Cyclohexylamine and aniline can be distinguished by Azo dye test. Because Cyclohexylamine is not the aromatic amine, it does not give azo dye test.

The reaction between acetaldehyde and benzaldehyde in the presence of NaOH is known as the Cannizzaro reaction. In this reaction, both aldehydes act as reducing agents and undergo disproportionation. However, the products formed depend on the reaction conditions.

Under normal Cannizzaro reaction conditions, where excess NaOH is used, the following products are formed:

Sodium Salt of Acetic Acid: The acetaldehyde (CH₃CHO) is reduced to form sodium salt of acetic acid (CH₃COONa) as one of the products.

Benzyl Alcohol: The benzaldehyde (C₆H₅CHO) is reduced to form benzyl alcohol (C₆H₅CH₂OH) as another product.

Therefore, in this specific reaction, two products are formed: sodium salt of acetic acid (CH₃COONa) and benzyl alcohol (C₆H₅CH₂OH).

It's important to note that the formation of these products can be influenced by reaction conditions such as the molar ratio of the reactants, concentration of NaOH, temperature, and reaction time. In some cases, additional products like benzoic acid or acetic acid may also be observed, depending on the specific reaction conditions.

Neoprene is a synthetic rubber that is composed primarily of chloroprene monomers. The bond present in neoprene is a covalent bond.

Covalent bonds are formed when atoms share electrons to achieve a stable electron configuration. In the case of neoprene, the chloroprene monomers undergo a polymerization process where the chloroprene molecules link together through covalent bonds, forming long chains.

p-Methyl benzene nitrile can be prepared from p-toluidine (p-methyl aniline) by the Sandmeyer reaction. In the Sandmeyer reaction, an aromatic primary amine is converted to a diazonium salt by treatment with sodium nitrite and hydrochloric acid at low temperature (273-278 K). The diazonium salt is then reacted with copper(I) cyanide (CuCN) or potassium cyanide (KCN) to replace the diazonium group with a cyanide group (CN).
(1) and (2) are not possible because aryl halides do not undergo nucleophilic substitution with cyanide ions easily.
(4) is not possible because the reaction of aniline with chloroform and potassium hydroxide is the Reimer
-Tiemann reaction, which introduces an aldehyde group (-CHO) at the ortho position.
Thus, the correct option is (3).

Step 1: Apply Graham's Law of Effusion. Graham's law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass.
Step 2: Calculate the relative rates of diffusion. Given that:
Molar mass of CH = 16 g/mol.
Molar mass of the other gas = 64 g/mol.

Step 3: Relate the rates to the volumes diffused over time. Using the given times for diffusion:
Step 4: Solve for the ratio . Since the ratio is equivalent to , the correct answer is confirmed to be option (4).

Step 1: Recall the energy ordering of atomic orbitals. Typically, the energy of orbitals increases with the principal quantum number , but due to the penetration and shielding effects, some orbitals like are lower in energy than .
Step 2: Identify the incorrect order. The correct energy order should be , making option (4) incorrect because it states .

Step 1: Understanding Swarts Reaction Swarts reaction is used to prepare alkyl fluorides by reacting alkyl bromides or chlorides with metal fluorides such as CoF , Hg F , or AgF.

Step 2: Identifying the Correct Reaction - Among the given reactions, only option (4) involves the replacement of Br with F using CoF , which follows the Swarts reaction.

In Kjeldahl's method, nitrogen in an organic compound is converted into ammonia ( ). This ammonia is absorbed in a known volume of standard sulfuric acid ( ). The unreacted sulfuric acid is then titrated against a standard solution of sodium hydroxide ( ).
Given data:

• Weight of organic compound = 0.2 g
• Volume of = 60 mL
• Molarity of = 0.1 M
• Volume of = 40 mL
• Molarity of = 0.1 M

Step 1: Calculate the milliequivalents of initial : Step 2: Calculate the milliequivalents of used for titration: Step 3: Calculate the milliequivalents of unused : Step 4: Calculate the milliequivalents of reacted with : Step 5: Milliequivalents of and Nitrogen: Equivalent weight of Nitrogen = 14 g/equivalent.
Step 6: Calculate the weight of Nitrogen: Step 7: Calculate the percentage of Nitrogen in the organic compound: Correct Answer: (3) 56

Step 1: Analyze the properties of alkali metal bicarbonates. Alkali metal bicarbonates typically form solids except for lithium bicarbonate ( ), which cannot be isolated as a solid under normal conditions and typically exists only in solution.
Step 2: Choose the correct option based on known chemical properties. Lithium bicarbonate's inability to form a solid is due to its highly soluble and unstable nature in solid form compared to other alkali metal bicarbonates like sodium ( ) or potassium ( ) bicarbonates.

The reaction proceeds in two steps.
Step 1: Dehydrohalogenation to form Propyne (A) 1,2-dibromopropane reacts with alcoholic KOH and heat, followed by , to undergo double dehydrohalogenation, resulting in the formation of propyne (compound A). The reaction is as follows: Thus, compound A is propyne.
Step 2: Hydration of Propyne to form Acetone (X) Propyne (A) undergoes hydration in the presence of and water at 333 K to form acetone (compound X). The reaction follows Markovnikov's rule. Thus, compound X is acetone (propan-2-one).
Step 3: Calculating Percentage of Carbon in Acetone (X) The molecular formula of acetone is . Molar mass of Acetone ( ) is calculated as: Mass of carbon in 1 mole of Acetone = g.
Percentage of carbon in Acetone (X): The percentage of carbon in compound X is approximately 62%.
Correct Answer: (2) 62

Step 1: Understanding the Reaction Sequence
1. First Reaction: Amide Formation
- The given reactant is acetic acid ( ).
- When reacted with ammonia (NH ) and heated ( ), it forms an amide (P): - So, is acetamide ( ).
2. Second Reaction: Hoffmann Bromamide Degradation - The Hoffmann degradation reaction involves treating an amide with Br /NaOH, which removes one carbon and forms a primary amine. - The product is methylamine ( ), which has one carbon less than in P.
Step 2: Identifying the Correct Answer
- (acetamide) contains two carbon atoms.
- (methylamine) contains one carbon atom, indicating the loss of one carbon.
- The final product is a primary amine.
Thus, the correct answer is: Option (2): A primary amine with one carbon less than in P

Step 1: Understanding the -R effect. Groups that withdraw electron density from a conjugated system by resonance show a negative resonance effect (-R). Such groups contain strongly electronegative elements or multiple bonds adjacent to the system.

Step 2: Identifying the correct groups.
- Formyl (-CHO), Cyano (-CN), and Nitro (-NO ) groups exhibit the -R effect.
- Alkoxy (-OR) and Amino (-NH ) groups show a positive resonance effect (+R).
Thus, the correct answer is .

Toluene reacts with KMnO /OH , followed by H O to give benzoic acid (Y). Benzoic acid dissolves in NaHCO due to the formation of sodium benzoate. When benzoic acid reacts with Br /Fe, it undergoes electrophilic substitution. The -COOH group is a deactivating and meta-directing group. So, the bromine atom will be substituted at the meta position. Thus, X is KMnO /OH , followed by H O and Z is m-bromobenzoic acid.

Analyzing the given statements:
Statement I: At constant pressure, physical adsorption decreases with increase in temperature. This is correct because physical adsorption is an exothermic process and its efficiency decreases as the temperature increases.
Statement II: Enthalpy of physisorption is higher than enthalpy of chemisorption. This is incorrect because chemisorption involves a chemical bond formation, which has higher enthalpy compared to physical adsorption.
Statement III: Gas masks used in coal mines consist of activated charcoal. This is correct because activated charcoal is used in gas masks due to its adsorption properties to remove harmful gases. Thus, the correct statements are I and III.

Step 1: Understanding the Reaction Sequence The given sequence involves the following transformations: 1. Ozonolysis Reaction: - The initial alkene undergoes ozonolysis in the presence of followed by reduction with Zn/H O. - This leads to the formation of an aldehyde and a ketone. 2. Addition of HBr in the Presence of Peroxides: - The anti-Markovnikov addition of HBr to the alkene forms a primary alkyl halide (X) via the free radical mechanism. 3. Formation of Grignard Reagent and Carboxylation: - The alkyl halide reacts with Mg in dry ether to form a Grignard reagent (RMgX). - When treated with CO , the Grignard reagent forms a carboxylate intermediate, which upon hydrolysis gives a primary carboxylic acid (Y).

Step 2: Identifying the Product (Y) - The final product is a primary carboxylic acid, corresponding to option (4).

Analysis of Gases:
Methane ( ) and Nitrous Oxide ( ) are well-known greenhouse gases. CFCs and Ozone ( ), although problematic for the ozone layer, also have greenhouse effects. is a primary greenhouse gas.
Identifying Non-Greenhouse Gases:
Nitrogen ( ) and Carbon Monoxide ( ) are not typically classified as greenhouse gases. is inert and does not absorb infrared radiation significantly, while 's contribution to the greenhouse effect is minimal compared to .

Step 1: Understanding the Reaction Sequence - The given reaction involves conversion of a substituted benzene with nitro (-NO ) and methyl (-CH ) groups to a brominated product.

Step 2: Identifying the Steps 1. Bromination:
- The presence of a methyl (-CH ) group directs bromine to the para position via electrophilic substitution. 2. Reduction of Nitro Group:
- The -NO group is reduced to an amine (-NH ) using reducing agents like Sn/HCl. 3. Carbylamine Reaction:
- The -NH group undergoes carbylamine reaction (using CHCl and KOH) to form an isocyanide (-NC).

Step 3: Conclusion
- Since this follows the sequence Bromination → Reduction → Carbylamine Reaction, the correct answer is option (1).

Step 1: Conversion of Chlorobenzene to Phenol
Chlorobenzene reacts with NaOH under high temperature (623 K) and pressure (300 atm) to form phenol.

Step 2: Nitration and Further Reaction
- The major product of nitration of chlorobenzene is p-nitrochlorobenzene.
- On treatment with NaOH (443 K, H ), it forms p-nitrophenol.

Step 1: Identifying the transformations.
- The first reaction involves the reduction of the ester to an aldehyde, which is best achieved using Diisobutylaluminum hydride (DIBAL-H) at low temperatures.
- The second reaction involves catalytic hydrogenation, which reduces the aldehyde to a primary alcohol.

Step 2: Assigning X and Y.
- The intermediate Y corresponds to the aldehyde, which is obtained by the partial reduction of the ester using DIBAL-H.
- The final product is a primary alcohol, which suggests the use of catalytic hydrogenation after the formation of the aldehyde.

Step 1: Identifying the nature of X.
- Potassium cyanide ( ) is an ionic compound that provides a nucleophilic cyanide ion (CN⁻).
- The cyanide ion attacks the alkyl halide via an SN2 mechanism, leading to the formation of an alkyl nitrile (R-CN).

Step 2: Understanding Y and its properties.
- The product Y is a nitrile ( ), which undergoes hydrolysis to give a carboxylic acid and ammonia (NH₃).
- Ammonia ( ) is a basic gas that turns red litmus paper blue, confirming the presence of a nitrile.

Step 1: Understanding the Reaction Mechanism - The given reaction is the anti-Markovnikov addition of HBr in the presence of peroxides. - This follows the free radical mechanism, leading to the addition of Br at the terminal carbon of the alkene.

Step 2: Identifying the Product Type - The resultant compound has a primary carbon attached to the bromine atom. - Since the halogen is attached to a primary carbon, the compound is a primary alkyl halide.

Step 1: Identifying the Carbonyl Compound X - Given that oxidation of gives a carboxylic acid ( ),
- This suggests that must be a ketone or aldehyde that can be oxidized to a carboxylic acid.

Step 2: Checking Possible Structures - The simplest possible ketone or aldehyde with is propanal or propanone.
- Since oxidation leads to a single carboxylic acid, the compound must be propanal (CH -CH -CHO).

Step 3: Identifying the Oxime of X
- The oxime is formed when aldehydes react with hydroxylamine (NH OH).
- This results in the formation of propanal oxime (CH -CH -CH=NOH).

Step 1: Identify the product 'X'. Given the reaction, 'X' is likely a boron compound where BF3 is reduced by LiAlH4. The exact structure of 'X' might depend on the specific conditions of the reaction, but it is unlikely to be diborane as initially assumed since the incorrect statement points to the absence of a B-B bond.
Step 2: Correct the understanding of 'X'. The reaction likely forms a complex where boron is bonded in a different configuration without B-B bonding, possibly a complex boron hydride or a coordination compound with aluminum and fluorine.
Step 3: Evaluate the statements.
(1) Likely True: The product is probably an electron-deficient hydride, consistent with boron's properties.
(2) False: 'X' does not have a B-B bond, making this statement incorrect.
(3) Likely True: If 'X' can react with ammonia, it might produce compounds like borazine under certain conditions, assuming it contains reactive boron-hydride units.
(4) Likely True: Hydrolysis of boron hydrides generally leads to the formation of boric acid, a weak monobasic acid.

Step 1: Consider the properties of Yb in forming hydrides. Ytterbium (Yb), as a rare earth element, is known to form complex metal hydrides that can display various hydrogen stoichiometries. The lattice structure of these hydrides can significantly differ from that of the pure metal due to the insertion of hydrogen atoms into the metal lattice.
Step 2: Understand the impact of hydrogen on metal lattices. When Yb absorbs hydrogen, it forms hydrides where the hydrogen atoms occupy interstitial sites within the Yb lattice. This can lead to an expansion and distortion of the lattice, which significantly changes the crystal structure from that of the metallic phase.
Step 3: Implications for . Given that for Yb suggests a high hydrogen content relative to ytterbium, the hydride lattice is likely to be different from the lattice of pure Yb due to these interstitial hydrogen atoms causing lattice expansion and potentially new crystallographic phases.

Understanding Conjugate Acids and Bases: In Brønsted-Lowry acid-base theory, a conjugate base is formed by the removal of a proton (H ) from an acid. Since is itself a base, we need to determine what would form if it were to lose a proton.
Step 1: Analyze the Possibility of OH Losing a Proton. OH losing a proton (which it doesn't possess in excess) would theoretically produce .
Step 2: Correct Misconception. However, is typically the conjugate base of . When acts as an acid and donates a proton, is what remains. Therefore, asking for the conjugate base of might seem unconventional, as is already a base. But in this context, further loss of a proton would indeed lead to , though this is a highly theoretical and not commonly observed scenario.