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Hydrocarbons · Question 15

JEE Main 2026 (Online) 23rd January Evening ShiftMCQHard📈 Topic in 32% papers+4−1
1

[figure] Identify (P)

  1. [figure]
  2. [figure]
  3. [figure]
  4. [figure]

Answer

(4) [figure]

Solution

Step 1: Zn, removes the two vicinal Br atoms (dehalogenation) a C=C bond is formed between the two carbons that carried Br.

Step 2: H+ of HBr adds to the double bond a 2° carbocation is formed next to either the ethyl-bearing or the methyl-bearing ring carbon.

A 1,2-hydride shift converts this 2° carbocation into a more stable 3° carbocation.

The 3° carbocation on the methyl-bearing carbon has 7 α-H atoms (3 of CH3 + 2 + 2 of the ring CH2 groups), while the one on the ethyl-bearing carbon has only 6 α-H atoms (2 of CH2CH3 + 2 + 2),

the carbocation on the carbon carrying CH3 is more stabilised by hyperconjugation.

Br- attacks this carbocation:

[figure]

(P) is 1-bromo-3-ethyl-1-methylcyclopentane, i.e. option (4).

Answer: (4)

JEE Main 2026 (Online) 23rd January Evening Shift

Keep practising — Hydrocarbons

2

Given below are two statements : In the light of the above statements, choose the most appropriate answer from the options given below

  1. Statement I is incorrect but Statement II is correct
  2. Both Statement I and Statement II are correct
  3. Both Statement I and Statement II are incorrect
  4. Statement I is correct but Statement II is incorrect
JEE Main 2026 (Online) 22nd January Morning Shift
3

The cycloalkene (X) on bromination consumes one mole of bromine per mole of (X) and gives the product (Y) in which C : Br ratio is 3:1. The percentage of bromine in the product (Y) is %. (Nearest integer) (Given : molar mass in g mol-1 H : 1, C : 12, O : 16, Br : 80)

JEE Main 2026 (Online) 22nd January Morning Shift
4

Consider the following reaction : [figure] The product Y formed is :

  1. 2-methylhex-3-yne
  2. 5-methylhex-2-yne
  3. Isopropylbut-1-yne
  4. 2-methylhex-2-yne
JEE Main 2026 (Online) 22nd January Evening Shift
5

The dibromo compound [P] (molecular formula: C9H10Br2) when heated with excess sodamide followed by treatment with dilute HCl gives [Q]. On warming [Q] with mercuric sulphate and dilute sulphuric acid yields [R] which gives positive Iodoform test but negative Tollens' test. The compound [P] is :

  1. [figure]
  2. [figure]
  3. [figure]
  4. [figure]
JEE Main 2026 (Online) 22nd January Evening Shift
6

Identify A in the following reaction. [figure]

  1. [figure]
  2. [figure]
  3. [figure]
  4. [figure]
JEE Main 2026 (Online) 21st January Morning Shift
7

[figure] 'P' is

  1. [figure]
  2. [figure]
  3. [figure]
  4. [figure]
JEE Main 2025 (Online) 8th April Evening Shift
8

Choose the correct set of reagents for the following conversion. [figure]

  1. Cl2/ anhy. AlCl3; Br2/Fe; alc. KOH
  2. Cl2/Fe; Br2/ anhy. AlCl3; aq. KOH
  3. Br2/Fe; Cl2, ; alc. KOH
  4. Br2/ anhy. AlCl3; Cl2, ; aq. KOH
JEE Main 2025 (Online) 8th April Evening Shift
9

Given below are two statements: In the light of the above statements, choose the correct answer from the options given below

  1. Statement I is false but Statement II is true
  2. Both Statement I and Statement II are false
  3. Statement I is true but Statement II is false
  4. Both Statement I and Statement II are true
JEE Main 2025 (Online) 7th April Morning Shift
10

Identify the structure of the final product (D) in the following sequence of the reactions : [figure] Total number of sp2 hybridised carbon atoms in product D is .

JEE Main 2025 (Online) 7th April Evening Shift
11

Predict the major product of the following reaction sequence: [figure]

  1. [figure]
  2. [figure]
  3. [figure]
  4. [figure]
JEE Main 2025 (Online) 4th April Morning Shift

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