Organic Chemistry

Step 1: Identify the structure.
The given structure is CH -CH-CH with a hydroxyl group at the 2-position. This indicates it is 3-methyl butan-2-ol. Step 2: Conclusion.
Hence, the correct answer is (C) 3-methyl butan-2-ol.

Step 1: Recall the Hunsdiecker reaction.
In the Hunsdiecker reaction, silver salts of carboxylic acids (RCOOAg) react with bromine in the presence of CCl to yield alkyl bromides (RBr) with the loss of one carbon atom.

Step 2: Analyzing the options.
- (A) Hunsdiecker reaction: Correct, this is exactly the given reaction.
- (B) Schmidt reaction: Converts carboxylic acids to amines using hydrazoic acid, not applicable here.
- (C) Hell-Volhard-Zelinsky reaction: Involves halogenation at the -position of carboxylic acids, different from the given case.
- (D) Tishchenko reaction: Forms esters from aldehydes, unrelated.

Step 3: Conclusion.
Thus, the given reaction is called the Hunsdiecker reaction.

Final Answer:

(i) Kolbe Reaction:
Phenol reacts with sodium hydroxide to form sodium phenoxide, which further reacts with carbon dioxide under pressure, followed by acidification to give salicylic acid.
(ii) Reimer-Tiemann Reaction:
Phenol reacts with chloroform in the presence of sodium hydroxide to give salicylaldehyde (ortho-hydroxybenzaldehyde).
(iii) Oxidation of Phenol:
Phenol undergoes oxidation with neutral ferric chloride ( ) or chromic acid to give para-benzoquinone.
(iv) Williamson Synthesis:
An alkoxide reacts with an alkyl halide to form an ether.
Example: (v) Industrial Preparation of Methanol:
Methanol is prepared industrially by the catalytic hydrogenation of carbon monoxide.
Conclusion:
Each reaction illustrates a fundamental organic transformation—carboxylation, formylation, oxidation, ether synthesis, and catalytic hydrogenation—all central to industrial and laboratory organic chemistry.

Step 1: Definition.
Acetylation is the process of introducing an acetyl functional group ( ) into a molecule. It usually involves the replacement of a hydrogen atom with an acetyl group.
Step 2: Types of Acetylation.
1.
Chemical Acetylation: Reaction of compounds with acetic anhydride or acetyl chloride. Example: Conversion of salicylic acid to acetylsalicylic acid (aspirin).
2.
Biological Acetylation: In living organisms, enzymes like acetyltransferases transfer acetyl groups, such as acetylation of histone proteins that regulate gene expression.
Conclusion:
Acetylation is important both in organic synthesis (e.g., formation of aspirin) and in biological systems (e.g., regulation of gene activity).

Step 1: Understanding Formalin.
Formalin is a common laboratory chemical that is used as a disinfectant and preservative. It is specifically an aqueous solution of formaldehyde gas in water.

Step 2: Composition of Formalin.
Formalin typically contains about 37–40% formaldehyde dissolved in water, along with a small amount of methanol to prevent polymerization.

Step 3: Analyzing the options.
- (A) Fluorescein: This is a dye, not related to formalin.
- (B) Formaldehyde: Correct. Formalin is an aqueous solution of formaldehyde.
- (C) Formic acid: This is a different organic acid, not formalin.
- (D) Acetic acid: This is vinegar’s main component, not related to formalin.

Step 4: Conclusion.
Hence, the correct answer is that formalin is an aqueous solution of formaldehyde.

Final Answer:

Step 1: Recall the conversion of amides to amines.
Amides are converted to amines (with one carbon less) by the Hofmann bromamide degradation reaction. In this reaction, amides react with bromine and potassium hydroxide (Br /KOH) to yield primary amines.

Step 2: Analyzing the options.
- (A) HCl/ZnCl : Used for Friedel–Crafts type reactions, not for converting amides to amines.
- (B) K Cr O /H SO : This is an oxidizing mixture, not used for this conversion.
- (C) NaOH/Ca(OH) : Commonly used in soda lime decarboxylation, not for amide-to-amine conversion.
- (D) Br /KOH: Correct. This reagent is used in Hofmann bromamide reaction to prepare amines from amides.

Step 3: Conclusion.
Thus, the correct reagent to prepare amines from amides is Br /KOH.

Final Answer:

Step 1: General chemical equation.
Acid-catalysed hydration of alkenes follows Markovnikov’s rule. Example with ethene: Step 2: Mechanism.
\begin{enumerate} \item
Protonation of alkene: (formation of carbocation) \item
Nucleophilic attack by water: \item
Deprotonation: \end{enumerate} Conclusion:
The final product is an alcohol, and the reaction is catalysed by an acid (usually ).

(I) Wurtz–Fittig Reaction:
Definition: The Wurtz–Fittig reaction involves the reaction of an aryl halide with an alkyl halide in the presence of sodium metal in dry ether to give alkyl-substituted aromatic hydrocarbons.
Equation:
Conclusion: It is used to prepare alkylbenzenes (like toluene).
(II) Sandmeyer Reaction:
Definition: The Sandmeyer reaction involves the replacement of the diazonium group (-N ) in an aromatic diazonium salt with a halogen (Cl, Br, CN) using copper(I) salts as catalysts.
Equation:
Conclusion: It is an important method for the preparation of aryl halides and aryl cyanides.

Step 1: Percentage composition.
Given: Step 2: Calculate empirical formula.
Dividing all by 1.16 (smallest): So, empirical formula = . Step 3: Molecular formula.
Empirical formula mass = .
Since molecular mass = 86, the molecular formula is the same: Step 4: Functional group tests.
\begin{itemize} \item Compound does not reduce Tollen’s reagent → Not an aldehyde. \item Forms addition compound with NaHSO → Presence of carbonyl group. \item Positive iodoform test → Presence of methyl ketone group ( ). \end{itemize} Thus, compound must be a ketone with structure . Step 5: Oxidation products.
On strong oxidation, compound gives ethanoic acid and propanoic acid. This indicates the structure is pentan-2-one. Step 6: Reaction equations.
(i) NaHSO addition: (ii) Iodoform test: (iii) Oxidation: Conclusion:
The possible structure of the compound is:

Step 1: Recall the carbylamine reaction.
In the carbylamine reaction, a primary amine (R–NH ) reacts with chloroform (CHCl ) and alcoholic KOH to form isocyanides (R–NC), which have a foul smell.

Step 2: Match with the given reaction.
The given equation: fits exactly with the carbylamine test.

Step 3: Analyzing options.
- (A) Coupling reaction: Involves diazonium salts, not applicable here.
- (B) Carbylamine reaction: Correct. Forms isocyanides (R–NC).
- (C) Hoffmann bromamide reaction: Converts amides to amines, not relevant here.
- (D) Schmidt reaction: Reaction of hydrazoic acid with carboxylic acids, different case.

Step 4: Conclusion.
Therefore, the given reaction is the carbylamine reaction.

Final Answer:

(i) Williamson’s Ether Synthesis:
It is a laboratory method for the preparation of ethers. In this method, an alkoxide ion reacts with a primary alkyl halide to produce an ether.
Example: This method is widely used for the synthesis of symmetrical and unsymmetrical ethers. (ii) Fermentation:
Fermentation is a biochemical process in which sugars such as glucose and sucrose are converted into ethanol and carbon dioxide by the action of enzymes (zymase) produced by yeast.
Fermentation is the basis of alcoholic beverages and bioethanol production. (iii) To obtain methyl alcohol from pyroligneous acid:
Pyroligneous acid is obtained by the destructive distillation of wood. It contains acetic acid, acetone, and methyl alcohol as impurities.
On neutralization and distillation, methyl alcohol is separated as the main product. Thus, wood spirit (methyl alcohol) is commercially obtained from pyroligneous acid. Conclusion:
These processes highlight important laboratory and industrial methods for preparing ethers, alcohols, and biofuels.

(i) \textit{n-Butyl chloride + alcoholic KOH:}
Dehydrohalogenation occurs, producing 1-butene.
(ii) Bromobenzene + Mg in dry ether:
Grignard reagent is formed.
(iii) Ethyl chloride + aqueous KOH:
Hydrolysis takes place, giving ethyl alcohol.
(iv) Methyl bromide + Na in dry ether:
Wurtz reaction occurs, forming ethane.
(v) Ethyl bromide + KCN (alc.):
Nucleophilic substitution occurs, forming propionitrile.
Conclusion:
These reactions show elimination (E2), substitution (SN1/SN2), Grignard reagent formation, and carbon–carbon bond formation, highlighting important pathways in organic synthesis.

Step 1: Presence of five –OH groups.
When glucose is acetylated with acetic anhydride, it forms penta-acetate, showing the presence of five –OH groups. Thus, glucose contains five hydroxyl groups. Step 2: Presence of –CHO group.
Glucose reacts with hydroxylamine (NH OH) to form oxime, and with hydrogen cyanide (HCN) to form cyanohydrin. These reactions prove the presence of the –CHO group in glucose. Step 3: Silver mirror test.
When glucose is warmed with Tollen’s reagent (ammoniacal AgNO ), the –CHO group is oxidised to –COOH, and metallic silver is deposited on the walls of the test tube as a shining silver mirror. Conclusion:
Glucose contains five alcoholic –OH groups and one aldehydic –CHO group. The aldehyde group is confirmed by Tollen’s reagent test, giving a silver mirror.

(A) Structure Analysis:

• Longest Chain: 6 Carbons containing the ketone group.
• Numbering: Start from the left to give the Ketone (functional group) the lowest number.
  • C1:
  • C2:
  • C3: (Ketone)
  • C4:
  • C5: (with a methyl substituent)
  • C6:
• Substituent: Methyl group at position 5.
• IUPAC Name: 5-Methylhexan-3-one

(B) Structure Analysis:

• Principal Group: Aldehyde ( ). Carbon of CHO is always C1.
• Chain: Propene chain.
• C1:
• C2:
• C3: (attached to Phenyl ring)
• Substituent: Phenyl group ( ) at C3.
• Unsaturation: Double bond starts at C2.
• IUPAC Name: 3-Phenylprop-2-enal (Also commonly known as Cinnamaldehyde).