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Chemistry
CH3CH2CH2OH xrightarrow[170° C] conc. H2SO4A xrightarrow[500° C]Cl2 B. A and B are
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Q. $CH_3CH_2CH_2OH \xrightarrow[170^°\, C] {conc. \,H_2SO_4}A \xrightarrow[500^°\, C]{Cl_2} B$. $A$ and $B$ are
Hydrocarbons
A
$A = CH_3CH_2CH_3$, $B = CH_3CH_2CH_2Cl$
21%
B
$A = CH_3CH = CH_2$, $B = CH_2ClCH = CH_2$
59%
C
$A = CH_2 = CH_2$, $B = CH_3CH_2Cl$
12%
D
$A = CH_3CH_2CH_3$, $B = CH_3CH = CH_2$
8%
Solution:
$CH_3CH_2CH_2OH \xrightarrow[170^°\, C]{conc. \,H_2SO_4} \underset{\text{'A'}}{CH_3CH = CH_2} \xrightarrow[500^°\, C]{Cl_2}\underset{\text{'B'}}{CH_2ClCH = CH_2}$