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Hydrocarbons

JEE Main 2025 (Online) 7th April Evening ShiftNumericalMedium📈 Topic in 29% papers+4
Ques. 1JEE Main 2025, 7 April (Evening Shift)

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 .

Answer

7

Solution

Step 1: PCl5 replaces the carbonyl oxygen by two Cl atoms (gem-dichloride):

A = Ph-CCl2-CH3

Step 2: Two equivalents of NaNH2 remove two molecules of HCl (double dehydrohalogenation) giving phenylacetylene; the third equivalent removes the acidic terminal H:

B = Ph-C C Na (sodium phenylacetylide)

Step 3: Acidification gives back the terminal alkyne:

C = Ph-C CH (phenylethyne)

Step 4: Hydroboration–oxidation of a terminal alkyne adds H and OH anti-Markovnikov; the enol formed tautomerises to an aldehyde:

D = Ph-CH2-CHO (phenylacetaldehyde) [figure]

sp2 carbons in D: 6 (benzene ring) + 1 (CHO carbon) = 7; the CH2 carbon is sp3.

Answer: 7

Keep practising — Hydrocarbons

Ques. 2JEE Main 2025, 4 April (Morning Shift)

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

  1. [figure]
  2. [figure]
  3. [figure]
  4. [figure]
Ques. 3JEE Main 2025, 4 April (Evening Shift)

Consider the following molecule(X). [figure] The major product formed when the given molecule (X) is treated with HBr (1 eq) is :

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

Which compound would give 3-methyl-6-oxoheptanal upon ozonolysis?

  1. [figure]
  2. [figure]
  3. [figure]
  4. [figure]
Ques. 5JEE Main 2025, 2 April (Morning Shift)

An optically active alkyl halide C4H9Br [A] reacts with hot KOH dissolved in ethanol and forms alkene [B] as major product which reacts with bromine to give dibromide [C]. The compound [C] is converted into a gas [D] upon reacting with alcoholic NaNH2. During hydration 18 gram of water is added to 1 mole of gas [D] on warming with mercuric sulphate and dilute acid at 333 K to form compound [E]. The IUPAC name of compound [E] is :

  1. Butan-2-ol
  2. But-2-yne
  3. Butan-2-one
  4. Butan-1-al
Ques. 6JEE Main 2025, 2 April (Evening Shift)

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. Statement I is correct but Statement II is incorrect
  4. Both Statement I and Statement II are incorrect
Ques. 7JEE Main 2025, 29 January (Evening Shift)

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 true but Statement II is false
  2. Both Statement I and Statement II are true
  3. Both Statement I and Statement II are false
  4. Statement I is false but Statement II is true
Ques. 8JEE Main 2025, 29 January (Evening Shift)

Isomeric hydrocarbons negative Baeyer's test (Molecular formula C9H12) The total number of isomers from above with four different non-aliphatic substitution sites is -

Ques. 9JEE Main 2025, 28 January (Morning Shift)

A molecule (P) on treatment with acid undergoes rearrangement and gives (Q). (Q) on ozonolysis followed by reflux under alkaline condition gives (R). The structure of (R) is given below. [figure] The structure of (P) is

  1. [figure]
  2. [figure]
  3. [figure]
  4. [figure]
Ques. 10JEE Main 2025, 28 January (Evening Shift)

Identify product [A], [B] and [C] in the following reaction sequence. CH3-C CH [H2]Pd/C [A] [(ii) Zn, H2O](i) O3 [B] + [C]

  1. [A]: CH3-CH=CH2, [B]: CH3CHO, [C]: CH3CH2OH
  2. [A]: CH3-CH=CH2, [B]: CH3CHO, [C]: HCHO
  3. [A]: CH3CH2CH3, [B]: CH3CHO, [C]: HCHO
  4. [A]: CH2=CH2, [B]: H3C- O \|C-CH3, [C]: HCHO
Ques. 11JEE Main 2025, 24 January (Morning Shift)

The product (A) formed in the following reaction sequence is [figure]

  1. CH3-CH2- OH |CH-CH2-NH2
  2. CH3-CH2- NH2 |CH-CH2-NH2
  3. CH3- CH3 | OH |C-CH2-NH2
  4. CH3- CH3 | NH2 |C-CH2-OH

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