What are the potential products of the reaction between acetylene and HOCl?

[Krushnaraj Pandya]

Homework Statement

What is the product of acetlyene and HOCl reaction?
I gathered that Cl acts as electrophile and so (HCOH)=CHCl is formed.
Now two things can happen, it could tautomerize to keto form since enol form is unstable, or another molecule of HOCl can add itself, if it adds again there are two ways of adding them, both chlorine atoms could end up on the same carbon or on adjacent carbons. How do we decide what will be the final product?

Homework Equations

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The Attempt at a Solution

given above

 

[symbolipoint]

This may be silly of me to ask, but could some conditions be made arranged that acetylene and that hypochlorous acid would have the opportunity to participate as reactants? One would imagine that acetylene is a gas and HOCl is dissolved in water. This LOOKS like the acid form of NaOCl, sodium hypochlorite, which would be expected to exist as aqueous solution.

 

[Krushnaraj Pandya]

This may be silly of me to ask, but could some conditions be made arranged that acetylene and that hypochlorous acid would have the opportunity to participate as reactants? One would imagine that acetylene is a gas and HOCl is dissolved in water. This LOOKS like the acid form of NaOCl, sodium hypochlorite, which would be expected to exist as aqueous solution.

H2OCl+ is formed in the presence of acetic acid, this is the attacking species

 

[TeethWhitener]

H2OCl+ is formed in the presence of acetic acid, this is the attacking species

I doubt this has to be the case. Ultimately you’re looking at the formation of a three-member ring featuring a chloronium. HOCl can likely do this on its own (in water, this is unstable to formation of chlorine gas). This proceeds quickly for the alkyne, but, as you’ve noted, after the first chlorohydrin addition, you have to contend with keto-enol tautomerism.

What happens here wil depend on the exact nature of the rates involved. Equilibrium heavily favors the aldehyde, so there’s likely a point at which you can stop the reaction and get the monochlorinated aldehyde product that is the tautomer of what you indicated in the OP. However, if you let the reaction continue running in the presence of xs HOCl, it will continue to react in the transient enol form.

To determine where the second chlorine will end up, remember that after the 3-member chloronium ring is formed, water adds as a nucleophile to the carbon with the largest partial positive charge (cf. epoxide ring opening).

 

[TeethWhitener]

Just to be clear about the very first step: in acidic water, HOCl is less important than free Cl₂ dissolved in water. The Cl₂ approaches perpendicular to the electron-rich triple bond and is polarized to give a partial positive charge on the chlorine nearest the alkyne moiety. A (roughly) concerted reaction gives the chloronium bridge and releases a chloride into solution.

 

[baldbrain]

Since it isn't explicitely mentioned that HOCl is in excess (assuming you have worded your question exactly 'as is' in your book), I think that it is ok to stop at the chloroacetaldehyde (major) product.
(Side question for OP: The enol + HOCl reaction product will be?...)
Additionally, chloroacetaldehyde isn't normally encountered in its anhydrous form, but rather occurs as a hydrate ClCH₂CH(OH)₂. This is a saturated compound that won't react with HOCl. Also, HOCl is a very weak acid that won't complete that intensely with the hydrate formation.

 

[Nik_2213]

Please beware, as acetylene is prone to exothermic reactions. Study the HAZMAT info and use due care.