In the treatment of HCl with acrolein 3-chloroprop-1-en-1-ol is the first product which is (my textbook says) unstable and turns into 3-chloropropanal.
Why is this 3-chloroprop-1-en-1-ol unstable? is −I effect of chlorine responsible?
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This earlier answer might provide some useful background – ron Aug 11 '16 at 19:04
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3-Chloroprop-1-en-1-ol is in equilibrium with the aldehyde, 3-chloropropanal (aka $\beta$-chloropropionaldehyde).
According to Organic Syntheses, this compound "is a very unstable substance which polymerizes rapidly, especially in the presence of traces of hydrochloric acid".
But I do not see it being reduced to 3-chloropropanol.
Reactivity/instability
In 3-chloropropanal, or more precisely in its enol form 3-chloropropenol, the chlorine atom is in allylic position (i.e. on a carbon atom adjacent to a $\ce{C=C}$ double bond) which makes it more reactive towards nucleophilic substitution. Another molecule of the enol can then substitute it and the resulting product is
$\ce{Cl-CH2-CH=CH-O-CH2-CH=CH-OH}$
which still has an allylic chlorine atom, so it can react further, hence the polymerization.
SteffX
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I have edited the question. Can you explain it more clearly. Your use of 'it' is puzzling me. – Mockingbird Aug 13 '16 at 22:20
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Enols are virtually always in equilibrium with their keto tautomer. In this case, the predominant compound you would expect to encounter is 3-chloropropanal. There are exceptions to this (1,3-diketones for example) but this is generally the case, particularly when acid or base is available to catalyze the reaction.
ChemProfMatt
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