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The relative reactivity of 1^(@): 2^(@):...

The relative reactivity of `1^(@): 2^(@):3^(@)H's` to chlorination is `1:3:8:5`. In the reaction
`CH_(3)CH_(2)CH_(2)underset("Excess)(CH_(3))+Cl_(2)rarrCH_(3)CH_(2)underset((A))CH_(2)CH_(2)Cl_(2)+CH_(2)underset((B))overset(Cl)overset(|)CHCH_(3)`
the percentage yields of the products `(A) and (B)` are expected to be

A

`48,52`

B

`28,72`

C

`99,4,0.6`

D

`35,65`

Text Solution

Verified by Experts

The correct Answer is:
B
`Coverset(1^(@))H_(3)overset(2^(@))CH_(2)Coverset(2^(@))H_(2)Coverset(1^(@))H_(3)`
There are `6` ( primary, `1^(@)`) equivalent `H` atoms leading to the formation of `1`-chlorobutane `(A) and (4)` (secondary, `2^(@)`) equivalent `H` atoms leading to the formation of `2`-chlorobutane `(B)`. The ratio of `1^(@) "to" 2^(@)H's is 3:2`. The amount of each isomers formed depends on its rate the formation, therefore, the ratio of amounts (i.e., relative amounts) is equal to the ratio of rates.
Relative amount of product
`((Relative),(reactivity))xx(("Number of"),("equivalent H's"))`
Relative amount of `A= (1)xx(6)=6`
Relative amount of `B=(4)xx(3.8)= 15.2`
Hence, the relative amounts of `A` to `B` are in the ratio `6//15.2`
`"Percentage yield"=("Relative amount of product")/("Sum of the relative amounts")xx100%`
`"Percentage of A"=(3)/(10.6)xx100%= 28%`
`"Percentage of B"=(7.6)/(10.6)xx100%= 72%`
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