
- At-Butyl ethyl ether
- B2,2-Dimethyl butane
- C2-Methyl pent-1-ene
- D2-Methyl prop-1-ene
View written solutionFree
Correct answer: D
The question text appears incomplete because the actual reaction scheme/reagents are missing. So the exact product cannot be determined with full certainty from the text alone.
However, based on the given options and the chapter context (haloalkanes and haloarenes), this is most likely a standard reaction involving a tert-butyl halide under conditions favoring elimination, whose major product is 2-methylprop-1-ene.
Let us analyze the options from that perspective.
1. Likely reaction pattern
A very common JEE question in this topic is of the type:
For a tertiary alkyl halide such as tert-butyl halide, substitution is hindered and under alcoholic base/heat, elimination is strongly favored.
The elimination gives:
This compound is 2-methylprop-1-ene (isobutene).
2. Checking the options
Option A: t-Butyl ethyl ether
This would be formed by substitution with an ethoxide/ethanol-type nucleophile under suitable conditions:
But for a tertiary halide, especially in elimination-favoring conditions, ether is usually not the major product.
So, A is unlikely as the major product.
Option B: 2,2-Dimethylbutane
This is an alkane and would require coupling/rearrangement, not the standard major outcome of a simple haloalkane reaction.
So, B is unlikely.
Option C: 2-Methyl pent-1-ene
This has six carbons, whereas tert-butyl derivatives generally lead to a four-carbon alkene on elimination. So unless a larger substrate is involved, this does not fit.
So, C is unlikely.
Option D: 2-Methyl prop-1-ene
This is exactly the alkene obtained by elimination from a tert-butyl halide:
So, D is the most plausible major product.
3. Final conclusion
The major product is:
So the correct option is:
4. Comparison with stored answer
Stored correct answer: D
Derived answer: D
Hence, they agree.
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