UPCATET SERIES

Chemistry

Organic Chemistry

14 previous year questions.

Volume: 14 Ques

Yield: Medium

High-Yield Trend

2025

2024

Chapter Questions
14 MCQs

PYQ 2024

medium

chemistry ID: upcatet-

Which of the following formula represents Alkene?

PYQ 2025

easy

chemistry ID: upcatet-

The compound which does not react with ammonical silver nitrate solution is:

2-methyl propanal

2-methyl cyclopentanone

3-methyl benzaldehyde

Methanoic acid

03

PYQ 2025

medium

chemistry ID: upcatet-

$The amine which is most basic in gas phase is:$

1

2

3

4

Official Solution

Correct Option: $(4)$

$The basicity of amines in the gas phase depends on their ability to donate a lone pair of electrons to a proton, forming . Unlike in aqueous solution, gas-phase basicity is not influenced by solvation effects. Step 1: Understand gas-phase basicity Alkyl groups () are electron-donating due to their inductive effect, increasing electron density on the nitrogen atom. More alkyl groups enhance the amine’s ability to stabilize the protonated form (), increasing basicity. The order of basicity in the gas phase is: Step 2: Analyze options - (A) : Primary amine, one alkyl group, less electron donation. - (B) : No alkyl groups, least basic. - (C) : Secondary amine, two alkyl groups, more basic than primary. - (D) : Tertiary amine, three alkyl groups, most basic due to maximum electron donation. Step 3: Conclusion In the gas phase, (tertiary amine) is the most basic, matching option (D).$

04

PYQ 2025

medium

chemistry ID: upcatet-

$Which of the following amines will give most easily gas on treatment with at 275 K:$

1

2

3

4

Official Solution

Correct Option: $(3)$

$The reaction of amines with and at low temperatures (e.g., 275 K, or 2°C) forms diazonium salts or other products, depending on the amine type. The evolution of gas is characteristic of diazonium salt formation and subsequent decomposition. Step 1: Reaction with - Primary aliphatic amines (): Form diazonium ions (), which are unstable and decompose to alcohols, releasing : - Primary aromatic amines (): Form stable diazonium salts () at low temperatures, which do not immediately release unless heated or reacted further: - Secondary amines (): Form N-nitrosoamines, no evolution: - Tertiary amines (): Aliphatic tertiary amines form ammonium salts or nitrosamines; aromatic tertiary amines may form p-nitroso derivatives, no . Step 2: Analyze options - (A) : Primary aliphatic amine, forms unstable diazonium ion, releases to form ethanol. - (B) : Secondary amine, forms N-nitrosomethylaniline, no . - (C) : Primary aromatic amine, forms stable benzenediazonium chloride at 275 K, which does not release immediately but can decompose with heat or reagents (e.g.,), releasing . - (D) : Tertiary aliphatic amine, forms trimethylammonium chloride or nitrosamine, no . Step 3: Interpret “most easily gas” The question likely assumes conditions where diazonium salts decompose to release . Aliphatic diazonium ions (from) are highly unstable and release readily at 275 K. Aromatic diazonium salts (from) are stable at 275 K but decompose under additional conditions (e.g., warming). Since aromatic diazonium salts are commonly tested for evolution in subsequent reactions, is often the intended answer in such questions. Step 4: Conclusion Assuming standard organic chemistry context, forms a diazonium salt most readily associated with evolution in follow-up reactions, matching option (C).$

05

PYQ 2025

medium

chemistry ID: upcatet-

$Electrophile used in Kolbe’s reaction is:$

1

$Carbondioxide$

2

$Dichlorocarbene$

3

$Carbommonoxide$

4

$Acyl carbocation$

Official Solution

Correct Option: $(1)$

$The Kolbe-Schmitt reaction involves the reaction of a phenoxide ion with carbon dioxide to form salicylic acid, a key step often confused with Kolbe’s electrolysis in some curricula. In this context, we identify the electrophile used. Step 1: Understand Kolbe’s reaction Kolbe’s electrolysis produces alkanes from carboxylate salts, but the Kolbe-Schmitt reaction is relevant here, where sodium phenoxide reacts with under heat and pressure to form salicylic acid. The electrophilic carbon in is attacked by the nucleophilic phenoxide ion. Step 2: Identify the electrophile - (A) Carbondioxide : has an electron-deficient carbon, acting as an electrophile in the Kolbe-Schmitt reaction, correct. - (B) Dichlorocarbene : Used in Reimer-Tiemann reaction, not Kolbe’s. - (C) Carbommonoxide : Not involved in Kolbe’s reactions. - (D) Acyl carbocation : Relevant to Friedel-Crafts acylation, not Kolbe’s. Step 3: Conclusion The electrophile in the Kolbe-Schmitt reaction is carbon dioxide, matching option (A).$

06

PYQ 2025

medium

chemistry ID: upcatet-

$Schiff’s base is formed by the reaction of aldehyde with:$

1

$Ammonia$

2

$Amines$

3

$Hydrazine$

4

$Phenyl hydrazine$

Official Solution

Correct Option: $(2)$

$A Schiff’s base, or imine, is characterized by a carbon-nitrogen double bond () formed through a condensation reaction. We need to identify the correct reagent that reacts with an aldehyde to produce this compound. Step 1: Reaction mechanism An aldehyde () reacts with a primary amine () to form a Schiff’s base: The amine’s nitrogen attacks the electrophilic carbonyl carbon, followed by dehydration to form the imine. Step 2: Analyze options - (A) Ammonia : Forms unstable intermediates like aldehyde ammonias, not Schiff’s bases. - (B) Amines : Primary amines form Schiff’s bases, correct. - (C) Hydrazine : Produces hydrazones (), a different compound. - (D) Phenyl hydrazine : Forms phenylhydrazones (), not Schiff’s bases. Step 3: Conclusion Schiff’s base is formed with primary amines, matching option (B).$

07

PYQ 2025

medium

chemistry ID: upcatet-

$Denaturation of alcohol is done by the help of:$

1

$+ Pyridine$

2

3

4

Official Solution

Correct Option: $(1)$

$Step 1: Understand denaturation Denaturation makes alcohol unfit for drinking by adding toxic or unpleasant substances while maintaining its industrial usefulness. Step 2: Standard denaturing agents The most common denaturing mixture is: This combination: Gives a blue color to alcohol Makes it extremely bitter Is difficult to separate Step 3: Why other options are incorrect : Too corrosive, destroys alcohol : Not an effective denaturant : Reacts with alcohol, forms byproducts Step 4: Mechanism of action The +Pyridine combination: Forms complex compounds with ethanol Creates persistent bad taste Maintains solvent properties for industrial use$

08

PYQ 2025

medium

chemistry ID: upcatet-

$When benzene diazonium chloride is treated with ethanol, the product obtained is:$

1

$Aniline$

2

$Phenetole$

3

$Benzene$

4

$p-Chloro phenol$

Official Solution

Correct Option: $(3)$

$Benzene diazonium chloride () is a reactive compound used in various transformations. Its reaction with ethanol involves a reduction process. Step 1: Reaction mechanism Ethanol () acts as a mild reducing agent, converting the diazonium group () to nitrogen gas and forming benzene: Ethanol is oxidized to acetaldehyde (), and the diazonium ion releases, yielding benzene (). Step 2: Analyze options - (A) Aniline () : Formed by strong reduction (e.g.,), not ethanol. - (B) Phenetole () : Requires phenol and ethanol under different conditions. - (C) Benzene () : Correct, as ethanol reduces the diazonium ion. - (D) p-Chloro phenol : Involves chlorination and phenolic conditions, not relevant. Step 3: Conclusion The product is benzene, matching option (C).$

09

PYQ 2025

medium

chemistry ID: upcatet-

$How many structural isomers does have:$

1

$66$

2

$7$

3

$8$

4

$9$

Official Solution

Correct Option: $(4)$

$Structural isomers are compounds with the same molecular formula but different connectivity of atoms. For, we need to find all possible isomers of heptane (an alkane,). Step 1: Understand alkane isomers Heptane () is a straight-chain alkane with 7 carbons, but branching creates isomers. The number of isomers increases with carbon count due to possible branching patterns. Step 2: List isomers systematically - n-Heptane: Straight chain (). - 2-Methylhexane: 6-carbon chain with a methyl group on carbon 2. - 3-Methylhexane: 6-carbon chain with a methyl group on carbon 3. - 2,2-Dimethylpentane: 5-carbon chain with two methyl groups on carbon 2. - 2,3-Dimethylpentane: 5-carbon chain with methyl groups on carbons 2 and 3. - 2,4-Dimethylpentane: 5-carbon chain with methyl groups on carbons 2 and 4. - 3,3-Dimethylpentane: 5-carbon chain with two methyl groups on carbon 3. - 3-Ethylpentane: 5-carbon chain with an ethyl group on carbon 3. - 2,2,3-Trimethylbutane: 4-carbon chain with three methyl groups (two on carbon 2, one on carbon 3). Total: 9 isomers. Step 3: Verify The number of structural isomers for is well-known to be 9, matching option (D).$

10

PYQ 2025

medium

chemistry ID: upcatet-

$Which of the following aldehydes does not form iodoform on heating with and alkali?$

1

2

3

4

Official Solution

Correct Option: $(3)$

$Step 1: Understand iodoform test The iodoform test identifies compounds containing: groups that can be oxidized to these structures. Step 2: Analyze each option (a) (methyl ketone) - Positive (has) (b) - Can form through halogenation (c) (propanal) - Negative (lacks) (d) - Can oxidize to Step 3: Reaction mechanism The test requires: Step 4: Conclusion Only propanal (option c) lacks the necessary group or oxidizable structure to give iodoform.$

11

PYQ 2025

medium

chemistry ID: upcatet-

$The group which shows and effect is:$

1

2

3

4

Official Solution

Correct Option: $(3)$

$Substituents on aromatic rings affect electron density via: - Inductive effect (): Electron withdrawal () or donation () through sigma bonds. - Resonance effect (): Electron withdrawal () or donation () through pi bonds. We need a group with both (electron-withdrawing via induction) and (electron-withdrawing via resonance). Step 1: Analyze each group - (A) : Chlorine is electronegative, causing a effect. It has lone pairs that can donate electrons via resonance () in aromatic systems, so not . - (B) : Alkoxy groups (e.g.,) are due to oxygen’s electronegativity but due to lone pair donation via resonance. - (C) : The nitro group is due to electronegative oxygens pulling electrons through sigma bonds. It is also because it withdraws electrons via resonance (e.g., resonance structures place positive charge on the aromatic ring). This matches both criteria. - (D) : The amide group has a nitrogen lone pair, leading to (electron donation via resonance). It may have a weak effect due to the electronegative oxygen in, but the resonance effect is dominantly . Step 2: Conclusion Only shows both and effects.$

12

PYQ 2025

medium

chemistry ID: upcatet-

$Statement (A): Alkyl halides are not soluble in water. Statement (B): Alkyl halides do not form H-bonds with water molecules through alkyl halide, but they are polar in nature. In light of the above statements, choose the most appropriate answer from the options given below:$

1

$Statement (A) is correct, B is incorrect$

2

$Statement (A) is incorrect, B is correct$

3

$Both are incorrect$

4

$Both statements are correct$

Official Solution

Correct Option: $(4)$

$Analysis of Statement (A): Alkyl halides () are generally not soluble in water because: They cannot form hydrogen bonds with water The energy required to break water's hydrogen bonds is not compensated by new interactions Analysis of Statement (B): Alkyl halides are polar due to the electronegativity difference between carbon and halogen, but: They cannot form hydrogen bonds with water as they lack or bonds The halogen (X) has lone pairs but cannot donate them effectively for H-bonding Key Observations: 1. Polarity order: (fluorides have some water solubility) 2. Solubility decreases with increasing alkyl chain length 3. Both statements correctly describe different aspects of alkyl halide properties Conclusion: Both statements (A) and (B) are factually correct and consistent with each other.$

13

PYQ 2025

medium

chemistry ID: upcatet-

$The given reaction is known as:$

1

$Rosenmund reaction$

2

$Stephen reaction$

3

$Etard reaction$

4

$Gatterman-Koch reaction$

Official Solution

Correct Option: $(2)$

$Step 1: Identify the reaction type The given reaction shows the conversion of a nitrile () to an aldimine () using followed by hydrolysis. This is characteristic of the Stephen reaction. Step 2: Understand the Stephen reaction mechanism The Stephen reaction involves: 1. Reduction of nitrile to imine using stannous chloride () in acidic medium 2. Hydrolysis of imine to aldehyde (though shown here as stopping at imine stage) Step 3: Compare with other options (a) Rosenmund reaction - Reduction of acyl chloride to aldehyde (c) Etard reaction - Partial oxidation of methyl group to aldehyde (d) Gatterman-Koch reaction - Formylation of aromatic compounds Step 4: Final confirmation The use of for nitrile reduction is unique to the Stephen reaction, making option (b) correct.$

14

PYQ 2025

medium

chemistry ID: upcatet-

$Given below are two statements: Statement I: Propene on treatment with diborane gives an addition product with the formula Statement II: Oxidation of with Hydrogen peroxide in the presence of NaOH gives propan-2-ol In the light of the above statements, choose the most appropriate answer from the options given below:$

1

$Statement I is correct but Statement II is incorrect$

2

$Statement I is incorrect but Statement II is correct$

3

$Both Statement I and Statement II are correct$

4

$Both Statement I and Statement II are incorrect$

Official Solution

Correct Option: $(3)$

$Step 1: Analyze Statement I The hydroboration of propene with diborane follows the reaction: This forms triisopropylborane, confirming Statement I is correct . Step 2: Analyze Statement II The oxidation of triisopropylborane with follows: This yields propan-2-ol, confirming Statement II is correct . Step 3: Verify both statements - Hydroboration-oxidation converts alkenes to alcohols via anti-Markovnikov addition - The mechanism involves: Formation of alkylborane (Statement I) Oxidation to alcohol (Statement II) Step 4: Conclusion Both statements accurately describe the hydroboration-oxidation process of propene.$

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Organic Chemistry

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Chapter Questions
14 MCQs