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.
A chiral compound with the structure CH3-CHBr-CH(OH)-CH3 will have how many stereoisomers?
Answer: B
The molecule has two chiral centers (carbons bearing Br and OH). Each chiral center can have R or S configuration, giving 2² = 4 possible stereoisomers (diastereomers and enantiomers).
Q.22Medium
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.23Medium
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.24Medium
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.25Medium
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.26Medium
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.27Medium
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.28Medium
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.29Medium
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.30Medium
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.31Medium
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.32Medium
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.33Medium
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.34Medium
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.35Medium
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.36Medium
In the Friedel-Crafts alkylation of benzene with 1-chloropropane and AlCl3, the major product is isopropylbenzene due to:
Answer: A
Primary carbocation rearranges via hydride shift to form more stable secondary carbocation, which then forms isopropylbenzene.
Q.37Medium
The pKa of phenol (C6H5OH) is approximately 10, while that of aliphatic alcohol is around 16. This difference is due to:
Answer: B
Phenoxide ion is stabilized by resonance with the benzene ring, making phenol more acidic than aliphatic alcohols.
Q.38Medium
Which of the following compounds will not give a positive Iodoform test?
Answer: D
Iodoform test requires either a methyl ketone (COCH3) or secondary alcohol with CH3 adjacent to CHOH. Ethanol has no such structure.
Q.39Medium
In the nitration of anisole (C6H5-O-CH3), the major product is:
Answer: A
Methoxy group is ortho-para directing and activating. Due to steric hindrance at para position, 2-nitroanisole is the major product.
Q.40Medium
The reaction of acetyl chloride with aniline preferentially gives N-acetylaniline rather than p-acetylaniline because:
Answer: A
The lone pair on nitrogen of aniline is more nucleophilic than the π electrons of benzene, making N-acylation the dominant pathway.
