The possible number of stereoisomers for 5-phenylpent-4-en-2-ol is .

Year	2026	2025	2024	Total
Questions	22	49	30	101

Q46 jee_main_2025_07_april_morning Quantitative Elemental Analysis

An organic compound weighing

500 mg

, produced

220 mg

of

CO₂

on complete combustion. The percentage composition of carbon in the compound is ______ %. (nearest integer) (Given molar mass in

g mol⁻¹

of

C: 12

,

O: 16

)

Numerical Answer. Answer: 12 to 12

Solution

Related Formula

% C = (12)/(44) \times Mass of CO₂ producedMass of organic compound taken \times 100

Core Logic

Given:

Mass of organic compound taken $= 500 mg = 500 \times 10⁻³ g$
Mass of $CO₂$ $\mathrm{CO}_2$ produced $= 220 mg = 220 \times 10⁻³ g$

Using the formula:

% C = (12)/(44) \times 220 \times 10⁻³500 \times 10⁻³ \times 100

% C = (12)/(44) \times (220)/(500) \times 100

% C = (12)/(44) \times 44 = 12 %

Thus, the percentage of carbon is $12$ $12$.

Pattern Recognition

Carbon dioxide has exactly $12/44 \approx 27.27%$ $12/44 \approx 27.27\%$ carbon by mass. Multiply the mass fraction of $CO₂$ $\mathrm{CO}_2$ ( $220/500 = 0.44$ $220/500 = 0.44$) by $12/44$ $12/44$ to directly get $0.12$ $0.12$ or $12%$ $12\%$.

Evaluation Rubric / Model Answer

A perfect step-by-step conversion of organic compound mass and combustion carbon dioxide mass to obtain a precise $12$ $12$ percent carbon composition.

Chapter Mix

Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

Q jee_main_2025_08_april_evening IUPAC Nomenclature

What is the correct IUPAC name of the following organic compound?

Cyclic substituted alkene organic molecule structure for Q35 — The molecule contains a five-membered carbon ring with one double bond, a hydroxyl substituent, and an ethyl group.

A. 4-Ethyl-1-hydroxycyclopent-2-ene
B. 1-Ethyl-3-hydroxycyclopent-2-ene
C. 1-Ethylcyclopent-2-en-3-ol
D. 4-Ethylcyclopent-2-en-1-ol

Solution

Core Logic

Let us apply official IUPAC priority indexing rules:

Principal Functional Group: The hydroxyl group ( $-OH$ $-\text{OH}$) possesses higher naming priority over double bonds and simple alkyl side chains. Thus, the carbon bearing the $-OH$ $-\text{OH}$ group is assigned position C-1.
Numbering Direction: We must number through the ring towards the double bond to assign it the lowest possible locant. Hence, the alkene carbons are given coordinates C-2 and C-3.
Locating Side Chains: Proceeding with this direction puts the ethyl group at position C-4.
The molecule contains a five-membered carbon ring with one double bond, a hydroxyl substituent, and an ethyl group.

Assembling the structural parts alphabetically:

Substituent: `4-Ethyl`
Parent root: `cyclopent-2-en`
Suffix: `1-ol`

Combined IUPAC format: 4-Ethylcyclopent-2-en-1-ol.

Pattern Recognition

Principal suffix priority hierarchy: $-OH > Double bond > Alkyl side-chain$ . Always fix the highest priority suffix at index 1 and head instantly towards the alkene bond to safely restrict locant numbers.

Chapter Mix

Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

Q27 jee_main_2025_08_april_evening Reactive Intermediates and Reagents

Match the LIST-I with LIST-II:

LIST-I	LIST-II
A. Carbocation	I. Species that can supply a pair of electrons.
B. C-Free radical	II. Species that can receive a pair of electrons.
C. Nucleophile	III. $sp²$ $sp^2$ hybridized carbon with empty p-orbital.
D. Electrophile	IV. $sp²/sp³$ $sp^2/sp^3$ hybridized carbon with one unpaired electron.

Choose the correct answer from the options given below:

A. $A-IV, B-II, C-III, D-I$ $\text{A-IV, B-II, C-III, D-I}$
B. $A-II, B-III, C-I, D-IV$ $\text{A-II, B-III, C-I, D-IV}$
C. $A-III, B-IV, C-II, D-I$ $\text{A-III, B-IV, C-II, D-I}$
D. $A-III, B-IV, C-I, D-II$ $\text{A-III, B-IV, C-I, D-II}$

Solution

Core Logic

Let us analyze each term carefully:

A. Carbocation: Features a positively charged trivalent carbon atom. It represents an $sp²$ $sp^2$ hybridized carbon with an empty unhybridized p-orbital.
Carbocation orbital hybridization diagram for Q27 - JEE Main 2025
B. Carbon Free Radical: Contains a trivalent carbon carrying a single unpaired lone electron. It typically exhibits $sp²$ $sp^2$ or $sp³$ $sp^3$ hybridization depending on structural environments.
Carbocation orbital hybridization diagram for Q27 - JEE Main 2025
C. Nucleophile: An electron-rich chemical species containing a lone pair or negative charge capable of donating/supplying a pair of electrons.
D. Electrophile: An electron-deficient chemical species possessing empty low-lying orbitals capable of accepting/receiving a pair of electrons.

Step 1: Alignment Matrix

Matching each item yields:

$A arrow III$ $\text{A} \rightarrow \text{III}$
$B arrow IV$ $\text{B} \rightarrow \text{IV}$
$C arrow I$ $\text{C} \rightarrow \text{I}$
$D arrow II$ $\text{D} \rightarrow \text{II}$

This sequence aligns flawlessly with Option (4).

Pattern Recognition

Nucleophiles donate ('nucleo-loving' = seeks positive sites with its electrons), Electrophiles accept ('electro-loving' = seeks electron density). Carbocations explicitly harbor a vacant p-orbital because of their positive charge configuration, making identification extremely swift.

Chapter Mix

Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

Q38 jee_main_2025_08_april_evening Quantitative Elemental Analysis

On complete combustion,

0.210 g

of an organic compound containing C, H, and O yielded

0.127 g

of

H₂O

and

0.307 g

of

CO₂

. The mass percentages of hydrogen and oxygen in the given organic compound respectively are:

A. 53.41, 39.6
B. 6.72, 53.41
C. 7.55, 43.85
D. 6.72, 39.87

Solution

Related Formula

Percentage of Hydrogen in organic analysis:

%H = (2)/(18) \times Mass of H₂OMass of Compound \times 100

Percentage of Carbon:

%C = (12)/(44) \times Mass of CO₂Mass of Compound \times 100

Percentage of Oxygen:

%O = 100 - (%C + %H)

Execution

Step 1: Compute the mass percent of Hydrogen:

%H = (2)/(18) \times (0.127)/(0.210) \times 100 = (0.254)/(3.78) \approx 6.72%

Step 2: Compute the mass percent of Carbon:

%C = (12)/(44) \times (0.307)/(0.210) \times 100 = (3.684)/(9.24) \approx 39.87%

Step 3: Deduce the remaining mass percent of Oxygen:

%O = 100 - (39.87 + 6.72) = 100 - 46.59 = 53.41%

Thus, the values of hydrogen and oxygen percentage are $6.72%$ $6.72\%$ and $53.41%$ $53.41\%$, matches with Option (2).

Pattern Recognition

Always focus on the order requested by the question stem. The query specifies 'hydrogen and oxygen respectively'. Option 2 and Option 4 both show these numbers but reversed—verifying the targeted sequence protects your score line.

Chapter Mix

Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

Q42 jee_main_2025_08_april_evening Qualitative Analysis of Functional Groups

Match the reagents in LIST-I with the corresponding chemical functional groups they detect in LIST-II:

LIST-I (Reagent)	LIST-II (Functional Group detected)
A. Sodium bicarbonate solution	I. double bond / unsaturation
B. Neutral ferric chloride	II. carboxylic acid
C. Ceric ammonium nitrate	III. phenolic - OH
D. Alkaline $KMnO₄$ $\text{KMnO}_4$	IV. alcoholic - OH

Choose the correct answer from the options given below:

A. $A-II, B-III, C-IV, D-I$ $\text{A-II, B-III, C-IV, D-I}$
B. $A-II, B-III, C-I, D-IV$ $\text{A-II, B-III, C-I, D-IV}$
C. $A-III, B-II, C-IV, D-I$ $\text{A-III, B-II, C-IV, D-I}$
D. $A-II, B-IV, C-III, D-I$ $\text{A-II, B-IV, C-III, D-I}$

Solution

Core Logic

Let us review the chemical basis for each qualitative test:

A. Sodium bicarbonate ( $NaHCO₃$ $\text{NaHCO}_3$) solution: Carboxylic acids are sufficiently acidic to decompose $NaHCO₃$ $\text{NaHCO}_3$, liberating carbon dioxide gas observed as vigorous effervescence. Therefore, $A arrow II$ $\text{A} \rightarrow \text{II}$.
B. Neutral ferric chloride ( $FeCl₃$ $\text{FeCl}_3$): Phenols react with neutral $FeCl₃$ $\text{FeCl}_3$ solution to form characteristic deeply colored violet coordination complexes. Therefore, $B arrow III$ $\text{B} \rightarrow \text{III}$.
C. Ceric ammonium nitrate (CAN): Alcohols react with CAN reagent to cause a distinct color shift to deep dark red due to complexation. Therefore, $C arrow IV$ $\text{C} \rightarrow \text{IV}$.
D. Alkaline $KMnO₄$ $\text{KMnO}_4$ (Baeyer's Reagent): Reacts readily via syn-hydroxylation across carbon-carbon double/triple bonds, resulting in decolored solutions alongside brown $MnO₂$ $\text{MnO}_2$ precipitates. This detects unsaturation. Therefore, $D arrow I$ $\text{D} \rightarrow \text{I}$.

Step 1: Assembly

Combining the validated relationships gives:

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

This maps perfectly to Option (1).

Pattern Recognition

Baeyer's test (alkaline $KMnO₄$ $\text{KMnO}_4$) always tests for alkenes/alkynes. $NaHCO₃$ $\text{NaHCO}_3$ is unique for acidic groups like carboxylic acids. Matching these two reliable pairs isolates the correct option without needing to review the entire table.

Chapter Mix

Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques Class 12 Chemistry: Alcohols, Phenols and Ethers

The possible number of stereoisomers for 5-phenylpent-4-en-2-ol is .

Solution & Explanation

Related Formula

Core Logic

Visual Mapping

Pattern Recognition

Chapter Mix

Reference Study Guides

More Organic Chemistry - Some Basic Principles and Techniques Previous-Year Questions — Page 8

Solution

Related Formula

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Solution

Core Logic

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Solution

Core Logic

Step 1: Alignment Matrix

Pattern Recognition

Chapter Mix

Solution

Related Formula

Execution

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Assembly

Pattern Recognition

Chapter Mix

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