Question #655ff

1 Answer
Ernest Z.
Nov 26, 2017

It may be an #"S"_text(N)"i"# mechanism.

Explanation:

#"S"_text(N)"i"# stands for #color(blue)("S")#ubstitution, #color(blue)("N")#ucleophilic, with #color(blue)("i")#nternal return.

Let's write the reaction as

#"C"_6"H"_13"CH(OH)CH"_3 + "COCl"_2 → "C"_6"H"_13"CHClCH"_3 + "HCl" + "CO"_2#

Here are the steps I suggest.

Step 1. Nucleophilic attack of the alcohol on the carbonyl group

Step 1

Step 2. Expulsion of #"Cl"^"-"# to form a protonated chlorocarbonate ion

Step 2

Step 3. Deprotonation by the expelled #"Cl"^"-"# ion

Step 3

Step 4. Departure of the leaving group

Step 4

The carbocation and the chlorocarbonate ion form an intimate ion pair that is held together tightly in space.

Step 5. Attack of #"Cl"#

Step 5

The #"Cl"# acts as a nucleophile, attacking the carbocation from the same face from which it was expelled, resulting in retention of configuration.

Step 6. Loss of #"CO"_2#

Step 6

This step may be simultaneous with Step 5.

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