#1 of 58
ETEA 2024
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The homolytic fission of C-H bond in an alkane results
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Correct Key: Option A
Diagnostic Explanation
Concept:Bond cleavage occurs in two ways: heterolytic (unequal sharing of electrons) and homolytic (equal sharing of electrons).
Solution:- In an alkane, the C-H bond consists of two shared electrons.
- When homolytic fission occurs (typically initiated by UV light or high heat), the bond breaks symmetrically.
- The carbon atom takes one electron, and the hydrogen atom takes the other electron.
- This produces a highly reactive, neutral species with an unpaired electron, known as an alkyl free radical.
Why other options are incorrect:- Carbanions (negative) and Carbocations (positive) are products of heterolytic fission, where both electrons go to one atom.
- Methylpropane is a completely separate stable molecule, not an intermediate species.
#2 of 58
DUHS 2024
Permalink
When an unsymmetrical alkene undergoes addition reactions, the negative part of attacking reagent is added to that double-bonded carbon atom which contains:
A
Highest number of chloride atoms
B
Lesser number of hydrogen atoms
C
Highest number of hydrogen atoms
D
Moderate number of hydrogen atoms
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:This question asks for the formal definition of Markovnikov's Rule for the electrophilic addition to unsymmetrical alkenes.
Solution:- Markovnikov's rule states that when an unsymmetrical reagent (like HX) adds to an unsymmetrical alkene, the positive part (\( \text{H}^+ \)) goes to the carbon with the greater number of hydrogen atoms.
- This ensures the formation of the most stable intermediate carbocation on the more highly substituted carbon.
- Consequently, the negative part of the reagent (e.g., \( \text{X}^- \)) must attach to the carbocation, which is the carbon atom containing the lesser number of hydrogen atoms.
- (Note: The source key erroneously selected C; however, B is the correct application of Markovnikov's Rule).
Why other options are incorrect:- Adding the negative part to the carbon with the highest number of hydrogen atoms (Option C) represents anti-Markovnikov addition, which only occurs under specific free-radical conditions (like HBr with peroxides).
#3 of 58
DUHS 2024
Permalink
The bond length between C-C in benzene is:
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Correct Key: Option B
Diagnostic Explanation
Concept:Because of resonance, the carbon-carbon bonds in benzene are intermediate in length and strength between a standard single bond and a standard double bond.
Solution:- A pure C-C single bond (alkane) has a length of 1.54 Å.
- A pure C=C double bond (alkene) has a length of 1.34 Å.
- In benzene, the pi electrons are completely delocalized over all six carbon atoms. This means every bond is functionally a "one-and-a-half" bond.
- X-ray diffraction shows all six C-C bonds in benzene are exactly equal at 1.397 Å (often cited as roughly 1.39 Å).
Why other options are incorrect:- 1.34 Å is the length of an isolated double bond.
- 1.54 Å (closest to 1.56 Å) is the length of an isolated single bond.
#4 of 58
DUHS 2024
Permalink
The density of benzene is (out of syllabus)
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Physical properties of aromatic hydrocarbons include their state, color, and density relative to water.
Solution:- Benzene is a colorless, highly flammable liquid with a sweet odor.
- Like most hydrocarbons, it is less dense than water (which has a density of 1.00 g/cm³).
- The experimentally determined density of pure liquid benzene at room temperature is approximately 0.876 g/cm³, which is standardly rounded to 0.88 g/cm³ in chemical literature.
Why other options are incorrect:- 0.80, 0.85, and 0.08 are factually incorrect values for the specific gravity/density of benzene.
#5 of 58
BUMHS 2024
Permalink
Natural gas consist mainly of
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Natural gas is a naturally occurring hydrocarbon gas mixture found deep underground, often in association with petroleum.
Solution:- While natural gas contains various low-molecular-weight alkanes, its primary and overwhelmingly dominant constituent is the simplest alkane: Methane (\( \text{CH}_4 \)).
- Methane typically makes up 70% to 90% of natural gas by volume.
- It is highly combustible and is the main energy source provided by municipal gas lines for heating and cooking.
Why other options are incorrect:- Ethane, Propane, and Butane are present in natural gas but only as minor components (usually less than 10-15% combined). They are often separated out as Natural Gas Liquids (NGLs).
#6 of 58
BUMHS 2024
Permalink
In alkane series, all the linkages between carbon atoms is ____ bonds
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Correct Key: Option A
Diagnostic Explanation
Concept:Hydrocarbons are classified into families based strictly on their degree of saturation, which is defined by the type of carbon-carbon bonds they contain.
Solution:- Alkanes (general formula \( \text{C}_n\text{H}_{2n+2} \)) are defined as fully saturated hydrocarbons.
- Because they are saturated, each carbon atom is sp3 hybridized and uses all its valence electrons to form independent bonds with four other atoms.
- Consequently, all the carbon-carbon linkages in an alkane must be strong, localized single (sigma) bonds.
Why other options are incorrect:- Double bonds define the alkene series.
- Triple bonds define the alkyne series.
- Quadruple bonds do not exist between carbon atoms in stable organic molecules.
#7 of 58
BUMHS 2024
Permalink
Benzene when reacts with chlorine, in presence of sunlight gives additional product. i.e.
D
Polyvinylchloride (PVC)
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:While benzene strongly prefers substitution to maintain its resonance stability, extreme conditions (like intense UV light) can force free-radical addition, breaking the aromatic ring.
Formula:$$ \text{C}_6\text{H}_6 + 3\text{Cl}_2 \xrightarrow{\text{h}\nu} \text{C}_6\text{H}_6\text{Cl}_6 $$
Solution:- When benzene is exposed to chlorine gas in the presence of strong sunlight (UV radiation), the pi cloud is broken via a free-radical mechanism.
- Three molecules of \( \text{Cl}_2 \) add entirely across the three double bonds.
- The actual chemical product is Hexachlorocyclohexane (\( \text{C}_6\text{H}_6\text{Cl}_6 \)), commonly known as Benzene Hexachloride (BHC).
- Note: The provided option C lists "Hexachlorobenzene" (\( \text{C}_6\text{Cl}_6 \)). While technically Hexachlorobenzene is a substitution product, in the context of many historical exam papers, this option is frequently used as a misnomer for the exhaustive chlorination product. Based on the options provided, it is the only answer reflecting the multi-chlorinated carbon ring structure.
Why other options are incorrect:- HCl is a byproduct of substitution, not an addition product.
- Chloroform is a single-carbon halocarbon.
- PVC is a polymer of vinyl chloride.
#8 of 58
NUMS 2024
Permalink
Identify the alkene which gives only one type of aldehyde upon ozonolysis:
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Correct Key: Option C
Diagnostic Explanation
Concept:Ozonolysis cleaves a \( \text{C}=\text{C} \) double bond, placing a \( \text{C}=\text{O} \) double bond on each resulting fragment. To yield only
one type of aldehyde, the alkene must be perfectly symmetrical and have exactly one hydrogen on each double-bonded carbon.
Solution:- A. 1-Pentene (\( \text{CH}_2=\text{CHCH}_2\text{CH}_2\text{CH}_3 \)): Asymmetrical. Yields formaldehyde and butanal (two different aldehydes).
- B. 2-Methylpropene (\( (\text{CH}_3)_2\text{C}=\text{CH}_2 \)): Yields acetone (a ketone) and formaldehyde.
- D. 2,3-Dimethyl-2-butene (\( (\text{CH}_3)_2\text{C}=\text{C}(\text{CH}_3)_2 \)): Symmetrical, but yields two molecules of acetone (a ketone, no aldehydes).
- C. 2-Butene (\( \text{CH}_3\text{CH}=\text{CHCH}_3 \)): Symmetrical. Cleavage exactly in the middle yields two identical molecules of acetaldehyde (\( \text{CH}_3\text{CHO} \)). This is the only option that produces exactly one type of aldehyde.
Why other options are incorrect:- They either produce a mixture of different compounds, or they produce ketones rather than aldehydes.
#9 of 58
NUMS 2024
Permalink
The most readily sulphonated compound is:
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:The rate of electrophilic aromatic substitution (like sulfonation) depends on the electron density of the aromatic ring. Electron-donating groups activate the ring (making it react faster), while electron-withdrawing groups deactivate it.
Solution:- Benzene: The baseline standard.
- Nitrobenzene: The \( -\text{NO}_2 \) group is highly electronegative and strongly withdraws electrons via resonance, heavily deactivating the ring.
- Chlorobenzene: Halogens withdraw electrons inductively, making the ring mildly deactivated compared to benzene.
- Toluene (Methylbenzene): The methyl group (\( -\text{CH}_3 \)) pushes electron density into the ring through inductive effects and hyperconjugation. This makes the ring significantly more electron-rich and thus most reactive/readily sulphonated among the choices.
Why other options are incorrect:- Benzene lacks the activating methyl group.
- Chlorobenzene and Nitrobenzene are deactivated and react much slower than pure benzene.
#10 of 58
PMC 2021
Permalink
Benzene does not show
B
Substitution receptions
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Correct Key: Option D
Diagnostic Explanation
Concept:Benzene undergoes substitution readily to preserve its resonance stability. It can undergo harsh addition and oxidation, but lacks a mechanism for basic elimination.
Solution:- Elimination reactions require saturated \( \text{sp}^3 \) carbons with adjacent leaving groups (like alkyl halides) to form a new pi bond.
- Benzene is already fully unsaturated and aromatic. Attempting to "eliminate" from it to create a triple bond in a six-membered ring would create unbearable ring strain (though transient arynes exist, they are extreme intermediates, not standard reactions).
- Therefore, chemically speaking in the context of standard pathways, benzene does not show elimination.
Why other options are incorrect:- Benzene easily undergoes electrophilic Substitution.
- Under extreme conditions, it can undergo Addition (with \( \text{Cl}_2 \) + UV, or \( \text{H}_2 \) + Ni) and Oxidation (with \( \text{V}_2\text{O}_5 \)).
#11 of 58
PMC 2021
Permalink
Which of the following is marsh gas
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Correct Key: Option D
Diagnostic Explanation
Concept:Marsh gas is a historical and common name for the gas primarily produced by the anaerobic bacterial decomposition of organic matter in wetlands and swamps.
Solution:- The simplest alkane, Methane (\( \text{CH}_4 \)), is the chief constituent of the biogas produced in marshy areas.
- Due to its origin, it was historically referred to as "marsh gas".
Why other options are incorrect:- Ethane, propane, and butane are heavier alkanes derived primarily from petroleum and natural gas reserves, not the primary product of surface-level swamp decomposition.
#12 of 58
PMC 2021
Permalink
Which of the following act as electrophile during nitration of benzene?
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Correct Key: Option D
Diagnostic Explanation
Concept:Nitration relies on the creation of a highly electrophilic, positively charged nitrogen species from nitric acid.
Formula:$$ \text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightleftharpoons \text{NO}_2^+ + \text{H}_3\text{O}^+ + 2\text{HSO}_4^- $$
Solution:- Concentrated sulfuric acid protonates nitric acid, forcing it to lose a water molecule.
- The resulting species is the nitronium ion, \( \text{NO}_2^+ \).
- This linear ion contains a central nitrogen with a full positive charge and empty orbitals, making it the aggressive electrophile that attacks the benzene pi cloud.
Why other options are incorrect:- \( \text{NO}^+ \) is the nitrosonium ion, used in nitrosation, not nitration.
- \( \text{NO}_3^+ \) and \( \text{HNO}_2^+ \) are not the stable active intermediates generated in standard nitrating mixtures.
#13 of 58
PMDC 2020
Permalink
Electrophile in sulphonation of benzene is
B
\( \text{H}_2\text{SO}_4 \)
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Electrophilic aromatic sulfonation relies on a neutral, heavily electron-deficient molecule rather than a positively charged ion as its primary electrophile.
Formula:$$ 2\text{H}_2\text{SO}_4 \rightleftharpoons \text{H}_3\text{O}^+ + \text{HSO}_4^- + \text{SO}_3 $$
Solution:- In concentrated or fuming sulfuric acid, an auto-ionization/dehydration equilibrium produces free sulfur trioxide (\( \text{SO}_3 \)).
- In \( \text{SO}_3 \), the highly electronegative oxygen atoms pull electron density intensely away from the central sulfur atom.
- This renders the sulfur atom highly electron-deficient, making the neutral \( \text{SO}_3 \) molecule a powerful electrophile capable of attacking the benzene ring.
Why other options are incorrect:- \( \text{H}_2\text{SO}_4 \) is the bulk solvent/reagent, not the active electrophile.
- \( \text{HSO}_4^- \) is an anion and acts as a base, not an electrophile.
#14 of 58
PMDC 2020
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Acetophenone can be formed by which of the following reaction of benzene?
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Acetophenone (phenyl methyl ketone, \( \text{C}_6\text{H}_5\text{COCH}_3 \)) features a carbonyl group directly attached to the benzene ring. Introducing a carbonyl-containing group is called acylation.
Solution:- To synthesize acetophenone, benzene must be reacted with acetyl chloride (\( \text{CH}_3\text{COCl} \)) in the presence of a Lewis acid (like \( \text{AlCl}_3 \)).
- The electrophile generated is the acylium ion (\( \text{CH}_3\text{C}^+=\text{O} \)).
- Because an acyl group (\( \text{R}-\text{CO}- \)) is being substituted onto the aromatic ring, the reaction is universally classified as Friedel-Crafts Acylation.
Why other options are incorrect:- Alkylation introduces simple alkyl groups (forming products like toluene or ethylbenzene).
- Halogenation introduces Cl, Br, etc.
- Nitration introduces an \( -\text{NO}_2 \) group.
#15 of 58
PMDC 2020
Permalink
When \( \text{CH}_3 \) is attached with the benzene ring, it makes the ring:
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Alkyl groups are electron-donating. A higher electron density on an aromatic ring enhances its ability to seek out and attack positive centers (electrophiles).
Solution:- The methyl group (\( -\text{CH}_3 \)) pushes electrons into the benzene ring via the positive inductive effect (+I) and hyperconjugation.
- This localized increase in pi-electron density makes the ring highly rich in negative charge.
- An electron-rich species that attacks electrophiles is by definition a good nucleophile. Thus, toluene is a stronger nucleophile than bare benzene.
Why other options are incorrect:- Electrophiles seek electrons; the electron-rich toluene ring repels other electrons, meaning it does not act as an electrophile.
- The ring is already a resonance hybrid; attaching \( \text{CH}_3 \) doesn't initiate that property.
- It actually becomes more reactive (less kinetically stable) toward substitution than plain benzene.
#16 of 58
SET 2019
Permalink
In Friedal Craft's Chlorination of Benzene, Iron III chloride acts as a catalyst, which is the initial reaction in the generation of an electrophile?
A
\( \text{Cl}_2 + \text{FeCl}_3 \rightarrow \text{Cl}^- + \text{FeCl}_4^+ \)
B
\( \text{Cl}_2 + \text{FeCl}_3 \rightarrow 2\text{Cl}^+ + \text{FeCl}_2^- \)
C
\( \text{Cl}_2 + \text{FeCl}_3 \rightarrow \text{Cl}^- + \text{FeCl}_3^+ \)
D
\( \text{Cl}_2 + \text{FeCl}_3 \rightarrow \text{Cl}^+ + \text{FeCl}_4^- \)
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Lewis acid catalysts facilitate electrophilic aromatic substitution by polarizing and breaking the halogen-halogen bond to create a strong electrophile.
Solution:- Iron(III) chloride (\( \text{FeCl}_3 \)) is an electron-deficient Lewis acid with empty d-orbitals.
- It accepts a pair of electrons from one of the chlorine atoms in \( \text{Cl}_2 \).
- This causes the Cl-Cl bond to break heterolytically. The chlorine atom attached to Fe leaves with the bonding electrons to form the complex anion \( \text{FeCl}_4^- \).
- The other chlorine is left deficient in electrons, forming the essential electrophile, \( \text{Cl}^+ \).
- The precise balanced generation equation is: \( \text{Cl}_2 + \text{FeCl}_3 \rightarrow \text{Cl}^+ + \text{FeCl}_4^- \).
Why other options are incorrect:- Equations proposing the formation of \( \text{Cl}^- \) alongside a positively charged Iron complex violate the fundamental behavior of Lewis acids, which accept (not donate) electron pairs to become negatively charged complexes.
#17 of 58
SET 2019
Permalink
Which of the options show all possible products of combustion of Butene?
A
\( \text{CO} + \text{CO}_2 + \text{H}_2\text{O} \)
B
\( \text{CO}_2 + \text{H}_2\text{O} \)
C
\( \text{C} + \text{CO} + \text{CO}_2 \)
D
\( \text{C} + \text{CO} + \text{CO}_2 + \text{H}_2\text{O} \)
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Combustion of hydrocarbons yields different products depending strictly on the availability of oxygen.
Solution:- Complete Combustion: With abundant oxygen, hydrocarbons oxidize entirely to Carbon Dioxide (\( \text{CO}_2 \)) and Water (\( \text{H}_2\text{O} \)).
- Incomplete Combustion: When oxygen is limited, the carbon does not fully oxidize. This produces toxic Carbon Monoxide (\( \text{CO} \)), unburned Carbon soot (\( \text{C} \)), and Water (\( \text{H}_2\text{O} \)).
- Since the question asks for all possible products of combustion (encompassing both complete and incomplete scenarios), the aggregate list must include all these species: Carbon (C), Carbon Monoxide (CO), Carbon Dioxide (\( \text{CO}_2 \)), and Water (\( \text{H}_2\text{O} \)).
Why other options are incorrect:- Other options are partial lists that omit either the solid carbon (soot) from severe incomplete combustion, or the water vapor which is invariably formed in all combustion of hydrocarbons.
#18 of 58
SET 2019
Permalink
In the reaction sequence given here: \( \text{H}_3\text{C}-\text{CH}_3 + \text{Br}_2 \xrightarrow{h\nu} \text{A} \xrightarrow{\text{Alc KOH}} \text{B} \). The end product is an unsaturated hydrocarbon. Identify the nature of reaction in the two steps.
A
Step I is a nucleophilic substitution and step II is elimination
B
Step I is addition and step II is nucleophilic substitution
C
Step I is free radical substitution and step II nucleophilic substitution
D
Step I is free radical substitution and step II is elimination
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Halogenation of alkanes in UV light proceeds via radicals, and reacting the resulting alkyl halide with alcoholic KOH triggers an elimination.
Solution:- Step I: Ethane (\( \text{H}_3\text{C}-\text{CH}_3 \)) reacts with \( \text{Br}_2 \) under UV light (\( h\nu \)). UV light initiates a homolytic cleavage, resulting in a free radical substitution that yields ethyl bromide (Compound A).
- Step II: Ethyl bromide is treated with alcoholic KOH. Alcoholic KOH is a strong base that removes a proton and the bromide ion from adjacent carbons. This process is elimination (specifically dehydrohalogenation), yielding ethene (Compound B), an unsaturated hydrocarbon.
Why other options are incorrect:- Step I is not nucleophilic substitution or addition, because alkanes only react via free radicals under these conditions.
- Step II is not nucleophilic substitution; substitution would require aqueous KOH to produce an alcohol, not the unsaturated hydrocarbon mentioned in the stem.
#19 of 58
SET 2019
Permalink
Homolysis of a covalent bond yields a very reactive species with incomplete octet in its valence shell. What is this species?
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Bond cleavage can be heterolytic (unequal sharing) or homolytic (equal sharing).
Solution:- In homolysis (homolytic fission), a covalent bond breaks symmetrically.
- Each atom involved in the bond retains exactly one electron from the shared pair.
- The resulting species possesses an unpaired electron, lacks a complete octet, is highly reactive, and is electrically neutral. This is the exact definition of a free radical.
Why other options are incorrect:- Complex ions and normal electrophiles/nucleophiles are typically formed via heterolytic fission, resulting in fully charged positive or negative species.
#20 of 58
NUMS 2019
Permalink
Order of reactivity of halogen toward alkane is
A
\( \text{F}_2 > \text{I}_2 > \text{Br}_2 > \text{Cl}_2 \)
B
\( \text{F}_2 > \text{Br}_2 > \text{Cl}_2 > \text{I}_2 \)
C
\( \text{F}_2 > \text{Cl}_2 > \text{Br}_2 > \text{I}_2 \)
D
\( \text{F}_2 > \text{Cl}_2 > \text{I}_2 > \text{Br}_2 \)
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:The reactivity of halogens in free-radical substitution of alkanes is dictated by the electronegativity of the halogen, the weakness of the X-X bond, and the high exothermicity of forming the H-X bond.
Solution:- Fluorine is incredibly reactive due to a weak F-F bond and the highly exothermic nature of the reaction. It often reacts explosively.
- Chlorine reacts rapidly under UV light.
- Bromine is less reactive and highly selective, reacting slower.
- Iodine is the least reactive; its reaction is endothermic, slow, and highly reversible.
- Therefore, the strictly descending order of reactivity follows the periodic group: \( \text{F}_2 > \text{Cl}_2 > \text{Br}_2 > \text{I}_2 \).
Why other options are incorrect:- Any other sequence breaks the periodic trend correlating reactivity with electronegativity and bond energies of the halogens.
#21 of 58
NUMS 2019
Permalink
Acetone can be obtained by ozonolysis of:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Ozonolysis cleaves the \( \text{C}=\text{C} \) double bond. A double bond attached to two methyl groups on one carbon yields a ketone (acetone) upon cleavage.
Formula:$$ (\text{CH}_3)_2\text{C}=\text{CH}_2 \xrightarrow{\text{O}_3, \text{Zn/H}_2\text{O}} (\text{CH}_3)_2\text{C}=\text{O} + \text{CH}_2\text{O} $$
Solution:- Iso-butene (2-methylpropene) has the structure \( (\text{CH}_3)_2\text{C}=\text{CH}_2 \).
- When the double bond is cleaved by ozone, the carbon bonded to the two methyl groups oxidizes to become a ketone: Acetone (\( \text{CH}_3-\text{CO}-\text{CH}_3 \)).
- The terminal \( \text{CH}_2 \) group oxidizes to formaldehyde (\( \text{HCHO} \)).
Why other options are incorrect:- 2-butene (\( \text{CH}_3\text{CH}=\text{CHCH}_3 \)) yields two molecules of acetaldehyde.
- 1-butene yields propionaldehyde and formaldehyde.
- 2-butyne would yield carboxylic acids upon oxidative cleavage.
#22 of 58
NUMS 2019
Permalink
Which catalyst is used in oxidation of benzene ring?
B
\( \text{Fe} + \text{Al}_2\text{O}_3 \)
C
\( \text{V}_2\text{O}_5 \)
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:The benzene ring is exceptionally stable and resists normal oxidizing agents (like \( \text{KMnO}_4 \) or \( \text{K}_2\text{Cr}_2\text{O}_7 \)). Harsh catalytic conditions are required to break the ring.
Solution:- To forcefully oxidize and cleave the aromatic benzene ring itself, it is heated strongly (around 450°C) with oxygen in the presence of a specific catalyst.
- Vanadium pentoxide (\( \text{V}_2\text{O}_5 \)) serves as the ideal catalyst for this severe oxidation, ultimately yielding maleic anhydride.
Why other options are incorrect:- \( \text{FeBr}_3 \) is a Lewis acid used for halogenation (substitution), not oxidation.
- Raney Ni is a powerful catalyst for reduction (hydrogenation) of benzene, not oxidation.
#23 of 58
NUMS 2019
Permalink
Which of the following compound react slower than benzene in electrophilic substitution reaction
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:The reactivity of substituted benzenes depends on whether the attached group pushes electrons into the ring (activating) or pulls electrons out (deactivating).
Solution:- Electrophilic substitution requires an electron-rich aromatic ring to attack a positive electrophile.
- The nitro group (\( -\text{NO}_2 \)) is a very strong electron-withdrawing group via both induction and resonance.
- It dramatically drains electron density from the pi system, making the ring heavily electron-deficient.
- Therefore, Nitrobenzene is strongly deactivated and reacts much slower than unsubstituted benzene.
Why other options are incorrect:- Phenol (\( -\text{OH} \)), Aniline (\( -\text{NH}_2 \)), and Toluene (\( -\text{CH}_3 \)) all contain electron-donating groups. They increase the ring's electron density, making them react faster than benzene.
#24 of 58
ETEA 2019
Permalink
A gas decolorizes alkaline \( \text{KMnO}_4 \) solution but does not give any PPT with ammonical \( \text{AgNO}_3 \)
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Chemical tests allow differentiation between saturated, double-bonded, and terminal triple-bonded hydrocarbons.
Solution:- Baeyer's Test (Alkaline \( \text{KMnO}_4 \)): Decolorization indicates the presence of a pi bond (unsaturation). Both alkenes and alkynes decolorize \( \text{KMnO}_4 \).
- Tollens' Test (Ammoniacal \( \text{AgNO}_3 \)): Formation of a precipitate indicates the presence of an acidic terminal alkyne (like ethyne).
- The unknown gas decolorizes \( \text{KMnO}_4 \) (meaning it is unsaturated) but does NOT form a precipitate with \( \text{AgNO}_3 \) (meaning it is not a terminal alkyne).
- Ethylene (ethene) is an alkene. It is unsaturated (passes Baeyer's test) but lacks acidic protons (fails Tollens' test).
Why other options are incorrect:- Methane and Ethane are saturated alkanes; they fail Baeyer's test entirely (will not decolorize \( \text{KMnO}_4 \)).
#25 of 58
ETEA 2019
Permalink
An olefin “X” on ozonolysis gives \( \text{CH}_3\text{CH}_2\text{COCH}_3 \) and \( \text{CH}_3\text{COCH}_3 \). The IUPAC name of X is:
C
2-3 Di methyl-2-pentene
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Ozonolysis cleaves a \( \text{C}=\text{C} \) double bond and replaces it with two \( \text{C}=\text{O} \) double bonds. The original alkene can be determined by mentally erasing the oxygens and stitching the two carbonyl carbons back together via a double bond.
Solution:- Product 1 is 2-butanone: \( \text{CH}_3\text{CH}_2-\text{C}(=\text{O})-\text{CH}_3 \).
- Product 2 is acetone: \( \text{CH}_3-\text{C}(=\text{O})-\text{CH}_3 \).
- Align the carbonyl groups: \( \text{CH}_3\text{CH}_2-\text{C}(\text{CH}_3)=\text{O} \) and \( \text{O}=\text{C}(\text{CH}_3)_2 \).
- Remove the oxygens and connect the carbons with a double bond: \( \text{CH}_3\text{CH}_2-\text{C}(\text{CH}_3)=\text{C}(\text{CH}_3)_2 \).
- Numbering the longest chain (5 carbons) to give the double bond and substituents the lowest numbers: 2,3-dimethyl-2-pentene.
Why other options are incorrect:- 2-butene and 2-pentene are straight-chain alkenes that would yield aldehydes, not ketones.
- 1-hexene is a terminal alkene that yields formaldehyde and pentanal.
#26 of 58
MDCAT 2019
Permalink
Treatment of ethene with cold sulphuric acid followed by reaction with boiling water yields:
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:The indirect hydration of alkenes involves intermediate formation of an alkyl hydrogen sulfate, which is then hydrolyzed by water to produce an alcohol.
Formula:$$ \text{CH}_2=\text{CH}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{CH}_3\text{CH}_2\text{OSO}_3\text{H} $$
$$ \text{CH}_3\text{CH}_2\text{OSO}_3\text{H} + \text{H}_2\text{O} \xrightarrow{\text{boil}} \text{CH}_3\text{CH}_2\text{OH} + \text{H}_2\text{SO}_4 $$
Solution:- Ethene reacts with cold concentrated \( \text{H}_2\text{SO}_4 \) via electrophilic addition to form ethyl hydrogen sulfate.
- When this intermediate is boiled with water, a hydrolysis reaction occurs, stripping the sulfate group and replacing it with a hydroxyl group (\( -\text{OH} \)).
- The final product is the two-carbon alcohol, Ethanol.
Why other options are incorrect:- Ethyne is an alkyne, representing a higher oxidation state.
- Ethanal is an aldehyde, which would require an oxidation step.
- Ethane requires direct catalytic hydrogenation (addition of \( \text{H}_2 \)), not hydration.
#27 of 58
MDCAT 2019
Permalink
Alkenes undergo:
A
Nucleophilic substitution
C
Electrophilic substitution
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:The characteristic reaction of a functional group is dictated by its electronic nature. Alkenes contain an electron-rich, easily accessible pi bond.
Solution:- Because the pi bond (\( \text{C}=\text{C} \)) has loosely held electrons situated above and below the molecular plane, it acts as a strong nucleophile.
- It instinctively attacks electron-deficient species (electrophiles).
- The pi bond breaks entirely, allowing new atoms to add to the carbon skeleton without any atoms leaving, thereby restoring stable single bonds.
- Consequently, the primary, characteristic reaction mechanism for alkenes is Electrophilic Addition.
Why other options are incorrect:- Nucleophiles are repelled by the high electron density of the pi bond, making nucleophilic attacks rare.
- Substitution implies removing an atom, which is less energetically favorable than simply adding across the easily broken pi bond. (Electrophilic substitution is typical for stable aromatic rings).
#28 of 58
MDCAT 2019
Permalink
Which of the following molecule shows cis – trans isomers?
A
\( \text{C}_2\text{HCl}_3 \)
B
\( \text{C}_2\text{H}_4 \)
C
\( \text{C}_2\text{H}_2\text{Cl}_2 \)
D
\( \text{C}_2\text{H}_2\text{Br}_2 \)
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Geometrical (cis-trans) isomerism requires restricted rotation (like a double bond) and for
each carbon atom of the double bond to be attached to two
different groups.
Solution:- Let's analyze \( \text{C}_2\text{H}_2\text{Br}_2 \) (1,2-dibromoethene): \( \text{CHBr}=\text{CHBr} \).
- Carbon 1 is attached to one H and one Br. Carbon 2 is attached to one H and one Br.
- Because both carbons have distinct groups, they can be arranged with the two Bromine atoms on the same side (cis) or on opposite sides (trans) of the double bond.
- Note: Option C (\( \text{C}_2\text{H}_2\text{Cl}_2 \)) can also show it if it is 1,2-dichloroethene, but based on the provided Answer Key, D was the specifically intended valid option (likely distinguishing from a 1,1-isomer not explicitly ruled out in other contexts). Both C and D mathematically can, but D is keyed.
Why other options are incorrect:- \( \text{C}_2\text{H}_4 \) (Ethene) has identical hydrogen atoms on both carbons.
- \( \text{C}_2\text{HCl}_3 \) (Trichloroethene) has two identical chlorine atoms on one of its carbons, making cis-trans impossible.
#29 of 58
MDCAT 2019
Permalink
In the reaction sequence:
\( \text{H}_3\text{C}-\text{CH}_2-\text{CH}_2-\text{Br} + \text{Alc.KOH} \rightarrow \text{C} \xrightarrow{\text{H}_2\text{SO}_4 / \text{H}_2\text{O}} \text{D} \)
Product D will be
A
Mixture of methanol and ethanol
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Elimination of a primary alkyl halide yields a terminal alkene. Hydration of that terminal alkene follows Markovnikov's rule, yielding a secondary alcohol.
Solution:- Step 1: 1-bromopropane (\( \text{H}_3\text{C}-\text{CH}_2-\text{CH}_2-\text{Br} \)) is treated with alcoholic KOH. This is a dehydrohalogenation (elimination) reaction, yielding the alkene propene (Compound C): \( \text{CH}_3-\text{CH}=\text{CH}_2 \).
- Step 2: Propene is hydrated using \( \text{H}_2\text{SO}_4 \) and water. This is an electrophilic addition of \( \text{H}_2\text{O} \).
- Following Markovnikov's rule, the H adds to the terminal carbon (\( \text{CH}_2 \)) and the OH adds to the central carbon (\( \text{CH} \)).
- This produces \( \text{CH}_3-\text{CH}(\text{OH})-\text{CH}_3 \), which is 2-propanol.
Why other options are incorrect:- 1-Propanol would be the product if anti-Markovnikov hydration (hydroboration-oxidation) were used instead of acid-catalyzed hydration.
- Propanoic acid requires an oxidation reaction.
#30 of 58
MDCAT 2019
Permalink
Which of the following reactions is used for the production of alcohols on industrial scale?
A
Hydrohalogenation of alkenes
B
Hydroxylation of alkenes
C
Hydrogenation of alkenes
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Industrial synthesis requires cheap, abundantly available precursors. Ethene from petroleum cracking and steam (water) fit this requirement perfectly.
Solution:- The direct hydration of alkenes (adding \( \text{H}_2\text{O} \) across a double bond) is the standard industrial method to synthesize large quantities of simple alcohols.
- For example, reacting ethene gas with high-pressure steam in the presence of a phosphoric acid catalyst yields vast amounts of industrial ethanol.
Why other options are incorrect:- Hydrohalogenation adds HX, yielding alkyl halides, not alcohols.
- Hydrogenation adds \( \text{H}_2 \), yielding alkanes.
- Hydroxylation uses expensive reagents (like \( \text{KMnO}_4 \) or \( \text{OsO}_4 \)) to form diols, which is not feasible for bulk mono-alcohol synthesis.
#31 of 58
MDCAT 2019
Permalink
Which derivative of benzene shows maximum reactivity in electrophilic substitution reactions?
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Ring reactivity is dictated by attached functional groups. Electron-donating groups (EDGs) activate the ring, while electron-withdrawing groups (EWGs) deactivate it.
Solution:- Methyl benzene (Toluene): The methyl group (\( -\text{CH}_3 \)) pushes electron density into the ring via the inductive effect and hyperconjugation. This makes the ring more electron-rich and highly reactive to electrophiles.
- Benzaldehyde, Benzoic acid, and Nitrobenzene: All contain potent EWGs (\( -\text{CHO} \), \( -\text{COOH} \), and \( -\text{NO}_2 \)) that drain electrons from the pi system through resonance, strongly deactivating the ring.
- Therefore, methyl benzene is the only activated ring listed, thus having maximum reactivity.
Why other options are incorrect:- They all contain deactivating meta-directing groups that significantly slow down electrophilic substitution compared to pure benzene.
#32 of 58
MDCAT 2018
Permalink
Reaction mechanism of alkanes with halogens is known as
C
Free radical substitution
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Alkanes are saturated hydrocarbons containing only strong sigma bonds. They lack pi electrons and are unreactive towards electrophiles or nucleophiles under standard conditions.
Solution:- In the presence of UV light or high heat, halogen molecules (like \( \text{Cl}_2 \)) undergo homolytic cleavage to form highly reactive free radicals.
- These radicals abstract a hydrogen atom from the alkane, and the resulting alkane radical reacts with another halogen molecule.
- Because a hydrogen atom is completely replaced by a halogen atom via radical intermediates, the overall mechanism is definitively named free radical substitution.
Why other options are incorrect:- Addition occurs only in unsaturated compounds (alkenes/alkynes).
- Elimination removes atoms to form double bonds; it does not replace them.
- Propagation is merely one step within the free radical mechanism, not the name of the overall reaction itself.
#33 of 58
MDCAT 2018
Permalink
Which compound is obtained by the elimination of bromopropane?
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Dehydrohalogenation (an elimination reaction) of an alkyl halide removes a hydrogen atom and a halogen atom from adjacent carbons to form an alkene.
Formula:$$ \text{CH}_3\text{CH}_2\text{CH}_2\text{Br} + \text{KOH}_{(\text{alc})} \rightarrow \text{CH}_3\text{CH}=\text{CH}_2 + \text{KBr} + \text{H}_2\text{O} $$
Solution:- Bromopropane is a 3-carbon alkyl halide.
- When subjected to an strong base (like alcoholic KOH), the bromine atom and a beta-hydrogen are eliminated.
- The 3-carbon skeleton remains intact, resulting in the formation of a double bond. The product is the 3-carbon alkene, propene.
Why other options are incorrect:- Butene and ethene have incorrect carbon chain lengths (4 and 2 carbons, respectively).
- Propane is a saturated alkane, which would be the product of reduction, not elimination.
#34 of 58
MDCAT 2018
Permalink
Bromination of alkene is shown in the following reaction. (Ethene + \( \text{Br}_2 \rightarrow \) 1,2-dibromoethane). This reaction is used for:
A
Identification of primary and secondary alcohols
D
Detection of double bond
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:The addition of molecular bromine (\( \text{Br}_2 \)) across a carbon-carbon double bond is a classic diagnostic chemical test.
Solution:- A solution of bromine in carbon tetrachloride (or water) has a distinct reddish-brown color.
- When reacted with an alkene, the bromine is rapidly consumed through an electrophilic addition reaction, forming a colorless vicinal dibromide (e.g., 1,2-dibromoethane).
- The immediate disappearance (decolorization) of the reddish-brown color provides visual confirmation of unsaturation. Therefore, it is used for the detection of a double bond.
Why other options are incorrect:- Alcohols, aldehydes, and ketones do not readily undergo addition reactions with bromine to decolorize it under these mild conditions.
#35 of 58
MDCAT 2018
Permalink
Which one of the following acts as an electrophile in the electrophilic substitution of benzene with bromine?
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Electrophilic aromatic substitution demands a powerful electrophile, normally generated via the interaction of the reagent with a Lewis acid catalyst.
Formula:$$ \text{Br}_2 + \text{FeBr}_3 \rightarrow \text{Br}^+ + \text{FeBr}_4^- $$
Solution:- When bromine (\( \text{Br}_2 \)) interacts with a Lewis acid (like \( \text{FeBr}_3 \) or \( \text{FeCl}_3 \)), the halogen bond is polarized and broken.
- This heterolytic cleavage produces the highly reactive bromonium ion, \( \text{Br}^+ \), which is electron-deficient.
- \( \text{Br}^+ \) is the actual electrophile that attacks the electron-rich pi cloud of the benzene ring.
Why other options are incorrect:- \( \text{FeCl}_4^- \) is a negatively charged counterion (a weak nucleophile), not an electrophile.
- \( \text{Fe}^{+3} \) and \( \text{Fe}^{+2} \) represent oxidation states of iron in the catalyst; they do not act as the carbon-attacking electrophile in the aromatic substitution step.
#36 of 58
MDCAT 2017
Permalink
The reaction of benzene with bromine in the presence of \( \text{FeBr}_3 \) follows the mechanism of ____ reaction.
B
Electrophilic substitution
C
Nucleophilic substitution
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Benzene is highly stable due to aromatic resonance. It resists reactions that destroy this stability (like addition) and prefers reactions that restore it.
Solution:- \( \text{FeBr}_3 \) generates the strong electrophile \( \text{Br}^+ \) from \( \text{Br}_2 \).
- The electron-rich benzene ring attacks the electrophile, temporarily breaking aromaticity.
- To rapidly regain its stable aromatic state, the intermediate expels a proton (\( \text{H}^+ \)), effectively substituting a hydrogen atom with a bromine atom.
- Therefore, the overriding mechanism is electrophilic substitution.
Why other options are incorrect:- Electrophilic addition would leave the ring saturated and non-aromatic, which is thermodynamically highly unfavorable.
- Nucleophilic reactions involve attack by electron-rich species, but benzene is itself electron-rich and strongly repels nucleophiles.
#37 of 58
MDCAT 2017
Permalink
Intermediate product formed when propanoyl chloride reacts with benzene is
D
\( \sigma \)-complex (Benzenonium ion) bearing the acyl group
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:In Friedel-Crafts acylation, the reaction proceeds through a distinct, non-aromatic cationic intermediate before restoring aromaticity.
Solution:- Propanoyl chloride reacts with a Lewis acid to form an acylium electrophile (\( \text{CH}_3\text{CH}_2\text{C}^+=\text{O} \)).
- When the benzene pi ring attacks this electrophile, a resonance-stabilized carbocation is formed within the ring.
- This intermediate—where one carbon becomes \( \text{sp}^3 \) hybridized holding both an H atom and the new acyl group—is called a \( \sigma \)-complex, arenium ion, or benzenonium ion.
Why other options are incorrect:- The acylium ion is the electrophile, but the actual intermediate product formed with benzene is the \( \sigma \)-complex.
- A phenyl radical belongs to free radical mechanisms, not electrophilic substitution.
- A simple primary carbocation is not formed in aromatic acylation.
#38 of 58
MDCAT 2017
Permalink
When benzene reacts with Acetyl chloride (\( \text{CH}_3\text{COCl} \)) in the presence of \( \text{AlCl}_3 \) acetophenone is formed. The electrophile in this reaction will be
A
\( \text{CH}_3\text{C}^+\text{O} \)
B
\( \text{C}^+\text{H}_3 \)
D
\( \text{CH}_3\text{COCl} \)
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:In Friedel-Crafts acylation, the Lewis acid catalyst serves specifically to strip a chloride ion from the acyl chloride, producing the active electrophile.
Formula:$$ \text{CH}_3\text{COCl} + \text{AlCl}_3 \rightarrow \text{CH}_3\text{C}^+=\text{O} + \text{AlCl}_4^- $$
Solution:- Acetyl chloride (\( \text{CH}_3\text{COCl} \)) loses \( \text{Cl}^- \) to \( \text{AlCl}_3 \).
- This generates the highly reactive, resonance-stabilized acylium ion: \( \text{CH}_3\text{C}^+\text{O} \).
- This acylium ion acts as the direct electrophile that attacks the benzene ring to yield acetophenone.
Why other options are incorrect:- \( \text{C}^+\text{H}_3 \) is a methyl carbocation, which would be the electrophile for Friedel-Crafts alkylation using methyl chloride, not acylation.
- \( \text{AlCl}_3 \) is the catalyst, not the attacking electrophile on the ring.
- \( \text{CH}_3\text{COCl} \) is the neutral starting reagent, not yet activated into an electrophile.
#39 of 58
MDCAT 2017
Permalink
Which of the following species are 3,5(meta) directing groups when second group is introduced into the benzene ring?
I = \( -\text{NH}_2 \)
II = \( -\text{CHO} \)
III = \( -\text{COOH} \)
IV = \( -\text{CH}_3 \)
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Substituents on a benzene ring dictate the position of subsequent electrophilic attacks. Electron-withdrawing groups (EWG) usually direct incoming electrophiles to the meta (3,5) position.
Solution:- Group I (\( -\text{NH}_2 \)) and Group IV (\( -\text{CH}_3 \)) are electron-donating groups (EDG). They activate the ring and are ortho/para directing.
- Group II (\( -\text{CHO} \)) and Group III (\( -\text{COOH} \)) contain carbonyl carbons directly attached to the ring. The oxygen atoms pull electron density away through resonance, making the ring electron-deficient, especially at the ortho and para positions.
- Consequently, \( -\text{CHO} \) and \( -\text{COOH} \) act as deactivating, meta-directing groups.
Why other options are incorrect:- Any option including I or IV is incorrect because amines and alkyl groups are ortho/para directors.
#40 of 58
MDCAT 2017
Permalink
For halogenation of benzene, which reagent is used:
A
\( \text{H}_2\text{SO}_4 \)
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Halogens alone are not electrophilic enough to break the aromaticity of benzene. A Lewis acid catalyst is strictly required.
Solution:- To substitute a halogen onto a benzene ring, a catalyst like \( \text{AlCl}_3 \), \( \text{FeCl}_3 \), or \( \text{FeBr}_3 \) is used.
- These Lewis acids possess empty d-orbitals that accept a lone pair from the halogen molecule (e.g., \( \text{Cl}_2 \)), polarizing the bond and forming the strongly electrophilic halonium ion (\( \text{Cl}^+ \)).
Why other options are incorrect:- \( \text{H}_2\text{SO}_4 \) and \( \text{HNO}_3 \) are used for nitration and sulfonation, not halogenation.
- HCl is a Brønsted acid and cannot effectively polarize molecular halogens for this reaction.
#41 of 58
MDCAT 2017
Permalink
Chlorination and Bromination mostly uses:
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Free radical halogenation of alkanes (chlorination and bromination) requires high-energy initiation to homolytically cleave the covalent halogen-halogen bond.
Solution:- The Cl-Cl and Br-Br bonds require a specific quantum of energy to undergo homolytic fission (breaking equally to form two free radicals).
- Ultraviolet (U.V.) light provides the exact frequency and sufficient energy required for this photochemical initiation step.
Why other options are incorrect:- Radiowaves and Infrared radiation are too low in energy to break covalent bonds (they cause nuclear spin transitions and molecular vibrations, respectively).
- Visible light does not possess sufficient energy to reliably initiate typical chlorination without specialized setups.
#42 of 58
MDCAT 2017
Permalink
Among the following, which one has electron withdrawing effect:
B
\( -\text{N}(\text{CH}_3)_2 \)
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Electron-withdrawing groups (EWG) usually possess atoms with a partial or full positive charge directly bonded to the ring, often due to multiple bonds with more electronegative atoms (like oxygen).
Solution:- The formyl group (\( -\text{CHO} \)) features a carbon atom double-bonded to a highly electronegative oxygen atom.
- The oxygen pulls electron density away from the carbon, leaving it with a partial positive charge. This carbon, in turn, withdraws pi electron density from the benzene ring via resonance.
- Hence, \( -\text{CHO} \) is a strong electron-withdrawing and ring-deactivating group.
Why other options are incorrect:- \( -\text{NH}_2 \) and \( -\text{N}(\text{CH}_3)_2 \) possess nitrogen atoms with a lone pair that can be donated into the ring via resonance, making them electron-donating groups (EDG).
- Iodine (\( -\text{I} \)) withdraws electrons inductively but donates them via resonance. It is considered weakly deactivating but ortho/para directing. However, compared to \( -\text{CHO} \), the aldehyde group is fundamentally the definitive EWG via strong resonance.
#43 of 58
ETEA 2016
Permalink
The correct order of the reactivity of hydrocarbon given below is
A
\( \text{C}_2\text{H}_4 > \text{C}_2\text{H}_2 > \text{C}_6\text{H}_6 \)
B
\( \text{C}_6\text{H}_6 > \text{C}_2\text{H}_4 > \text{C}_2\text{H}_2 \)
C
\( \text{C}_2\text{H}_2 > \text{C}_2\text{H}_4 > \text{C}_6\text{H}_6 \)
D
\( \text{C}_2\text{H}_2 > \text{C}_6\text{H}_6 > \text{C}_2\text{H}_4 \)
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Reactivity of hydrocarbons towards electrophilic attack is based on the availability and energy of their pi-electrons.
Solution:- Ethene (\( \text{C}_2\text{H}_4 \), an alkene): Has one loosely held pi bond whose electron density sits above and below the plane. It is highly accessible to electrophiles, making it the most reactive.
- Ethyne (\( \text{C}_2\text{H}_2 \), an alkyne): Possesses two pi bonds, but the sp-hybridized carbons pull the electron density closer and tighter, making the pi electrons less available than in alkenes. It is less reactive than ethene toward electrophiles.
- Benzene (\( \text{C}_6\text{H}_6 \)): Contains a cyclic, delocalized pi electron system. This resonance creates extraordinary thermodynamic stability, making it highly unreactive toward addition reactions compared to both alkenes and alkynes.
- Therefore, the correct reactivity sequence is \( \text{C}_2\text{H}_4 > \text{C}_2\text{H}_2 > \text{C}_6\text{H}_6 \).
Why other options are incorrect:- Any order placing benzene or ethyne before ethene misrepresents the resonance stability of benzene and the tight sp-orbital electronegativity of alkynes.
#44 of 58
ETEA 2016
Permalink
Select meta directing group of the following?
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Meta-directing groups withdraw electron density from the benzene ring, typically via resonance, possessing an atom with a partial positive charge directly attached to the ring.
Solution:- The cyano group (\( -\text{C}\equiv\text{N} \)) consists of a carbon atom triple-bonded to a highly electronegative nitrogen.
- The nitrogen pulls electron density, placing a partial positive charge on the carbon. This carbon subsequently withdraws electrons from the benzene ring through resonance.
- This makes \( -\text{CN} \) a strong deactivating, meta-directing group.
Why other options are incorrect:- \( -\text{OH} \), \( -\text{NR}_2 \), and \( -\text{OR} \) all have atoms (O or N) directly attached to the ring containing lone pairs. These lone pairs are donated into the ring via resonance, rendering them activating, ortho/para directing groups.
#45 of 58
MDCAT 2015
Permalink
Order of reactivity of alkenes with hydrogen halide is
A
\( \text{HBr} > \text{HI} > \text{HCl} \)
B
\( \text{HI} > \text{HCl} > \text{HBr} \)
C
\( \text{HF} > \text{HI} > \text{HCl} \)
D
\( \text{HI} > \text{HBr} > \text{HCl} \)
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:The reactivity of hydrogen halides (HX) towards alkenes is determined by the bond dissociation energy of the H-X bond.
Solution:- Electrophilic addition requires the H-X bond to break in order to donate a proton (\( \text{H}^+ \)) to the alkene.
- As you move down Group 17, atomic size increases, making the H-X bond longer and significantly weaker.
- Because the H-I bond is the weakest, HI donates a proton most readily, making it the most reactive. HCl has a stronger, shorter bond and is least reactive.
- Thus, the correct reactivity order is: \( \text{HI} > \text{HBr} > \text{HCl} \).
Why other options are incorrect:- Any sequence placing HBr or HF as the most reactive is fundamentally flawed because HF has the highest bond dissociation energy and is highly unreactive towards alkenes in this context.
#46 of 58
MDCAT 2015
Permalink
Which one of the following is a powerful electrophile used to attack on the electrons of benzene ring?
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Aromatic rings are electron-rich and require a strongly electron-deficient species (an electrophile) to initiate substitution.
Solution:- During the halogenation (chlorination) of benzene, neutral \( \text{Cl}_2 \) is not powerful enough to disrupt the aromatic pi cloud.
- A Lewis acid catalyst (like \( \text{FeCl}_3 \)) polarizes and breaks the Cl-Cl bond to generate the chloronium ion, \( \text{Cl}^+ \).
- \( \text{Cl}^+ \) has a complete positive charge and an empty orbital, making it a powerful electrophile that can successfully attack the benzene ring.
Why other options are incorrect:- \( \text{Cl}_2 \) is relatively weak and neutral.
- \( \text{FeCl}_4^- \) is a negatively charged complex ion (a nucleophile/base), not an electrophile.
- \( \text{FeCl}_2 \) is a salt of Iron(II), not the active attacking species.
#47 of 58
MDCAT 2014
Permalink
Addition of unsymmetrical reagent to an unsymmetrical alkene is governed by:
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Regioselectivity in the electrophilic addition of alkenes is guided by the stability of the intermediate carbocation.
Solution:- When an unsymmetrical reagent (like HX) adds to an unsymmetrical alkene, the reaction strictly follows Markovnikov's rule.
- The rule states that the positive part of the reagent (usually \( \text{H}^+ \)) attaches to the double-bonded carbon bearing the greater number of hydrogen atoms.
- This ensures the formation of the most highly substituted, and thus most stable, intermediate carbocation.
Why other options are incorrect:- Cannizzaro and Aldol reactions are specific to aldehydes/ketones, not alkenes.
- Kirchhoff's rules govern electrical circuits and thermodynamics, not organic reaction regiochemistry.
#48 of 58
MDCAT 2013
Permalink
What is the product formed when propene reacts with HBr?
A
\( \text{CH}_3-\text{CH}_2-\text{CH}_2\text{Br} \)
B
\( \text{CH}_3-\text{CH}(\text{Br})-\text{CH}_3 \)
C
\( \text{BrCH}_2-\text{CH}=\text{CHBr} \)
D
\( \text{CH}_3-\text{C}(\text{Br})_2-\text{CH}_3 \)
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:The electrophilic addition of an unsymmetrical hydrogen halide (HBr) to an unsymmetrical alkene (propene) follows Markovnikov's rule.
Solution:- Propene is \( \text{CH}_3-\text{CH}=\text{CH}_2 \).
- According to Markovnikov's rule, the electrophile (\( \text{H}^+ \)) adds to the \( \text{sp}^2 \) carbon with the greater number of hydrogen atoms to form the most stable intermediate carbocation.
- \( \text{H}^+ \) adds to \( \text{CH}_2 \), yielding a stable secondary carbocation (\( \text{CH}_3-\text{C}^+\text{H}-\text{CH}_3 \)).
- The bromide ion (\( \text{Br}^- \)) then attacks the central carbon, resulting in 2-bromopropane: \( \text{CH}_3-\text{CH}(\text{Br})-\text{CH}_3 \).
Why other options are incorrect:- Option A (1-bromopropane) would require anti-Markovnikov addition (only occurs with HBr in the presence of peroxides).
- Option C and D imply multiple additions or substitutions, which does not happen with a single equivalent of HBr on an alkene.
#49 of 58
MDCAT 2013
Permalink
The introduction of an alkyl group in benzene takes place in the presence of \( \text{AlCl}_3 \) and:
A
\( \text{R}-\text{C}(=\text{O})-\text{OH} \)
B
\( \text{R}-\text{C}(=\text{O})-\text{Cl} \)
C
\( \text{R}-\text{Cl} \)
D
\( \text{R}-\text{C}(=\text{O})-\text{O}-\text{R} \)
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Friedel-Crafts alkylation requires an alkylating agent to generate an alkyl carbocation (\( \text{R}^+ \)) when reacted with a Lewis acid catalyst like \( \text{AlCl}_3 \).
Solution:- To attach a standard alkyl group (\( \text{R} \)) to benzene, you must start with an alkyl halide.
- The reagent \( \text{R}-\text{Cl} \) (an alkyl chloride) reacts with \( \text{AlCl}_3 \) to form an \( \text{R}^+ \) electrophile.
- This electrophile attacks the benzene ring, completing the alkylation.
Why other options are incorrect:- \( \text{R}-\text{C}(=\text{O})-\text{Cl} \) (acyl chloride) is used for Friedel-Crafts acylation.
- \( \text{R}-\text{C}(=\text{O})-\text{OH} \) (carboxylic acid) and ester derivatives do not readily undergo Friedel-Crafts reactions under simple \( \text{AlCl}_3 \) conditions.
#50 of 58
MDCAT 2012
Permalink
In the reaction of ethene with bromine the intermediate formed is
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Electrophilic addition of halogens to alkenes does not proceed through a standard open carbocation, but rather a stable three-membered cyclic intermediate.
Solution:- When ethene (\( \text{CH}_2=\text{CH}_2 \)) reacts with \( \text{Br}_2 \), the pi electrons attack the electrophilic bromine.
- Instead of a full positive charge resting on one carbon, the bromine atom shares its lone pairs to bridge both carbons, forming a three-membered ring with a positive charge on bromine.
- This intermediate is called the cyclic bromonium ion.
Why other options are incorrect:- A primary carbocation is highly unstable and would lead to rearrangement or unstereospecific addition; the cyclic intermediate explains the strict anti-addition of halogens.
- Free radicals are involved in UV-catalyzed substitution, not cold electrophilic addition.
- Carbanions are negatively charged species, impossible to form in electrophilic attack.
#51 of 58
MDCAT 2012
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Ethene on polymerization, give the product polyethene, this reaction may be called as
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Polymerization reactions are broadly classified based on whether the monomer molecules simply add together or if a small byproduct molecule is eliminated.
Solution:- Ethene (\( \text{CH}_2=\text{CH}_2 \)) has a carbon-carbon double bond.
- During polymerization, the pi bond breaks, and the ethene monomers link together in a continuous chain without the loss of any atoms.
- Because nothing is lost and the monomers simply "add" to one another, this is strictly classified as an addition polymerization reaction.
Why other options are incorrect:- Condensation involves the loss of a small molecule (like water) when monomers join (e.g., nylon or polyester).
- Substitution replaces an atom or group, which doesn't build a polymer chain here.
- Pyrolysis is thermal decomposition (cracking) of large molecules into smaller ones.
#52 of 58
MDCAT 2012
Permalink
The introduction of R - C(=O) group in benzene is called
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Reactions introducing specific functional groups onto an aromatic ring are named based on the functional group being attached.
Solution:- The \( \text{R}-\text{C}(=\text{O})- \) group is generally known as an acyl group.
- Attaching this group to a benzene ring (typically via a Friedel-Crafts reaction with an acyl halide and \( \text{AlCl}_3 \)) is therefore formally termed acylation.
Why other options are incorrect:- Alkylation introduces a simple alkyl group (\( \text{R}- \)), not an acyl group.
- Carbonyl reduction is the process of removing oxygen/adding hydrogen to a carbonyl group (e.g., Clemmensen reduction).
- Formylation strictly refers to adding a formyl group (\( \text{H}-\text{C}(=\text{O})- \)), which is a specific, narrower subset of acylation.
#53 of 58
MDCAT 2011
Permalink
Hydrogenation of unsaturated oils is done by using
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Catalytic hydrogenation involves the addition of \( \text{H}_2 \) across double bonds (\( \text{C}=\text{C} \)) in unsaturated oils to convert them into saturated solid fats (like vegetable ghee).
Solution:- This industrial process requires a transition metal catalyst to lower the activation energy by adsorbing hydrogen gas and the alkene onto its surface.
- Finely divided nickel (Ni) is widely used industrially because it provides a large surface area and is cost-effective at high temperatures (around 200°C).
Why other options are incorrect:- Iron (Fe) is a catalyst for the Haber process (ammonia synthesis).
- Vanadium pentoxide (\( \text{V}_2\text{O}_5 \)) is used in the Contact process to oxidize \( \text{SO}_2 \) to \( \text{SO}_3 \).
- Copper is generally not active enough for direct alkene hydrogenation.
#54 of 58
MDCAT 2011
Permalink
The substitution of –H group by \( -\text{NO}_2 \) group in benzene is called
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:The nomenclature of electrophilic aromatic substitution reactions is based on the functional group being introduced onto the benzene ring.
Solution:- When a nitro group (\( -\text{NO}_2 \)) replaces a hydrogen atom on the aromatic ring, the process is definitively termed nitration.
Why other options are incorrect:- Ammonolysis involves cleavage by ammonia.
- Sulphonation introduces a sulfonic acid group (\( -\text{SO}_3\text{H} \)).
- Reduction of benzene would add hydrogen atoms (e.g., forming cyclohexane), rather than substituting them with a nitro group.
#55 of 58
MDCAT 2011
Permalink
Benzene in presence of \( \text{AlCl}_3 \) gives acetophenone when reacts with
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Friedel-Crafts acylation attaches an acyl group (\( \text{R}-\text{C}=\text{O} \)) to a benzene ring using an acyl halide and a Lewis acid catalyst.
Formula:$$ \text{C}_6\text{H}_6 + \text{CH}_3\text{COCl} \xrightarrow{\text{AlCl}_3} \text{C}_6\text{H}_5\text{COCH}_3 + \text{HCl} $$
Solution:- Acetophenone (\( \text{C}_6\text{H}_5\text{COCH}_3 \)) specifically requires an acetyl group (\( \text{CH}_3\text{CO}- \)).
- Reacting benzene with acetyl chloride (\( \text{CH}_3\text{COCl} \)) in the presence of anhydrous \( \text{AlCl}_3 \) generates the highly reactive acylium ion (\( \text{CH}_3\text{C}^+=\text{O} \)), which then attacks the ring to form acetophenone.
Why other options are incorrect:- Acetic acid and ethanoic acid (same compound) are carboxylic acids, which do not undergo Friedel-Crafts acylation directly without being converted to an acyl chloride or anhydride first.
- Ethyl benzene is a hydrocarbon product of Friedel-Crafts alkylation, not an acylating agent.
#56 of 58
MDCAT 2010
Permalink
When hydrogen atom is removed from benzene, group left is called:
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:The removal of a hydrogen atom from an aromatic hydrocarbon ring generates an aryl group.
Formula:$$ \text{C}_6\text{H}_6 \xrightarrow{-\text{H}} \text{C}_6\text{H}_5- $$
Solution:- Benzene has the molecular formula \( \text{C}_6\text{H}_6 \).
- When one hydrogen atom is abstracted, the remaining fragment is \( \text{C}_6\text{H}_5- \).
- This specific radical/group is formally named the phenyl group.
Why other options are incorrect:- An alkyl group is derived from an alkane by removing a hydrogen.
- A benzyl group is \( \text{C}_6\text{H}_5\text{CH}_2- \), formed by removing a hydrogen from the methyl group of toluene.
- A methyl group is \( \text{CH}_3- \), derived strictly from methane.
#57 of 58
MDCAT 2010
Permalink
The introduction of \( \text{NO}_2 \) group in the benzene ring is called nitration. The nitration of benzene takes place when it is heated with a 1:1 mixture of -----------at 50-55°C
A
Conc. \( \text{HNO}_3 \) and Conc. \( \text{H}_2\text{SO}_4 \)
B
Conc. \( \text{HNO}_3 \) and Conc. \( \text{HCl} \)
C
Conc. \( \text{HNO}_3 \) and Conc. Acetic acid
D
Conc. \( \text{HNO}_3 \) and Conc. \( \text{H}_3\text{PO}_4 \)
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Electrophilic aromatic substitution (nitration) requires a strong acid catalyst to generate the highly reactive nitronium ion (\( \text{NO}_2^+ \)) from nitric acid.
Formula:$$ \text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightleftharpoons \text{NO}_2^+ + \text{H}_3\text{O}^+ + 2\text{HSO}_4^- $$
Solution:- A 1:1 mixture of concentrated nitric acid (\( \text{HNO}_3 \)) and concentrated sulfuric acid (\( \text{H}_2\text{SO}_4 \)) is known as a nitrating mixture.
- Sulfuric acid acts as a stronger acid, protonating nitric acid and causing it to dehydrate, yielding the essential \( \text{NO}_2^+ \) electrophile.
Why other options are incorrect:- HCl, acetic acid, and phosphoric acid are not strong enough or suitable dehydrating agents to efficiently generate the nitronium ion for benzene nitration under these standard conditions.
#58 of 58
MDCAT 2010
Permalink
Ethyne molecule is formed when two carbon atoms joined together to form a sigma bond only:
C
\( \text{sp}^2-\text{sp}^2 \) overlap
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:In alkynes (like ethyne), the carbon atoms are sp hybridized, featuring linear geometry and a triple bond consisting of one sigma (\( \sigma \)) bond and two pi (\( \pi \)) bonds.
Solution:- Each carbon in ethyne (\( \text{HC}\equiv\text{CH} \)) utilizes sp hybridization.
- The primary sigma bond directly between the two carbon atoms is formed by the head-to-head overlap of an sp orbital from one carbon with an sp orbital from the other carbon.
Why other options are incorrect:- sp-s overlap forms the C-H sigma bonds in ethyne, not the C-C bond.
- \( \text{sp}^2-\text{sp}^2 \) overlap is characteristic of the C-C double bond in alkenes like ethene.
- 2py-2py (and 2pz-2pz) sideways overlaps form the pi (\( \pi \)) bonds, not the sigma bond.
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