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.
In the oxidation of primary alcohols using PCC (pyridinium chlorochromate), the primary alcohol is converted to:
Answer: B
PCC is a mild oxidizing agent that oxidizes primary alcohols to aldehydes without further oxidation to carboxylic acids (unlike acidic KMnO4 or K2Cr2O7).
Q.122Medium
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.123Easy
The coupling constant (J) in 1H-NMR for vicinal coupling (3J) typically ranges from:
Answer: B
Vicinal coupling (3J) between protons separated by 3 bonds typically has values of 6-18 Hz, with typical values around 7-8 Hz for anti-periplanar and 2-5 Hz for gauche conformations.
Q.124Medium
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.125Medium
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.126Medium
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.127Easy
In the Williamson ether synthesis, the reactivity order for nucleophilic substitution by alkoxide ions (RO-) follows which pattern?
Answer: A
In Williamson ether synthesis, the alkoxide ion attacks via SN2 mechanism, which prefers primary alkyl halides due to less steric hindrance. Tertiary substrates don't react due to steric hindrance.
Q.128Hard
A compound with molecular formula C6H12O shows a broad O-H stretch in IR at 3300 cm⁻¹ and no C=O peak. The 1H-NMR shows a singlet at δ 3.3 ppm. The compound is likely:
Answer: B
The singlet at δ 3.3 ppm suggests the OH is on a quaternary carbon (no neighboring H, hence no coupling). 2,3-dimethyl-2-butanol fits: the OH is on a quaternary carbon, explaining the singlet and broad O-H in IR.
Q.129Hard
In the hydroboration-oxidation of alkenes, the reaction is stereospecific because:
Answer: B
Hydroboration occurs through a concerted mechanism involving a four-membered ring transition state, resulting in syn-addition of BH across the double bond.
Q.130Medium
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.131Easy
In the ozonolysis of 2-methylbut-2-ene, the number of organic products formed is:
Answer: B
2-methylbut-2-ene: (CH3)2C=CH-CH3. Ozonolysis cleaves the C=C to give (CH3)2C=O (acetone) and CH3-CHO (acetaldehyde). Two different organic products are formed.
Q.132Easy
The compound that will show geometrical isomerism is:
Answer: B
CH3-CH2-CH=CH-CH3 (pent-2-ene) shows geometrical isomerism because the C=C has two different groups on each carbon. Option (a) is butane with symmetric substituents; (c) has geminal methyls; (d) is hexene with symmetric groups.
Q.133Easy
In the polymer synthesis by condensation polymerization, the type of linkage formed when dicarboxylic acids react with diols is:
Answer: B
When dicarboxylic acids (HOOC-R-COOH) react with diols (HO-R'-OH), ester bonds form between the carboxyl and hydroxyl groups, creating polyester polymers with repeating ester linkages.
Q.134Hard
A compound C5H10O2 with no C=O in IR but shows broad O-H stretch. 1H-NMR exhibits signals at δ 3.8 and δ 4.7 ppm. The compound is most likely:
Answer: B
No C=O in IR rules out ketone/aldehyde. The signals at δ 3.8 (OCH2) and δ 4.7 suggest acetal carbons (characteristic of dioxolane ring). 2,2-dimethyl-1,3-dioxolane fits C5H10O2 with protected diol functionality.
Q.135Medium
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.136Medium
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.137Medium
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.138Easy
Which of the following compounds will undergo SN2 reaction most readily?
Answer: B
SN2 reaction requires easy access to the carbon bearing the leaving group. Ethyl bromide (primary alkyl halide) has minimal steric hindrance and is highly reactive in SN2 reactions.
Q.139Easy
The IUPAC name of the compound with structure CH3-CH(OH)-CH2-CHO is:
Answer: B
The longest carbon chain contains 4 carbons with the aldehyde group (CHO) at position 1. The hydroxyl group is at position 3, giving 3-hydroxybutanal.
Q.140Easy
In the nitration of benzene using HNO3/H2SO4, the electrophile is:
Answer: B
H2SO4 protonates HNO3 to form H2NO3+, which loses water to generate the nitronium ion (NO2+), the actual electrophile in electrophilic aromatic substitution.
