Organic Chemistry

Step 1: Recall the carbylamine reaction. Primary amines on heating with chloroform (CHCl ) and alcoholic KOH form isocyanides.
Step 2: Aniline is a primary aromatic amine.
Step 3: When aniline reacts with CHCl and KOH, it forms phenyl isocyanide, which has an extremely foul smell.
Step 4: The reaction is:
Step 5: Therefore, the reagent ‘X’ is chloroform (CHCl ).

Step 1: Recall the meaning of heterolytic fission. In heterolytic bond cleavage, the shared pair of electrons is unequally divided between the two atoms.
Step 2: One atom takes both electrons and becomes a negatively charged ion (anion).
Step 3: The other atom loses the shared electrons and becomes a positively charged ion (cation).
Step 4: Therefore, heterolytic fission always produces one cation and one anion.
Step 5: Hence, the correct option is (B).

Step 1: Recall that the acidic nature of phenols depends on the stability of the phenoxide ion formed after loss of H .
Step 2: Electron-withdrawing groups increase acidity by stabilising the phenoxide ion, while electron-donating groups decrease acidity.
Step 3: In o-nitrophenol, the –NO group is a strong electron-withdrawing group, which greatly increases acidity.
Step 4: In o-cresol, the –CH group is electron-donating, which decreases acidity compared to phenol.
Step 5: Therefore, the relative acidic strength is:
Step 6: This relationship is correctly represented by option (B).

Step 1: Write the structure of 2-methyl-2-butene:
Step 2: Ozonolysis cleaves the C=C double bond and converts each carbon of the double bond into a carbonyl compound.
Step 3: The carbon atom attached to two CH groups forms a ketone:
Step 4: The other carbon atom attached to one CH group and one hydrogen forms an aldehyde:
Step 5: Therefore, the products of ozonolysis are CH CHO and CH COCH .

Step 1: Determine the mass of carbon in CO . The molar mass of CO is 44 g/mol, and carbon contributes 12 g/mol.
Step 2: Use the ratio of masses to find the mass of carbon:
Step 3: Calculate the percentage of carbon in the compound:
Step 4: Correct the mass of compound: given is 0.2 g
Step 5: Check the calculation: mass of carbon in CO2 = g. Percentage = ? Wait, mass of compound = 0.2 g (given), so

Step 1: Recall directing effects in electrophilic aromatic substitution. Substituents on a benzene ring influence the position of incoming electrophiles:
Ortho/para directors — donate electron density to the ring.
Meta directors — withdraw electron density from the ring.
Step 2: Classify the given substituents.
: Strongly electron-donating, ortho/para-directing
: Electron-donating (hyperconjugation), ortho/para-directing
: Deactivating but donates by resonance, ortho/para-directing
: Strongly electron-withdrawing ( and effects), meta-directing
Step 3: Identify the meta-directing group. Only the group withdraws electron density from the ring sufficiently to direct substitution predominantly to the meta position.
Hence, the correct answer is

Step 1: Secondary amine with strong H SO undergoes oxidation coupling to give hydrazine derivative. = tetramethylhydrazine → matches first part.
Step 2: Primary methyl amine under acidic Hg oxidation forms = dimethyl hydroxyl amine.
Step 3: Option (A) states exactly this pair. Hence → (A).

Step 1: The reagent set is / dil.\;H SO \) in presence of water and heat. This combination is used for oxymercuration–demercuration of terminal alkynes.
Step 2: Terminal alkyne attached to aromatic ring undergoes via Markovnikov hydration.
Step 3: In the given molecule, the side chain is attached to benzene adjacent to a carbonyl group. After hydration the triple bond converts into a methyl ketone ring closure.
Step 4: The resulting system behaves as if the alkyne portion forms a six-membered saturated ring with → cyclohexanone fused to benzene.
Step 5: Match with options → only (D) represents this transformation.

Step 1: Write the structure of 3-methylpentane: Carbon types: - C and C – equivalent primary (2 sites) - C and C – equivalent secondary (2 sites) - C – tertiary (1 site) - branch methyl – primary (1 site)
Step 2: Non-equivalent substitution positions: 1. on terminal primary → 1 product 2. on secondary → 1 product 3. on tertiary → 1 product 4. on branch methyl → 1 product Total constitutionally = 4.
Step 3: Secondary carbon substitution creates a chiral centre. Each such gives two enantiomers. There are 2 secondary positions (equivalent) → 2×2 = 4 stereoisomers.
Step 4: Add non-chiral others (primary + tertiary + branch = 2).
Step 5: Overall monochloro derivatives = 4 + 2 = 6. Hence → (C).

Step 1: Hyperconjugation is interaction of C–H σ bond with adjacent empty/partially filled π orbital.
Step 2: Therefore type = σ–π only.
Step 3: π–π is resonance; p–p is covalent sigma. Hence → (A).

Step 1: Calculate the moles of sulphuric acid used. Step 2: Use the reaction between ammonia and sulphuric acid. From the equation: Step 3: Calculate the mass of nitrogen present. Each mole of contains 1 mole of nitrogen. Step 4: Calculate the percentage of nitrogen in the soil. Final Answer:

Step 1: Recall the metal reactivity series. The reactivity series of metals (from most reactive to least reactive) is: Step 2: Identify the given metals in the series.

Potassium (K) — most reactive
Magnesium (Mg) — highly reactive
Zinc (Zn) — moderately reactive
Iron (Fe) — less reactive
Step 3: Arrange them in increasing order of reactivity (least to most). Step 4: Match with the given options. This order corresponds to option (D).

Step 1: Recall the melting points of the given substances.

Germanium: Melting point — solid at ordinary temperature
Gallium: Melting point — melts near room temperature
Gold: Melting point — solid
Galena (PbS): High melting point — solid
Step 2: Identify the substance liquid at ordinary temperature. Since gallium melts slightly above room temperature, it can exist as a liquid at ordinary conditions. Step 3: Final conclusion. The substance that is liquid at ordinary temperature is .

Step 1: Recall the rule for nature of salt solutions.

Salt of strong acid + strong base → neutral solution
Salt of strong acid + weak base → acidic solution
Salt of weak acid + strong base → basic solution
Step 2: Analyze each option.

Sodium chloride (NaCl) Formed from HCl (strong acid) and NaOH (strong base) → neutral solution
Copper sulphate (CuSO ) Formed from H SO (strong acid) and Cu(OH) (weak base) → acidic solution
Ferric chloride (FeCl ) Formed from HCl (strong acid) and Fe(OH) (weak base) → acidic solution
Sodium acetate (CH COONa) Formed from CH COOH (weak acid) and NaOH (strong base) → basic solution ✔
Step 3: Final conclusion. The salt that makes the solution basic when dissolved in water is sodium acetate.