obtained by chlorination of n-butane, will be

Question

obtained by chlorination of n-butane, will be (A) meso form (B) racemic mixture (C) d-form (D) I-form

Tardigrade Admin · Accepted Answer

Chlorination of n-butane takes place via free radical formation i.e., Cl 2 xrightarrow h v Cl bullet+ Cl bullet CH 3- CH 2- CH 2- CH 3 xrightarrow Cl 2 / hv <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/5b608b4c4435b95770c6051befa254c7-.png" /> Cl bullet may attack on either side and give a racemic mixture of 2-chlorobutane which contain 50 % d form and 50 % l-form.

Q. obtained by chlorination of n-butane, will be

meso form

racemic mixture

d-form

I-form

Chlorination of $n$ -butane takes place via free radical formation i.e., $C l_{2} h v C l^{∙} + C l^{∙}$
$C H_{3} - C H_{2} - C H_{2} - C H_{3} C l_{2} / h v$

$C l^{∙}$ may attack on either side and give a racemic mixture of $2$ -chlorobutane which contain $50% d$ form and $50% l$ -form.

Q. obtained by chlorination of n-butane, will be

meso form

racemic mixture

d-form

I-form

Chlorination of n-butane takes place via free radical formation i.e., Cl2​hv​Cl∙+Cl∙ CH3​−CH2​−CH2​−CH3​Cl2​/hv​ Cl∙ may attack on either side and give a racemic mixture of 2-chlorobutane which contain 50%d form and 50%l-form.

Chlorination of $n$ -butane takes place via free radical formation i.e., $C l_{2} h v C l^{∙} + C l^{∙}$
$C H_{3} - C H_{2} - C H_{2} - C H_{3} C l_{2} / h v$

$C l^{∙}$ may attack on either side and give a racemic mixture of $2$ -chlorobutane which contain $50% d$ form and $50% l$ -form.