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Q.

An alkane with mol.mass $= 86$ on bromination gives only two monobromo derivatives (excluding stereoisomers). The alkane is

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a

${(C H_{3})}_{4} C$

b

${(C H_{3})}_{2} C H C H_{2} C H_{2} C H_{3}$

c

${(C H_{3})}_{3} C - C H_{2} C H_{3}$

d

${(C H_{3})}_{2} C H - C H {(C H_{3})}_{2}$

answer is C.

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Detailed Solution

The alkane with mol. Mass 86 is 2,3-Dimethylbutane, i.e., Option 3 has only two types of hydrogens and hence forms two monobromo derivatives

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