Chemistry carries the highest scoring potential in JEE for anyone who keeps the three branches separate in revision. This set covers physical chemistry numericals, organic reaction mechanisms and named reactions, and inorganic chemistry including periodic trends, chemical bonding and coordination compounds. Organic questions show the mechanism arrow by arrow, so the reasoning transfers to reactions you have not seen before.
Which of the following compounds will undergo SN2 reaction most readily?
Answer: B
SN2 reaction requires easy access to the carbon bearing the leaving group. Ethyl bromide (primary alkyl halide) has minimal steric hindrance and is highly reactive in SN2 reactions.
Q.42Easy
The IUPAC name of the compound with structure CH3-CH(OH)-CH2-CHO is:
Answer: B
The longest carbon chain contains 4 carbons with the aldehyde group (CHO) at position 1. The hydroxyl group is at position 3, giving 3-hydroxybutanal.
Q.43Easy
In the nitration of benzene using HNO3/H2SO4, the electrophile is:
Answer: B
H2SO4 protonates HNO3 to form H2NO3+, which loses water to generate the nitronium ion (NO2+), the actual electrophile in electrophilic aromatic substitution.
Q.44Easy
Which functional group shows a strong absorption around 1700-1750 cm⁻¹ in IR spectrum?
Answer: C
The C=O stretch of carbonyl compounds (aldehydes, ketones, carboxylic acids, esters) appears characteristically around 1700-1750 cm⁻¹ in IR spectrum.
Q.45Easy
Grignard reagent (RMgX) reacts with water to give:
Answer: C
Grignard reagents are strong nucleophiles and strong bases. They readily abstract a proton from water, producing an alkane (R-H) and magnesium halide salt.
Q.46Medium
The reactivity order of alkyl halides in SN1 reactions is:
Answer: B
SN1 proceeds through carbocation formation. Tertiary carbocations are most stable (hyperconjugation and inductive effects), followed by secondary, then primary.
Q.47Medium
In the addition of HBr to propene, the major product is 2-bromopropane because of:
Answer: B
Markovnikov's rule states that in addition to unsymmetrical alkenes, the hydrogen adds to the carbon with more hydrogen atoms, and the addendum to the carbon with fewer hydrogens, forming the more stable carbocation intermediate.
Q.48Medium
Which of the following will give a positive Tollens test?
Answer: B
Tollens test detects aldehydes. Benzaldehyde contains an aldehyde group (-CHO) and will be oxidized to benzoate ion, giving a positive test (silver mirror).
Q.49Medium
The rate-determining step in an E1 elimination reaction is:
Answer: B
E1 elimination occurs in two steps: slow carbocation formation followed by fast deprotonation. The carbocation formation is the rate-determining step.
Q.50Medium
In the bromination of toluene with Br2/FeBr3, the major product is:
Answer: C
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.
Q.51Medium
Which statement about the aldol condensation is correct?
Answer: B
Aldol condensation produces a β-hydroxy carbonyl compound (aldol) initially, which can further dehydrate under heating to form an α,β-unsaturated carbonyl compound.
Q.52Medium
The product formed by the hydroboration-oxidation of 1-butene is:
Answer: A
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).
Q.53Medium
Identify the compound with molecular formula C6H12 that shows geometrical isomerism:
Answer: B
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.
Q.54Medium
When 2-methylpropene reacts with cold dilute KMnO4, the product is:
Answer: B
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).
Q.55Medium
In the oxidation of primary alcohols using K2Cr2O7/H2SO4, the final product is:
Answer: B
K2Cr2O7 in acidic medium is a strong oxidizing agent that oxidizes primary alcohols first to aldehydes, then further oxidizes the aldehyde to carboxylic acids.
Q.56Medium
The stereochemistry of an SN2 reaction is:
Answer: B
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.
Q.57Medium
Which of the following is the correct order of acidity for carboxylic acids?
Answer: A
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.
Q.58Medium
In the reaction of acetylene with HgSO4/H2SO4, the product is:
Answer: A
This is the hydration of alkynes using Hg2+ catalyst. Acetylene (HC≡CH) undergoes hydration to form acetaldehyde (CH3CHO) via enol intermediate, which tautomerizes.
Q.59Medium
The Wittig reaction converts a carbonyl compound into:
Answer: B
The Wittig reaction uses a phosphonium ylide to convert carbonyl compounds (aldehydes and ketones) to alkenes. It's valuable for forming C=C double bonds with defined positions.
Q.60Easy
In the Friedel-Crafts alkylation of benzene with alkyl halides, the catalyst used is:
Answer: B
AlCl3 is the standard Lewis acid catalyst for Friedel-Crafts alkylation. It activates the alkyl halide to form a carbocation, which then attacks the aromatic ring.
