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NEET CHEMISTRYHydrocarbonsMedium

Question

The correct IUPAC name of the given compound is:

A

3-Ethyl-4-ethenylheptane

B

3-Ethyl-4-propylhex-5-ene

C

3-(1-Ethyl propyl) hex-1-ene

D

4-Ethyl-3-propylhex-1-ene

Step-by-Step Solution

  1. Structure Deduction: The options describe a hydrocarbon with a double bond. Comparing Options 3 and 4 suggests the structure involves a 6-carbon chain containing a double bond, with ethyl and propyl groups attached. The probable structure is CH2=CH−CH(CH2CH2CH3)−CH(CH2CH3)−CH2CH3CH_2=CH-CH(CH_2CH_2CH_3)-CH(CH_2CH_3)-CH_2CH_3.
  2. Longest Chain Selection:
  • The parent chain must contain the double bond (C=CC=C).
  • There are two possible 6-carbon chains containing the double bond:
  • Path A (Straight): C1−C2−C3−C4−C5−C6C_1-C_2-C_3-C_4-C_5-C_6. This chain has two substituents: a propyl group at C-3 and an ethyl group at C-4.
  • Path B (Turning into the propyl group): C1−C2−C3−C(Pr)C_1-C_2-C_3-C(Pr). This chain also has 6 carbons but results in only one complex substituent (1-ethylpropyl) at C-3.
  • Rule: If two chains have the same length, the one with the greater number of substituents is chosen as the parent chain [NCERT 11th, Ch 12, Sec 12.5.2; Source 83, 84]. Thus, Path A is selected.
  1. Numbering: Numbering starts from the end giving the double bond the lowest locant. C1C_1 is the double-bonded carbon (=CH2=CH_2). This makes the double bond position 1.
  2. Naming: Substituents are listed alphabetically: Ethyl before Propyl.
  • Position 4: Ethyl
  • Position 3: Propyl
  • Parent: Hex-1-ene
  • Full Name: 4-Ethyl-3-propylhex-1-ene.
  1. Analysis of Other Options:
  • Option 2 is incorrect because numbering must give the double bond the lowest number (1-ene, not 5-ene).
  • Option 3 is incorrect because it chooses the chain with fewer substituents.

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.

CHEMISTRYHydrocarbonscorrectcompound

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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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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)
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C.(iv) > (iii) > (ii) > (i)
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A.X = 2–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)
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A.$60^{\circ}$
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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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