Home Subjects JEE Chemistry Organic Chemistry

JEE Chemistry
Organic Chemistry

Chemistry questions for JEE Main — Physical, Organic, Inorganic Chemistry.

100 Q 5 Topics Take Test
Advertisement
Difficulty: All Easy Medium Hard 61–70 of 100
Topics in JEE Chemistry
Q.61 Medium Organic Chemistry
A chiral compound with the structure CH3-CHBr-CH(OH)-CH3 will have how many stereoisomers?
A 2
B 4
C 8
D 16
Correct Answer:  B. 4
EXPLANATION

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).

Take Test
Q.62 Medium Organic Chemistry
In the conversion of benzene to benzoic acid through the Kolbe reaction, the carboxylic acid is ultimately derived from:
A The phenolic oxygen
B CO2 fixation on the aromatic ring
C Oxidation of a side chain
D Displacement of a leaving group on the ring
Correct Answer:  B. CO2 fixation on the aromatic ring
EXPLANATION

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.

Take Test
Q.63 Medium Organic Chemistry
The rate of E1 elimination reaction depends on:
A Concentration of both substrate and base
B Concentration of substrate only
C Concentration of base only
D Temperature but not concentrations
Correct Answer:  B. Concentration of substrate only
EXPLANATION

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.

Take Test
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:
A 2-methylbutane-1,4-diol
B 2,2-dimethyl-1,3-dioxolane
C Cyclopentanone
D Tetrahydrofurfuryl alcohol
Correct Answer:  B. 2,2-dimethyl-1,3-dioxolane
EXPLANATION

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.

Take Test
In the polymer synthesis by condensation polymerization, the type of linkage formed when dicarboxylic acids react with diols is:
A Peptide bond
B Ester linkage
C Ether linkage
D Glycosidic bond
Correct Answer:  B. Ester linkage
EXPLANATION

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.

Take Test
The compound that will show geometrical isomerism is:
A CH3-CH=CH-CH3
B CH3-CH2-CH=CH-CH3
C CH2=C(CH3)2
D CH3-CH2-CH=CH-CH2-CH3
Correct Answer:  B. CH3-CH2-CH=CH-CH3
EXPLANATION

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.

Take Test
In the ozonolysis of 2-methylbut-2-ene, the number of organic products formed is:
A 1
B 2
C 3
D 4
Correct Answer:  B. 2
EXPLANATION

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.

Take Test
Q.68 Medium Organic Chemistry
Which statement about aldol condensation is correct?
A Both carbonyl compounds must be aldehydes
B It requires a strong acid catalyst
C A compound with α-hydrogen can serve as the enolate donor
D The reaction is irreversible under all conditions
Correct Answer:  C. A compound with α-hydrogen can serve as the enolate donor
EXPLANATION

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.

Take Test
In the hydroboration-oxidation of alkenes, the reaction is stereospecific because:
A It involves a free radical mechanism
B It proceeds through a four-membered ring transition state with syn-addition
C The oxidation step is stereospecific
D The alkene must be planar
Correct Answer:  B. It proceeds through a four-membered ring transition state with syn-addition
EXPLANATION

Hydroboration occurs through a concerted mechanism involving a four-membered ring transition state, resulting in syn-addition of BH across the double bond.

Take Test
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:
A 1-hexanol
B 2,3-dimethyl-2-butanol
C Tetrahydropyran
D 1,2-hexanediol
Correct Answer:  B. 2,3-dimethyl-2-butanol
EXPLANATION

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.

Take Test
IGET
iget AI
Online · Ask anything about exams
Hi! 👋 I'm your iget AI assistant.

Ask me anything about exam prep, MCQ solutions, study tips, or strategies! 🎯
UPSC strategy SSC CGL syllabus Improve aptitude NEET Biology tips