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.
RMgX + CO2 → RMgX-O2C- → RCOO- MgX+ → RCOOH (after acidification). This is a classic method for carboxylic acid synthesis.
Q.62Medium
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.63Medium
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.
Q.64Medium
The compound 2,2-dimethyl-1-propanol when heated with concentrated H2SO4 gives the major product as:
Answer: B
This is a dehydration reaction where the OH group leaves, and a rearrangement occurs via hydride shift to form the more stable tertiary carbocation, leading to 2-methylbut-2-ene (more substituted alkene, Zaitsev's rule).
Q.65Medium
Which isomer of dimethylbenzene (xylene) will show only 2 signals in 1H-NMR?
Answer: C
Para-xylene has high symmetry: all four aromatic protons are equivalent (one signal) and all six methyl protons are equivalent (one signal), giving only 2 total signals in 1H-NMR.
Q.66Medium
In the preparation of phenol from cumene (isopropylbenzene), the intermediate cumene hydroperoxide is cleaved in acidic conditions to give:
Answer: B
The cumene hydroperoxide undergoes an acid-catalyzed rearrangement (Hock rearrangement) where the isopropyl group rearranges to give phenol and acetone as the cleavage products.
Q.67Medium
Which of the following carboxylic acids is most acidic?
Answer: B
CF3-COOH is most acidic because fluorine is highly electronegative and strongly withdraws electron density through inductive effects, stabilizing the conjugate base carboxylate ion.
Q.68Medium
The reaction of benzene with excess Cl2 in the presence of FeCl3 followed by hydrolysis gives:
Answer: B
Chlorobenzene formed initially is more deactivated than benzene, but further chlorination can occur at ortho/para positions under excess Cl2 conditions, yielding a mixture of polychlorobenzenes.
Q.69Medium
Which statement about aldol condensation is correct?
Answer: C
Aldol condensation requires a compound with α-hydrogens that can form an enolate/enol as the nucleophilic component, and a carbonyl compound as the electrophile. Ketones can also be used.
Q.70Medium
The rate of E1 elimination reaction depends on:
Answer: B
E1 elimination is a two-step mechanism where the rate-determining step is the formation of carbocation (unimolecular). The rate depends only on substrate concentration, following first-order kinetics.
Q.71Medium
In the conversion of benzene to benzoic acid through the Kolbe reaction, the carboxylic acid is ultimately derived from:
Answer: B
In the Kolbe carboxylation of phenols, CO2 is directly incorporated into the aromatic ring ortho to the hydroxyl group under high temperature and pressure with alkali, forming salicylic acid derivatives.
Q.72Medium
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.73Medium
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.74Medium
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.75Medium
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.76Medium
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.77Medium
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.78Medium
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.79Medium
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.80Medium
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.
