Consider the following reactions A i C H 3 M B r ( i i ) H 3 O + B C u 573 K 2 - m e t h y l - 2 - b u t e n e . The mass percentage of carbon in A is _____________.

Compound A is Percentage carbon in compound A= 12 × 4 48 + 16 + 8 × 100 = 66.67

Chemistry - Carbonyl compounds Question with Solution | TestHub

ChemistryCarbonyl compoundsMiscellaneous/MixedHard2 minPYQ_2020

ChemistryHardnumerical

Consider the following reactions $A \frac{(i) C H_{3} M B r}{(i i) H_{3} O^{+}} B \frac{C u}{573 K} 2 - m e t h y l - 2 - b u t e n e$ . The mass percentage of carbon in $A$ is _____________.

Answer:

66.67

Solution:

Compound A is
Percentage carbon in compound A= $(\frac{12 \times 4}{48 + 16 + 8} \times 100) = 66.67$

Stream:JEESubject:ChemistryTopic:Carbonyl compoundsSubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: PYQ_2020

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