Basic Organic Principles and Practical Organic Chemistry (JEE Advanced)

Question 1

Which one of the following molecules has all its carbon atoms in the same state of hybridisation, and is also linear? Exactly one option is correct.

Accepted Answer

D. Assign hybridisation from the number of σ bonds plus lone pairs on each carbon (4 → sp³, 3 → sp², 2 → sp).
(A) Propadiene CH₂=C=CH₂: terminal C sp², central C sp. Mixed.
(B) But-1-en-3-yne CH₂=CH–C≡CH: two sp² and two sp carbons. Mixed.
(C) Cyclohexa-1,4-diene: four sp² and two sp³ (CH₂) carbons. Mixed.
(D) Dicyanoacetylene N≡C–C≡C–C≡N: every carbon forms two σ bonds and two π bonds, so all four a

Question 2

Which one of the following compounds can exist as a pair of cis–trans (geometrical) isomers? Exactly one option is correct.

Accepted Answer

C. Key idea: in a cumulene the π bonds on either side of an sp carbon use two mutually perpendicular p orbitals.
(A) Penta-2,3-diene CH₃CH=C=CHCH₃ (2 cumulated C=C): the plane of H–C2–CH₃ is perpendicular to the plane of H–C4–CH₃. With the ends perpendicular there is no 'same side'/'opposite side' relationship, so no cis–trans isomers. (Its two stereoisomers are mirror images, i.e. enantiomers, not g

Question 3

Consider the following phenols in water at 25 °C: (I) 4-nitrophenol, (II) 3-nitrophenol, (III) 3-methoxyphenol, (IV) phenol, (V) 4-methylphenol. The correct decreasing order of acid strength is ______

Accepted Answer

A. Acidity rises when the phenoxide ion is stabilised.
(I) 4-NO₂: strong –I and –R; the negative charge is delocalised onto the nitro oxygens from the para position. Most acidic (pKₐ ≈ 7.15).
(II) 3-NO₂: –I only (no resonance delocalisation from meta). Less acidic than para (pKₐ ≈ 8.36) but much more than phenol.
(III) 3-OCH₃: the +R effect of OCH₃ does not reach the meta position, so only its –I eff

Basic Organic Principles and Practical Organic Chemistry: questions with solutions

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