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NEET CHEMISTRYHydrocarbonsEasy

Question

The reaction of HBr with propene in the presence of peroxide gives:

A

3-bromopropane

B

Allyl bromide

C

n-propyl bromide

D

Isopropyl bromide

Step-by-Step Solution

When an unsymmetrical alkene like propene reacts with HBr\text{HBr} in the presence of a peroxide, the addition takes place contrary to Markovnikov's rule. This phenomenon is known as the peroxide effect, Kharash effect, or anti-Markovnikov addition . The reaction proceeds via a free radical chain mechanism, resulting in the attachment of the bromide radical to the doubly bonded carbon atom that possesses more hydrogen atoms. Consequently, the major product formed is 1-bromopropane (CH3-CH2-CH2Br\text{CH}_3\text{-CH}_2\text{-CH}_2\text{Br}), which is commonly known as n-propyl bromide .

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.

CHEMISTRYHydrocarbonsreactionpropenepresenceperoxide

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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)
MediumSolve

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
MediumSolve

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
MediumSolve

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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