JEE Main 20265 AprilMorning Shift
- Exam
- JEE Main
- Year
- 2026
- Date
- 5 April
- Shift
- Morning Shift
- Subject
- Chemistry
- Chapter
- Hydrocarbons
JEE Main 2026 · 5 April · Morning Shift · Chemistry · Hydrocarbons
"P" is a hydrocarbon of molecular formula: C _8 H _ 14 . On...
"P" is a hydrocarbon of molecular formula: CH. On ozonolysis, "P" forms "Q". "Q" on treatment with alkali under reflux condition produces "R", which on treatment with I/NaOH gives a yellow precipitate. Acidification of the solution gives "S". The structure of "S" is given below:
The correct structure of "P" is
The correct structure of "P" is
Options
- A
![Formation of Na_4[Fe(CN)_5 NOS] , a purple coloured complex formed by addition of sodium nitroprusside in sodium](/api/proxy-image?url=https%3A%2F%2Ffirebasestorage.googleapis.com%2Fv0%2Fb%2Fquantrexacademy-app.firebasestorage.app%2Fo%2Fquestions%252Ffigs%252Fpyq%252Fjee_main_2026_april_watermarked%252F05APRS1__q64_d2.png%3Falt%3Dmedia&clean=1&v=qxfig138)
- B
![Formation of Na_4[Fe(CN)_5 NOS] , a purple coloured complex formed by addition of sodium nitroprusside in sodium](/api/proxy-image?url=https%3A%2F%2Ffirebasestorage.googleapis.com%2Fv0%2Fb%2Fquantrexacademy-app.firebasestorage.app%2Fo%2Fquestions%252Ffigs%252Fpyq%252Fjee_main_2026_april_watermarked%252F05APRS1__q64_d3.png%3Falt%3Dmedia&clean=1&v=qxfig138)
- C
![Formation of Na_4[Fe(CN)_5 NOS] , a purple coloured complex formed by addition of sodium nitroprusside in sodium](/api/proxy-image?url=https%3A%2F%2Ffirebasestorage.googleapis.com%2Fv0%2Fb%2Fquantrexacademy-app.firebasestorage.app%2Fo%2Fquestions%252Ffigs%252Fpyq%252Fjee_main_2026_april_watermarked%252F05APRS1__q64_d4.png%3Falt%3Dmedia&clean=1&v=qxfig138)
- D
![Formation of Na_4[Fe(CN)_5 NOS] , a purple coloured complex formed by addition of sodium nitroprusside in sodium](/api/proxy-image?url=https%3A%2F%2Ffirebasestorage.googleapis.com%2Fv0%2Fb%2Fquantrexacademy-app.firebasestorage.app%2Fo%2Fquestions%252Ffigs%252Fpyq%252Fjee_main_2026_april_watermarked%252F05APRS1__q64_d5.png%3Falt%3Dmedia&clean=1&v=qxfig138)
Correct answer
D.
Step-by-step solution
The structure of "S" is 2-methylcyclopent-1-ene-1-carboxylic acid.
Since "S" is obtained by the acidification of the product from the iodoform reaction of "R", "R" must contain a methyl ketone group (). The iodoform reaction converts the group into a group alongside the formation of a yellow precipitate of iodoform (). Therefore, the structure of "R" is 1-acetyl-2-methylcyclopentene.
"R" is formed by the intramolecular aldol condensation of "Q". Performing a retro-aldol condensation on "R" by cleaving the double bond in the cyclopentene ring yields octane-2,7-dione as "Q" ().
"Q" is obtained by the ozonolysis of "P". To determine the structure of "P", we perform retro-ozonolysis by removing the oxygen atoms from the two carbonyl groups of "Q" and connecting the corresponding carbon atoms with a double bond. Connecting the C2 and C7 carbons forms a six-membered ring with methyl groups at the double bond.
Thus, "P" is 1,2-dimethylcyclohexene.
Answer:
Since "S" is obtained by the acidification of the product from the iodoform reaction of "R", "R" must contain a methyl ketone group (). The iodoform reaction converts the group into a group alongside the formation of a yellow precipitate of iodoform (). Therefore, the structure of "R" is 1-acetyl-2-methylcyclopentene.
"R" is formed by the intramolecular aldol condensation of "Q". Performing a retro-aldol condensation on "R" by cleaving the double bond in the cyclopentene ring yields octane-2,7-dione as "Q" ().
"Q" is obtained by the ozonolysis of "P". To determine the structure of "P", we perform retro-ozonolysis by removing the oxygen atoms from the two carbonyl groups of "Q" and connecting the corresponding carbon atoms with a double bond. Connecting the C2 and C7 carbons forms a six-membered ring with methyl groups at the double bond.
Thus, "P" is 1,2-dimethylcyclohexene.
Answer: