Write the chemical reactions for the following: I. Chlorobenzene is heated with fuming H SO II. Ethyl bromide is heated with silver acetate
Official Solution
Correct Option: (1)
1. Chlorobenzene is heated with fuming H SO : When chlorobenzene is heated with fuming sulfuric acid (H SO ), a sulfonation reaction takes place, where a sulfonic group (-SO H) is introduced into the benzene ring at the position of the chlorine atom. The reaction is as follows: In this reaction, chlorobenzene reacts with fuming H SO to form 2-chlorobenzenesulfonic acid (C H ClSO ). 2. Ethyl bromide is heated with silver acetate: When ethyl bromide (C H Br) is heated with silver acetate (AgOAc), an esterification reaction occurs, leading to the formation of ethyl acetate (C H OAc) and the release of silver bromide (AgBr). The reaction is as follows: In this reaction, ethyl bromide reacts with silver acetate to form ethyl acetate and silver bromide.
02
PYQ 2023
medium
chemistryID: maharash
What is the action of the following reagents on ethylamine? I. Chloroform and caustic potash II. Nitrous acid
Official Solution
Correct Option: (1)
1. Chloroform and Caustic Potash on Ethylamine: When ethylamine ( ) reacts with chloroform ( ) in the presence of caustic potash (KOH), a reaction occurs forming an isocyanide (also known as an isonitrile). The reaction is: The product is ethyl isocyanide ( ). 2. Nitrous Acid on Ethylamine: When ethylamine reacts with nitrous acid ( ), it undergoes diazotization to form a diazonium salt. The reaction is: The product is ethyl diazonium chloride.
03
PYQ 2023
medium
chemistryID: maharash
Write Dow process for preparation of Phenol. What is the action of bromine water on phenol? Give reason: Group 16th elements have lower ionization enthalpy compared to group 15th elements. Write two uses of dioxygen.
Official Solution
Correct Option: (1)
Step 1: Dow Process for Preparation of Phenol. The Dow process involves the chlorination of benzene to form benzene hexachloride (also known as "Lindane") which is then treated with sodium hydroxide to produce phenol:
This yields phenol ( ). Step 2: Action of Bromine Water on Phenol. When phenol reacts with bromine water, it undergoes electrophilic substitution to form 2,4,6-tribromophenol:
The bromine atoms are added to the ortho and para positions of the phenol ring. Step 3: Ionization Enthalpy Comparison. Group 16 elements have lower ionization enthalpy compared to group 15 elements because:
- Group 16 elements have a lower effective nuclear charge due to the additional electron in the p-orbital, making the outer electrons easier to remove.
- Group 15 elements, being nitrogen, phosphorus, arsenic, etc., have a half-filled p-orbital which is more stable, resulting in higher ionization enthalpy. Step 4: Uses of Dioxygen. - Respiration: Dioxygen is essential for respiration in most aerobic organisms.
- Combustion: Dioxygen supports combustion and is used in various industrial processes like steelmaking.
04
PYQ 2023
medium
chemistryID: maharash
Silver crystallizes in fcc structure. If edge length of unit cell is 400 pm, calculate density of silver (Atomic mass of Ag = 108). Write a note on Halform reaction.
Official Solution
Correct Option: (1)
Step 1: Formula for Density of a Cubic Unit Cell. The density ( ) of a crystal is given by the formula:
Where:
- is the number of atoms per unit cell (for fcc structure, ),
- is the molar mass of the substance (108 g/mol for silver),
- is Avogadro's number ( atoms/mol),
- is the volume of the unit cell. Step 2: Volume of the Unit Cell. For fcc structure, the edge length is related to the volume of the unit cell:
Where . The volume of the unit cell is:
Step 3: Calculate the Density. Substitute the values into the density formula:
Thus, the density of silver is . Step 4: Halform Reaction. The Haloform reaction is a chemical reaction where a haloform (CHCl3, CHBr3, or CHI3) is produced when a methyl group is attached to a carbonyl group, typically in the presence of halogens. For example, when chloroform is reacted with a base, it forms chloroform and a carboxylate ion:
05
PYQ 2023
medium
chemistryID: maharash
Explain Aldol condensation of ethanal.
Official Solution
Correct Option: (1)
The Aldol Condensation is a reaction in which an enolate ion attacks the carbonyl carbon of an aldehyde or ketone to form a β-hydroxy aldehyde or ketone, which then undergoes dehydration to form an α,β-unsaturated carbonyl compound. Step 1: Aldol Addition Reaction. When ethanal (acetaldehyde, ) undergoes the Aldol addition, the enolate ion formed from one molecule of ethanal reacts with another molecule of ethanal:
This forms β-hydroxybutanal, also known as an Aldol addition product. Step 2: Dehydration to form Aldol Condensation Product. The β-hydroxy aldehyde undergoes elimination of water, resulting in the formation of an α,β-unsaturated aldehyde:
This product, crotonaldehyde, is the result of the Aldol condensation.
06
PYQ 2023
medium
chemistryID: maharash
Write chemical reactions for the following conversions: i. Acetic acid into acetic anhydride ii. Acetic acid into ethyl alcohol Write IUPAC name and structure of methylphenylamine.
Official Solution
Correct Option: (1)
i. Acetic Acid into Acetic Anhydride: Acetic acid can be converted into acetic anhydride by dehydration in the presence of a dehydrating agent such as phosphorus pentachloride ( ):
This forms acetic anhydride. ii. Acetic Acid into Ethyl Alcohol: Acetic acid can be reduced to ethyl alcohol (ethanol) using reducing agents such as lithium aluminium hydride ( ):
This is the reduction of acetic acid to ethanol. iii. IUPAC Name and Structure of Methylphenylamine: The IUPAC name of methylphenylamine is N-methylaniline. Its structure is as follows:
07
PYQ 2023
medium
chemistryID: maharash
Explain the dehydrohalogenation reaction of 2-chlorobutane. Write use and environmental effect of CFC.
Official Solution
Correct Option: (1)
Step 1: Dehydrohalogenation of 2-Chlorobutane. The dehydrohalogenation reaction involves the elimination of a hydrogen atom and a halogen atom (chlorine) from a compound. In 2-chlorobutane, the reaction occurs in the presence of a strong base like potassium hydroxide ( ) to form 2-butene:
This follows an E2 elimination mechanism, where both the base and substrate react simultaneously. Step 2: Use and Environmental Effect of CFCs. Chlorofluorocarbons (CFCs) were once widely used as refrigerants, solvents, and in aerosol propellants. However, their use has been phased out due to their role in ozone layer depletion. CFCs break down ozone molecules in the stratosphere, leading to increased UV radiation reaching Earth, which causes skin cancer, cataracts, and other health issues.
08
PYQ 2023
medium
chemistryID: maharash
Convert the following: i. Ethyl alcohol into ethyl acetate
ii. Phenol into benzene
iii. Diethyl ether into ethyl chloride
Official Solution
Correct Option: (1)
i. Ethyl Alcohol into Ethyl Acetate: To convert ethyl alcohol (ethanol) into ethyl acetate, you need to perform an esterification reaction. This is done by reacting ethanol with acetic acid ( ) in the presence of sulfuric acid ( ):
ii. Phenol into Benzene: Phenol can be converted into benzene via reduction. The most common method is by using zinc dust in a reduction reaction:
iii. Diethyl Ether into Ethyl Chloride: To convert diethyl ether to ethyl chloride, you can perform a substitution reaction using hydrochloric acid ( ):
09
PYQ 2023
medium
chemistryID: maharash
Write the chemical reaction for the preparation of glucose from sucrose. Write structure of D-ribose.
Official Solution
Correct Option: (1)
1. Preparation of Glucose from Sucrose: Sucrose (a disaccharide) can be hydrolyzed to give glucose and fructose. This is a simple hydrolysis reaction that requires an acid or enzyme for catalysis. The chemical reaction is: Sucrose ( ) breaks down into glucose ( ) and fructose ( ) under acidic conditions. 2. Structure of D-Ribose: D-Ribose is a five-carbon monosaccharide (pentose sugar) with an aldehyde group. Its structure is: The structure shows the hydroxyl groups on the second, third, and fourth carbon atoms, and an aldehyde group on the first carbon.
10
PYQ 2023
medium
chemistryID: maharash
Write the structure of the product formed when chlorobenzene is treated with sodium metal in the presence of dry ether.
Official Solution
Correct Option: (1)
When chlorobenzene is treated with sodium metal in dry ether, the reaction follows the Wurtz-Fittig coupling mechanism. This leads to the formation of biphenyl ( ). The reaction is:
Thus, the product formed is biphenyl with the structure:
11
PYQ 2023
medium
chemistryID: maharash
Ozonolysis of 2, 3 dimethyl but-2-ene, followed by decomposition by Zn dust and water gives:
1
acetaldehyde
2
propionaldehyde and acetone
3
acetaldehyde and butyraldehyde
4
acetalaldehyde and butyraldehyde
Official Solution
Correct Option: (2)
Step 1: Analyze the ozonolysis reaction. Ozonolysis of an alkene involves the addition of ozone ( ), followed by cleavage of the double bond. In this case, ozonolysis of 2, 3 dimethyl but-2-ene will break the carbon-carbon double bond. Step 2: Products of ozonolysis. The ozonolysis of 2, 3 dimethyl but-2-ene will produce two aldehydes:
After decomposition with Zn dust and water, the products will be acetaldehyde and acetone. Thus, the correct answer is (B).
12
PYQ 2023
medium
chemistryID: maharash
The product of the following reaction is:
1
2
3
4
Official Solution
Correct Option: (1)
Step 1: Analyze the reaction. The given reaction involves the reduction of an aldehyde ( ) with , followed by hydrolysis with water. Lithium aluminum hydride ( ) is a strong reducing agent that reduces aldehydes to primary alcohols. Step 2: Determine the product. Upon reduction of the aldehyde group and subsequent hydrolysis, the product is a primary alcohol:
Thus, the correct answer is (A).
13
PYQ 2023
medium
chemistryID: maharash
The allylic halide, among the following is:
1
2
3
4
Official Solution
Correct Option: (2)
Step 1: Understanding Allylic Halides. Allylic halides are compounds where a halogen is attached to a carbon atom adjacent to a double bond (the allylic position). The structure is an example of an allylic halide. Step 2: Analyzing the Options. - (A) : This is not an allylic halide because it doesn't have a halogen attached to the allylic carbon. - (B) : Correct. This is an allylic halide because the halogen is attached to the carbon next to the double bond. - (C) : This is not an allylic halide because it is an aryl halide. - (D) : This is not an allylic halide because the halogen is not directly attached to an allylic carbon. Thus, the correct answer is (B).
14
PYQ 2024
easy
chemistryID: maharash
Why are amines basic in nature? Among dimethylamine (pK = 3.27) and diethylamine (pK = 3.0), which one is more basic?
Official Solution
Correct Option: (1)
Amines exhibit basic properties due to the presence of a lone pair of electrons on the nitrogen atom, which allows them to act as Lewis bases by accepting protons . Comparatively, a lower value indicates a stronger base. Therefore, diethylamine is a stronger base than dimethylamine .
15
PYQ 2024
hard
chemistryID: maharash
Write the preparation of (a) diethyl ether and (b) ethyl cyanide from ethyl bromide.
Ethyl bromide reacts with magnesium in dry ether to form an ethyl magnesium bromide (Grignard reagent).
17
PYQ 2024
hard
chemistryID: maharash
Silver Nitrite
Official Solution
Correct Option: (1)
When ethyl bromide reacts with silver nitrite ( ), ethyl nitrite is formed due to the ambident nature of the nitrite ion.
18
PYQ 2024
medium
chemistryID: maharash
HF, HCl, HBr, HI (thermal stability)
Official Solution
Correct Option: (1)
Increasing Order of Thermal Stability: Thermal stability of hydrogen halides (HX) decreases down the group due to decreasing bond strength (H-F) being the strongest and (H-I) the weakest.
19
PYQ 2024
medium
chemistryID: maharash
MF, MCl, MBr, MI (ionic character)
Official Solution
Correct Option: (1)
Order of Decreasing Ionic Character:
The ionic character of compounds formed between metals (M) and halogens is largely determined by the electronegativity difference between the two elements. As you move down the halogen group, their electronegativity decreases, which in turn reduces the ionic character of the bond they form with metals. Consequently, the ionic character diminishes from fluoride to iodide.
20
PYQ 2024
easy
chemistryID: maharash
Arrange the following in the increasing order of the property mentioned:
(i) HClO, HClO₂, HClO₃, HClO₄ (acidic strength)
(ii) MF, MCl, MBr, MI (ionic character)
(iii) HF, HCl, HBr, HI (thermal stability)
Official Solution
Correct Option: (1)
21
PYQ 2024
easy
chemistryID: maharash
What is the action of the following on ethyl bromide ? (a) silver nitrite (b) Mg in dry ether (c) alcoholic sodium hydroxide
Official Solution
Correct Option: (1)
22
PYQ 2024
medium
chemistryID: maharash
Among the following, the vinylic halide is ____________.
1
R - CHX - R
2
3
4
Official Solution
Correct Option: (2)
Vinylic halides are compounds in which the halogen is directly bonded to a doubly bonded carbon (alkene carbon). - Contains a halogen on a saturated carbon, making it an alkyl halide. - (c) Represents an aryl halide (halogen attached to a benzene ring). - (d) Contains a halogen on an allylic position, making it an allylic halide. - Contains a halogen directly bonded to a doubly bonded carbon, making it a vinylic halide. Thus, the correct answer is .
23
PYQ 2025
hard
chemistryID: maharash
Write salient features of SN mechanism. What is the action of following reagents on bromomethane : (i) bromobenzene (ii) mercurous fluoride
Official Solution
Correct Option: (1)
Salient Features of SN Mechanism
It is a single-step, concerted mechanism where bond breaking and bond formation occur simultaneously. It follows second-order kinetics, i.e., the rate is proportional to the concentration of both the substrate and the nucleophile: Rate = k[Alkyl Halide][Nucleophile]. The attack of the nucleophile occurs from the side opposite to the leaving group (backside attack). It results in a complete inversion of the stereochemical configuration, known as Walden inversion. The reactivity order for alkyl halides is governed by steric hindrance: CH X>1 >2 >3 .
Reactions of Bromomethane
[(i)] Bromobenzene (in presence of Na/dry ether): This is the Wurtz-Fittig reaction. Bromomethane reacts with bromobenzene in the presence of sodium metal in dry ether to form toluene. [(ii)] Mercurous fluoride (Hg F ): This is a halogen exchange reaction known as the Swarts reaction, used for the synthesis of alkyl fluorides. Bromomethane reacts with mercurous fluoride to form fluoromethane.
24
PYQ 2025
easy
chemistryID: maharash
Identify ‘A’ and ‘B’ in the following reaction and rewrite the complete reaction :
Official Solution
Correct Option: (1)
Step 1: First reaction (Propene to A)
This is the addition of HBr to propene in the presence of peroxide. This follows the anti-Markovnikov's rule (Kharasch effect), where the negative part of the addendum (Br ) attaches to the carbon atom with more hydrogen atoms.
So, A is 1-bromopropane.
Step 2: Second reaction (A to B)
This is a nucleophilic substitution reaction where 1-bromopropane reacts with alcoholic potassium cyanide (KCN). The cyanide ion (CN ) displaces the bromide ion (Br ).
So, B is Butanenitrile.
Complete Reaction:
25
PYQ 2025
medium
chemistryID: maharash
Write the structural formula of N, N-dimethylethanamine.
Official Solution
Correct Option: (1)
The name indicates an ethanamine parent structure (CH CH NH ). The "N, N-dimethyl" prefix means that two methyl groups (CH ) replace the two hydrogens on the nitrogen atom. This results in a tertiary amine.
26
PYQ 2025
medium
chemistryID: maharash
Write the reagents used for the reduction of carbonyl group in Clemmensen's reduction.
Official Solution
Correct Option: (1)
The reagents for the Clemmensen reduction are Zinc amalgam (Zn-Hg) and concentrated Hydrochloric acid (conc. HCl).
27
PYQ 2025
medium
chemistryID: maharash
Write the IUPAC name of isoprene.
Official Solution
Correct Option: (1)
Structure and IUPAC Name of Isoprene
The molecular structure of isoprene is given as:
Step 1: Identify the Longest Chain
The longest continuous carbon chain that includes both double bonds contains 4 carbon atoms. Hence, the parent hydrocarbon is buta-1,3-diene.
Step 2: Numbering the Chain
Numbering starts from the end that gives the double bonds the lowest locants. Therefore, the double bonds are at carbons 1 and 3.
Step 3: Identify Substituents
A methyl group ( ) is attached to the second carbon atom of the chain.
Step 4: Combine the Name
Putting it all together:
Final Answer:
The IUPAC name of isoprene is
28
PYQ 2025
medium
chemistryID: maharash
The highest acidic compound among the following is _______.
1
o-Hydroxybenzoic acid % Placeholder for image
2
o-Aminobenzoic acid % Placeholder for image
3
Benzoic acid % Placeholder for image
4
p-Methoxybenzoic acid % Placeholder for image
Official Solution
Correct Option: (1)
Step 1: Analyze the effect of substituents on the acidity of benzoic acid.
Acidity is determined by the stability of the conjugate base (carboxylate anion, -COO ). Electron-withdrawing groups (EWGs) stabilize the anion and increase acidity. Electron-donating groups (EDGs) destabilize the anion and decrease acidity.
Step 2: Evaluate each compound.
- Benzoic acid: This is our reference compound.
- o-Hydroxybenzoic acid (Salicylic acid): The -OH group is an EWG via its inductive effect (-I) but an EDG via resonance (+R). However, at the ortho position, the conjugate base is significantly stabilized by intramolecular hydrogen bonding between the -OH proton and the -COO group. This is a powerful stabilizing effect known as the "ortho effect," making it much more acidic than benzoic acid.
- o-Aminobenzoic acid: The -NH group is a strong EDG via resonance (+R), which destabilizes the conjugate base and decreases acidity.
- p-Methoxybenzoic acid: The -OCH group at the para position is a strong EDG via resonance (+R), destabilizing the conjugate base and making it less acidic than benzoic acid.
Step 3: Compare the acidities.
Due to the strong stabilization from intramolecular hydrogen bonding, o-hydroxybenzoic acid is the most acidic compound among the choices.
29
PYQ 2025
medium
chemistryID: maharash
Which of the following is a secondary amine?
1
Cyclohexylamine
2
Isopropylamine
3
Diphenylamine
4
N, N-Dimethylaniline
Official Solution
Correct Option: (3)
Step 1: Define the classes of amines.
Amines are classified based on the number of alkyl or aryl groups directly bonded to the nitrogen atom.
- Primary (1 ) amine: One carbon group attached to nitrogen (general formula R-NH ).
- Secondary (2 ) amine: Two carbon groups attached to nitrogen (general formula R -NH).
- Tertiary (3 ) amine: Three carbon groups attached to nitrogen (general formula R -N). Step 2: Classify each option.
- Cyclohexylamine: The nitrogen is attached to one cyclohexyl group. It is a primary amine (C H -NH ).
- Isopropylamine: The nitrogen is attached to one isopropyl group. It is a primary amine ((CH ) CH-NH ).
- Diphenylamine: The nitrogen is attached to two phenyl groups. It is a secondary amine ((C H ) NH).
- N, N-Dimethylaniline: The nitrogen is attached to one phenyl group and two methyl groups, for a total of three carbon groups. It is a tertiary amine. Therefore, diphenylamine is the secondary amine.
30
PYQ 2025
medium
chemistryID: maharash
Explain Cannizzaro’s reaction with the help of benzaldehyde. Write the reaction for the conversion of cyclohexene to adipic acid.
Official Solution
Correct Option: (1)
Cannizzaro's Reaction
The Cannizzaro reaction is a redox disproportionation reaction given by aldehydes that do not have an -hydrogen atom. When heated with a concentrated alkali solution (like 50% NaOH), one molecule of the aldehyde is reduced to an alcohol, and another molecule is oxidized to the salt of a carboxylic acid. Example with Benzaldehyde: Benzaldehyde (C H CHO) has no -hydrogen. When two molecules of benzaldehyde are treated with concentrated NaOH, one is reduced to benzyl alcohol, and the other is oxidized to sodium benzoate.
Conversion of Cyclohexene to Adipic Acid
This conversion is achieved by the strong oxidation of cyclohexene. The double bond is cleaved, and the carbons of the double bond are oxidized to carboxylic acid groups. A common oxidizing agent for this is alkaline potassium permanganate (KMnO ) followed by acidification, or hot concentrated nitric acid (HNO ).
31
PYQ 2025
medium
chemistryID: maharash
Write the structural formula and IUPAC name of the alcohol having molecular formula C H O which does not undergo oxidation under normal condition.
Official Solution
Correct Option: (1)
An alcohol that does not undergo oxidation under normal conditions must be a tertiary alcohol. This is because a tertiary alcohol has no hydrogen atom attached to the carbinol carbon (the carbon atom bonded to the -OH group).
For the molecular formula C H O, the tertiary alcohol is 2-methylpropan-2-ol.
Structural Formula: IUPAC Name:
The longest carbon chain containing the -OH group has three carbons (propane). The -OH group is on carbon 2, and a methyl group is also on carbon 2.
Therefore, the name is 2-methylpropan-2-ol.
32
PYQ 2025
medium
chemistryID: maharash
Write the reaction for the preparation of : (i) acetaldehyde by Rosenmund reaction. (ii) benzaldehyde by Gatterman-Koch formylation.
Official Solution
Correct Option: (1)
(i) Rosenmund Reaction for Acetaldehyde
This reaction involves the catalytic hydrogenation of an acid chloride to an aldehyde. The catalyst is poisoned to prevent further reduction of the aldehyde to an alcohol.
(ii) Gatterman-Koch Formylation for Benzaldehyde
This reaction introduces a formyl group (-CHO) onto an aromatic ring. Benzene is treated with carbon monoxide and hydrogen chloride in the presence of a catalyst mixture of anhydrous aluminum chloride and cuprous chloride.
33
PYQ 2025
medium
chemistryID: maharash
Write the reactions for the action of following reagents on phenol : (i) Nitrating mixture (ii) Zinc dust What is the action of phosphorous pentachloride on ethyl methyl ether?
Official Solution
Correct Option: (1)
Reactions on Phenol
[(i)] Nitrating mixture (conc. HNO + conc. H SO ): The -OH group of phenol is highly activating. Reacting phenol with a nitrating mixture results in the substitution of nitro groups at the ortho and para positions, forming 2,4,6-trinitrophenol (Picric acid). [(ii)] Zinc dust: When phenol is heated with zinc dust, it undergoes dehydroxylation (removal of the -OH group) to form benzene.
Action of PCl on Ethyl Methyl Ether
When ethyl methyl ether (CH OC H ) is heated with phosphorus pentachloride (PCl ), the ether linkage breaks, and both alkyl groups are converted into their corresponding alkyl chlorides.
34
PYQ 2025
easy
chemistryID: maharash
What is the action of : (i) Liquid bromine in acetic acid on anisole. (ii) Soda-lime on sodium acetate?
Official Solution
Correct Option: (1)
(i) Action of liquid bromine in acetic acid on anisole: Anisole ( ) undergoes electrophilic aromatic substitution. The methoxy group ( ) is an ortho-para directing group. Therefore, bromination gives a mixture of ortho- and para-bromoanisole.
(ii) Action of soda-lime on sodium acetate: Sodium acetate ( ) undergoes decarboxylation when heated with soda lime ( ), giving methane.
Final Answer:
(i) Anisole gives ortho- and para-bromoanisole. (ii) Sodium acetate with soda lime gives methane (decarboxylation reaction).
35
PYQ 2026
medium
chemistryID: maharash
Contrary to electrophilic addition reactions observed in alkenes, the aldehydes and ketones undergo nucleophilic addition reactions. (True/False)
Official Solution
Correct Option: (1)
Concept: Different functional groups undergo different types of reactions depending on electron distribution and bond polarity. Explanation:
Alkenes contain a carbon–carbon double bond ( ) rich in electron density. Therefore, they attract electrophiles and undergo electrophilic addition reactions. Aldehydes and Ketones:
They contain a polar carbonyl group ( ). Oxygen is more electronegative, making the carbonyl carbon partially positive. This electrophilic carbon is attacked by nucleophiles. Hence, aldehydes and ketones undergo nucleophilic addition reactions. Example: (HCN adds via nucleophilic addition) Conclusion: The given statement is correct.
36
PYQ 2026
medium
chemistryID: maharash
Lower aliphatic amines are soluble in water while higher amines are essentially insoluble in water. (True/False)
Official Solution
Correct Option: (1)
Concept: Solubility of organic compounds in water depends on their ability to form hydrogen bonds and the size of the hydrophobic hydrocarbon chain. Explanation:
Amines contain a nitrogen atom with a lone pair of electrons. This allows formation of hydrogen bonds with water molecules. Lower aliphatic amines (e.g., methylamine, ethylamine) have small alkyl groups, so hydrogen bonding dominates → high solubility. Higher Amines:
As the alkyl chain length increases, the non-polar hydrocarbon portion increases. Hydrophobic effect reduces interaction with water. Thus, higher amines become less soluble or insoluble. Conclusion: The statement is correct.
37
PYQ 2026
medium
chemistryID: maharash
Amines behave as ________ due to presence of an unshared pair of electron on Nitrogen atom.
Official Solution
Correct Option: (1)
Concept: Amines are organic derivatives of ammonia ( ) in which one or more hydrogen atoms are replaced by alkyl or aryl groups. The nitrogen atom in amines contains a lone pair of electrons. Explanation:
Nitrogen atom in amines has an unshared (lone) pair of electrons. This lone pair can accept a proton ( ). Hence, amines behave as Lewis bases or Brønsted bases. Example Reaction: This shows proton acceptance, confirming basic nature. Conclusion: Due to the presence of a lone pair on nitrogen, amines exhibit basic properties.
38
PYQ 2026
medium
chemistryID: maharash
Bromo, iodo and polychloro derivatives of hydrocarbons are heavier than water. (True/False)
Official Solution
Correct Option: (1)
Concept: The density of organic compounds depends on the atoms present in the molecule. Halogen atoms like chlorine, bromine, and iodine have higher atomic masses compared to hydrogen. Explanation:
Hydrocarbons contain mainly carbon and hydrogen, so they are usually lighter than water. When halogens (Cl, Br, I) replace hydrogen atoms, molecular mass increases significantly. Bromo and iodo compounds especially have high densities due to heavy halogen atoms. Polychloro derivatives (e.g., chloroform, carbon tetrachloride) are also denser than water. Examples:
Chloroform ( ) – denser than water Carbon tetrachloride ( ) – sinks in water Bromoform ( ) – very dense liquid Conclusion: Hence, the statement is correct.
39
PYQ 2026
medium
chemistryID: maharash
In addition of bromine in to an alkene resulting in disappearance of reddish brown colour of bromine constitutes an important method for the detection of ________ in a molecule.
Official Solution
Correct Option: (1)
Concept: Bromine test is a common qualitative test used in organic chemistry to detect unsaturation in hydrocarbons. Explanation:
Bromine in carbon tetrachloride ( ) has a reddish-brown colour. When an alkene is added, bromine undergoes an addition reaction across the double bond. This reaction removes the colour of bromine solution. Reaction:
Conclusion: Disappearance of bromine colour indicates the presence of a double bond or unsaturation in the molecule.
