Chlorinating specifically one carbon on a symmetrical alcohol

Question

The final target molecule is 1-chloro-2-amino propane.

The amination step would be simple via reductive amination of 1-chloro-iso-propanone.

Obtaining 1-chloro-iso-propanone would be from nucleophillic addition-elimination of 1-chloro-iso-propanol by H2O under acidic conditions (or oxidation using potassium permanganate but no need for such a reagent when a simpler route available).

However, assuming you started with only iso-propanol and you wanted to chlorinate only one of the two terminal carbons, how would you do this? I understand $\ce{SOCl2}$ would of course be the chlorinating reagent, but with no blocking group it would simply attack both carbons. Attempting to add a blocking group would block both carbons also as they are symmetrical and identical.

Answer 1

It might be overkill for your purposes, but there is a nice reaction that coverts secondary alcohols to $\alpha$-chloroketones in a one-pot synthesis, using trichloroisocyanuric acid (for instance, they got 75% yield of 2-chlorocyclohexanone from cyclohexanol). The paper is 'Direct conversion of alcohols to $\alpha$-chloro aldehydes and $\alpha$-chloro ketones' (Jing, Daniliuc, Studer 2014, Org. Lett. 16 (18) pp. 4932-4935).

Direct conversion of primary and secondary alcohols into the corresponding α-chloro aldehydes and α-chloro ketones using trichloroisocyanuric acid, serving both as stoichiometric oxidant and α-halogenating reagent, is reported. For primary alcohols, TEMPO has to be added as an oxidation catalyst, and for the transformation of secondary alcohols (TEMPO-free protocol), MeOH as an additive is essential to promote chlorination of the intermediary ketones.

Trichloroisocyanuric acid has the advantages that it's inexpensive, widely available and easy to handle.

Answer 2

If you start from iso-propanol, you can first perform dehydration using concentrated $\ce{H_2SO_4}$ and heat. You will get prop-1-ene.

Then you can use 1 equivalent of $\ce{HOCl}$ (hypochlorous acid) to facilitate chlorohydrin formation. Thus, you will have 1-chloropropan-2-ol. Then, you can simply partially oxidise the secondary alcohol group to a ketone by using pyridinium chlorochromate (PCC) or pyridinium dichromate (PDC), or by performing Swern oxidation, etc. That will be a better way to get 1-chloropropan-2-one (i.e. chloroacetone). After that you can perform reductive amination.

Answer 3

Alternatively, you can use acetone in place of 2-propanol directly. The selective chlorination on a single carbon can be achieve in one step by using a dichloroemethane solution of sulfuryl chloride ($\ce{SO2Cl2}$) in the presence of methanol (see reference below). Then, using a one of the methods suggested by others here, you may convert resulting 1-chloroprop-2-one (e.g., the reductive amination) to 2-amino-1-chloropropane.

For your convenience, I have put the abstract below:

Cyclohexanone reacts with sulfuryl chloride in sulfur dioxide, ether, tetrahydrofuran, dioxane, and tetraglyme to give mixtures of mono- and dichlorocyclohexanones. However, when the reaction was carried out in sulfur dioxide solution containing a slight excess of methanol, only monochlorocyclohexanone was obtained in high yield. Almost as good selectivities are obtained in methylene chloride solution in the presence of methanol. A similar procedure with acetone affords monochloroacetone cleanly. While phenols do not react with sulfuryl chloride in methylene chloride even under reflux, the chlorination can be induced by the addition of various organic “bases”, resulting in selective monochlorination.

Reference:

Sulfuryl chloride as a reagent for selective chlorination of symmetrical ketones and phenols: D. Masilamani, and M. M. Rogic, J. Org. Chem., 1981, 46(22), 4486–4489 (https://pubs.acs.org/doi/abs/10.1021/jo00335a033).

Answer 4

Alternatively you can oxidise iso-propanol by any of the methods mentioned above to acetone, directly chlorinate with $\ce{Cl2}$ to chloroacetone then do the reductive amination to 1-chloro-2-aminopropane. It should be noted that this material can form the aziridine fairly easily so should be kept cool. Indeed I would favour doing the reductive amination by forming the benzyl imine then hydrogenating in the presence of $\ce{HCl}$ so that the hydrochloride is formed directly.