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 exhibits optical isomerism?
Answer: B
2-bromobutane has a chiral center (C2 with four different groups: Br, H, CH3, and C2H5), exhibiting optical isomerism. Other options lack a true chiral center.
Q.2Easy
The IUPAC name of (CH3)3C-CH2-OH is:
Answer: B
The structure has 5 carbons with main chain having OH at position 1 and two methyl groups at position 2. IUPAC name is 2,2-dimethyl-1-propanol.
Q.3Easy
Which reagent is used to distinguish between primary and secondary alcohols?
Answer: B
Lucas reagent (HCl + ZnCl2) forms turbidity with primary alcohols immediately, secondary alcohols after 5 minutes, and tertiary alcohols instantly with dehydration.
Q.4Medium
In the Williamson ether synthesis, the reactivity of alkyl halides follows the order:
Answer: A
Williamson ether synthesis follows SN2 mechanism, which is favored by primary alkyl halides due to minimal steric hindrance. Reactivity: Primary > Secondary > Tertiary.
Q.5Medium
Which of the following compounds will show aldol condensation?
Answer: C
Aldol condensation requires an α-hydrogen. Acetaldehyde has α-hydrogens and undergoes aldol condensation. Benzaldehyde and formaldehyde lack α-hydrogens.
Q.6Medium
The oxidation of primary alcohol to aldehyde can be selectively done using:
Answer: B
PCC oxidizes primary alcohols to aldehydes without further oxidation to carboxylic acids. KMnO4 and Na2Cr2O7 further oxidize aldehydes to carboxylic acids.
Q.7Medium
Which mechanism is followed in the hydrolysis of alkyl halides under basic conditions?
Answer: B
Basic hydrolysis of alkyl halides follows SN2 mechanism (bimolecular nucleophilic substitution) with OH- as nucleophile, especially for primary halides.
Q.8Medium
In the nitration of benzene, the attacking species is:
Answer: A
Nitration involves formation of NO2+ (nitronium ion) from HNO3 in presence of H2SO4. NO2+ acts as the electrophile in electrophilic aromatic substitution.
Q.9Medium
The dehydration of 2-methylbutan-2-ol gives primarily:
Answer: B
Dehydration follows Zaitsev's rule, producing the most substituted (stable) alkene. 2-methylbut-2-ene is the major product with the double bond at the tertiary position.
Q.10Medium
Which compound will not undergo Cannizzaro reaction?
Answer: C
Cannizzaro reaction occurs with aldehydes lacking α-hydrogens. Acetaldehyde has α-hydrogens, so it undergoes aldol condensation instead of Cannizzaro reaction.
Q.11Medium
The rate of SN1 reaction is independent of:
Answer: B
SN1 is a unimolecular reaction with rate = k[R-X]. It is first-order with respect to alkyl halide only, independent of nucleophile concentration.
Q.12Hard
In Friedel-Crafts alkylation, the reactivity of alkyl halides follows:
Answer: B
F-C alkylation proceeds via carbocation formation. Tertiary carbocations are most stable, followed by secondary, then primary. Reactivity: Tertiary > Secondary > Primary.
Q.13Medium
The acidity order of carboxylic acids is:
Answer: A
Formic acid is most acidic (Ka = 1.8×10-4) due to favorable +I effect of hydrogen. Acidity decreases with increasing alkyl chain length due to +I effect: Formic > Acetic > Propionic.
Q.14Hard
Which of the following will give a positive iodoform test?
Answer: B
Iodoform test requires CH3CO- group. Pentan-2-ol (CH3CH2CH(OH)CH3) can be oxidized to CH3COCH2CH3 which contains the required methyl ketone group.
Q.15Easy
The most stable conformation of ethane is:
Answer: B
In staggered conformation, bonds are maximally separated reducing electron-electron repulsion and torsional strain. Staggered is most stable with ~12 kJ/mol lower energy than eclipsed.
Q.16Medium
Grignard reagent reacts with CO2 to give:
Answer: C
RMgX + CO2 → RMgX-O2C- → RCOO- MgX+ → RCOOH (after acidification). This is a classic method for carboxylic acid synthesis.
Q.17Hard
The compound that will undergo decarboxylation easily is:
Answer: C
β-Ketoacids (RCOCH2COOH) undergo easy decarboxylation through a six-membered transition state, producing ketones. This is more facile than decarboxylation of regular carboxylic acids.
Q.18Medium
The esterification reaction is catalyzed by:
Answer: B
Esterification (Fischer esterification) requires acid catalyst (H2SO4 or HCl) which protonates the carbonyl oxygen, increasing electrophilicity and facilitating nucleophilic attack by alcohol.
Q.19Easy
Which set of reagents can convert benzene to toluene in one step?
Answer: A
Friedel-Crafts alkylation using CH3Cl with AlCl3 as Lewis acid catalyst converts benzene to toluene. This is the standard industrial method for alkylbenzene synthesis.
Q.20Medium
In the bromination of benzene, the role of FeBr3 is to:
Answer: B
FeBr3 is a Lewis acid that polarizes Br2, generating Br+ (bromonium ion) which acts as the electrophile in electrophilic aromatic substitution of benzene.
