Draw bond-line formulas of all dichloro derivatives that might be formed when 1-chloro-2,2,3,3,-tetramethylpentan e is allowed to react with Cl 2 under UV irradiation. For each structure, indicate, with an asterisk, any stereocenters that might be present. Predcit the percentage of each product using the relative reactivities 3 0 = 5.3, 2 0 = 3.6, 1 0 = 1

Explanation:

This is a halogenation reaction i.e substitution or replacement of a single or more than a single hydrogen atom in the organic alkane compound with the halogen(here it is chlorine).

The chlorination of 1-chloro-2,2,3,3-tetramethylpentane under UV light resulted in the formation of five (5) dichloro derivatives which are shown in the image attached below.

Also, the compounds containing a stereocenter (i.e a location within the compound composing of various substituents in which the interchangeability of these substituents has the tendency of resulting into a stereoisomer) are indicated with an asterisk in the image below.

From the image below:

compound 1 ⇒  1,1-dichloro-2,2,3,3-tetramethylpentane = 2° C

∴

The given relative reactivity rate for 2°  = 3.6x

For compound 2 ⇒  1,4-dichloro-2,2,3,3-tetramethylpentane = 2°  = 3.6x

For compound 3 ⇒ 1,5-dichloro-2,2,3,3-tetramethylpentane = 1°  = 1x

For compound 4 ⇒ 1-chloro-2-chloromethyl-2,3,3-trimethylpentane

= 1°  = 1x

For compound 5 ⇒ 1-chloro-3-chloromethyl-2,2,3-trimethylpentane

= 1°  = 1x

As such, we have:

2(3.6x) + 3(1x) = 100

7.2x + 3x = 100

10.2x = 100

x = 100/10.2

x = 9.803°

∴

For compound (1) = 3.6(9.803) = 35.3%

For compound (2) = 3.6(9.803) = 35.3%

For compound (3) = 1(9.803) = 9.803°%

For compound (4) = 1(9.803) = 9.803°%

For compound (5) = 1(9.803) = 9.803°%

neutrons are in the middle of the atom and the electrons, which are much smaller.