Question

The major product expected when (r)-3-methylhexane reacts with bromine in the presence of light is _________. the major product expected when (r)-3-methylhexane reacts with bromine in the presence of light is _________. (s)-3-bromo-3-methylhexane a mixture of (r)- and (s)-3-bromo-3-methylhexane where the (s) stereoisomer would predominate (r)-3-bromo-3-methylhexane a racemic mixture of (r)- and (s)-3-bromo-3-methylhexane

Answer 1

Answer: -

a racemic mixture of (R)- and (S)-3-bromo-3-methylhexane

Explanation: -

In free radical halogenation, the radical intermediate formed is sp2 hybridization with a trigonal planar geometry.

Thus attack of the bromine can occur from both the top direction as well as the down direction that is from either side of the molecule.

Thus when when (r)-3-methylhexane reacts with bromine in the presence of light, a racemic mixture of (R)- and (S)-3-bromo-3-methylhexane

Answer 2

A racemic mixture of (r) and (s) - 3 - bromo - 3 - methylhexane

Step-by-step explanation:

In the picture you have the mechanism that is taking place to produce the (r) and (s). Now, what happens here?, when the bromine it's in presence of light, the Br2, split to form to bromine ion. One of this bromine, substract one atom of hydrogen in the molecule of hexane. Now, it will substract the atom of hydrogen that can leave the most stable carbon. The stability of the carbon goes from 3° > 2° > 1°, so in this case, it will go to the hydrogen that has the methyl (Carbon 3). This can be done for either direction (Upward or downward), and then, the remaining bromine goes there as the new electrophyle forming the racemic mixture.

The products are optically inactive and also, they are chiral in nature as the central carbon contains four different groups bonded to it.

In the picture, you have the mechanism: