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Question
Minimum no. of carbon atoms required for an alkane to exhibit positional isomerism is:
Minimum no. of carbon atoms required for an alkane to exhibit positional isomerism is:
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Solution
The correct option is B 4
For positional isomerism chain of atleast 4 carbons is required in alkane.H3C−CH2−CH2−CH3⟶CH3−CH|−CH3
CH3
For positional isomerism chain of atleast 4 carbons is required in alkane.
H3C−CH2−CH2−CH3⟶CH3−CH|−CH3
CH3
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