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NEET CHEMISTRYHydrocarbonsMedium

Question

The product(s) that are formed when the given compound is treated with Br2Br_2 in the presence of FeBr3FeBr_3 are:

A

(Structure Missing)

B

(Mixture of Isomers)

C

(Structure Missing)

D

(Structure Missing)

Step-by-Step Solution

  1. Reagents and Reaction Type: The reaction uses Bromine (Br2Br_2) in the presence of Ferric Bromide (FeBr3FeBr_3). FeBr3FeBr_3 acts as a Lewis acid catalyst to generate the electrophile, the bromonium ion (Br+Br^+) [NCERT 12th, Ch 10, Sec 10.5; NCERT 11th, Ch 13, Sec 13.5.5]. This indicates an Electrophilic Aromatic Substitution reaction. Br2+FeBr3→Br++[FeBr4]−Br_2 + FeBr_3 \rightarrow Br^+ + [FeBr_4]^-
  2. Substrate Analysis: Although the specific structure of the 'given compound' is missing from the input, the reaction conditions and the probable answer format ('and', implying a mixture) suggest the substrate is a mono-substituted benzene ring containing an ortho-para directing group (e.g., −CH3-CH_3, −OCH3-OCH_3, −Cl-Cl, etc.).
  3. Directive Influence: Ortho-para directing groups increase electron density at the ortho (2, 6) and para (4) positions via resonance or inductive effects. Consequently, the electrophile (Br+Br^+) attacks these positions, resulting in a mixture of o-bromo and p-bromo derivatives [NCERT 12th, Ch 10, Sec 10.5; NCERT 11th, Ch 13, Sec 13.5.6].
  4. Conclusion: The products formed would be a mixture of the ortho and para isomers of the brominated compound. (e.g., if the reactant was Toluene, the products would be o-Bromotoluene and p-Bromotoluene).

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Hydrocarbons. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYHydrocarbonsproductsformedcompoundtreatedpresence

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Phenylacetylene undergoes a reaction with dilute $\text{H}_2\text{SO}_4$ in the presence of $\text{HgSO}_4$ to yield:

A.Option 1 (Structure missing)
B.Option 2 (Structure missing)
C.Option 3 (Structure missing)
D.Option 4 (Structure missing)
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The reaction among the following that proceeds through an electrophilic substitution reaction is:

A.(Option 1 Structure Missing)
B.(Option 2 Structure Missing)
C.4
D.(Option 4 Structure Missing)
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A.(i) > (ii) > (iii) > (iv)
B.(ii) > (iv) > (i) > (iii)
C.(iv) > (iii) > (ii) > (i)
D.(iv) > (ii) > (i) > (iii)
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A.X = 2–Butyne; Y = 3–Hexyne
B.X = 2-Butyne; Y = 2-Hexyne
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Match the bond line structures of hydrocarbons given in List-I with the corresponding boiling points given in List-II. List-II (Boiling point in K): (i) 300.9 (ii) 282.5 (iii) 309.1 (iv) 341.9 Choose the correct answer from the options given below:

A.(a)-(i), (b)-(iv), (c)-(iii), (d)-(ii)
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A.$60^{\circ}$
B.$0^{\circ}$
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A.0
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The correct statement regarding the comparison of staggered and eclipsed conformations of ethane is:

A.The eclipsed conformation of ethane is more stable than staggered conformation because eclipsed conformation has no torsional strain.
B.The eclipsed conformation of ethane is more stable than staggered conformation even though the eclipsed conformation has a torsional strain.
C.The staggered conformation of ethane is more stable than eclipsed conformation because staggered conformation has no torsional strain.
D.The staggered conformation of ethane is less stable than eclipsed conformation because staggered conformation has a torsional strain.
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