In the following series of reactions identify the major products A & B respectively:
Electrophilic Aromatic Substitution
A.
Electrophilic Aromatic Substitution
B.
Electrophilic Aromatic Substitution
C.
Electrophilic Aromatic Substitution
D.
Electrophilic Aromatic Substitution
Solution & Explanation
### Related Formula
Orienting effects in electrophilic aromatic substitution:
- Bromine (-Br$-Br$) is ortho/para-directing (para-dominated due to steric hindrance).
- Sulfonic acid group (-SO_3H$-SO_3H$) is a strong deactivating, meta-directing group.
Electrophilic Aromatic Substitution
### Core Logic
Analyzing the first step:
- Sulfonation of bromobenzene with mathrmSO_3/mathrmH_2mathrmSO_4$\mathrm{SO}_3/\mathrm{H}_2\mathrm{SO}_4$ yields 4-bromobenzenesulfonic acid as the major product (A) due to steric hindrance at the ortho-position.
### Step 1: Determine orientation for the second step
In 4-bromobenzenesulfonic acid, we have two substituents:
- -Br$-Br$ (ortho/para director)
- -SO_3H$-SO_3H$ (meta director)
The positions meta to the deactivating -SO_3H$-SO_3H$ group correspond to the positions ortho to the -Br$-Br$ group. Both directing effects align on the same position (carbon-3/carbon-5). Since -Br$-Br$ is activating relative to -SO_3H$-SO_3H$, it controls the orientation.
### Step 2: Identify Product B
Halogenation with mathrmBr_2/mathrmFe$\mathrm{Br}_2/\mathrm{Fe}$ introduces a bromine atom ortho to the existing bromine atom (meta to -SO_3H$-SO_3H$):
textProduct B = text3,4-dibromobenzenesulfonic acid$$\text{Product B} = \text{3,4-dibromobenzenesulfonic acid}$$
This matches Option (2).
### Pattern Recognition
When an activating group (-Br$-Br$) and a deactivating group (-SO_3H$-SO_3H$) compete on a benzene ring, the orienting influence of the activating group wins. Position ortho to the bromine atom is favored over meta positions of the sulfonic acid group.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Hydrocarbons
Class 12 Chemistry: Haloalkanes and Haloarenes
More Hydrocarbons Previous-Year Questions — Page 2
Q47jee_main_2025_29_jan_eveningAromatic Hydrocarbons and Isomerism
Isomeric hydrocarbons giving negative Baeyer's test have the molecular formula C_9H_12$C_{9}H_{12}$.
The total number of isomers from above with exactly four different non-aliphatic substitution sites is ________.
Numerical Answer.Answer: 2 to 2
Solution
### Core Logic
A negative Baeyer's test confirms that the hydrocarbon structural isomers contain no aliphatic alkene or alkyne unsaturations, establishing that they are purely aromatic benzene derivatives with side alkyl chains.
Aromatic Hydrocarbons and Isomerism diagram for Q47 - JEE Main 2025 Evening
To find structures possessing four distinct ring positions available for electrophilic substitution, we examine the symmetries of specific tri-substituted configurations. There are exactly 2$2$ such structural isomers satisfying these spatial conditions.
### Pattern Recognition
Negative test = aromatic ring constraint. Calculate positional substitution patterns meticulously to ensure ring symmetry matches the required counts.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Hydrocarbons
Q42jee_main_2025_03_april_morningOzonolysis of Alkenes
Which compound would give 3-methyl-6-oxoheptanal upon ozonolysis ?
A. Structure (1)
B. Structure (2)
C. Structure (3)
D. Structure (4)
Solution
### Core Logic
Let us reconstruct the original alkene from the given ozonolysis product fragments. Write down the line-structure formula for 3-methyl-6-oxoheptanal:
textO=textCH-textCH2-textCH(CH*3)-textCH*2-textCH*2-textC(O)-textCH*3$$
\text{O}=\text{CH}-\text{CH}2-\text{CH(CH}*3)-\text{CH}*2-\text{CH}*2-\text{C(O)}-\text{CH}*3$$
Remove both carbonyl oxygen atoms and link carbon-1 directly to carbon-6 with a double bond. This cyclizes into a 6-membered ring structure: 1,4-dimethylcyclohexene. Alkene cyclization path for Q42 - JEE Main 2025 Morning
### Step 1: Ozonolysis Verification
Performing reductive ozonolysis (O_3 / Zn, H_2O$O_3 / Zn, H_2O$) cleaves the internal endocyclic double bond of 1,4-dimethylcyclohexene, perfectly regenerating the acyclic keto-aldehyde compound.
### Pattern Recognition
Shortcut: Count the carbons in the main chain product (7text carbons total$7\text{ carbons total}$). Ozonolysis of option (2) creates an open chain containing a terminal aldehyde group on one end and a methyl ketone group on the other.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Hydrocarbons
Q38jee_main_2025_04_april_morningProperties of Benzene
Benzene is treated with oleum to produce compound (X) which when further heated with molten sodium hydroxide followed by acidification produces compound (Y). The compound Y is treated with zinc metal to produce compound (Z). Identify the structure of compound (Z) from the following options:
A.textStructure 1$\text{Structure 1}$
B.textStructure 2$\text{Structure 2}$
C.textStructure 3$\text{Structure 3}$
D.textStructure 4$\text{Structure 4}$
Solution
### Core Logic
Let's map out this complete aromatic synthesis pathway:
1. **Benzene + Oleum:** Sulfonation steps take place to form Benzene Sulfonic Acid (C_6H_5SO_3H$C_6H_5SO_3H$, Compound X).
2. **Fusion with molten NaOH$NaOH$ followed by H^+$H^+$ activation:** The sulfonic group is displaced, passing through a sodium phenoxide intermediate to yield Phenol (C_6H_5OH$C_6H_5OH$, Compound Y).
3. **Phenol + Zinc dust distillation:** Phenol undergoes clean deoxygenation reduction when heated with Zinc metal, stripping the hydroxyl group away to reform **Benzene** (Compound Z).
### Pattern Recognition
Zinc dust distillation is a highly reliable reduction tool designed explicitly to strip phenolic hydroxyl groups away, leaving a clean unsubstituted aromatic ring behind.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Hydrocarbons
Class 12 Chemistry: Alcohols, Phenols and Ethers
Predict the major product of the following reaction sequence:
The sequence pathways specify Br2/hv free radical substitution, alcoholic KOH elimination, and HBr/peroxide addition.
A.textProduct 1$\text{Product 1}$
B.textProduct 2$\text{Product 2}$
C.textProduct 3$\text{Product 3}$
D.textProduct 4$\text{Product 4}$
Solution
### Core Logic
Let's analyze the steps of the reaction sequence:
1. **Step 1 (Br_2 / hnu$Br_2 / h\nu$):** Light-induced free radical substitution targeted at the most stable tertiary position, producing 1-bromo-1-methylcyclohexane.
2. **Step 2 (Alcoholic KOH, Delta$KOH, \Delta$):** Dehydrohalogenation occurs via an E2 mechanism. Following Saytzeff's rule, elimination favors the formation of the more highly substituted, stable alkene: 1-methylcyclohexene.
3. **Step 3 (HBr / R-O-O-R, hnu$HBr / R-O-O-R, h\nu$):** Radical hydrobromination across the unsymmetrical alkene. The presence of peroxide shifts addition toward the **Anti-Markovnikov** path, placing the bromine atom cleanly at the less-substituted secondary carbon to yield **1-bromo-2-methylcyclohexane**.
### Pattern Recognition
Combining Saytzeff elimination with a peroxide-promoted HBr$HBr$ addition allows you to reposition functional groups from highly substituted tertiary carbons to adjacent secondary positions.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Hydrocarbons
Class 12 Chemistry: Haloalkanes and Haloarenes
Q41jee_main_2025_07_april_eveningOzonolysis and Stereochemistry
The number of optically active products obtained from the complete ozonolysis of the given compound is: [cite: 364, 365]
The image details a symmetrically structured long-chain polyene containing multiple internal alkene bonds and chiral carbon sites.
A.2$2$
B.0$0$
C.1$1$
D.4$4$
Solution
### Related Formula
textR_1text-CH=textCH-R_2 xrightarrow[textZn / H_2textO]textO_3 textR_1text-CHO + textR_2text-CHO$$\text{R}_1\text{-CH}=\text{CH-R}_2 \xrightarrow[\text{Zn / H}_2\text{O}]{\text{O}_3} \text{R}_1\text{-CHO} + \text{R}_2\text{-CHO}$$
### Core Logic
Complete oxidative cleavage of all double bonds via reductive ozonolysis breaks the molecule into smaller fragments:
textCH_3text-CH=textCH-CH(textCH_3)text-CH=textCH-CH(textCH_3)text-CH=textCH-CH_3$$\text{CH}_3\text{-CH}=\text{CH-CH}(\text{CH}_3)\text{-CH}=\text{CH-CH}(\text{CH}_3)\text{-CH}=\text{CH-CH}_3$$
Let's trace the fragmentation logic mapping visually:
The image details a symmetrically structured long-chain polyene containing multiple internal alkene bonds and chiral carbon sites.
### Step 1: Tracking Fragment Structures
The chemical reaction outputs two major molecular product types:
1. textCH3text-CHO$\text{CH}3\text{-CHO}$ (Acetaldehyde): Optically inactive as it lacks a chiral carbon.
2. textOHC-CH(textCH_3)text-CHO$\text{OHC-CH}(\text{CH}_3)\text{-CHO}$ (2-methylpropanedial): Let's inspect the substituted central carbon. It is bonded to: a hydrogen atom (-textH$-\text{H}$), a methyl group (-textCH_3$-\text{CH}_3$), and two identical formyl groups (-textCHO$-\text{CHO}$).
Because two of the groups are identical (-textCHO$-\text{CHO}$), this molecule does not have a chiral center and is entirely optically inactive.
### Step 2: Total Summation
Since every single product formed is achiral, the number of optically active products is zero.
### Pattern Recognition
Symmetry check shortcut: When a symmetrical dialkene is cleaved, it yields symmetric fragments. The central carbon is attached to identical flanking aldehyde units post-cleavage, destroying any prior asymmetry and leaving 0 optically active compounds.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Hydrocarbons
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
We Map Every Repeating Question in Competitive Exams.
Say goodbye to generic mock test fatigue. RankBit uses smart analysis to group past exam questions into their foundational Repeating Question Types. Find chapter weightage, track repeating questions, and score higher with targeted practice.
Select Your Target Exam
Choose an exam track below to find formulas per chapter and patterns.
Syncing Exam Intelligence
Mapping formulas and patterns across all tracks…
PATH A — FULL LENGTH PRACTICE
Full Mock Test Hub
Simulate real NTA exam conditions with fully tracked mocks. Time yourself against past papers.
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
.svg)
.svg)
.svg)
.svg)
.svg)
)
 "alkene chain, stereochemistry, ozonolysis reactant")
