Identify A in the following reaction.
A chemical reaction scheme converting an unknown compound A into different cyclic structures using specific reagents.
A.
A chemical reaction scheme converting an unknown compound A into different cyclic structures using specific reagents.
B.
A chemical reaction scheme converting an unknown compound A into different cyclic structures using specific reagents.
C.
A chemical reaction scheme converting an unknown compound A into different cyclic structures using specific reagents.
D.
A chemical reaction scheme converting an unknown compound A into different cyclic structures using specific reagents.
Solution & Explanation
### Core Logic
Let's analyze the given reaction sequences.
1. Reaction with 2H_2 / Pt$2H_2 / Pt$: Compound A undergoes hydrogenation with 2 moles of hydrogen, suggesting the presence of two double bonds.
2. Reaction with mathrmKMnO_4 / Delta$\mathrm{KMnO_4} / \Delta$: Compound A undergoes oxidative cleavage to give a substituted dicarboxylic acid (cyclohexane-1,2-dicarboxylic acid) and oxalic acid (HOOC-COOH$HOOC-COOH$).
The structure of A must have a diene system that, when cleaved completely at the double bonds, yields these specific acid fragments.
A chemical reaction scheme converting an unknown compound A into different cyclic structures using specific reagents.
By matching the cleavage points, compound A is identified as 1,2,3,4,4a,5,8,8a-octahydronaphthalene derivative with two isolated double bonds in the ring.
A chemical reaction scheme converting an unknown compound A into different cyclic structures using specific reagents.
### Pattern Recognition
Hot KMnO_4$KMnO_4$ cleaves double bonds entirely, converting =C-H$=C-H$ into -COOH$-COOH$. The formation of oxalic acid indicates a -CH=CH-$-CH=CH-$ fragment caught between two cleavable double bonds in a ring system.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: Hydrocarbons
Keywords:#Reaction scheme#KMnO4#H2/Pt
More Hydrocarbons Previous-Year Questions — Page 5
Q84jee_main_2024_31_jan_eveningHalogenation of Alkanes (Isomers)
Number of isomeric products formed by mono-chlorination of 2-methylbutane in presence of sunlight is ________
Numerical Answer.Answer: 6 to 6
Solution
### Core Logic
The structure of 2-methylbutane is CH_3-CH(CH_3)-CH_2-CH_3$CH_3-CH(CH_3)-CH_2-CH_3$. It has four different types of hydrogen atoms, which can be substituted to form structural isomers:
1) 1-chloro-2-methylbutane: Chlorination at terminal CH_3$CH_3$ near branch. Yields a chiral center at C2 (2 enantiomers).
2) 2-chloro-2-methylbutane: Chlorination at the tertiary carbon (1 achiral product).
3) 2-chloro-3-methylbutane: Chlorination at the CH_2$CH_2$ group. Yields a chiral center at C2 (2 enantiomers).
4) 1-chloro-3-methylbutane: Chlorination at the far terminal CH_3$CH_3$. No chiral center (1 achiral product).
Halogenation of Alkanes (Isomers) diagram for Q84 - JEE Main 2024 Evening
### Step 1: Counting Stereoisomers
Total isomeric products = 2 (from 1st) + 1 (from 2nd) + 2 (from 3rd) + 1 (from 4th) = 6$2 (from 1st) + 1 (from 2nd) + 2 (from 3rd) + 1 (from 4th) = 6$.
### Pattern Recognition
When asked for "isomeric products" in halogenation without specifying "structural isomers", you must count stereoisomers (enantiomers) as well.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: Hydrocarbons
Class 12 Chemistry: Haloalkanes and Haloarenes
Number of alkanes obtained on electrolysis of a mixture of CH_3COONa$CH_3COONa$ and C_2H_5COONa$C_2H_5COONa$ is
Numerical Answer.Answer: 3 to 3
Solution
### Core Logic
Kolbe's electrolytic method generates free radicals at the anode, which then combine to form alkanes.
The given mixture yields two types of carboxylate radicals which decarboxylate to form alkyl radicals:
CH_3COONa rightarrow dotCH_3$CH_3COONa \rightarrow \dot{C}H_3$C_2H_5COONa rightarrow dotC_2H_5$C_2H_5COONa \rightarrow \dot{C}_2H_5$
These radicals can couple in three different ways:
1. Cross coupling: dotCH_3 + dotC_2H_5 rightarrow CH_3-CH_2-CH_3$\dot{C}H_3 + \dot{C}_2H_5 \rightarrow CH_3-CH_2-CH_3$ (Propane)
2. Self-coupling 1: dotCH_3 + dotCH_3 rightarrow CH_3-CH_3$\dot{C}H_3 + \dot{C}H_3 \rightarrow CH_3-CH_3$ (Ethane)
3. Self-coupling 2: dotC_2H_5 + dotC_2H_5 rightarrow CH_3-CH_2-CH_2-CH_3$\dot{C}_2H_5 + \dot{C}_2H_5 \rightarrow CH_3-CH_2-CH_2-CH_3$ (Butane)
Thus, a total of 3 different alkanes are formed.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: Hydrocarbons
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