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Hinsberg’s reagent is benzenesulfonyl chloride. Its formula is:
Hence, the correct answer is:

Aniline is strongly activating and highly reactive. During direct nitration, the oxidizing acidic medium can cause oxidation and side reactions, producing tarry products. Therefore, nitration of aniline gives tarry oxidation products.
Hence, the correct answer is:

Among the given carbohydrates:

• sucrose, maltose, and lactose are soluble sugars,
• amylose is a component of starch,
• amylopectin is a branched polysaccharide and is insoluble in water.

Hence, the correct answer is:

The one-letter codes are: So the correct pair is:
Hence, the correct answer is:

HVZ stands for Hell-Volhard-Zelinsky reaction. In this reaction, a carboxylic acid having an -hydrogen is converted into an -halogen substituted acid. So HVZ reaction involves:
Hence, the correct answer is:

The actual matching table is not fully visible in the parsed text, but the official answer key marks option (D) as correct.
Hence, the correct answer is:

The structure image itself is not visible in the parsed text, but the official answer key marks option (D) as correct. So the IUPAC name is:
Hence, the correct answer is:

The reaction scheme image/details are not fully visible in the parsed text, but the official answer key for this paper marks option (E) as correct. So the products are:
Hence, the correct answer is:

In aqueous medium, aliphatic amines are generally more basic than aromatic amines because the lone pair on nitrogen in aromatic amines is partly delocalized into the benzene ring. Among the options, N-ethylethanamine is a secondary aliphatic amine and is strongly basic.
Hence, the correct answer is:

Dodecane is: On catalytic cracking, a long-chain alkane breaks into a smaller alkane and a smaller alkene. Check option (C): Adding them: So this satisfies both carbon and hydrogen balance and matches the expected cracking pattern.
Hence, the correct answer is:

2-chloro-1-phenyl butane undergoes dehydrohalogenation with alcoholic KOH or EtOK/EtOH. The major alkene formed is the more stable alkene: Now HBr adds to this alkene according to Markovnikov’s rule. The proton adds in such a way that the more stable carbocation is formed, which is the benzylic carbocation. Then bromide attacks that carbon. So the major product formed is:
Hence, the correct answer is:

Organohalogen compounds are generally more polar than the corresponding hydrocarbons because of the polar carbon-halogen bond. They also have higher molecular mass, which increases van der Waals forces. So their boiling points are higher mainly because of:

• dipole-dipole interactions
• van der Waals forces

Hence, the correct answer is:

Kolbe’s method, also called Kolbe electrolysis, involves the electrolysis of aqueous solutions of sodium or potassium salts of carboxylic acids. In this reaction, the carboxylate ion loses carbon dioxide at the anode and forms an alkyl radical, which then combines to form an alkane. So the defining feature of Kolbe’s method is:
Hence, the correct answer is:

In Lassaigne’s test, sulfur in sodium fusion extract forms sulfide ion. When treated with sodium nitroprusside, the sulfide ion gives a violet or blood-red coloured complex: Therefore, the coloured species formed is:
Hence, the correct answer is:

Nitrogen estimation in organic compounds is done by:

• Dumas method
• Kjeldahl method

Carius method is used for halogens and sulfur estimation. Therefore, the correct pair is:
Hence, the correct answer is:

Step 1: Understanding threshold energy.
Threshold energy refers to the minimum energy required to eject an electron from an atom during the photoelectric effect.
Step 2: Analyze the elements.
Among the elements K (Potassium), Na (Sodium), Mg (Magnesium), and Li (Lithium), the threshold energy is highest for the element with the smallest atomic size and highest ionization energy.
Step 3: Ionization energy trend.
Ionization energy generally increases across a period (left to right) and decreases down a group (top to bottom) in the periodic table.
Step 4: Determine the trend.
In the given elements:
- Li (Lithium) has the highest ionization energy (and therefore the highest threshold energy).
- K (Potassium) has the lowest ionization energy (and therefore the lowest threshold energy).

Step 5: Conclusion.
Therefore, Li has the maximum threshold energy, and K has the minimum threshold energy.

Concept:
Many organic compounds have specific industrial uses.
Step 1: Check correct pairs.
- Esters of benzoic acid are used in perfumery. - Ethanoic acid is used in food industry (vinegar, preservatives). - Methanoic acid is used in electroplating. - Sodium benzoate is a food preservative.
Step 2: Check incorrect pair.
Hexanedioic acid (adipic acid) is mainly used in manufacture of nylon-6,6, not detergents.
Step 3: Final answer.

Hence, correct option is (B).

Concept:
Lower means stronger base. In aqueous medium, basic strength of aliphatic amines depends on +I effect and solvation.
Step 1: Compare types of amines.
- Primary amines: methanamine, ethanamine - Secondary amine: N-ethylethanamine - Other substituted amines
Step 2: Aqueous phase trend.
In water, secondary amines are usually strongest because: - Two alkyl groups increase electron density on nitrogen. - Solvation is still effective. General trend:
Step 3: Choose strongest base.
N-ethylethanamine is a secondary amine. Hence it has lowest .
Step 4: Final answer.

Hence, correct option is (D).

Concept:
Organic reactions are identified from standard reagents and their known actions.
Step 1: Match oxidation of secondary alcohol.
Secondary alcohols are oxidized to ketones by chromic anhydride.
Step 2: Match dehydration reaction.
Secondary alcohols give alkenes on dehydration using phosphoric acid at high temperature.
Step 3: Match reduction of ketone.
Ketones are reduced to secondary alcohols by sodium borohydride.
Step 4: Match oxidation of phenol.
Phenol is oxidized to benzoquinone using acidified sodium dichromate.
Step 5: Final matching.

Hence, correct option is (B).

Concept:
Nucleotides join together in DNA and RNA chains through phosphate linkages.
Step 1: Understand bonding positions.
The phosphate group connects: - carbon of one sugar - carbon of next sugar
Step 2: Name of linkage.
Since one phosphate forms two ester bonds with two sugars, it is called:
Step 3: Final answer.

Hence, correct option is (E).

Concept:
Alkyl halides react with to form nitriles, and nitriles on catalytic hydrogenation give primary amines.
Step 1: Use final molecular formula.
Given final product: This corresponds to phenethylamine: which is 2-phenyl ethanamine.
Step 2: Find nitrile precursor B.
Reduction of nitrile: So precursor must be: This is phenylethane nitrile.
Step 3: Find starting compound A.
Nitrile is formed from corresponding benzyl halide using ethanolic : Thus A is chlorophenyl methane (benzyl chloride).
Step 4: Final matching.

Hence, correct option is (D).

Concept:
Common uses of haloalkanes and haloarenes.
Step 1: Match each compound with known use.
Methylene chloride is used in cleaning and degreasing metals. Chloroform is used as a solvent for iodine. Iodoform is used as an antiseptic. Carbon tetrachloride was used in manufacture of propellents and refrigerants.
Step 2: Final matching.

Hence, correct option is (A).

Concept:
Ethers react with HI to undergo cleavage of the C-O bond.
Step 1: Given ether.
This is benzyl phenyl ether.
Step 2: Site of cleavage.
The bond breaks at alkyl side (benzyl carbon), because aryl-O bond is resistant to cleavage.
Step 3: Products formed.
Benzyl part gives benzyl iodide: Phenoxy part gives phenol after protonation:
Step 4: Final answer.

Hence, correct option is (E).