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.
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.82Medium
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.83Medium
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.84Medium
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.85Medium
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.86Medium
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.87Medium
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.88Medium
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.89Medium
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.90Medium
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.91Medium
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.
Q.92Medium
In the reaction of phenylmagnesium bromide (Grignard reagent) with CO2 followed by hydrolysis, the main product is:
Answer: B
Grignard reagent attacks CO2 to form a salt, which upon hydrolysis gives benzoic acid.
Q.93Medium
In the Clemmensen reduction of a ketone, the reducing agent used is:
Answer: A
Clemmensen reduction uses zinc amalgam with concentrated HCl to convert C=O to CH2, unlike Wolff-Kishner which uses hydrazine.
Q.94Medium
The major product when 2-methylbut-3-en-1-ol undergoes PCC oxidation is:
Answer: A
PCC (Pyridinium chlorochromate) selectively oxidizes primary alcohols to aldehydes without further oxidation, and doesn't affect C=C.
Q.95Medium
The compound that will show maximum hydrogen bonding is:
Answer: B
Amides have both N-H (hydrogen bond donor) and C=O (hydrogen bond acceptor), allowing formation of strong intermolecular hydrogen bonds.
Q.96Medium
Which of the following alkenes will undergo hydroboration-oxidation to give a secondary alcohol as the major product?
Answer: A
Hydroboration-oxidation follows anti-Markovnikov's rule and gives Markovnikov's hydration product after oxidation. 1-methylcyclohexene gives secondary alcohol, while pent-1-ene gives primary and 2-methylbut-2-ene gives tertiary alcohol.
Q.97Medium
The reaction of phenol with excess bromine in water produces:
Answer: A
Phenol is highly activated towards electrophilic aromatic substitution due to the electron-donating -OH group. Bromine can add at all three ortho and para positions with excess bromine, giving 2,4,6-tribromophenol.
Q.98Medium
In the dehydration of 2-methylbutan-2-ol, the major product is:
Answer: B
Dehydration of 2-methylbutan-2-ol follows Zaitsev's rule, producing the most stable (most substituted) alkene. 2-methylbut-2-ene is a trisubstituted alkene and is the major product.
Q.99Medium
The addition of HBr to propene in the presence of peroxides follows:
Answer: B
Peroxides initiate free radical mechanism (Kharasch effect). In free radical addition, HBr adds anti-Markovnikov to propene, giving 1-bromopropane as the major product. The Br radical adds first to the terminal carbon.
Q.100Medium
In the nitration of benzene with HNO₃/H₂SO₄, the rate-determining step involves attack by:
Answer: A
The HNO₃/H₂SO₄ mixture generates NO₂⁺ (nitronium ion), which is the electrophile attacking the benzene ring in the rate-determining step. This is a classic electrophilic aromatic substitution.
