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A hydrocarbon on ozonolysis gives one mole each of compounds A and B. Both A and B on reduction with sodium borohydride gives compounds C and D belonging to same homologous series. C has no position isomer whereas D has position isomer. A and B do not belong to the same homologous series. However, both A and B on oxidation with acidified KMnO4 give same compound. Identify the hydrocarbon.

A
3- Methyl-2-butene
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B
2,3-Dimethyl-1-butene
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C
2-Methyl-2-butene
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D
3,4-Dimethyl-1-butene
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Solution

The correct option is C 2-Methyl-2-butene
Ozonolysis of alkenes form ketone or aldehydes. Their reduction with NaBH4 gives alcohols. C has no position isomer whereas D has position isomer.Thus C is a primary alcohol and D is a secondary alcohol. A is an aldehyde and B is a ketone. And the compound is 2methyl2butene
CH3(CH3)C=CHCH3(O3)CH3(CH3)C=O[A]+O=CHCH3[B]NaBH4CH3(CH3)CHOH[C]+HOCH2CH3[D]KMnO4CH3(CH3)CHOH+HOOCCH3

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