Organic Chemistry

Step 1: Understanding the stability of free radicals.
The stability of free radicals increases with the number of alkyl groups attached to the carbon center. The more alkyl groups, the greater the electron donation, stabilizing the radical.

Step 2: Analyzing the options.
(A) R-CH : This is a primary radical and is less stable due to fewer electron-donating groups.
(B) R -C•: This is a tertiary radical, which is the most stable due to the three electron-donating alkyl groups.
(C) CH •: This is a methyl radical, which is a primary radical and less stable.
(D) R -C•H: This is a secondary radical and more stable than a primary radical, but less stable than a tertiary radical.

Step 3: Conclusion.
The correct answer is (B) R -C•, as tertiary radicals are the most stable.

Step 1: Understanding the hydrogenation process.
In the hydrogenation process, ethene (C H ) reacts with hydrogen (H ) to form ethane (C H ). The balanced equation for the reaction is: This indicates that 1 mole of ethene reacts with 1 mole of hydrogen to form 1 mole of ethane.
Step 2: Calculation.
The molar mass of ethane (C H ) is 30 g/mol. To prepare 6 g of ethane, we need: Since the molar ratio of ethene to ethane is 1:1, we need 0.2 moles of ethene.
Step 3: Conclusion.
The correct answer is (A) 0.2 mole.

Step 1: Understanding the reaction with HCl and propane.
When HCl reacts with propane in the presence of sodium peroxide (Na O ), a free radical substitution reaction occurs, leading to the formation of alkyl chlorides. The major product in this case is 1-chloropropane due to the primary carbon atom being more susceptible to radical attack.
Step 2: Analyzing the options.
(A) 1-Chloropropane: This is correct. The reaction leads to the formation of 1-chloropropane.
(B) 1, 2-dichloropropane: This is incorrect. The major product is not a dihalide.
(C) 2-Chloropropane: This is incorrect. The reaction preferentially forms 1-chloropropane, not 2-chloropropane.
(D) 2, 2-dichloropropane: This is incorrect. The reaction does not lead to a double chlorination at the same carbon atom.
Step 3: Conclusion.
The correct answer is (A) 1-Chloropropane.

Step 1: Understanding (+R) effect.
The (+R) effect, or the +M effect, involves the donation of electron density through resonance, which is typically seen in groups that have lone pairs on heteroatoms (such as -OH, -NH , -OCH ).
Step 2: Analyzing the options.
(A) -NH : The amino group (-NH ) shows the (+R) effect due to the lone pairs on nitrogen.
(B) -NHCO R: The -NHCO group shows the (+R) effect due to nitrogen's lone pairs.
(C) -CN: The -CN group is electron-withdrawing through resonance (not donating), so it does not show the (+R) effect.
(D) -NR: Similar to -NH , -NR shows the (+R) effect due to lone pairs on nitrogen.
Step 3: Conclusion.
The correct answer is (C) -CN, as it does not show the (+R) effect.

Step 1: Understanding the reactions.
- The first reaction involves the substitution of chlorine with a hydroxyl group in an alcohol, forming a primary alcohol.
- The second reaction with HBr (in the presence of peroxide) leads to a radical halogenation that gives an alkyl halide.
- The final reaction with KCN in alcohol results in the formation of a nitrile group by nucleophilic substitution of the halide with the cyanide ion.
Step 2: Analyzing the options.
(A) CH -CH -CH -CN: This is the correct answer. The final product is a nitrile (CN) group attached to a three-carbon chain.
(B) CH -CH -CH -Br: This is incorrect, as the final product should contain a nitrile group, not a bromine atom.
(C) CH -CH -CH -CN: This is incorrect, as the structure shown is not consistent with the reactions described.
(D) CH -CH =CH : This is incorrect. The final product is not an alkene, but a nitrile.
Step 3: Conclusion.
The correct answer is (A) CH -CH -CH -CN, which is the nitrile product formed.

Step 1: Understanding the reaction type.
In the Swarts reaction, alkyl halides (usually bromides) react with metallic fluorides to form alkyl fluorides. This reaction is a method of preparing fluorinated organic compounds.
Step 2: Analyzing the options.
(A) Sandmeyer reaction: The Sandmeyer reaction involves the substitution of halides in aromatic compounds using copper salts, not the preparation of alkyl fluorides.
(B) Wurtz reaction: The Wurtz reaction involves the coupling of alkyl halides using sodium metal, not fluorination.
(C) Swarts reaction: This is the correct reaction. The Swarts reaction involves the preparation of alkyl fluorides from alkyl bromides and metallic fluorides.
(D) Finkelstein reaction: The Finkelstein reaction involves the substitution of halides using alkali halides, not metallic fluorides.
Step 3: Conclusion.
The correct answer is (C) Swarts reaction.

Step 1: Understanding dicarboxylic acids.
Dicarboxylic acids contain two carboxyl groups (-COOH) per molecule. To identify which one is not a dicarboxylic acid, we examine the number of carboxyl groups in each option.
Step 2: Analyzing the options.
(A) Succinic acid: Succinic acid is a dicarboxylic acid with two carboxyl groups.
(B) Acrylic acid: Acrylic acid has only one carboxyl group, making it a monocarboxylic acid, not a dicarboxylic acid.
(C) Malonic acid: Malonic acid is a dicarboxylic acid with two carboxyl groups.
(D) Phthalic acid: Phthalic acid is a dicarboxylic acid with two carboxyl groups.
Step 3: Conclusion.
The correct answer is (B) Acrylic acid, as it is not a dicarboxylic acid.

Step 1: Understanding sulphapyridine.
Sulphapyridine is a type of sulfonamide antibiotic, used to treat bacterial infections. It works by inhibiting the growth of bacteria.

Step 2: Analyzing the options.
(A) Antibiotic: Correct — Sulphapyridine is an antibiotic.
(B) Tranquilizer: Incorrect — Sulphapyridine is not a tranquilizer.
(C) Antihistamine: Incorrect — It is not an antihistamine.
(D) Analgesic: Incorrect — Sulphapyridine is not an analgesic.

Step 3: Conclusion.
The correct answer is (A) Antibiotic, as sulphapyridine is used as an antibiotic.

Step 1: Understanding electrophoresis.
Electrophoresis is a technique used to separate particles based on their charge by applying an electric field. It is used to determine the charge on colloidal particles.

Step 2: Analyzing the options.
(A) to count number of particles in colloidal dispersions: Incorrect — Electrophoresis is not used to count particles.
(B) for stability of colloids: Incorrect — While electrophoresis can be used to study colloidal properties, it is not directly used to determine stability.
(C) to determine charge on colloidal particles: Correct — Electrophoresis is used to determine the charge on colloidal particles by observing their movement in an electric field.
(D) to determine size of colloidal particles: Incorrect — Electrophoresis does not directly measure the size of particles.

Step 3: Conclusion.
The correct answer is (C) to determine charge on colloidal particles, as electrophoresis is used for this purpose.

Step 1: Understanding phantom atoms.
A phantom atom is an atom that forms part of the structure but is not directly involved in bonding, such as the nitrogen in methyl cyanide ( ).
Step 2: Analyzing the options.
(A) Acetaldehyde: Acetaldehyde has a carbonyl group, with no phantom atoms.
(B) Methyl cyanide: This contains a phantom atom (the nitrogen).
(C) n-propyl alcohol: This does not contain any phantom atoms, as the structure is fully bonded.
(D) Propionic acid: Propionic acid contains a carboxyl group but no phantom atoms.

Step 3: Conclusion.
The correct answer is (C) n-propyl alcohol, which does not contain any phantom atoms.

Step 1: Define resonance.
Resonance occurs in molecules having conjugated -electron systems or lone pair participation.
Step 2: Analyze the options.
Aniline, nitrobenzene, and phenol all contain benzene rings where delocalisation of -electrons occurs.
Step 3: Identify exception.
Cyclohexane is a saturated compound with only -bonds and no delocalised electrons.
Step 4: Conclusion.
Therefore, cyclohexane does not exhibit resonance.

Step 1: Applying the Nernst equation.
The standard free energy change can be calculated using the relation: where is the number of moles of electrons transferred, is Faraday's constant (96,485 C/mol), and is the cell potential. In this reaction, 2 moles of electrons are transferred. Therefore,
Step 2: Conclusion.
The standard free energy change is (A) -231.6 kJ.

Step 1: Understanding dihydric phenol.
Dihydric phenols are compounds that contain two hydroxyl groups (-OH) attached to the benzene ring. Among the options, resorcinol (1,3-dihydroxybenzene) is a dihydric phenol.

Step 2: Analyzing the options.
(A) Resorcinol: Correct — Resorcinol is a dihydric phenol, with two hydroxyl groups at positions 1 and 3 of the benzene ring.
(B) m-Cresol: m-Cresol is a monohydric phenol, containing one hydroxyl group.
(C) Phloroglucinol: Phloroglucinol contains three hydroxyl groups, making it a trihydric phenol, not dihydric.
(D) Pyrogallol: Pyrogallol also has three hydroxyl groups and is a trihydric phenol.

Step 3: Conclusion.
The correct answer is (A) Resorcinol.

Step 1: Understanding zone refining.
Zone refining is a technique used to purify metals, especially semiconductors like germanium. The process involves heating a small region of the metal and moving this molten zone along the metal. Impurities move to the cooler part of the metal and can be removed.

Step 2: Analyzing the options.
(A) Germanium: Correct — Germanium is commonly purified using the zone refining method.
(B) Tin: Tin is not typically purified using zone refining; it is usually refined by other methods.
(C) Copper: Copper is generally purified using electrolysis, not zone refining.
(D) Bismuth: Bismuth is also not refined using zone refining.

Step 3: Conclusion.
The correct answer is (A) Germanium.

Step 1: Understanding the aliphatic series.
Vitamins belonging to the aliphatic series contain long carbon chains or structures with straight or branched chains, with no aromatic rings. Among the options, Vitamin C (ascorbic acid) belongs to the aliphatic series as it has an aliphatic structure with a hydroxyl group.

Step 2: Analyzing the options.
(A) Vitamin C: Correct — Vitamin C is a water-soluble vitamin with an aliphatic structure.
(B) Vitamin A: Vitamin A has a structure with an aromatic ring, so it does not belong to the aliphatic series.
(C) Vitamin K: Vitamin K has a quinone structure, which is aromatic, not aliphatic.
(D) Vitamin B complex: The B vitamins are a group of compounds that include both aliphatic and non-aliphatic structures, but Vitamin C is the only one fully aliphatic.

Step 3: Conclusion.
The correct answer is (A) Vitamin C.

Step 1: Understanding salvarsan.
Salvarsan, also known as arsphenamine, is an organoarsenic compound that contains arsenic (As), nitrogen (N), and oxygen (O) but not phosphorus (P). It was historically used as a treatment for syphilis.

Step 2: Analyzing the options.
(A) P: Correct — Phosphorus is not present in salvarsan.
(B) O: Oxygen is present in salvarsan.
(C) N: Nitrogen is present in salvarsan.
(D) As: Arsenic is a key component of salvarsan.

Step 3: Conclusion.
The element that is not present in salvarsan is (A) P (Phosphorus).

Step 1: Understanding ester linkages in polymers.
Ester linkages are formed by the reaction between a carboxylic acid group and an alcohol group. Some polymers, like PHBV and Dacron, contain ester linkages. However, Nylon-2-Nylon-6 is a polyamide, not a polyester, and therefore does not contain ester linkages.

Step 2: Analyzing the options.
(A) Dextrin: Dextrin is a polysaccharide, but it does not contain ester linkages.
(B) PHBV: PHBV (Polyhydroxybutyrate-co-hydroxyvalerate) is a polyester containing ester linkages.
(C) Dacron: Dacron (Polyethylene terephthalate) is a polyester and contains ester linkages.
(D) Nylon-2-Nylon-6: Nylon-2-Nylon-6 is a polyamide, which contains amide linkages, not ester linkages.

Step 3: Conclusion.
The polymer that does not contain ester linkages is (D) Nylon-2-Nylon-6, as it is a polyamide.

Step 1: Understanding acidity in alcohols.
The acidity of alcohols is influenced by the electron-donating or electron-withdrawing effects of substituents attached to the carbon atom bonded to the hydroxyl group. The more electron-withdrawing the group, the more acidic the alcohol.

Step 2: Analyzing the options.
(A) (CH ) C-OH: This alcohol is a tertiary alcohol, where the hydroxyl group is attached to a carbon surrounded by three methyl groups. Methyl groups are electron-donating, making the alcohol less acidic.
(B) (CH ) CH-OH: This is a secondary alcohol with electron-donating effects from two methyl groups. It is less acidic than CH OH.
(C) CH OH: This is a primary alcohol with no additional electron-donating groups. It is more acidic than tertiary or secondary alcohols.
(D) CH -CH -OH: This is also a primary alcohol, but the ethyl group is slightly more electron-donating than the methyl group in CH OH, making it slightly less acidic.

Step 3: Conclusion.
The most acidic alcohol is (C) CH OH, as it is a primary alcohol without electron-donating substituents.