What is the sterostructure of the monochlobutanes obtained form the free radical chlorination of butane?
(i) `(+)-2-`Chlorbutane
(ii) `(-)-2-`Chlorobutane
(ii) `(+-)-2-`Chlorobutane
(iv) Achiral`-1-`Chlorobutane

Options

A. (i),(ii),(iii),(iv)
B. `(i),(iv)
C. `(ii),(iv)
D. `(iii),(iv)`

Answer

D

Explanation

The two products are achiral `1-`chlorobutane and racemic `2-`chlorobutane:
`CH_(3)CH_(2)CH_(2)CH_(3)overset(CI_(2))underset(hv)rarrunderset(1-"Chlorobutane"("achiral"))(CH_(3)CH_(2)CH_(2)CH_(2)CI+CH_(3))underset(("racemic mixture"))underset(2-"Chlorobutane")underset(CI)underset(|)(CHCH_(2)CH_(3))`
The intermediate from the first propagation step is the radical `CH_(3)CHCH_(2)CH_(3)`(for `2-`"chlorobutane"). The radical `C` is `sp^(2)` hybridized and its three bonds are coplanar, with the odd electron in a `p`-orbital at right angles to this plane. In the second propagation step, the radical can remove a `CI` atom form `CI_(2)` on either side of the plane to given one product that is `R` and the other `S`. These attacks occur with equal probability, resulting in equal amounts of `R` and `S` (i.e. a racemic form).

The flat radical is said to have enantiotopic faces beacuse bonding of a fourth ligand to one or the other face gives one or the other enantiomer.
Nite that racemic `2-`chlorobutane would also be obtained if the sec-buty radical were not flat but pyramidal beacuse abstraction of either `2^(@)` hydrogen is equally likely, giving equal amounts of enantiomeric pyramidal radicals, each of which reacts equally radialy with chlorine, thus giving equal amounts of enantiomeric sec-`BuCI` molecules (i.e. the racemic mdification). Moreover, a pyramidal radical undergoes rapid inversion and lose configuration.