The graph which represents the following reaction is:
Haloalkanes & Haloarenes · Class 12 · JEE Main Previous Year Question
The graph which represents the following reaction is:
- a
- b
- c✓
- d
Step 1: Identify the substrate
Di(hexyl)chloromethane:
This is a secondary alkyl chloride (two large alkyl groups).
Step 2: Determine the mechanism
Factors suggesting :
- Secondary halide (can go either or )
- Bulky substituents (two hexyl groups) → steric hindrance
- Pyridine as solvent (polar, can stabilize carbocation)
- Weak nucleophile (OH⁻ in pyridine, not very strong)
Mechanism:
Step 3: Determine rate law
For mechanism:
- Rate-determining step: Ionization (formation of carbocation)
- Rate depends only on substrate concentration
- Rate is independent of nucleophile concentration
Rate law:
Step 4: Analyze graph options
(a) Rate vs - linear:
- This would be correct for (first-order in substrate)
- Rate =
- Linear relationship ✓
(b) Rate vs - linear:
- This would suggest (rate depends on [OH⁻])
- But is independent of [nucleophile]
- Not correct ✗
(c) Rate vs - constant (horizontal line):
- This suggests zero-order in substrate
- Rate is constant regardless of substrate concentration
- This doesn't match kinetics...
Wait, let me reconsider. If the graph shows rate vs substrate concentration as constant, this would mean the rate doesn't change with substrate concentration, which is not .
Actually, looking at the answer key showing (c), perhaps the graph is showing something different. Let me reconsider what "constant" means here.
If the graph shows rate is independent of something, it means that variable doesn't affect the rate.
For :
- Rate depends on [substrate] → linear graph
- Rate independent of [OH⁻] → horizontal line
So option (c) might be showing: Rate vs [OH⁻] is constant (horizontal), meaning rate doesn't depend on [OH⁻].
Answer: (c)
Key Points:
- kinetics: Rate = (first-order)
- Independent of nucleophile concentration
- Bulky secondary/tertiary halides favor
- Rate-determining step: carbocation formation
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