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A sample of 1.79 mg of a compound (molar mass 90 g $m o l^{- 1}$ ) generates 1.34 mL of gas when

treated with ${CH}_{3} MgBr$ . The number of active hydrogen in the compound is two.

The IUPAC name of the compound Question Image is 5 – isopropyl – 3 –

Methyl octane

The number of trichloro derivatives of cyclopentane is three

The number of monochloro derivatives of isohexane is five

answer is A.

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Detailed Solution

(A) 1.79 gm – 1340 ml

90 gm – 90 × 1340 ml

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No of moles of methane = $\frac{90 \times 1340}{1.79 \times 22400} = 3$

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