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.
NO2+ (nitronium ion) is an electrophile with positive charge, not a nucleophile. All others have lone pairs or negative charge.
Q.182Easy
The Williamson ether synthesis involves reaction of an alkoxide with:
Answer: B
Williamson ether synthesis uses SN2 reaction between alkoxide (R-O-) and alkyl halide (R'-X) to form R-O-R'.
Q.183Medium
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.184Medium
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.185Easy
Which of the following shows the most stable carbocation?
Answer: C
Tertiary carbocation (CH3)3C+ is most stable due to maximum hyperconjugation and inductive stabilization from three alkyl groups.
Q.186Hard
In the acetoacetic ester synthesis, the initial condensation is between acetoacetic ester and alkyl halide in the presence of:
Answer: A
Sodium ethoxide creates the enolate of acetoacetic ester, which then undergoes SN2 with alkyl halide for alkylation.
Q.187Medium
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.188Medium
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.189Easy
In the ozonolysis of but-2-ene, the products formed are:
Answer: A
But-2-ene (CH₃-CH=CH-CH₃) on ozonolysis gives two identical fragments, each being acetaldehyde (CH₃CHO). The double bond cleaves symmetrically.
Q.190Easy
Which mechanism is involved in the reaction of tert-butyl bromide with aqueous NaOH?
Answer: A
tert-Butyl bromide is a tertiary alkyl halide. Tertiary substrates preferentially undergo SN1 mechanism due to stable tertiary carbocation formation. SN2 is hindered by steric effects.
Q.191Medium
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.192Easy
Which of the following compounds will give a positive Tollens test?
Answer: B
Tollens test is positive for aldehydes. Propanal (CH₃CH₂CHO) is an aldehyde and gets oxidized to carboxylic acid, reducing Ag⁺ to Ag (silver mirror). Acetone is a ketone, ether and alcohol do not give positive test.
Q.193Medium
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.194Medium
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.195Easy
Which of the following is the correct IUPAC name for the compound CH₃-CHBr-CH₂-CH₃?
Answer: A
The parent chain is butane (4 carbons). Numbering from the end nearest to substituent gives Br at position 2. IUPAC name is 2-bromobutane. Option D is a common name.
Q.196Medium
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.
Q.197Hard
Which intermediate is formed when acetyl chloride reacts with benzene in the presence of AlCl₃?
Answer: A
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.
Q.198Easy
Which of the following is the correct order of atomic radius?
Answer: B
Atomic radius increases down a group in the periodic table. As we go from Na to K to Rb to Cs, the number of shells increases, leading to increased atomic radius.
Q.199Medium
The ionization energy of nitrogen is higher than oxygen because:
Answer: B
Nitrogen (1s² 2s² 2p³) has a half-filled 2p orbital which provides extra stability. Removing an electron from this configuration requires more energy than from oxygen's configuration.
Q.200Medium
Which compound shows the highest ionic character?
Answer: B
Ionic character depends on electronegativity difference and charge density. MgO has the highest charge density (smallest cation with 2+ charge) and highest electronegativity difference.
