Organic Chemistry

In R-S configuration, priority sequence is decided by the atoms directly attached to the chiral carbon arranged in decreasing atomic number.
From the given groups, sulphur (S) has the highest atomic number i.e., 16 therefore, it gets highest priority.

The functional group that has electron donating or positive inductive effect is Since, the positive inductive effect is due to the presence of an electron donating groups within the polar bond. Thus, option (c) is correct as rest of the three groups are electron withdrawing in nature.

Step 1: First reaction (dehalogenation).
Propylene dibromide on treatment with Zn/alcohol undergoes dehalogenation to form propene (A).
Step 2: Addition of HBr.
Propene reacts with HBr (Markovnikov addition) to form 2-bromopropane (B).
Step 3: Wurtz reaction.
2-Bromopropane reacts with sodium in dry ether to undergo coupling, forming 2,3-dimethylbutane.
Step 4: Conclusion.
Hence, the final product is 2,3-dimethylbutane.

Step 1: Understanding the reaction.
The balanced reaction shows that 2 moles of KClO decompose to give 3 moles of O . We need to calculate the volume of oxygen gas produced at standard temperature and pressure (STP). At STP, 1 mole of any ideal gas occupies 22.4 L.

Step 2: Calculating the volume.
The volume of oxygen produced is directly proportional to the amount of KClO used. The given reaction shows that 2 moles of KClO produce 3 moles of O . Therefore, for every 2 moles of KClO , 3 moles of O are produced, which corresponds to:
Step 3: Conclusion.
Thus, the correct answer is (D) 67.2 L, as this is the volume of oxygen gas produced at STP.

Step 1: Understanding the chlorinating property of \text{PCl_5.}
Phosphorus pentachloride (\text{PCl}_5) is a chlorinating agent that reacts with various substances to replace or add chlorine atoms. The reaction with tin (Sn) is a classic example of its chlorinating property.

Step 2: Analyzing the options.
(A) : This reaction shows the hydrolysis of \text{PCl}_5, not its chlorinating property.
(B) : This is the formation of \text{PCl}_5 from phosphorus and chlorine, not proof of its chlorinating property.
(C) : This shows decomposition of \text{PCl}_5, but not its chlorinating property.
(D) : Correct — This reaction demonstrates the chlorinating property of \text{PCl}_5, where it chlorinates tin to form SnCl₄ and PCl₃.

Step 3: Conclusion.
The correct answer is (D) as it directly demonstrates the chlorinating ability of phosphorus pentachloride.

Step 1: Identifying the structure.
The compound is a cyclopentane ring with a chlorine ( ) attached to the 4th carbon, a methyl group ( ) attached to the 2nd carbon, and a hydroxyl group ( ) attached to the 4th carbon. The IUPAC name should reflect the correct positions of these groups.
Step 2: Analyzing the options.
(A) 3-chloro-5-methyl cyclopentanol: Incorrect — The position of the chlorine and methyl groups is wrong.
(B) 1-chloro-3-methyl cyclopentan-4-ol: Incorrect — The positions of the chlorine and hydroxyl groups are incorrect.
(C) 4-chloro-2-methyl cyclopentan-ol: Correct — This is the correct IUPAC name based on the structure provided.
(D) 4-chloro-2-hydroxy-1-methylcyclopentane: Incorrect — The hydroxyl group is incorrectly named as hydroxy, and the methyl group is not on the correct position.
Step 3: Conclusion.
The correct answer is (C) 4-chloro-2-methyl cyclopentan-ol.

Step 1: Understanding the first order reaction.
For a first-order reaction, the integrated rate law is given by: By converting the natural logarithm to base 10, we get the equation: This equation represents a straight line with a slope of .
Step 2: Conclusion.
Thus, the slope of the graph of vs. time is equal to .

Step 1: Understanding the reaction sequence.
- The first step involves the conversion of ethanol (CH -CH -OH) into an alkyl chloride, CH -CH -Cl, by reaction with PCl . - The second step uses alcoholic KOH, which leads to a dehydrohalogenation reaction, producing an alkene. - The final step involves treating the alkene with concentrated sulfuric acid (H SO ), which causes the alkene to undergo an elimination reaction, forming a hydrocarbon.

Step 2: Analyzing the options.
(A) CH -CH -OH: This is the starting alcohol and does not fit the final product of the reaction.
(B) (CH ) CH-CH : This is incorrect as the final product is an alkane, not this structure.
(C) CH -CH=CH : This is an alkene intermediate, but the final product is an alkane.
(D) CH -CH -OH: Correct — This is the alkane product formed after the elimination step.

Step 3: Conclusion.
The correct answer is (D) CH -CH -OH.

Step 1: Analyzing the compound.
The compound contains an ethyl group (Et) and a methyl group (Me). The structure consists of an ethyl group (C H ) and a methyl group (CH ) attached to a carbon-carbon chain. The molar mass of the compound is calculated by adding the atomic masses of the constituent atoms.
Step 2: Calculation.
For the ethyl group (Et): For the methyl group (Me): For the entire compound: Step 3: Conclusion.
The correct molar mass of the compound is 128 g/mol, so the correct answer is (A).

Step 1: Understanding Gabriel phthalimide synthesis.
Gabriel phthalimide synthesis is a method used to prepare primary amines. It involves the reaction of phthalimide with an alkyl halide to form a phthalimide salt, which upon hydrolysis gives the corresponding amine.
Step 2: Analyzing the options.
(A) Ethylamine: Ethylamine can be synthesized using Gabriel phthalimide synthesis, as it involves an alkyl halide (ethyl bromide) and phthalimide.
(B) Sec-butylamine: Sec-butylamine can also be synthesized in a similar manner using sec-butyl bromide.
(C) Aniline: Aniline cannot be synthesized using Gabriel phthalimide synthesis, as it is a primary amine with an aromatic group. The synthesis works only for aliphatic amines.
(D) Isopropylamine: Isopropylamine can be synthesized using Gabriel phthalimide synthesis, using isopropyl bromide as the alkyl halide.
Step 3: Conclusion.
The correct answer is (C) Aniline, as it cannot be prepared by Gabriel phthalimide synthesis due to its aromatic nature.

Step 1: Understanding the reaction.
The reaction with iodine and sodium hydroxide (NaOH + I ) is known as the iodoform test, which forms a yellow precipitate of CHI in the presence of methyl ketones or compounds with the group CH CO-.

Step 2: Analyzing the options.
(A) Acetophenone: Acetophenone reacts with iodine and NaOH to form a yellow precipitate of iodoform.
(B) Benzaldehyde: Benzaldehyde does not have the CH CO- group, so it does not give the yellow precipitate.
(C) Acetone: Acetone reacts with iodine and NaOH to give the yellow precipitate of iodoform.
(D) Acetaldehyde: Acetaldehyde reacts with iodine and NaOH to give the yellow precipitate of iodoform.

Step 3: Conclusion.
The correct answer is (B) Benzaldehyde.

Step 1: Understanding the reaction.
Propane nitrile reacts with ethyl magnesium iodide (a Grignard reagent) in the presence of dry ether to form an imine complex. Upon acid hydrolysis, this imine complex forms a ketone.

Step 2: Analyzing the options.
(A) 2-Pentanone: This is incorrect as it does not match the reaction outcome.
(B) Butanone: This is incorrect. The expected product is not butanone.
(C) Propanone: This is incorrect. The structure of the final product does not match propanone.
(D) 3-Pentanone: Correct — 3-Pentanone is the product formed after hydrolysis of the imine complex.

Step 3: Conclusion.
The correct answer is (D) 3-Pentanone.

Step 1: Naming the compound.
The given compound is 1-Chloro-2, 2-dimethyl propane. In common nomenclature, this compound is known as neo-pentyl chloride. The name "neo-pentyl" comes from the structure with a five-carbon chain and a chloromethyl group attached to the first carbon.

Step 2: Analyzing the options.
(A) neo-pentyl chloride: Correct — This is the common name for 1-Chloro-2, 2-dimethyl propane.
(B) isopropyl chloride: This is a different compound.
(C) n-pentyl chloride: This is not the correct name for the compound.
(D) isopentyl chloride: This is another name for a different compound.

Step 3: Conclusion.
The correct answer is (A) neo-pentyl chloride.

Step 1: Understanding the reaction.
The reaction involves the hydrolysis of an amide (C H C(O)NH ) in the presence of water and hydrochloric acid. Hydrolysis of amides produces a carboxylic acid and ammonia.
Step 2: Analyzing the options.
(A) C H C(O)OH: This is the correct product. Hydrolysis of the amide group results in the formation of a carboxylic acid, phenyl acetic acid (C H C(O)OH).
(B) C H C(NH ): This is incorrect, as it represents an amine group, which is not formed in this reaction.
(C) C H C(Cl): This is incorrect, as no chlorination is involved in this reaction.
(D) C H C(H): This is incorrect, as no reduction to an aldehyde is involved.
Step 3: Conclusion.
The correct product is (A) C H C(O)OH, phenyl acetic acid.

Step 1: Understanding the carbylamine test.
The carbylamine test is used to detect primary amines. In this test, the amine reacts with chloroform (CHCl ) and sodium hydroxide (NaOH) to form an isocyanide (carbylamine), which typically gives a strong odor. Secondary and tertiary amines do not give this test.

Step 2: Analyzing the options.
(A) Ethylamine: Ethylamine is a primary amine and will give a positive carbylamine test.
(B) Sec. butylamine: Sec. butylamine is a secondary amine and will not give the carbylamine test.
(C) Isopropylamine: Isopropylamine is a secondary amine and will not give the carbylamine test.
(D) Dimethylamine: Dimethylamine is a secondary amine and will not give the carbylamine test.

Step 3: Conclusion.
The correct answer is (D) Dimethylamine.

Step 1: Understanding Solvay's Process.
In the Solvay process, sodium carbonate (Na CO ) is produced by reacting sodium chloride (NaCl) with calcium carbonate (CaCO ) and carbon dioxide (CO ) at high temperature. Carbon dioxide is bubbled through the brine solution to form sodium bicarbonate (NaHCO ), which is then heated to produce sodium carbonate.

Step 2: Analyzing the options.
(A) CO (g): Correct. Carbon dioxide is the gas that is bubbled through the brine solution in the Solvay process.
(B) N (g): Incorrect. Nitrogen is an inert gas and does not play a role in this process.
(C) NO (g): Incorrect. Nitrogen dioxide is not involved in the Solvay process.
(D) O (g): Incorrect. Oxygen is not used in the Solvay process.

Step 3: Conclusion.
The correct answer is (A) CO (g), as carbon dioxide is bubbled through the brine solution in the Solvay process.

Step 1: Understanding the reaction.
The reaction involves the oxidation of phenol with sodium dichromate ( ) in the presence of sulfuric acid, leading to the formation of benzoquinone. This is a common reaction where phenol undergoes oxidation to produce benzoquinone.
Step 2: Analyzing the options.
(A) Benzaldehyde: Incorrect. Benzaldehyde is not formed in this reaction.
(B) Salicylic acid: Incorrect. Salicylic acid is not produced by this oxidation reaction.
(C) Benzoquinone: Correct — Benzoquinone is the product of the oxidation of phenol with sodium dichromate.
(D) Benzoic acid: Incorrect. Benzoic acid is not the product of this reaction, which involves the formation of a quinone group.
Step 3: Conclusion.
The correct answer is (C) Benzoquinone, as it is the product obtained in this oxidation reaction of phenol.

Step 1: Understand phantom atom.
A phantom atom is an atom which is not shown explicitly in the molecular formula but is assumed to be present to complete valency, especially in resonance structures.
Step 2: Analyze acetic acid structure.
In acetic acid ( ), resonance structures of the carboxyl group involve delocalisation of electrons, where one oxygen behaves as if a hydrogen is shared or shifted, giving rise to the concept of a phantom atom.
Step 3: Eliminate other options.
The other compounds do not show such resonance-induced phantom atom behaviour.
Step 4: Conclusion.
Hence, contains a phantom atom.

Step 1: Identify carbolic acid.
Carbolic acid is phenol with molecular formula .
Step 2: Count the sigma bonds.
In the benzene ring, there are 6 sigma bonds and 5 sigma bonds.
The bond contributes 1 sigma bond and the bond contributes 1 sigma bond.
Step 3: Total sigma bonds.

Step 4: Conclusion.
Thus, carbolic acid contains a total of sigma bonds.

Step 1: Identify the reaction.
The given reaction is an example of the Wurtz–Fittig reaction, where an aryl halide reacts with an alkyl halide in the presence of sodium and dry ether.
Step 2: Possible products formed.
In this reaction, three types of products can be formed:
Alkyl–alkyl coupling gives n-butane, aryl–aryl coupling gives diphenyl, and aryl–alkyl coupling gives ethylbenzene.
Step 3: Analyze the options.
Toluene requires a methyl group attached to benzene, which cannot be formed from bromoethane under these conditions.
Step 4: Conclusion.
Therefore, toluene is not obtained in this reaction.

Step 1: Boiling points and intermolecular forces.
Boiling point is determined by the strength of intermolecular forces. Alcohols like propan-1-ol have hydrogen bonding, which results in a higher boiling point compared to aldehydes (propanal), ethers (methoxy ethane), and alkanes (propane), which have weaker intermolecular forces.
Step 2: Analyzing the options.
(A) Propanal: Propanal is an aldehyde and has weaker intermolecular forces compared to alcohols, so it has a lower boiling point than propan-1-ol.
(B) Methoxy ethane: This is an ether, which has weaker dipole-dipole interactions compared to alcohols.
(C) Propane: Propane is an alkane and has the weakest intermolecular forces (London dispersion forces), so it has the lowest boiling point.
(D) Propan-1-ol: This is an alcohol and has the highest boiling point due to hydrogen bonding.
Step 3: Conclusion.
The correct answer is (D) Propan-1-ol.

Step 1: Understanding Werner's theory.
According to Werner's theory, the primary valence of a metal ion refers to its oxidation state, and the secondary valence refers to the coordination number, i.e., the number of ligands attached to the metal ion. In , the coordination number is 6, meaning the secondary valence is 6. However, because the complex ion has a charge of , the primary valence or oxidation state of Co is .
Step 2: Analyzing the options.
(A) 5: Incorrect — This is not the secondary valence.
(B) 4: Incorrect — The secondary valence is 6 in this complex.
(C) 3: Correct — The secondary valence refers to the coordination number of 3.
(D) 6: Correct — The coordination number is 6. But we needed the secondary valence.

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

Step 1: Formula for radius in fcc unit cell.
In the face-centered cubic (fcc) structure, the relationship between edge length and atomic radius is: Where . Solving for :
Step 2: Conclusion.
The radius of the silver atom is (B) 111.91 pm.

Step 1: Identifying the -CN Group.
The -CN group (cyano group) is an organic functional group consisting of a carbon triple-bonded to a nitrogen atom. Among the listed antihistamines, cimetidine contains the -CN group. Cimetidine is a H -receptor antagonist used for the treatment of ulcers and gastroesophageal reflux disease (GERD).

Step 2: Analyzing the options.
(A) Dimetapp: Incorrect. Dimetapp does not contain a -CN group. It is primarily used for treating allergies and cold symptoms.
(B) Cimetidine: Correct. Cimetidine contains a -CN group in its chemical structure.
(C) Terfenadine: Incorrect. Terfenadine does not contain a -CN group. It is an antihistamine used for allergic conditions.
(D) Ranitidine: Incorrect. Ranitidine does not contain a -CN group. It is similar to cimetidine but lacks the cyano group.

Step 3: Conclusion.
The antihistamine that contains the -CN group is Cimetidine, corresponding to option (B).

Step 1: Understanding resonance.
Resonance occurs when a molecule can be represented by two or more valid Lewis structures that differ only in the placement of electrons. Cyclohexane does not exhibit resonance as it has a simple, single structure with no delocalized electrons.
Step 2: Analyzing the options.
(A) Cyclohexane: Correct — Cyclohexane does not exhibit resonance because its structure is a stable, single form with no delocalized electrons.
(B) Phenol: Incorrect. Phenol exhibits resonance due to the delocalization of the lone pair of electrons on the oxygen atom.
(C) Aniline: Incorrect. Aniline exhibits resonance with the lone pair of electrons on the nitrogen atom delocalizing into the benzene ring.
(D) Nitro ethane: Incorrect. Nitro ethane exhibits resonance due to the delocalization of electrons in the nitro group.
Step 3: Conclusion.
The correct answer is (A) Cyclohexane, as it does not exhibit resonance.

Step 1: Understanding interhalogen compounds.
Interhalogen compounds are compounds formed by the combination of two halogens. They can take the form XX'Y, where X and X' are halogens, and Y can be another halogen.

Step 2: Analyzing the options.
(A) Br and F: This combination forms BrF, which is an interhalogen compound but not of the form XX'Y.
(B) Cl and F: This combination forms ClF, but this is not of the form XX'Y.
(C) I and F: This combination forms IF and IF , both of which are interhalogen compounds of the type XX'Y.
(D) I and Cl: This combination forms ICl, which is an interhalogen compound, but not in the XX'Y form.

Step 3: Conclusion.
The correct answer is (C) I and F, which forms interhalogen compounds of the type XX'Y.

Step 1: Understanding tetradentate ligands.
A tetradentate ligand is a ligand that can form four bonds with a metal center. Triethylene tetramine is a common tetradentate ligand that can coordinate with metal centers through four nitrogen atoms.

Step 2: Analyzing the options.
(A) Ethylene diamine tetracetato: This ligand is bidentate, meaning it can form two bonds, not four.
(B) Triethylene tetramine: This is the correct answer as it is a tetradentate ligand.
(C) Dimethyl glyoximato: This is a bidentate ligand, not tetradentate.
(D) Oxalato: This is also a bidentate ligand, not tetradentate.

Step 3: Conclusion.
The correct answer is (B) Triethylene tetramine.

Step 1: Understanding the polymers.
Styrene-butadiene rubber (SBR) is a synthetic rubber made from the copolymerization of styrene and butadiene. It is widely used in the automotive industry, particularly for making tires.

Step 2: Analyzing the options.
(A) Buna-N: This is a copolymer of butadiene and acrylonitrile, not styrene.
(B) PHBV: This is a biopolymer made from 3-hydroxybutyrate and 3-hydroxyvalerate.
(C) Butyl rubber: Butyl rubber is made from isobutene and isoprene, not styrene and butadiene.
(D) SBR: This is the correct option; styrene-butadiene rubber is formed from the copolymerization of styrene and butadiene.

Step 3: Conclusion.
The correct answer is (D) SBR, as it is obtained from styrene and 1-3-butadiene.

Step 1: Understanding p character.
In acetylene (C H ), the carbon atoms undergo sp hybridization, where 50% of the orbital character comes from the p orbitals. This results in a 50% p character of the hybrid orbitals.

Step 2: Analyzing the options.
(A) Propene: Propene involves sp hybridization, but it does not exhibit 50% p character in its hybridization.
(B) Acetylene: This is the correct answer, as acetylene has a 50% p character in its sp hybrid orbitals.
(C) Methane: Methane involves sp hybridization, which has no p character.
(D) Ethane: Ethane also involves sp hybridization and does not exhibit p character.

Step 3: Conclusion.
The correct answer is (B) Acetylene, as it contains 50% p character of hybrid orbitals in carbon atoms.

Step 1: Understanding allylic alcohols.
Allylic alcohols are alcohols where the hydroxyl group (-OH) is attached to a carbon that is adjacent to a double bond, i.e., the -OH group is on a carbon in an alkene.

Step 2: Analyzing the options.
(A) 2-Phenyl propane -2-ol: This compound is not an allylic alcohol because the -OH group is not adjacent to a double bond.
(B) 2-Methylbut -3-en-2-ol: This is the correct option. The -OH group is on the second carbon, which is adjacent to the double bond at position 3, making it an allylic alcohol.
(C) Propane -1, 2, 3 - triol: This is a triol, not an allylic alcohol. The -OH groups are attached to non-adjacent carbons.
(D) Propane -1, 3-diol: This compound is not an allylic alcohol because the -OH groups are on non-adjacent carbons, with no double bond involved.

Step 3: Conclusion.
The correct answer is (B) 2- Methylbut -3-en-2-ol, as it is the only allylic alcohol.

Step 1: Understanding the concept of effective atomic number.
The effective atomic number (EAN) is the total number of electrons surrounding the metal ion. In , the Zn ion is surrounded by four NH molecules. Each NH donates 2 electrons, and the Zn ion contributes 2 electrons. The total effective atomic number is:
Step 2: Conclusion.
The effective atomic number of Zn is 36.

Step 1: Understanding the reaction.
The Mendius reduction involves the reduction of alkyl cyanides (nitriles) to primary amines when treated with sodium and ethanol. This is an important method in organic synthesis for preparing amines.

Step 2: Analyzing the options.
(A) Wolff-Kishner reduction: This reaction involves the reduction of carbonyl compounds to alkanes, not the reduction of nitriles.
(B) Hell-Vohlard-Zelinsky reaction: This is a halogenation reaction, not related to the reduction of nitriles.
(C) Mendius reduction: This is the correct name for the reduction of alkyl cyanides to primary amines.
(D) Clemensen reduction: This is a reduction of carbonyl compounds to alkanes using zinc and hydrochloric acid, not related to nitriles.

Step 3: Conclusion.
The correct answer is (C) Mendius reduction.

Step 1: Understanding the van't Hoff factor.
The van't Hoff factor is the number of particles the solute dissociates into. For K [Fe(CN) ], the dissociation can be represented as: Hence, the total number of ions formed is 4 (3 K ions and 1 [Fe(CN) ] ).
Step 2: Applying the degree of dissociation.
The degree of dissociation , so the van't Hoff factor is:
Step 3: Conclusion.
The van't Hoff factor is 3.34.

Step 1: Understanding boiling points.
Boiling points of organic compounds are influenced by factors such as molecular weight, polarity, and hydrogen bonding. Alcohols generally have higher boiling points than alkanes and amines due to hydrogen bonding.

Step 2: Analyzing the options.
(A) :} This is an alcohol, and alcohols have higher boiling points due to hydrogen bonding.
(B) : This is an alkane, which typically has a lower boiling point than alcohols.
(C) :} This is an amine, and while amines have higher boiling points than alkanes, they typically have lower boiling points than alcohols.
(D) :} This is also an amine with a lower boiling point than the alcohol.

Step 3: Conclusion.
The correct answer is (A) }, as alcohols generally have the highest boiling points due to hydrogen bonding.