Question #00280

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Question #00280

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Answer 1 · 2016-09-27T03:55:34

Let the chlorination reaction of Ethane $(C_{2} H_{6})$ $(C_2H_6)$ with chlorine replaces n atoms from its molecule producing chloro ethane of formula $C_{2} H_{6 - n} C l_{n}$ $C_2H_(6-n)Cl_n$

Considering atomic masses as

$C = 12 \frac{g}{mol}$ $C=12g/"mol"$

$H = 1 \frac{g}{mol}$ $H=1g/"mol"$

$C l = 35.5 \frac{g}{mol}$ $Cl=35.5g/"mol"$

We get the molar mass of chloroethane as

$2 \times 12 + (6 - n) \times 1 + n \times 35.5$ $2xx12+(6-n)xx1+nxx35.5$
$= 30 + 34.5 n$ $=30+34.5n$

And % Cl in the compound as calculated from its molecular formula $C_{2} H_{6 - n} C l_{n}$ $C_2H_(6-n)Cl_n$ will be

$= \frac{35.5 \times n \times 100}{30 + 34.5 n}$ $=(35.5xxnxx100)/(30+34.5n)$

Equating this with the given percentage we can write

$\frac{35.5 \times n \times 100}{30 + 34.5 n} = 71.7$ $(35.5xxnxx100)/(30+34.5n)=71.7$

$\Rightarrow \frac{35.5 \times n \times 100}{71.7} = (30 + 34.5 n)$ $=> (35.5xxnxx100)/71.7=(30+34.5n)$

$\Rightarrow 49.5 n = (30 + 34.5 n)$ $=> 49.5n=(30+34.5n)$

$\Rightarrow 49.5 n - 34.5 n = 30$ $=> 49.5n-34.5n=30$

$\Rightarrow n = \frac{30}{15} = 2$ $=>n=30/15=2$

(A) Hence the molecular formula of chloroethane is $C_{2} H_{4} C l_{2}$ $C_2H_4Cl_2$

(B) The condensed structural formula and name of the isomers

$C H_{3} C H C l_{2} \to 1, 1 - dichloroethane$ $CH_3CHCl_2->1,1-"dichloroethane"$

$C H_{2} C l C H_{2} C l \to 1, 2 - dichloroethane$ $CH_2ClCH_2Cl->1,2-"dichloroethane"$

(C) Between the two isomers the major product will be
$C H_{3} C H C l_{2} \to 1,1-dichloroethane$ $CH_3CHCl_2->"1,1-dichloroethane"$

The reaction follows free radical mechanism.

(1) $C l \cdot$ $Cl*$ free radical is first produced by homolytic fission of $C l - C l σ b o n d$ $Cl-Cl" "sigmabond$ under the influence of light or heat energy.

(2)This $C l \cdot$ $Cl*$ free radical extracts $H$ $H$ atom from $C H_{3} C H_{3}$ $CH_3CH_3$ molecule forming $C H_{3} C H_{2} \cdot$ $CH_3CH_2*$ free radical

(3) $C H_{3} C H_{2} \cdot$ $CH_3CH_2*$ free radical then combines with $C l \cdot$ $Cl*$ free radical to form $C H_{3} C H_{2} C l$ $CH_3CH_2Cl$ ,monochloroethane.

(4) Now the extraction of $H -atom$ $H"-atom"$ from $C H_{3} C H_{2} C l$ $CH_3CH_2Cl$ by $C l \cdot$ $Cl*$ again occurs with two possibilities forming $C H_{3} C H C l \cdot$ $CH_3CHCl*$ or $\cdot C H_{2} C H_{2} C l$ $*CH_2CH_2Cl$

(5) The first one $C H_{3} C H C l \cdot$ $color(blue)(CH_3CHCl*)$ is more stable free radical as it has got more $α H$ $alpha"H$ in respect of free radical providing higher hyper conjugative effect. So the first one which is formed in higher concentration in the reaction mixture,will combine with $C l \cdot$ $Cl*$ free radical more and thus the major product will be

$C H_{3} C H C l_{2} \to 1,1-dichloroethane$ $color(red)(CH_3CHCl_2->"1,1-dichloroethane")$

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