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JEE Chemistry

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

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Difficulty: All Easy Medium Hard 141–150 of 225
Topics in JEE Chemistry
Q.141 Medium Organic Chemistry
Which of the following is the correct order of acidity for carboxylic acids?
A Formic > Acetic > Propionic
B Acetic > Formic > Propionic
C Propionic > Acetic > Formic
D All are equally acidic
Correct Answer:  A. Formic > Acetic > Propionic
EXPLANATION

As alkyl chain length increases, the electron-donating effect of the alkyl group increases, destabilizing the conjugate base carboxylate ion. Thus, formic acid (no alkyl group) is most acidic.

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Q.142 Medium Organic Chemistry
The stereochemistry of an SN2 reaction is:
A Retention of configuration
B Complete inversion of configuration
C Racemization
D Random stereochemistry
Correct Answer:  B. Complete inversion of configuration
EXPLANATION

SN2 is a one-step bimolecular mechanism where the nucleophile attacks from the back side of the carbon bearing the leaving group, resulting in complete (Walden) inversion of configuration.

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Q.143 Medium Organic Chemistry
In the oxidation of primary alcohols using K2Cr2O7/H2SO4, the final product is:
A An aldehyde
B A carboxylic acid
C A ketone
D An ester
Correct Answer:  B. A carboxylic acid
EXPLANATION

K2Cr2O7 in acidic medium is a strong oxidizing agent that oxidizes primary alcohols first to aldehydes, then further oxidizes the aldehyde to carboxylic acids.

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Q.144 Medium Organic Chemistry
When 2-methylpropene reacts with cold dilute KMnO4, the product is:
A Acetone and methanol
B 2-methylpropane-1,2-diol
C Propanoic acid
D 2-methylpropanol
Correct Answer:  B. 2-methylpropane-1,2-diol
EXPLANATION

Cold dilute KMnO4 causes hydroxylation of alkenes to form vicinal diols via syn addition. 2-methylpropene forms 2-methylpropane-1,2-diol (geminal diol arrangement on same carbon after rearrangement).

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Q.145 Medium Organic Chemistry
Identify the compound with molecular formula C6H12 that shows geometrical isomerism:
A Cyclohexane
B hex-3-ene
C 2-methylpent-2-ene
D Hexane
Correct Answer:  B. hex-3-ene
EXPLANATION

hex-3-ene (CH3CH2CH=CHCH2CH3) has different groups on each carbon of the double bond, allowing cis-trans (E-Z) isomerism. Cyclohexane and alkanes have no double bonds, and 2-methylpent-2-ene is not a correct formula for C6H12.

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Q.146 Medium Organic Chemistry
The product formed by the hydroboration-oxidation of 1-butene is:
A 1-butanol
B 2-butanol
C Butanal
D Butanone
Correct Answer:  A. 1-butanol
EXPLANATION

Hydroboration-oxidation follows anti-Markovnikov's rule with syn addition. The OH adds to the less substituted carbon (primary), giving 1-butanol (butan-1-ol).

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Q.147 Medium Organic Chemistry
Which statement about the aldol condensation is correct?
A It requires an aldehyde and ketone in equal amounts
B It produces a β-hydroxy carbonyl compound as the intermediate product
C It only works with aromatic aldehydes
D It requires acidic conditions exclusively
Correct Answer:  B. It produces a β-hydroxy carbonyl compound as the intermediate product
EXPLANATION

Aldol condensation produces a β-hydroxy carbonyl compound (aldol) initially, which can further dehydrate under heating to form an α,β-unsaturated carbonyl compound.

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Q.148 Medium Organic Chemistry
In the bromination of toluene with Br2/FeBr3, the major product is:
A 2-bromotoluene only
B 4-bromotoluene only
C ortho and para-bromotoluene (major)
D meta-bromotoluene only
Correct Answer:  C. ortho and para-bromotoluene (major)
EXPLANATION

The methyl group (-CH3) is an alkyl group, which is an electron-donating group that activates the benzene ring and is ortho/para-directing in electrophilic aromatic substitution.

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Q.149 Medium Organic Chemistry
The rate-determining step in an E1 elimination reaction is:
A Removal of hydrogen by base
B Formation of the carbocation
C Protonation of the alkene
D Formation of the alkyl halide
Correct Answer:  B. Formation of the carbocation
EXPLANATION

E1 elimination occurs in two steps: slow carbocation formation followed by fast deprotonation. The carbocation formation is the rate-determining step.

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Q.150 Medium Organic Chemistry
Which of the following will give a positive Tollens test?
A Acetone
B Benzaldehyde
C Diethyl ketone
D tert-Butyl alcohol
Correct Answer:  B. Benzaldehyde
EXPLANATION

Tollens test detects aldehydes. Benzaldehyde contains an aldehyde group (-CHO) and will be oxidized to benzoate ion, giving a positive test (silver mirror).

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