(C) Different fractions of crude oil in petroleum industry
(III) Distillation at reduced pressure
(D) Chloroform-Aniline mixture
(IV) Steam distillation
Choose the correct answer from the options given below:
A.(A)-(IV), (B)-(III), (C)-(II), (D)-(I)
B.(A)-(I), (B)-(II), (C)-(III), (D)-(IV)
C.(A)-(III), (B)-(IV), (C)-(I), (D)-(II)
D.(A)-(II), (B)-(I), (C)-(IV), (D)-(III)
Solution & Explanation
### Core Logic
Evaluating standard NCERT laboratory purification matches:
- **(A) Aniline from aniline-water mixture:** Aniline is steam volatile and immiscible with water, so it is separated via **Steam distillation (IV)**.
- **(B) Glycerol from spent-lye in soap industry:** Glycerol decomposes at or below its boiling point, hence it is separated via **Distillation at reduced pressure (III)**.
- **(C) Different fractions of crude oil:** Separated using their small differences in boiling points via **Fractional distillation (II)**.
- **(D) Chloroform-Aniline mixture:** Separated due to a substantial boiling point difference via **Simple distillation (I)**.
### Pattern Recognition
Match key words directly: Glycerol/spent-lye always links to reduced pressure (vacuum distillation). Crude oil always couples to fractional columns. Aniline + water implies steam injection.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Some Basic Principles of Organic Chemistry
Keywords:#purification of organic compounds#JEE Main 2025 Evening Q34#glycerol distillation reduced pressure#steam distillation aniline water
More Some Basic Principles of Organic Chemistry Previous-Year Questions — Page 10
Q34jee_main_2025_24_jan_morningQuantitative Analysis of Organic Compounds
Given below are two statements I and II.
Statement I: Dumas method is used for estimation of "Nitrogen" in an organic compound.
Statement II: Dumas method involves the formation of ammonium sulphate by heating the organic compound with conc mathrmH_2mathrmSO_4$\mathrm{H}_2\mathrm{SO}_4$
In the light of the above statements, choose the correct answer from the options given below
A. Both Statement I and Statement II are true.
B. Statement I is false but Statement II is true
C. Both Statement I and Statement II are false.
D. Statement I is true but Statement II is false
Solution
### Core Logic
Statement I is fully accurate: Dumas method is standardly applied for estimating elemental nitrogen across structural compounds.
Statement II is incorrect: The reaction leading to ammonium sulphate generation by intense thermal heating alongside concentrated mathrmH_2mathrmSO_4$\mathrm{H}_2\mathrm{SO}_4$ describes the **Kjeldahl method**, not the Dumas strategy. The Dumas process instead relies on burning carbonaceous compounds explicitly with copper oxide to convert nitrogen cleanly into free gas (N_2$N_2$).
### Pattern Recognition
Dumas method collects element gas N_2$N_2$ via volumetric analysis; Kjeldahl maps digestions via standard (mathrmNH_4)_2mathrmSO_4$(\mathrm{NH}_4)_2\mathrm{SO}_4$ pathways.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q31jee_main_2025_28_jan_eveningPurification of Organic Compounds
The purification method based on the following physical transformation is :
textSolid xrightarrow[(textX)]textHeat textVapour xrightarrow[(textX)]textCool textSolid$$\text{Solid} \xrightarrow[(\text{X})]{\text{Heat}} \text{Vapour} \xrightarrow[(\text{X})]{\text{Cool}} \text{Solid}$$
A. Sublimation
B. Distillation
C. Crystallization
D. Extraction
Solution
### Related Formula
Direct phase transition without passing through an intermediate liquid phase defines sublimation:
textSolid rightleftharpoons textVapour$$\text{Solid} \rightleftharpoons \text{Vapour}$$
### Core Logic
The schematic diagram represents a solid turning directly into vapor on heating, which then reverts back to a solid phase upon cooling. This distinct behavior isolates sublimable solids from non-sublimable impurities.
### Step 1: Identification
This transformation perfectly defines the laboratory purification process known as **Sublimation**.
### Pattern Recognition
Look for the skipping of the liquid state entirely: Solid rightarrow$\rightarrow$ Vapour rightarrow$\rightarrow$ Solid. Common examples include camphor, naphthalene, benzoic acid, and iodine.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q44jee_main_2025_28_jan_eveningIsomerism
Given below are two statements:
Statement (I): Oxacyclobutane and prop-2-en-1-ol are isomeric compounds.
Statement (II): Propan-1-amine and N-methylethanamine are functional group isomers.
In the light of the above statements, choose the correct answer from the options given below :
A. Both Statement I and Statement II are false
B. Both Statement I and Statement II are true
C. Statement I is true but Statement II is false
D. Statement I is false but Statement II is true
Solution
### Related Formula
Isomers share an identical molecular formula but differ in structural arrangement or functional groups:
textSame M_f neq textSame structural layout$$\text{Same } M_f \neq \text{Same structural layout}$$
### Core Logic
Evaluating the structural parameters:
- **Statement I**: Oxacyclobutane (a cyclic ether) and prop-2-en-1-ol (an unsaturated alcohol) both possess the molecular formula C_3H_6O$C_3H_6O$. They are functional/ring-chain isomers, so Statement I is true.
- **Statement II**: Propan-1-amine (1^circ$1^\circ$ amine) and N-methylethanamine (2^circ$2^\circ$ amine) both share the molecular formula C_3H_9N$C_3H_9N$. Because primary, secondary, and tertiary amines contain different functional groups, they act as functional group isomers. Thus, Statement II is true.
### Step 1: Conclusion Match
Since both structural statements are valid, both Statement I and Statement II are true.
Skeletal representations for the specified organic isomers
### Pattern Recognition
Always remember that 1^circ$1^\circ$, 2^circ$2^\circ$, and 3^circ$3^\circ$ amines are classified as *different functional groups* in IUPAC nomenclature. Consequently, structural shifts between them with a constant carbon count represent functional group isomerism.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
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 rightarrow$\rightarrow$ Nucleophile
* mathrmPhSH$\mathrm{PhSH}$: Contains lone pairs on sulfur rightarrow$\rightarrow$ Nucleophile
* (mathrmH_3mathrmC)_2mathrmS$(\mathrm{H}_3\mathrm{C})_2\mathrm{S}$: Contains lone pairs on sulfur rightarrow$\rightarrow$ Nucleophile
* mathrmH_2mathrmC=mathrmCH_2$\mathrm{H}_2\mathrm{C}=\mathrm{CH}_2$: Contains a nucleophilic pi$\pi$-bond rightarrow$\rightarrow$ Nucleophile
* stackrelominusmathrmOmathrmH$\stackrel{\ominus}{\mathrm{O}}\mathrm{H}$: Negatively charged with lone pairs rightarrow$\rightarrow$ Nucleophile
* mathrmH_3mathrmO^oplus$\mathrm{H}_3\mathrm{O}^{\oplus}$: Electron deficient, positively charged oxygen cannot donate electrons rightarrow$\rightarrow$ 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
Match List-I with List-II.
Choose the correct answer from the options given below:
IUPAC Nomenclature of Organic Compounds
A. (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
B. (A)-(III), (B)-(II), (C)-(I), (D)-(IV)
C. (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
D. (A)-(II), (B)-(III), (C)-(I), (D)-(IV)
Solution
### 1. RELATED FORMULA
IUPAC Rules select the longest principal carbon chain and number it to give substituents the lowest possible locants.
### 2. EXECUTION
**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 rightarrow$\rightarrow$ 3-Ethyl-5-methylheptane (II) .
* (B) C expanded yields a 7 carbon main chain with two methyl groups at carbon-4 rightarrow$\rightarrow$ 4,4-Dimethylheptane (III) .
* (C) 5-carbon diene numbered from the left double bond side rightarrow$\rightarrow$ 2-Methyl-1,3-pentadiene (IV) .
* (D) 5-carbon alkene starting from the double bond end rightarrow$\rightarrow$ 4-Methylpent-1-ene (I) .
Therefore, matching sequence: (A)-(II), (B)-(III), (C)-(IV), (D)-(I).
### 3. PATTERN RECOGNITION
Expanding compressed groupings such as mathrm(C_3H_7)_2$\mathrm{(C_3H_7)_2}$ prevents errors regarding parent chain carbon counts.
### 4. EVALUATION RUBRIC / MODEL ANSWER
null
### 5. CHAPTER MIX
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
More Some Basic Principles of Organic Chemistry Questions — jee_main_2025_04_april_evening
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