Cis-but-2-ene has higher dipole moment than trans-but-2-ene. (D) 2-methylbut-2-ene shows two geometrical isomers. (E) Trans-isomer has lower melting point than cis isomer. Choose the correct answer from the options given below : JEE Main 2025 (Online) 22nd January Morning Shift Options: A. (A) and (E) Only B. (A), (D) and (E) Only C. (C), (D) and (E) Only D. (B) and (C) Only Ans: B Solution:Read More
FAQs on JEE Main Previous Year Questions (2025): General Organic Chemistry (GOC)
| 1. What is General Organic Chemistry (GOC) and why is it important for JEE preparation? |  |
Ans. General Organic Chemistry (GOC) refers to the foundational concepts and principles that govern organic compounds and reactions. It includes topics such as hybridization, resonance, acidity and basicity, functional groups, and stereochemistry. Understanding GOC is crucial for JEE preparation as it forms the basis for more complex organic chemistry topics and helps students solve problems related to reaction mechanisms and organic synthesis effectively.
| 2. How does hybridization affect the structure and reactivity of organic compounds? | |
Ans. Hybridization involves the mixing of atomic orbitals to form new hybrid orbitals that influence the geometry and bonding of organic molecules. For example, sp³ hybridization leads to a tetrahedral structure (e.g., in methane, CH₄), while sp² hybridization results in a trigonal planar structure (e.g., in ethene, C₂H₄). The type of hybridization affects the compound's reactivity; for instance, sp² hybridized compounds are generally more reactive in electrophilic addition reactions compared to sp³ compounds.
| 3. What role does resonance play in determining the stability of organic molecules? | |
Ans. Resonance is the phenomenon where a molecule can be represented by two or more valid Lewis structures, which contribute to the overall hybrid structure. The greater the number of resonance structures, the more stable the molecule tends to be due to delocalization of electrons. For example, benzene (C₆H₆) has multiple resonance forms, which explains its remarkable stability compared to other unsaturated hydrocarbons.
| 4. How do acidity and basicity concepts apply to organic compounds? | |
Ans. Acidity and basicity in organic chemistry are influenced by the stability of the conjugate base and the nature of functional groups. For instance, carboxylic acids (RCOOH) are generally more acidic than alcohols (R-OH) because the carboxylate ion (RCOO⁻) formed after deprotonation is stabilized by resonance. Understanding these concepts is essential for predicting reaction outcomes and the behavior of different organic molecules in various chemical environments.
| 5. What are functional groups, and why are they significant in organic chemistry? | |
Ans. Functional groups are specific groups of atoms within molecules that impart characteristic chemical properties and reactivity patterns. Examples include hydroxyl (-OH), carbonyl (C=O), and carboxyl (-COOH) groups. The presence of functional groups determines the classification, reactivity, and behavior of organic compounds, making them crucial for understanding organic reactions and synthesis in the context of JEE preparation.
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