Assuming reactivity ratio 1 : 4 : 5 for chlorination with 1 ∘ H : 2 ∘ H : 3 ∘ H respectively. The percentage of 2 ∘ alkyl chloride is

1 ∘ product = 9 × 1 = 9 2 ∘ product = 6 × 4 = 24 3 ∘ product = 1 × 5 = 5 9 + 24 + 5 = 38 % of 2 ∘ product = 38 24 × 100 = 63.1578 ≈ 63.16

Chemistry - Hydrocarbon Question with Solution | TestHub

ChemistryHydrocarbonCHEMICAL REACTIONS OF ALKANESMedium2 minQB

ChemistryMediuminteger

Assuming reactivity ratio $1 : 4 : 5$ for chlorination with $1^{\circ} H : 2^{\circ} H : 3^{\circ} H$ respectively. The percentage of $2^{\circ}$ alkyl chloride is

Answer:

Solution:

$1^{\circ}$ product $= 9 \times 1 = 9$

$2^{\circ}$ product $= 6 \times 4 = 24$

$3^{\circ}$ product $= 1 \times 5 = 5$

$9 + 24 + 5 = 38$

$%$ of $2^{\circ}$ product $= \frac{24}{38} \times 100 = 63.1578 \approx 63.16$

Stream:JEESubject:ChemistryTopic:HydrocarbonSubtopic:CHEMICAL REACTIONS OF ALKANES

⏱ 2mℹ️ Source: QB

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