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NEET CHEMISTRYHydrocarbonsMedium

Question

Which of the following is a free radical substitution reaction?

A

Benzene with Br2/AlCl3\text{Br}_2/\text{AlCl}_3

B

Acetylene with HBr\text{HBr}

C

Methane with Br2/hν\text{Br}_2/h\nu

D

Propene with HBr/(C6H5COO)2\text{HBr}/(\text{C}_6\text{H}_5\text{COO})_2

Step-by-Step Solution

The reaction of methane with halogens like bromine (Br2\text{Br}_2) in the presence of ultraviolet light (hνh\nu) or at high temperature is a free radical substitution reaction . The mechanism involves three steps: initiation (homolysis of the Br-Br\text{Br-Br} bond to generate free radicals), propagation, and termination .

  • Benzene with Br2/AlCl3\text{Br}_2/\text{AlCl}_3: This is an electrophilic substitution reaction where AlCl3\text{AlCl}_3 acts as a Lewis acid catalyst to generate the electrophile (Br+\text{Br}^+) .
  • Acetylene with HBr\text{HBr}: This is an electrophilic addition reaction across the triple bond .
  • Propene with HBr/(C6H5COO)2\text{HBr}/(\text{C}_6\text{H}_5\text{COO})_2: This is a free radical addition reaction, also known as the peroxide effect or anti-Markovnikov addition .

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.

CHEMISTRYHydrocarbonsfollowingradicalsubstitutionreaction

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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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The order of decreasing reactivity towards an electrophilic reagent for the following compounds is: (i) Benzene (ii) Toluene (iii) Chlorobenzene (iv) Phenol

A.(i) > (ii) > (iii) > (iv)
B.(ii) > (iv) > (i) > (iii)
C.(iv) > (iii) > (ii) > (i)
D.(iv) > (ii) > (i) > (iii)
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X and Y in the above-mentioned reaction are respectively:

A.X = 2–Butyne; Y = 3–Hexyne
B.X = 2-Butyne; Y = 2-Hexyne
C.X = 1-Butyne; Y = 2-Hexyne
D.X = 1-Butyne; Y = 3–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)
B.(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
C.(a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
D.(a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
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The dihedral angle of the least stable conformer of ethane is:

A.$60^{\circ}$
B.$0^{\circ}$
C.$120^{\circ}$
D.$180^{\circ}$
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Phenanthrene is an aromatic compound and follows Hückel's $(4n+2)\pi$ electron rule. The value of $n$ is:

A.0
B.1
C.2
D.3
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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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