The steam volatile compounds among the following are: Choose the correct answer from the options given below:

Steam volatile compounds isomers for Q44 - JEE Main 2025 Morning
Four different disubstituted benzene structures are indexed to show spatial isomer distributions.

Solution & Explanation

### Related Formula textIntramolecular H-bonding implies textLower boiling point implies textSteam 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 group right next to an -mathrmNO_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 .
Hydrogen bonding configurations for volatile compounds
Four different disubstituted benzene structures are indexed to show spatial isomer distributions.
Hydrogen bonding configurations for volatile compounds
Four different disubstituted benzene structures are indexed to show spatial isomer distributions.
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
Hydrogen bonding configurations for volatile compounds
Four different disubstituted benzene structures are indexed to show spatial isomer distributions.

More Organic Chemistry - Some Basic Principles and Techniques Previous-Year Questions — Page 15

Q65 jee_main_2024_30_jan_morning Aromaticity
Which of the following molecule/species is most stable?
  • A.
  • B.
  • C. textOption 3
  • D. textOption 4

Solution

### Core Logic Stability of cyclic carbocations can be determined using Huckel's rule for aromaticity. A species is exceptionally stable if it is aromatic. Aromaticity requires the system to be cyclic, planar, fully conjugated, and possess (4n + 2) pi electrons. ### Step 1: Analyze Option 1 The tropylium cation (Option 1) is a 7-membered ring with 3 double bonds and a positive charge in continuous conjugation.
Aromaticity solution diagram for Q65 - JEE Main 2024 Morning
Aromaticity solution diagram for Q65 - JEE Main 2024 Morning
Number of pi electrons = 6. Since 6 satisfies (4n+2) for n=1, it is aromatic and therefore highly stable. ### Pattern Recognition Tropylium ion (C_7H_7^+) is a classic example of a stable aromatic carbocation. It frequently appears in stability comparison questions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q79 jee_main_2024_30_jan_morning Qualitative Analysis of Organic Compounds
The Lassiagne's extract is boiled with dil. HNO_3 before testing for halogens because,
  • A. textAgCN is soluble in HNO_3
  • B. textSilver halides are soluble in HNO_3
  • C. Ag_2Stext is soluble in HNO_3
  • D. Na_2Stext and NaCN are decomposed by HNO_3

Solution

### Core Logic In Lassaigne's test for halogens, we add AgNO_3 to form a precipitate of silver halide (AgX). However, if the organic compound also contains Nitrogen or Sulphur, the Lassaigne's extract will contain NaCN or Na_2S. ### Step 1: Reason for adding HNO3 These ions (CN^- and S^2-) would also react with AgNO_3 to form precipitates (AgCN - white, Ag_2S - black), which would interfere with the test for halogens. Boiling the extract with concentrated/dilute HNO_3 decomposes the cyanide and sulphide to HCN and H_2S gases, which escape, thus removing the interference. NaCN + HNO_3 rightarrow NaNO_3 + HCN uparrow Na_2S + 2HNO_3 rightarrow 2NaNO_3 + H_2S uparrow ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q86 jee_main_2024_30_jan_morning Chromatography
On a thin layer chromatographic plate, an organic compound moved by 3.5text cm, while the solvent moved by 5text cm. The retardation factor of the organic compound is ________ times 10^-1
Numerical Answer. Answer: 7 to 7

Solution

### Related Formula R_f = fractextDistance travelled by compoundtextDistance travelled by solvent ### Step 1: Substitution and calculation R_f = frac3.55 R_f = 0.7 R_f = 7 times 10^-1 ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q75 jee_main_2024_31_jan_evening Purification of Organic Compounds
The fragrance of flowers is due to the presence of some steam volatile organic compounds called essential oils. These are generally insoluble in water at room temperature but are miscible with water vapour in vapour phase. A suitable method for the extraction of these oils from the flowers is -
  • A. text1. crystallisation
  • B. text2. distillation under reduced pressure
  • C. text3. distillation
  • D. text4. steam distillation

Solution

### Core Logic Steam distillation technique is applied to separate substances which are steam volatile and are immiscible with water. Since the essential oils in flowers are steam volatile and insoluble in water, steam distillation is the perfect technique for their extraction. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q66 jee_main_2024_31_jan_morning Reaction Intermediates
A species having carbon with sextet of electrons and can act as electrophile is called
  • A. textcarbon free radical
  • B. textcarbanion
  • C. textcarbocation
  • D. textpentavalent carbon

Solution

### Core Logic
Reaction Intermediates diagram for Q66 - JEE Main 2024 Morning
Reaction Intermediates diagram for Q66 - JEE Main 2024 Morning
A carbocation has three bonds and an empty p-orbital, yielding a sextet (6) of electrons in its valence shell. Due to its electron deficiency, it acts as a strong electrophile. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
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