Organic Chemistry

Step 1: Type of reaction.
This is an example of a Friedel–Crafts alkylation reaction. It involves the introduction of an alkyl group into an aromatic ring using an alkyl halide and a Lewis acid catalyst like anhydrous AlCl .
Step 2: Reaction.
Step 3: Product formed.
The main product is ethylbenzene.
Step 4: Mechanism overview.
- AlCl reacts with ethyl bromide to generate the ethyl carbocation (C H ).
- This carbocation then attacks the benzene ring, forming ethylbenzene after deprotonation.
Step 5: Importance.
This reaction is an important synthetic route for alkyl-substituted aromatic compounds.

Step 1: Definition.
A vinyl halide is a compound in which the halogen atom is directly attached to a carbon atom of a double bond (C=C).
Step 2: Analysis of options.
(A) CH =CH–CHBrCH : Here Br is attached to an alkyl group, not directly to the double bond. Wrong.
(B) CH =CH–Br: Correct, because Br is directly bonded to the carbon of the C=C double bond.
(C) CH =CH–CH CH Cl: This is an allyl halide, not vinyl.
(D) HC≡C–Br: This is a haloalkyne, not vinyl halide.
Step 3: Conclusion.
The correct answer is (B) CH =CH–Br.

Step 1: Solubility rule.
Polar compounds with hydrogen bonding are highly soluble in water. Non-polar compounds are insoluble.
Step 2: Analyze options.
(A) CH OH: Polar, forms H-bonds → soluble.
(B) CHCl : Weak polarity, limited solubility.
(C) CCl : Non-polar, insoluble.
(D) CS : Non-polar, insoluble.
Step 3: Conclusion.
The correct answer is (A) CH OH.

Step 1: Understanding ether.
An ether is an organic compound in which an oxygen atom is bonded to two alkyl or aryl groups (R–O–R’).
Step 2: Analysis of options.
(A) R–C=O: Represents carbonyl group (aldehydes or ketones), not ether.
(B) R–C=O: Same as above, carbonyl compound.
(C) R–O–R: Correct, this is the general structure of ethers.
(D) HO–C=O: Represents carboxylic acid group, not ether.
Step 3: Conclusion.
The correct answer is (C) R–O–R.

Step 1: Definition of Carbyl Amine Reaction.
The carbyl amine reaction is a chemical reaction where a primary amine reacts with an aldehyde or ketone, resulting in the formation of an isocyanide (carbyl amine). This reaction is used to test the presence of primary amines.
Step 2: Reaction Example.
For example, when phenylamine (aniline) reacts with chloroform (CHCl₃) and a strong base (like NaOH), an isocyanide (carbyl amine) is formed, and the reaction is characterized by a strong, foul odor.

Step 1: Understanding the structure of the compound.
The given compound is , where an amine group is attached to a carbon chain. Step 2: Analysis of options.
(A) Primary amine: A primary amine is an amine where the nitrogen atom is attached to one carbon atom. In this case, the nitrogen is attached to a single carbon chain, making it a primary amine.
(B) Secondary amine: A secondary amine has two alkyl groups attached to the nitrogen atom. This is not the case here.
(C) Tertiary amine: A tertiary amine has three alkyl groups attached to the nitrogen atom. This is not applicable here.
(D) Quaternary salt: A quaternary salt involves a positively charged nitrogen atom attached to four groups, which is not the case here.
Step 3: Conclusion.
The correct answer is (A) because the given compound is a primary amine.

Step 1: Understanding the structure of Butan-2-ol.
Butan-2-ol has the structure , with the hydroxyl group attached to the second carbon of the butane chain. Since the hydroxyl group is attached to a secondary carbon (a carbon that is bonded to two other carbons), it is classified as a secondary alcohol. Step 2: Analysis of options.
(A) Primary alcohol: This is incorrect because the hydroxyl group is attached to a secondary carbon, not a primary carbon.
(B) Secondary alcohol: Correct — The hydroxyl group is attached to a secondary carbon, making this a secondary alcohol.
(C) Tertiary alcohol: This is incorrect because the hydroxyl group is not attached to a tertiary carbon.
(D) Dihydric alcohol: This is incorrect because Butan-2-ol is a monohydric alcohol, containing only one hydroxyl group.
Step 3: Conclusion.
The correct answer is (B) because Butan-2-ol is a secondary alcohol, with the hydroxyl group attached to a secondary carbon.

Step 1: Aldol Condensation.
Aldol condensation is a reaction in which an aldehyde or ketone reacts with another aldehyde or ketone to form a β-hydroxy aldehyde or ketone, which then undergoes dehydration to form an α,β-unsaturated carbonyl compound. This reaction is catalyzed by a base or an acid.
Example of Aldol Condensation:
Consider the reaction between acetaldehyde molecules: In this reaction, acetaldehyde (CH₃CHO) undergoes aldol condensation, leading to the formation of crotonaldehyde (CH₃CH=CHCHO).
Step 2: Cannizzaro's Reaction.
The Cannizzaro reaction is a type of disproportionation reaction in which non-enolizable aldehydes (those without an α-hydrogen) undergo a base-catalyzed redox reaction to produce an alcohol and a carboxylate ion. The aldehyde is simultaneously reduced and oxidized.
Example of Cannizzaro's Reaction:
Consider the reaction between two molecules of benzaldehyde: In this reaction, one molecule of benzaldehyde is reduced to benzyl alcohol (C₆H₅CH₂OH), and the other is oxidized to benzoate ion (C₆H₅COO⁻).
Step 3: Conclusion.
- Aldol condensation involves the formation of a β-hydroxy aldehyde or ketone followed by dehydration to yield an α,β-unsaturated compound.
- Cannizzaro’s reaction is a disproportionation reaction where non-enolizable aldehydes undergo reduction and oxidation in the presence of a base.

Step 1: Identifying and naming the compounds.
- (i) CH₃-CH₂-COOH: This is propanoic acid (also known as propionic acid), a simple carboxylic acid with a three-carbon chain. The IUPAC name is propanoic acid.
- (ii) CH₂=COOH: This compound is an α,β-unsaturated carboxylic acid, commonly known as acrylic acid. The IUPAC name is prop-2-enoic acid.
- (iii) ClCH₂-COOH: This compound is a chlorinated carboxylic acid, commonly known as chloroacetic acid. The IUPAC name is chloroethanoic acid.
- (iv) CH₃-CH=CH-COOH: This compound is a conjugated diene with a carboxyl group. The IUPAC name is 2-propenoic acid.
- (v) CH₃-CO-CH₂-COOH: This compound is a diketone derivative of acetic acid, also known as acetoacetic acid. The IUPAC name is 3-oxobutanoic acid.
Step 2: Conclusion.
- (i) propanoic acid
- (ii) prop-2-enoic acid
- (iii) chloroethanoic acid
- (iv) 2-propenoic acid
- (v) 3-oxobutanoic acid

Step 1: Understanding alcohol classification.
Alcohols are classified based on the number of alkyl groups attached to the carbon that holds the hydroxyl group (-OH):
- Primary alcohols have the hydroxyl group attached to a carbon atom that is bonded to only one alkyl group.
- Secondary alcohols have the hydroxyl group attached to a carbon atom that is bonded to two alkyl groups.
- Tertiary alcohols have the hydroxyl group attached to a carbon atom that is bonded to three alkyl groups.
Step 2: Distinguishing tests.
1. Oxidation test:
- Primary alcohols can be oxidized to aldehydes and then to carboxylic acids (e.g., using potassium dichromate).
- Secondary alcohols can be oxidized to ketones but cannot be further oxidized to carboxylic acids.
- Tertiary alcohols do not undergo oxidation easily because the carbon bearing the -OH group is attached to three other carbon atoms, making it sterically hindered.
2. Reaction with Lucas reagent (a mixture of concentrated HCl and ZnCl₂):
- Primary alcohols react slowly with Lucas reagent.
- Secondary alcohols react moderately with Lucas reagent.
- Tertiary alcohols react rapidly with Lucas reagent, leading to the formation of an alkyl chloride.
3. Color reaction with ceric ammonium nitrate (Ce(NH₄)₂(NO₃)₆):
- Primary alcohols give a yellow to red color change when tested with ceric ammonium nitrate.
- Secondary alcohols show a weaker color change compared to primary alcohols.
- Tertiary alcohols do not give any color change.
Step 3: Conclusion.
Primary, secondary, and tertiary alcohols can be distinguished by their reaction with oxidation agents, Lucas reagent, and ceric ammonium nitrate. The number of alkyl groups attached to the carbon with the -OH group determines their classification.

Step 1: Understanding the molecular formula.
The general formula for saturated monocarboxylic acids is , where represents the number of carbon atoms. This formula represents a chain of carbon atoms with a single carboxyl group (-COOH) attached. Step 2: Analysis of options.
(A) : This is the formula for saturated alkanes, not carboxylic acids.
(B) : This is the formula for alcohols, not carboxylic acids.
(C) : Correct — This is the correct formula for saturated monocarboxylic acids, where the carboxyl group (-COOH) is attached to a carbon chain.
(D) : This is similar to option (C) and also represents a general formula for carboxylic acids.
Step 3: Conclusion.
The correct answer is (C) because the general molecular formula for saturated monocarboxylic acids is .

Step 1: Understanding the reactions.
Formic acid and formaldehyde can be distinguished using Tollen's reagent. Tollen's reagent reacts with aldehydes to form a silver mirror, but does not react with formic acid, which is a carboxylic acid. Step 2: Analysis of options.
(A) Benedict solution: Benedict's solution is used to detect reducing sugars, not for distinguishing between formic acid and formaldehyde.
(B) Fehling solution: Fehling's solution is used for reducing sugars and aldehydes, but Tollen's reagent is more specific for distinguishing aldehydes like formaldehyde.
(C) Tollen's reagent: Correct — Tollen's reagent reacts with aldehydes like formaldehyde, forming a silver mirror, but does not react with formic acid.
(D) Sodium bicarbonate: Sodium bicarbonate is used to test for acids, but it does not specifically distinguish between formic acid and formaldehyde.
Step 3: Conclusion.
The correct answer is (C) because Tollen's reagent is used to distinguish between formic acid and formaldehyde.

Step 1: Understanding acetamide.
Acetamide is an amide with both acidic and basic properties. It can act as a weak acid due to the presence of the carbonyl group, and as a base due to the nitrogen atom in the amide group. Step 2: Analysis of options.
(A) Acidic: Acetamide is not strongly acidic.
(B) Alkaline: Acetamide is not strongly alkaline.
(C) Amphoteric: This is correct because acetamide has both acidic and basic properties.
(D) Neutral: Acetamide is not neutral; it exhibits amphoteric behavior.
Step 3: Conclusion.
The correct answer is (C) because acetamide is amphoteric, showing both acidic and basic characteristics.

Step 1: Understanding basicity of amines.
The basicity of amines depends on the electron-donating effect of the alkyl group attached to the nitrogen. Alkyl groups increase the electron density on nitrogen, making the amine more basic. Step 2: Analysis of options.
(A) : Aniline (C₆H₅NH₂) has a phenyl group, which is electron-withdrawing, thus making it less basic than alkyl amines.
(B) Another form of : Same reasoning as (A), still less basic.
(C) : Ethylamine (C₂H₅NH₂) is more basic because the ethyl group is electron-donating.
(D) Another form of : Same reasoning as (C), still more basic.
Step 3: Conclusion.
The correct answer is (C) because (ethylamine) is the most basic due to the electron-donating effect of the ethyl group.

Step 1: Understanding acyl chloride formation.
Acyl chlorides are typically formed by reacting carboxylic acids with chlorinating agents like , , and . These reagents react with acetic acid to form acyl chloride. Step 2: Analysis of options.
(A) : Phosphorus pentachloride reacts with acetic acid to form acyl chloride.
(B) : Phosphorus trichloride also reacts with acetic acid to form acyl chloride.
(C) : Thionyl chloride reacts with acetic acid to form acyl chloride.
(D) : Chlorine gas does not react with acetic acid to form acyl chloride.
Step 3: Conclusion.
The correct answer is (D) because chlorine gas does not react with acetic acid to form acyl chloride.

Step 1: Understand the structure of the compound.
The given compound is CH COOC H , which is an ester formed by the reaction of acetic acid (CH COOH) and ethanol (C H OH).
Step 2: Analyze the options.
- Methyl propanoate (Option A) is incorrect because the ester is formed from acetic acid and ethanol. - Ethyl ethanoate (Option B) is the correct IUPAC name for CH COOC H , where "ethyl" comes from the ethanol part and "ethanoate" refers to the acetic acid part. - Acetotheane (Option C) is not a valid name. - Ethoxyethane (Option D) refers to an ether, not an ester.
Step 3: Apply to the options.
The correct IUPAC name is (B) Ethyl ethanoate.
Final Answer:

Step 1: Understand the structure of the compound.
The given compound is CH -C-NH , where an amine group (NH ) is attached to a carbonyl group (C). This is a simple primary amine, as there is only one alkyl group (CH ) attached to the nitrogen atom.
Step 2: Analyze the options.
- Primary amine (Option A) is correct because the amine group is attached to one carbon atom. - Secondary amine (Option B) would have two carbon atoms attached to the nitrogen. - Tertiary amine (Option C) would have three carbon atoms attached to the nitrogen. - Ammonium salt (Option D) would have a positively charged nitrogen atom, which is not the case here.
Step 3: Apply to the options.
The correct answer is (A) Primary amine.
Final Answer:

Step 1: Understand the Carbymine reaction.
The Carbymine reaction is a reaction of primary amines with chloroform (CHCl ) in the presence of alcoholic KOH to form an isocyanide (carbamine).
Step 2: Analyze the options.
- Hydrolysis is the reaction of a compound with water to break it down. It does not involve chloroform. - Reduction refers to the gain of electrons or decrease in oxidation state, which is not applicable here. - Wurtz reaction involves the coupling of two alkyl halides in the presence of sodium metal and is unrelated to amines or chloroform. - Carbymine reaction is the correct reaction, as it involves the formation of an isocyanide when a primary amine reacts with chloroform and KOH.
Step 3: Apply to the options.
The correct answer is (D) Carbymine reaction.
Final Answer:

Step 1: Understand dipole moment.
The dipole moment is a measure of the separation of charge in a molecule. Molecules with symmetrical structures where the dipoles cancel out have zero dipole moment.
Step 2: Analyze the options.
- CH Cl, CH Cl , and CHCl all have asymmetrical structures, so their dipoles do not cancel out, and they have non-zero dipole moments. - CCl has a symmetrical tetrahedral structure, and the dipoles of the four C-Cl bonds cancel each other out, resulting in a zero dipole moment.
Step 3: Apply to the options.
The compound with zero dipole moment is (B) CCl .
Final Answer: