Consider the following reactions: A xrightarr | vedclass

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(D) is formed by the reaction of $A$ with $CH_3MgBr$. Since $B$ gives $2	ext{-methyl-2-butene}$ $(CH_3-C(CH_3)=CH-CH_3)$ upon heating with $Cu$ at $5

Vedclass · Accepted Answer

$66.67$

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(D) is formed by the reaction of $A$ with $CH_3MgBr$. Since $B$ gives $2	ext{-methyl-2-butene}$ $(CH_3-C(CH_3)=CH-CH_3)$ upon heating with $Cu$ at $573 	ext{ K}$,$B$ must be a tertiary alcohol,specifically $2	ext{-methyl-2-butanol}$ $(CH_3-C(OH)(CH_3)-CH_2-CH_3)$. Therefore,$A$ must be butanone $(CH_3-CO-CH_2-CH_3)$. The molecular formula of $A$ is $C_4H_8O$. The molar mass of $A = (4 	imes 12) + (8 	imes 1) + 16 = 72 	ext{ g/mol}$. The mass of carbon in $A = 4 	imes 12 = 48 	ext{ g}$. The mass percentage of carbon $= \frac{48}{72} 	imes 100 = 66.67\%$.

Reactant	$K_{eq.}$
$PhCHO$	$a$
$CH_3CH_2COCH_3$	$b$
$PhCOCH_3$	$c$
$CH_3CHO$	$d$

Consider the following reactions:
$A$ $\xrightarrow[(i) CH_3MgBr]{(ii) H_3O^{+}} B$ $\xrightarrow[573 \text{ K}]{Cu} 2\text{-methyl-2-butene}$
The mass percentage of carbon in $A$ is:

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Consider the following reactions: $A$ $\xrightarrow[(i) CH_3MgBr]{(ii) H_3O^{+}} B$ $\xrightarrow[573 \text{ K}]{Cu} 2\text{-methyl-2-butene}$ The mass percentage of carbon in $A$ is: