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JEE Main 2025 Jan 29 Shift 1, Chemistry Q5: Introduction to Nomenclature

JEE Main2025Jan 29, Shift 1Chemistry
Q.

Match List-I with List-II.

List-I (Structure)List-II (IUPAC Name)
(A) (I) 4-Methylpent-1-ene
(B) (II) 3-Ethyl-5-methylheptane
(C) (III) 4,4-Dimethylheptane
(D) (IV) 2-Methyl-1,3-pentadiene

Choose the correct answer from the options given below:

  1. A

    (A)-(III), (B)-(II), (C)-(IV), (D)-(I)

  2. B

    (A)-(III), (B)-(II), (C)-(I), (D)-(IV)

  3. C

    (A)-(II), (B)-(III), (C)-(IV), (D)-(I)

  4. D

    (A)-(II), (B)-(III), (C)-(I), (D)-(IV)

Solution

Number each chain along the longest carbon backbone and apply the lowest-locant rule for substituents.

(A) The longest chain has 7 carbons (heptane). Numbering gives an ethyl group at C-3 and a methyl group at C-5: 3-Ethyl-5-methylheptane (II).

(B) : the longest chain runs through two propyl arms (7 C atoms). The quaternary C carries two methyls; numbering from the end nearest the substituents places both methyls on C-4: 4,4-Dimethylheptane (III).

(C) A 5-carbon diene with a methyl branch on C-2 and double bonds at 1,3: 2-Methyl-1,3-pentadiene (IV).

(D) A 5-carbon chain with terminal and a methyl at C-4: 4-Methylpent-1-ene (I).

Hence (A)-(II), (B)-(III), (C)-(IV), (D)-(I).

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