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Organic Chemistry: Some Basic Principles and Techniques

4 solved JEE Chemistry previous year questions on Organic Chemistry: Some Basic Principles and Techniques, each with the correct answer and a full explanation.

In Friedel-Crafts alkylation of aniline, one gets: [JEE Main 2022, 29 Jun (Shift 2)]

aalkylated product with ortho and para substitution.

bsecondary amine after acidic treatment

can amide product.

dPositively charged nitrogen at benzene ring.

✓ Correct answer: d) Positively charged nitrogen at benzene ring.

ExplanationThe friedel-crafts alkylation of aniline occurs as: Aniline shows an acid-base reaction with the , a lewis acid. aniline is a Lewis base, while acts as a lewis acid.

With respect to the compounds , choose the incorrect statement(s).

aThe acidity of compound is due to delocalization in the conjugate base.

bThe conjugate base of compound IV is aromatic.

cCompound II becomes more acidic, when it has a substituent.

dThe acidity of compounds follows the order I

✓ Correct answer: d) The acidity of compounds follows the order I

ExplanationOption D is incorrect because the given order of acidity is not accurate.The acidity of a compound depends on the stability of its conjugate base.In compound I, the conjugate base (carbanion) is highly stabilized by resonance over three benzene rings, making it the most acidic.Compound IV forms a conjugate base that is aromatic and thus gains extra stability, so it is also relatively acidic.Next comes compound V (a terminal alkyne), where the negative charge is stabilized due to the high s-character of the sp-hybridized carbon.Compound II (benzene) is much less acidic, and compound III (an alkane) is the least acidic.Therefore, the exact order given in option D is not correctly justified, making D the incorrect statement.

In the given structure, number of sp and hybridized carbon atoms present respectively are : [JEE Main 2025, 24 Jan (Shift 2)]

a3 and 6

b4 and 6

c3 and 5

d4 and 5

✓ Correct answer: c) 3 and 5

Explanationsp and hybridized carbon atoms in the structure can be shown as: No. of sp hybridized carbon-3 No. of hybridized carbon-5

Methyl group attached to a positively charged carbon atom stabilizes the carbocation due to [Re-NEET 2024]

a-I-effect

belectromeric effect

chyperconjugation

dmesomeric effect

✓ Correct answer: c) hyperconjugation

ExplanationIn hyperconjugation, the σ C–H bonds adjacent to the positively charged carbon overlap with the empty p-orbital, delocalizing the positive charge and stabilizing the carbocation.

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