Question 1

Arrange the following carbanions in decreasing order of basicity: (A) CH3CH2−, (B) CH2=CH−, and (C) HC≡C−.

Accepted Answer

(A) > (B) > (C). Carbanion basicity decreases with increasing s-character: CH3CH2- (sp3) > CH2=CH- (sp2) > HC#C- (sp), i.e. A > B > C (option 3). The stored order is reversed.

Question 2

In acid-catalyzed hydration of alkenes, all alkenes except ethene generally give rise to which type of alcohol product?

Accepted Answer

either a secondary alcohol or a tertiary alcohol. Acid-catalysed (Markovnikov) hydration puts OH on the more substituted carbon, giving a secondary or tertiary alcohol for all alkenes except ethene (which gives the primary ethanol). So the answer is 'secondary or tertiary', not 'primary and secondary'.

Question 3

Which of the following organic compounds has the same hybridization as the carbon atom in its combustion product, CO2?

Accepted Answer

Ethyne. Ethyne has sp hybridization, which is the same as the carbon atoms in carbon dioxide (CO2). In CO2, the carbon atom forms two double bonds with oxygen, requiring sp hybridization to accommodate the linear geometry.

Question 4

What is formed when CH3MgX reacts with CH3C≡CH?

Accepted Answer

CH4. Propyne has an acidic terminal C-H, so CH3MgX acts as a base: CH3C#CH + CH3MgX -> CH3C#C-MgX + CH4. The evolved product is methane, not 2-butyne.

Question 5

Which of the following reagent sets can be used to convert propene into 1-propanol?

Accepted Answer

B2H6, then H2O2 and OH−. Converting propene to the primary 1-propanol requires anti-Markovnikov hydration: hydroboration-oxidation (B2H6 then H2O2/OH-). The stored MgSO4/NaBH4 route does not do this.

Question 6

When ethyne (H–C≡CH) is treated first with (1) NaNH2 in liquid NH3 and then with (2) CH3CH2Br, the product X is formed. If X is again subjected to the same two-step sequence, the product Y obtained is:

Accepted Answer

X = 1-Butyne; Y = 3-Hexyne. The correct option is right because treating ethyne with NaNH2 generates a terminal alkyne, which in this case is 1-butyne. When 1-butyne is further treated with CH3CH2Br, a nucleophilic substitution occurs, leading to the formation of 3-hexyne.

Question 7

An alkyl chloride A with molecular formula C4H9Cl is treated with sodium in dry ether. The hydrocarbon formed on this reaction yields only one monochloro product when chlorinated. Identify A.

Accepted Answer

tert-Butyl chloride. Wurtz coupling doubles the carbon skeleton. tert-Butyl chloride gives (CH3)3C-C(CH3)3 (2,2,3,3-tetramethylbutane), in which all 18 hydrogens are equivalent, so chlorination yields only one monochloro product.

Question 8

When toluene is treated with chlorine in the presence of FeCl3, product 'X' is formed. If the same hydrocarbon is chlorinated in sunlight, product 'Y' is obtained. Identify 'X' and 'Y'.

Accepted Answer

X = o- and p-chlorotoluene, Y = trichloromethylbenzene. With Cl2/FeCl3 (ionic conditions) toluene undergoes ring substitution giving o- and p-chlorotoluene (X). With Cl2 in sunlight (free-radical) the methyl side chain is chlorinated, ultimately to C6H5CCl3, trichloromethylbenzene (Y).

Question 9

When benzene is treated with n-propyl chloride in the presence of anhydrous AlCl3, the major product formed is:

Accepted Answer

Isopropylbenzene. The reaction involves the Friedel-Crafts alkylation, where n-propyl chloride reacts with benzene in the presence of AlCl3, leading to the formation of a more stable carbocation. The n-propyl carbocation can rearrange to form the more stable isopropyl carbocation, resulting in isopropylbenzene as the major product.

Question 10

SLV-3 was powered by which type of propellant system?

Accepted Answer

solid. India's SLV-3 (Satellite Launch Vehicle) was a four-stage rocket in which every stage used SOLID propellant. Therefore it is a solid-propellant launch vehicle.

Question 11

The number of types of bonds between two carbon atoms in calcium carbide is

Accepted Answer

One sigma, two pi. In calcium carbide, the carbon atoms are connected by a triple bond, which consists of one sigma bond and two pi bonds. The sigma bond is formed by the head-on overlap of orbitals, while the two pi bonds result from the side-to-side overlap of p orbitals.

Question 12

Which of the following alkenes shows optical isomerism?

Accepted Answer

3-methyl-1-pentene. 3-methyl-1-pentene has a double bond between the first and second carbon atoms, which creates a chiral center when the substituents on the double bond are different, allowing for the existence of optical isomers. In contrast, the other options do not possess the necessary chiral centers to exhibit optical isomerism.

Question 13

When 2-methylbutane is treated with bromine under sunlight, the principal product formed is

Accepted Answer

2-bromo-2-methylbutane. The correct option is 2-bromo-2-methylbutane because the reaction involves free radical bromination, which favors the formation of the most stable radical. In this case, the tertiary radical formed at the second carbon is more stable than any secondary or primary radicals, leading to the predominant formation of the product.

Question 14

When ethylbenzene is oxidized with KMnO4, which compound is produced?

Accepted Answer

benzoic acid. Oxidation of ethylbenzene with KMnO4 leads to the formation of benzoic acid, as the oxidation process converts the ethyl group into a carboxylic acid functional group.

Question 15

Which of the following reactions produces 2,2-dibromopropane?

Accepted Answer

CH3–C≡CH + 2HBr →. The reaction of propyne (CH3–C≡CH) with two equivalents of HBr adds bromine across the triple bond, resulting in the formation of 2,2-dibromopropane, as both bromine atoms attach to the same carbon atom.

Question 16

When toluene is treated with chlorine in the presence of FeCl3, the major products formed are

Accepted Answer

o- and p-chlorotoluenes. The presence of FeCl3 as a catalyst facilitates the electrophilic aromatic substitution reaction, where chlorine substitutes hydrogen atoms on the aromatic ring of toluene. Due to the electron-donating methyl group, the substitution occurs predominantly at the ortho and para positions, leading to the formation of o- and p-chlorotoluenes.

Question 17

3-Methyl-pent-2-ene on reaction with HBr in presence of peroxide forms an addition product. The number of possible stereoisomers for the product is:

Accepted Answer

Four. HBr with peroxide gives anti-Markovnikov addition: Br adds to C2, giving 2-bromo-3-methylpentane (CH3-CHBr-CH(CH3)-CH2CH3). Both C2 and C3 are stereocentres and are different, so 2^2 = 4 stereoisomers are possible.

Question 18

The trans-alkenes are formed by the reduction of alkynes with:

Accepted Answer

Na/liq. NH3. The reduction of alkynes to trans-alkenes is effectively achieved using sodium in liquid ammonia, which facilitates the formation of the trans configuration through a radical mechanism.

Question 19

The major product of the following reaction is:
CH3C≡CH (i) DCl (excess) / (ii) D2

Accepted Answer

CH3CD(Cl)CHD(Cl). The reaction involves the addition of DCl to the alkyne, resulting in the formation of a deuterated vinyl halide. The excess DCl leads to the substitution of both hydrogen atoms with deuterium and chlorine, yielding the product CH3CD(Cl)CHD(Cl), which is the most stable and substituted product.

Question 20

The trans-alkenes are formed by the reduction of alkynes with:
(1) H2 - Pd/C, BaSO4
(2) NaBH4
(3) Na/liq. NH3
(4) Sn - HCl

Accepted Answer

Na/liq. NH3. The reduction of alkynes to trans-alkenes is effectively achieved using sodium in liquid ammonia, which facilitates the formation of the trans isomer through a unique mechanism that avoids the formation of the more stable cis isomer.

Question 21

Which one of the following alkenes when treated with HCl yields majorly an anti Markovnikov product?

Accepted Answer

F3C-CH=CH2. Ionic HCl addition follows the more stable carbocation. The strongly electron-withdrawing CF3 destabilizes a cation on the adjacent carbon, so the positive charge forms on the terminal CH2 instead, giving the anti-Markovnikov product. Hence F3C-CH=CH2.

Question 22

The minimum number of moles of O2 required for complete combustion of 1 mole of propane and 2 moles of butane is ______.

Accepted Answer

18. To completely combust propane (C3H8), 5 moles of O2 are needed, and for butane (C4H10), 13 moles of O2 are required. Therefore, the total for 1 mole of propane and 2 moles of butane is 5 + (2 x 13) = 18 moles of O2.

Question 23

The major product formed in the following reaction is:
CH3CH = CHCH(CH3)2
HBr →

Accepted Answer

CH3CH2CH2C(Br)(CH3)2. The correct option is the result of Markovnikov's rule, where HBr adds across the double bond, leading to the more stable carbocation. In this case, the bromine atom attaches to the more substituted carbon, resulting in the formation of the branched product.

Question 24

The major product (Y) in the following reactions is: CH3–CH(CH3)–C≡CH
(i) HgSO4, H2SO4
(ii) C2H5MgBr, H2O
(iii) Conc. H2SO4/Δ

Accepted Answer

CH3–C(CH3)=C(CH3)–C2H5. HgSO4/H2SO4 gives the methyl ketone (CH3)2CH-CO-CH3; C2H5MgBr/H2O gives the tertiary alcohol (CH3)2CH-C(OH)(CH3)-C2H5; conc. H2SO4 dehydrates by Zaitsev toward the isopropyl CH to give the tetrasubstituted alkene CH3-C(CH3)=C(CH3)-C2H5.

Question 25

An unsaturated hydrocarbon X on ozonolysis gives A. Compound A when warmed with ammonical silver nitrate forms a bright silver mirror along the sides of the test tube. The unsaturated hydrocarbon X is:

Accepted Answer

HCC-CH2-CH3. Oxidative ozonolysis of the terminal alkyne HC=C-CH2CH3 gives HCOOH (formic acid) plus propanoic acid. Formic acid contains a -CHO group and reduces Tollens' reagent, producing the silver mirror. The symmetrical alkene/alkyne options give only ketones or acetic acid, which do not.

Question 26

Metallic sodium does not react normally with:

Accepted Answer

But-2-yne. But-2-yne is a stable alkyne that does not readily react with metallic sodium under normal conditions, as it lacks the necessary functional groups or reactivity to engage in a reaction with sodium.

Question 27

The number of acyclic structural isomers (including geometrical isomers) for pentene are ____

Accepted Answer

6. Structural isomers: pent-1-ene, pent-2-ene, 2-methylbut-1-ene, 3-methylbut-1-ene, 2-methylbut-2-ene (5). Pent-2-ene has cis and trans forms, adding 1 -> total 6.

Question 28

Which of the following reagent is used for the following reaction ? CH3CH2CH3 → CH3CH2CHO

Accepted Answer

Molybdenum oxide. Direct oxidation of propane (CH3CH2CH3) to propanal (CH3CH2CHO) is achieved by catalytic oxidation over molybdenum oxide (MoO3-based catalyst), which selectively abstracts hydrogen to give the aldehyde.

Question 29

Excess of isobutane on reaction with Br2 in presence of light at 125°C gives which one of the following, as the major product?

Accepted Answer

CH3-C(Br)(CH3)-CH3. The major product is CH3-C(Br)(CH3)-CH3 because the reaction involves free radical bromination, which favors the formation of the most stable tertiary radical. Isobutane has a tertiary carbon that can stabilize the radical formed during the reaction, leading to the preferential formation of this product.

Question 30

In CH2 = C = CH – CH3 molecule, the hybridization of carbon 1, 2, 3 and 4 respectively are:

Accepted Answer

sp2, sp, sp2, sp3. In CH2=C=CH-CH3, C1 (=CH2) is sp2, C2 (central cumulene carbon with two double bonds) is sp, C3 (=CH-) is sp2, and C4 (CH3) is sp3. So the order is sp2, sp, sp2, sp3.

Question 31

An organic compound 'A' C4H8 on treatment with KMnO4/H+ yields compound 'B' C3H6O. Compound 'A' also yields 'B' on ozonolysis. Compound 'A' is:

Accepted Answer

2-Methylpropene. 2-Methylpropene is an alkene that can undergo oxidation with KMnO4 to form a ketone, which matches the formula of compound 'B' (C3H6O). Additionally, ozonolysis of 2-Methylpropene also yields the same compound 'B', confirming it as the correct structure for compound 'A'.

Question 32

Given below are two statements. Statement I: The presence of weaker π-bonds make alkenes less stable than alkanes. Statement II: The strength of the double bond is greater than that of carbon-carbon single bond. In the light of the above statements, choose the correct answer from the optio

Accepted Answer

Both Statement I and Statement II are correct.. Both statements accurately describe the stability of alkenes compared to alkanes. Alkenes have weaker π-bonds, making them less stable than alkanes, which have stronger σ-bonds, while the strength of the double bond is indeed greater than that of a single bond.

Question 33

The product formed in the following reaction

(CH3)2C=CH2 + H-C(CH3)3
H?

Accepted Answer

CH3-C(H)(CH3)-CH2-C(CH3)2-CH3. The correct option represents the product of the reaction where the alkene undergoes an addition reaction with the alkyl group from the second reactant, resulting in a branched structure that maintains the integrity of the carbon skeleton while adding the substituents appropriately.

Question 34

Cl· + CH4 → A + B
A and B in the above atmospheric reaction step are:

Accepted Answer

CH3· and HCl. The reaction involves the chlorine radical (Cl·) abstracting a hydrogen atom from methane (CH4), resulting in the formation of a methyl radical (CH3·) and hydrochloric acid (HCl). This process is characteristic of radical chain reactions where a radical species reacts with a stable molecule to produce new radicals and other products.

Question 35

Two isomers 'A' and 'B' with molecular formula C4H8 give different products on oxidation with KMnO4 / H+ results in effervescence of a gas and gives ketone. The compound 'A' is -

Accepted Answer

2-methyl propene. 2-methyl propene is a branched alkene that, upon oxidation with KMnO4, produces a ketone and releases gas due to the formation of a stable product. In contrast, the other isomers either do not yield a ketone or do not produce gas during the reaction.

Question 36

But-2-yne is reacted separately with one mole of Hydrogen as shown below
B —Na / liq NH3→ CH3–C≡C–CH3 —Pd/C, Δ→ A
+H2
(A) A is more than soluble than B.
(B) The boiling point & melting point of A are higher and lower than B respectively.
(C) A is more polar than B because dipole moment of

Accepted Answer

B, C & D only. Options B, C, and D are incorrect because A is a more stable alkane with higher boiling and melting points than B, which is an alkyne. Additionally, A is less polar than B due to its symmetrical structure, resulting in a zero dipole moment, and the addition of Br2 is more favorable to the more reactive alkyne B than the alkane A.

Question 37

Q.61 The major product 'P' formed in the given reaction is

Given compound: a naphthalene derivative bearing an ethyl group (CH2CH3) on one ring, a side chain ester group (–CH2–CH2–COOCH3) on the other side, and an exocyclic CH=CH2 substituent. Reagents: (i) alk.KMnO4, Δ (ii) H3O+

Accepted Answer

A naphthalene derivative with COOH groups at three positions: one on the upper left ring carbon, one on the lower left ring carbon, and one on the right side-chain position. The correct option describes the major product formed after oxidation of the naphthalene derivative, where the alkali permanganate oxidizes the ethyl group and the ester to carboxylic acids, resulting in COOH groups at three distinct positions on the naphthalene structure.

Question 38

In the following reaction ‘X’ is
CH3(CH2)4CH3 Anhy. AlCl3, X major product

Accepted Answer

CH3CH(CH3)(CH2)2CH3. The correct option represents a branched alkane formed through the alkylation of the starting compound with the aluminum chloride catalyst, which facilitates the formation of a more stable tertiary carbocation, leading to the major product.

Question 39

When a hydrocarbon A undergoes combustion in the presence of air, it requires 9.5 equivalents of oxygen and produces 3 equivalents of water. What is the molecular formula of A?

Accepted Answer

C8H6. The combustion of a hydrocarbon can be represented by the equation CxHy + O2 → CO2 + H2O. Given that 9.5 equivalents of oxygen are required and 3 equivalents of water are produced, we can deduce the ratio of carbon to hydrogen in the hydrocarbon. The correct molecular formula C8H6 indicates that for every 8 carbon atoms, there are 6 hydrogen atoms, which aligns with the stoichiometry of the combustion reaction.

Question 40

Number of σ and π bonds present in ethylene molecule is respectively:

Accepted Answer

5 and 1. Ethylene (CH2=CH2) has 4 C-H sigma bonds and 1 C-C sigma bond (total 5 sigma) plus 1 pi bond. So the answer is 5 and 1, not 4 and 1.

Question 41

An optically active alkyl halide C4H9Br [A] reacts with hot KOH dissolved in ethanol and forms alkene [B] as major product which reacts with bromine to give dibromide [C]. The compound [C] is converted as a gas [D] upon reacting with alcoholic NaNH2. During hydration 1 gram of gas [D] is a

Accepted Answer

Butan-2-one. The correct option is Butan-2-one because the reaction sequence indicates that the alkene formed is a symmetrical compound that, upon hydration with mercuric sulfate and dilute acid, leads to the formation of a ketone. In this case, the addition of water to the double bond of but-2-yne results in butan-2-one, which is the only ketone among the options provided.

Question 42

Given below are two statements:
Statement I: Neopentane forms only one monosubstituted derivative.
Statement II: Melting point of neopentane is higher than n-pentane.
In the light of the above statements, choose the most appropriate answer from the options given below:
(1) Statement I is i

Accepted Answer

Both Statement I and Statement II are correct. Statement I is correct because neopentane has a symmetrical structure that allows for only one unique monosubstituted derivative. Statement II is also correct as neopentane has a higher melting point than n-pentane due to its compact structure, which leads to stronger van der Waals forces.

Question 43

Given below are two statements:
Statement (I): On nitration of m-xylene with HNO3, H2SO4 followed by oxidation, 4-nitrobenzene-1, 3-dicarboxylic acid is obtained as the major product.
Statement (II): CH3 group is o/p-directing while -NO2 group is m-directing.
In the light of the above stat

Accepted Answer

Both Statement I and Statement II are true. Both statements are accurate; the methyl group in m-xylene directs electrophilic substitution to the ortho and para positions, while the nitro group directs it to the meta position. Consequently, the nitration of m-xylene followed by oxidation indeed leads to the formation of 4-nitrobenzene-1,3-dicarboxylic acid as the major product.

Question 44

Which of these will not react with acetylene ?

Accepted Answer

NaOH. NaOH does not react with acetylene because it is a strong base and acetylene is a relatively stable alkyne that does not undergo reactions with bases under normal conditions.

Question 45

What is the product when acetylene reacts with hypochlorous acid ?

Accepted Answer

Cl2CHCHO. The reaction between acetylene and hypochlorous acid leads to the formation of a compound where chlorine atoms are added to the carbon chain, resulting in the product Cl2CHCHO, which contains both the aldehyde functional group and two chlorine substituents.

Question 46

On mixing a certain alkane with chlorine and irradiating it with ultraviolet light, it forms only one monochloroalkane. This alkane could be

Accepted Answer

neopentane. Neopentane has a symmetrical structure that allows for only one unique position for chlorine substitution, resulting in a single monochloroalkane product when reacted with chlorine under UV light.

Question 47

Butene-1 may be converted to butane by reaction with

Accepted Answer

Pd/H2. Reducing the C=C of butene-1 to butane requires addition of H2 over a metal catalyst, i.e. Pd/H2 (catalytic hydrogenation). Sn-HCl, Zn-Hg, and Zn-HCl do not hydrogenate isolated alkenes.

Question 48

2 methylbutane on reacting with bromine in the presence of sunlight gives mainly

Accepted Answer

2 - bromo - 2 - methylbutane. 2-methylbutane, (CH3)2CH-CH2-CH3, has a tertiary hydrogen at C2. Because bromination strongly favors the tertiary position, the main product is 2-bromo-2-methylbutane.

Question 49

Acid catalyzed hydration of alkenes except ethene leads to the formation of

Accepted Answer

secondary or tertiary alcohol. Acid-catalyzed hydration follows Markovnikov's rule, placing OH on the more substituted carbon, so for alkenes other than ethene it yields secondary or tertiary alcohols. Stored answer 'primary alcohol' is wrong; the correct answer is secondary or tertiary alcohol.

Question 50

Which of the following reactions will yield 2,2-dibromopropane?

Accepted Answer

CH3−C≡CH + 2HBr →. The reaction of propyne (CH3−C≡CH) with two equivalents of HBr adds bromine across the triple bond, resulting in the formation of 2,2-dibromopropane. This is because the addition of HBr occurs in a Markovnikov fashion, leading to the formation of the more stable carbocation intermediate.

JEE Main Chemistry: Hydrocarbons questions with solutions

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