Substituted cyclohexane molecule labeled as compound P.
A.1-Ethyl-5, 5-dimethylcyclohexane
B.3-Ethyl-1,1-dimethylcyclohexane
C.1-Ethyl-3, 3-dimethylcyclohexane
D.1,1-Dimethyl-3-ethylcyclohexane
Solution & Explanation
### Core Logic
Number the ring to give substituents the lowest possible locants. Setting locant 1 at the carbon carrying the two methyl groups provides a locant list of (1,1,3)$(1,1,3)$, whereas setting it at the ethyl-bearing carbon gives (1,3,3)$(1,3,3)$. The set (1,1,3)$(1,1,3)$ wins by the lowest locant rule. Then arrange alphabetically: 3-ethyl precedes 1,1-dimethyl. Hence, the correct systematic tag is 3-Ethyl-1,1-dimethylcyclohexane. Substituted cyclohexane molecule labeled as compound P.
### Pattern Recognition
Lowest locant grouping set takes ultimate priority before checking alphabetical organization rules.
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Keywords:#IUPAC name substituted cyclohexane#JEE Main 2024 Morning Q77#Lowest locant rule organic#Systematic chemical notation#IUPAC name#cyclohexane indexing#alphabetical priority
More Organic Chemistry - Some Basic Principles and Techniques Previous-Year Questions — Page 10
Qjee_main_2025_29_jan_morningNucleophiles and Electrophiles
Total number of nucleophiles from the following is :-
mathrm N H _ 3, mathrm P h S H, (mathrm H _ 3 mathrm C) _ 2 mathrm S, mathrm H _ 2 mathrm C = mathrm C H _ 2, stackrel ominus mathrm O mathrm H, mathrm H _ 3 mathrm O ^ oplus, (mathrm C H _ 3) _ 2 mathrm C O, > = mathrm N C H _ 3$\mathrm {N H} _ {3}, \mathrm {P h S H}, (\mathrm {H} _ {3} \mathrm {C}) _ {2} \mathrm {S}, \mathrm {H} _ {2} \mathrm {C} = \mathrm {C H} _ {2}, \stackrel {\ominus} {\mathrm {O}} \mathrm {H}, \mathrm {H} _ {3} \mathrm {O} ^ {\oplus}, (\mathrm {C H} _ {3}) _ {2} \mathrm {C O}, > = \mathrm {N C H} _ {3}$
A. 5
B. 4
C. 7
D. 6
Solution
### Related Formula
Nucleophiles are electron-rich species containing lone pairs of electrons or pi$\pi$-bonds that can donate an electron pair to an electrophilic center.
### Core Logic
Let us examine each species:
* mathrmNH_3$\mathrm{NH}_3$: Contains a lone pair on nitrogen
ightarrow$
ightarrow$ Nucleophile
* mathrmPhSH$\mathrm{PhSH}$: Contains lone pairs on sulfur
ightarrow$
ightarrow$ Nucleophile
* (mathrmH_3mathrmC)_2mathrmS$(\mathrm{H}_3\mathrm{C})_2\mathrm{S}$: Contains lone pairs on sulfur
ightarrow$
ightarrow$ Nucleophile
* mathrmH_2mathrmC=mathrmCH_2$\mathrm{H}_2\mathrm{C}=\mathrm{CH}_2$: Contains a nucleophilic pi$\pi$-bond
ightarrow$
ightarrow$ Nucleophile
* stackrelominusmathrmOmathrmH$\stackrel{\ominus}{\mathrm{O}}\mathrm{H}$: Negatively charged with lone pairs
ightarrow$
ightarrow$ Nucleophile
* mathrmH_3mathrmO^oplus$\mathrm{H}_3\mathrm{O}^{\oplus}$: Electron deficient, positively charged oxygen cannot donate electrons
ightarrow$
ightarrow$ Electrophile
* (mathrmCH_3)_2mathrmCO$(\mathrm{CH}_3)_2\mathrm{CO}$: Carbonyl carbon is electrophilic
* >=mathrmNCH_3$>=\mathrm{NCH}_3$: Imine carbon is electrophilic
Thus, the total number of nucleophiles is 5
### Pattern Recognition
Neutral molecules with lone pairs (N, S) or alkenes/alkynes with available pi$\pi$-electrons operate as good nucleophiles, along with full anions.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Qjee_main_2025_29_jan_morningIUPAC Nomenclature of Organic Compounds
(IV) 2-Methyl-1,3-pentadiene
Choose the correct answer from the options given below:
A. \text{(A)-(III), (B)-(II), (C)-(IV), (D)-(I)}
B. \text{(A)-(III), (B)-(II), (C)-(I), (D)-(IV)}
C. \text{(A)-(II), (B)-(III), (C)-(IV), (D)-(I)}
D. \text{(A)-(II), (B)-(III), (C)-(I), (D)-(IV)}
Solution
### Related Formula
IUPAC Rules select the longest principal carbon chain and number it to give substituents the lowest possible locants.
### Core Logic
Let's systematic decode each name :
* (A) Longest chain contains 7 carbons (heptane) with an ethyl group at position 3 and a methyl group at position 5
ightarrow$
ightarrow$ 3-Ethyl-5-methylheptane (II) .
* (B) C expanded yields a 7 carbon main chain with two methyl groups at carbon-4
ightarrow$
ightarrow$ 4,4-Dimethylheptane (III) .
* (C) 5-carbon diene numbered from the left double bond side
ightarrow$
ightarrow$ 2-Methyl-1,3-pentadiene (IV) .
* (D) 5-carbon alkene starting from the double bond end
ightarrow$
ightarrow$ 4-Methylpent-1-ene (I) .
Therefore, matching sequence: (A)-(II), (B)-(III), (C)-(IV), (D)-(I).
### Pattern Recognition
Expanding compressed groupings such as mathrm(C_3H_7)_2$\mathrm{(C_3H_7)_2}$ prevents errors regarding parent chain carbon counts.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Qjee_main_2025_29_jan_morningPurification of Organic Compounds - Steam Distillation
The steam volatile compounds among the following are:
(A) Four different disubstituted benzene structures are indexed to show spatial isomer distributions.
(B) Four different disubstituted benzene structures are indexed to show spatial isomer distributions.
(C) Four different disubstituted benzene structures are indexed to show spatial isomer distributions.
(D) Four different disubstituted benzene structures are indexed to show spatial isomer distributions.
Choose the correct answer from the options given below:
A. (B) and (D) only
B. (A) and (C) only
C. (A) and (B) only
D. (A), (B) and (C) only
Solution
### Related Formula
textIntramolecular H-bonding implies textLower boiling point implies textSteam Volatile$\text{Intramolecular H-bonding} \implies \text{Lower boiling point} \implies \text{Steam Volatile}$
### Core Logic
Purification via steam distillation requires a high relative vapor pressure at the boiling point of water. Let us assess the isomers :
* (A) o-Nitrophenol contains an -mathrmOH$-\mathrm{OH}$ group right next to an -mathrmNO_2$-\mathrm{NO}_2$ group, allowing for strong intramolecular hydrogen bonding. This minimizes external interactions, lowering the boiling point and making it steam volatile .
* (B) o-Nitroaniline similarly stabilizes itself via internal intramolecular hydrogen bonding between the amine and nitro components, making it steam volatile .
* (C) & (D) The para isomers form extensive intermolecular networks with neighboring molecules, which significantly elevates their boiling points and prevents steam volatility .
Hence, compounds (A) and (B) are steam volatile, matching option (3).
### Pattern Recognition
ortho-substituted functional networks form self-contained internal loops through hydrogen bonding, preventing external pairing and maximizing volatility.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Class 11 Chemistry: Chemical Bonding and Molecular Structure
Q48jee_main_2025_29_jan_morningSigma and Pi Bonds
The sum of sigma (sigma)$(\sigma)$ and mathrmpi(pi)$\mathrm{\pi}(\pi)$ bonds in Hex-1,3-dien-5-yne is ________.
Numerical Answer.Answer: 15 to 15
Solution
### Related Formula
textSingle bond = 1sigma, quad textDouble bond = 1sigma + 1pi, quad textTriple bond = 1sigma + 2pi$\text{Single bond} = 1\sigma, \quad \text{Double bond} = 1\sigma + 1\pi, \quad \text{Triple bond} = 1\sigma + 2\pi$
### Core Logic
Let us map out the complete structural bond geometry of Hex-1,3-dien-5-yne :
mathrmH_2C=CH-CH=CH-Cequiv CH$\mathrm{H_2C=CH-CH=CH-C\equiv CH}$
Counting the individual bonds :
* Carbon-Hydrogen (mathrmC-H$\mathrm{C-H}$) single bonds = 2 + 1 + 1 + 1 + 1 = 6 \, sigma$= 2 + 1 + 1 + 1 + 1 = 6 \, \sigma$ bonds
* Carbon-Carbon single, double, and triple linkages:
* mathrmC_1=mathrmC_2
ightarrow 1sigma + 1pi$\mathrm{C}_1=\mathrm{C}_2
ightarrow 1\sigma + 1\pi$
* mathrmC_2-mathrmC_3
ightarrow 1sigma$\mathrm{C}_2-\mathrm{C}_3
ightarrow 1\sigma$
* mathrmC_3=mathrmC_4
ightarrow 1sigma + 1pi$\mathrm{C}_3=\mathrm{C}_4
ightarrow 1\sigma + 1\pi$
* mathrmC_4-mathrmC_5
ightarrow 1sigma$\mathrm{C}_4-\mathrm{C}_5
ightarrow 1\sigma$
* mathrmC_5equivmathrmC_6
ightarrow 1sigma + 2pi$\mathrm{C}_5\equiv\mathrm{C}_6
ightarrow 1\sigma + 2\pi$
Total values :
textTotal sigma text bonds = 6 + 5 = 11$\text{Total } \sigma \text{ bonds} = 6 + 5 = 11$textTotal pi text bonds = 1 + 1 + 2 = 4$\text{Total } \pi \text{ bonds} = 1 + 1 + 2 = 4$textSum (sigma + pi) = 11 + 4 = 15$\text{Sum } (\sigma + \pi) = 11 + 4 = 15$
### Pattern Recognition
Do not skip implicit mathrmC-H$\mathrm{C-H}$ single bonds when scanning skeletal formulas, as each counts as a full sigma$\sigma$ bond.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q73jee_main_2024_01_february_morningFission of a Covalent Bond
Ionic reactions with organic compounds proceed through:
(A) Homolytic bond cleavage
(B) Heterolytic bond cleavage
(C) Free radical formation
(D) Primary free radical
(E) Secondary free radical
Choose the correct answer from the options given below:
A.text(A) only$\text{(A) only}$
B.text(C) only$\text{(C) only}$
C.text(B) only$\text{(B) only}$
D.text(D) and (E) only$\text{(D) and (E) only}$
Solution
### Core Logic
Bond fission in organic chemistry can happen in two ways:
1. Homolytic cleavage: The shared pair of electrons gets distributed equally, leading to free radical formation.
2. Heterolytic cleavage: The shared pair of electrons goes completely to one of the bonded atoms, leading to the formation of positive (cation) and negative (anion) ions.
### Step 1: Match Definition
Since ionic reactions involve ions, they inherently proceed through heterolytic bond cleavage where ionic intermediates (carbocations, carbanions, or leaving groups) are formed.
### Pattern Recognition
Ionic reactions rightarrow$\rightarrow$ Ions formed rightarrow$\rightarrow$ Heterolytic cleavage.
Radical reactions rightarrow$\rightarrow$ Free radicals formed rightarrow$\rightarrow$ Homolytic cleavage.
### 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_2024_27_jan_morning
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