From the following, the least stable structure is :
A.textOption 1$\text{Option 1}$
Resonance Effects
B.textOption 2$\text{Option 2}$
Resonance Effects
C.textOption 3$\text{Option 3}$
Resonance Effects
D.textOption 4$\text{Option 4}$
Resonance Effects
Solution & Explanation
### Core Logic
In structure 3, there are positive formal charges on two adjacent atoms (Oxygen and the Carbon adjacent to it). Like charges on adjacent atoms cause extreme electrostatic repulsion, making the structure highly unstable.
Resonance Effects diagram for Q62 - JEE Main 2026 Morning
### Pattern Recognition
Rules of resonance stability:
1. Complete octets are more stable.
2. More covalent bonds = more stable.
3. Least charge separation is more stable.
4. Like charges on adjacent atoms create massive destabilization (Least stable scenario).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Keywords:#least stable resonance structure#JEE Main 2026 Morning Q62#Organic Chemistry - Some Basic Principles and Techniques JEE Main 2026#Resonance Effects JEE Main 2026
More Organic Chemistry - Some Basic Principles and Techniques Previous-Year Questions — Page 4
Which of the following is the correct IUPAC name of given organic compound (X)?
The image shows structural representation of compound X with a double bond and a bromine substituent.
### Core Logic
To determine the IUPAC name of the compound shown in The image shows structural representation of compound X with a double bond and a bromine substituent.:
1. Identify the principal functional group, which is the double bond (alkene).
2. Find the longest carbon chain containing the double bond:
mathrmC1(H_2Br) - C2(CH_3) = C3(H) - C4(H_3)$$\mathrm{C1(H_2Br) - C2(CH_3) = C3(H) - C4(H_3)}$$
The longest chain has 4$4$ carbons, which means the parent alkane is butane, and with a double bond it's "but-2-ene".
3. Number the chain from the end that gives lower locants to the double bond. Starting from left or right both give the double bond at position 2$2$. However, starting from left gives substituent locants as 1$1$ (for bromo) and 2$2$ (for methyl), whereas starting from right gives substituent locants as 3$3$ and 4$4$.
4. Hence, correct numbering is:
- textC1$\text{C1}$: bonded to Bromine (-textBr$-\text{Br}$)
- textC2$\text{C2}$: bonded to Methyl (-textCH_3$-\text{CH}_3$)
- textC3$\text{C3}$: alkene carbon
- textC4$\text{C4}$: terminal methyl group
The image shows structural representation of compound X with a double bond and a bromine substituent.
Combining these rules, the name is: **1-Bromo-2-methylbut-2-ene**.
### Pattern Recognition
Double bond takes precedence over halogen substituent in numbering direction. If double bond is symmetrical (at position 2$2$ in a 4$4$-carbon chain), use the substituent positions to break the tie, choosing lowest possible locants (1$1$ and 2$2$ vs 3$3$ and 4$4$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Class 12 Chemistry: Haloalkanes and Haloarenes
An organic compound weighing 500mathrm\ mg$500\mathrm{\ mg}$, produced 220mathrm\ mg$220\mathrm{\ mg}$ of mathrmCO_2$\mathrm{CO}_2$ on complete combustion. The percentage composition of carbon in the compound is ______ %. (nearest integer)
(Given molar mass in mathrmg\ mol^-1$\mathrm{g\ mol}^{-1}$ of mathrmC: 12$\mathrm{C}: 12$, mathrmO: 16$\mathrm{O}: 16$)
Numerical Answer.Answer: 12 to 12
Solution
### Related Formula
\% mathrmC = frac1244 times fractextMass of mathrmCO_2 text producedtextMass of organic compound taken times 100$$\% \mathrm{C} = \frac{12}{44} \times \frac{\text{Mass of } \mathrm{CO}_2 \text{ produced}}{\text{Mass of organic compound taken}} \times 100$$
### Core Logic
Given:
- Mass of organic compound taken = 500 text mg = 500 times 10^-3 text g$= 500 \text{ mg} = 500 \times 10^{-3} \text{ g}$
- Mass of mathrmCO_2$\mathrm{CO}_2$ produced = 220 text mg = 220 times 10^-3 text g$= 220 \text{ mg} = 220 \times 10^{-3} \text{ g}$
Using the formula:
\% mathrmC = frac1244 times frac220 times 10^-3500 times 10^-3 times 100$$\% \mathrm{C} = \frac{12}{44} \times \frac{220 \times 10^{-3}}{500 \times 10^{-3}} \times 100$$\% mathrmC = frac1244 times frac220500 times 100$$\% \mathrm{C} = \frac{12}{44} \times \frac{220}{500} \times 100$$\% mathrmC = frac1244 times 44 = 12 \%$$\% \mathrm{C} = \frac{12}{44} \times 44 = 12 \%$$
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 mathrmCO_2$\mathrm{CO}_2$ (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
Q27jee_main_2025_08_april_eveningReactive 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. sp^2$sp^2$ hybridized carbon with empty p-orbital.
D. Electrophile
IV. sp^2/sp^3$sp^2/sp^3$ hybridized carbon with one unpaired electron.
Choose the correct answer from the options given below:
### Core Logic
Let us analyze each term carefully:
* **A. Carbocation**: Features a positively charged trivalent carbon atom. It represents an sp^2$sp^2$ 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 sp^2$sp^2$ or sp^3$sp^3$ 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:
* textA rightarrow textIII$\text{A} \rightarrow \text{III}$
* textB rightarrow textIV$\text{B} \rightarrow \text{IV}$
* textC rightarrow textI$\text{C} \rightarrow \text{I}$
* textD rightarrow textII$\text{D} \rightarrow \text{II}$
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.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
What is the correct IUPAC name of the following organic compound?
{{Q_IMG1}}
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:
1. **Principal Functional Group**: The hydroxyl group (-textOH$-\text{OH}$) possesses higher naming priority over double bonds and simple alkyl side chains. Thus, the carbon bearing the -textOH$-\text{OH}$ group is assigned position **C-1**.
2. **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**.
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: -textOH > textDouble bond > textAlkyl side-chain$-\text{OH} > \text{Double bond} > \text{Alkyl side-chain}$. Always fix the highest priority suffix at index 1 and head instantly towards the alkene bond to safely restrict locant numbers.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
On complete combustion, 0.210 text g$0.210 \text{ g}$ of an organic compound containing C, H, and O yielded 0.127 text g$0.127 \text{ g}$ of textH_2textO$\text{H}_2\text{O}$ and 0.307 text g$0.307 \text{ g}$ of textCO_2$\text{CO}_2$. 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:
\%textH = frac218 times fractextMass of H_2OtextMass of Compound times 100$$\%\text{H} = \frac{2}{18} \times \frac{\text{Mass of } H_2O}{\text{Mass of Compound}} \times 100$$
Percentage of Carbon:
\%textC = frac1244 times fractextMass of CO_2textMass of Compound times 100$$\%\text{C} = \frac{12}{44} \times \frac{\text{Mass of } CO_2}{\text{Mass of Compound}} \times 100$$
Percentage of Oxygen:
\%textO = 100 - (\%textC + \%textH)$$\%\text{O} = 100 - (\%\text{C} + \%\text{H})$$
### Execution
Step 1: Compute the mass percent of Hydrogen:
\%textH = frac218 times frac0.1270.210 times 100 = frac0.2543.78 approx 6.72\%$$\%\text{H} = \frac{2}{18} \times \frac{0.127}{0.210} \times 100 = \frac{0.254}{3.78} \approx 6.72\%$$
Step 2: Compute the mass percent of Carbon:
\%textC = frac1244 times frac0.3070.210 times 100 = frac3.6849.24 approx 39.87\%$$\%\text{C} = \frac{12}{44} \times \frac{0.307}{0.210} \times 100 = \frac{3.684}{9.24} \approx 39.87\%$$
Step 3: Deduce the remaining mass percent of Oxygen:
\%textO = 100 - (39.87 + 6.72) = 100 - 46.59 = 53.41\%$$\%\text{O} = 100 - (39.87 + 6.72) = 100 - 46.59 = 53.41\%$$
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.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
More Organic Chemistry - Some Basic Principles and Techniques Questions — jee_main_2026_21_jan_morning
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