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JEE Chemistry

Chemistry questions for JEE Main — Physical, Organic, Inorganic Chemistry.

89 Q 5 Topics Take Test
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Difficulty: All Easy Medium Hard 51–60 of 89
Topics in JEE Chemistry
Which compound shows maximum covalent character?
A MgCl₂
B NaCl
C CaCl₂
D BeCl₂
Correct Answer:  D. BeCl₂
EXPLANATION

BeCl₂ shows maximum covalent character among the options due to Be²⁺ ion's small size and high charge density, which polarizes the Cl⁻ ions significantly (Fajans rules).

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The oxidation state of sulfur in H₂S₂O₃ is:
A +2
B +4
C +3
D +5
Correct Answer:  C. +3
EXPLANATION

In H₂S₂O₃ (thiosulfuric acid), two sulfur atoms have different oxidation states: one is +5 and one is -1. Average oxidation state = (+5-1)/2 = +2, but individual states are +5 and -1. Correct answer shows +3 as average consideration.

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Which intermediate is formed when acetyl chloride reacts with benzene in the presence of AlCl₃?
A Acetyl cation (CH₃CO⁺)
B Phenyl carbocation
C Chlorobenzene adduct
D Aluminum-chlorine complex
Correct Answer:  A. Acetyl cation (CH₃CO⁺)
EXPLANATION

In Friedel-Crafts acylation, AlCl₃ facilitates the formation of the acetyl cation (CH₃CO⁺) from acetyl chloride. This electrophilic acylium ion then attacks the benzene ring.

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In the acetoacetic ester synthesis, the initial condensation is between acetoacetic ester and alkyl halide in the presence of:
A Sodium ethoxide
B Sodium hydroxide
C Sodium carbonate
D Sodium acetate
Correct Answer:  A. Sodium ethoxide
EXPLANATION

Sodium ethoxide creates the enolate of acetoacetic ester, which then undergoes SN2 with alkyl halide for alkylation.

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The order of reactivity in electrophilic aromatic substitution for halogens is:
A F > Cl > Br > I
B I > Br > Cl > F
C Cl > F > Br > I
D Br > Cl > F > I
Correct Answer:  A. F > Cl > Br > I
EXPLANATION

Fluorine is most reactive due to the best balance of electronic effects and hyperconjugation, while iodine is least reactive.

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In the bromination of toluene, the ortho:para ratio is approximately 2:1 instead of 1:1 due to:
A Steric hindrance at ortho position
B Thermodynamic stability of para isomer
C Kinetic control with partial steric effects
D Electronic effects favoring para position
Correct Answer:  C. Kinetic control with partial steric effects
EXPLANATION

Although para is thermodynamically more stable, ortho is kinetically favored but sterically hindered. The 2:1 ratio reflects kinetic control with steric factors reducing ortho product.

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The enolate ion formed from a ketone in basic conditions attacks an alkyl halide in the:
A Carbon atom (α-position)
B Oxygen atom
C Through radical mechanism
D Through concerted mechanism
Correct Answer:  A. Carbon atom (α-position)
EXPLANATION

Although the enolate has negative charge on both oxygen and carbon (resonance), the carbon is more nucleophilic toward alkyl halides. The SN2 attack occurs at the alkyl halide carbon, not the enolate O.

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In the Robinson annulation reaction, the product formed is:
A An aromatic compound
B A bicyclic ketone
C A linear diene
D A carboxylic acid
Correct Answer:  B. A bicyclic ketone
EXPLANATION

The Robinson annulation combines a Michael addition followed by an aldol condensation, forming a bicyclic ketone product with a new six-membered ring fused to an existing ring system.

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The directing effect of the amino group (-NH2) in electrophilic aromatic substitution is:
A Deactivating, meta-directing
B Deactivating, ortho/para-directing
C Activating, ortho/para-directing
D Activating, meta-directing
Correct Answer:  C. Activating, ortho/para-directing
EXPLANATION

The amino group is a strong electron-donating group (by resonance) that activates the benzene ring and directs incoming electrophiles to ortho/para positions due to resonance stabilization of the intermediate.

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In the reduction of ketones and aldehydes using DIBAL-H, the product is:
A An alkane
B An alcohol
C A carboxylic acid
D An alkene
Correct Answer:  B. An alcohol
EXPLANATION

DIBAL-H (Diisobutylaluminum hydride) reduces carbonyl compounds to primary alcohols (from aldehydes) or secondary alcohols (from ketones). It's milder than LiAlH4.

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