[jena]

Hi,

My question is:

Prop-2-en-1-ol + HBr then Mg then H2O = ?

My Answer: 1-propanol   

Is this answer correct?

Thank You  

[Winga]

Remember, there is an acidic proton in 2-bromopropan-1-ol (major).

[Cedric]

For me the answer looks allright

[Dude]

I agree with Winga.  The likely major product of HBr and C=C-C-OH would be 2-bromopropanol resulting from the electrophilic addition of HBr.  There is no way to initiate a Grignard reaction in the presence of hydrogens more acidic than ammonia.  You would have to protect the alcohol group before reaction.

[Winga]

But I want to know will Grignard reagent be formed if Mg is just added to 2-bromo-1-propanol or no reaction.

[movies]

The Grignard would perhaps form, but it would almost instantaneously quench itself by deprotonating the alcohol.

However, I think that the answer may be something different from what has been suggested so far.  Has anyone else had any other ideas for what might be happening?

[Dude]

No reaction.

Movies,
There should be no way to induce E1 type dehydration or bromine substitution of the alcohol if that is where you are going.  The addition reaction should be much quicker, eliminating the possibility of a somewhat stabilized allyl cation.  No matter what is postulated from the first step, there is still the problem of getting an alcohol since the Grignard reagent would only extract the hydrogen.

[jdurg]

Yeah, the initial addition of HBr to the Prop-2-en-1-ol would create 2-bromopropanol.  That's almost without question.  However, you simply cannot get a Gringard reagent with the hydroxyl group sitting there.  As soon as one would form, it would immediately deprotonate the alcohol leaving you with CH3-CH2-CH2-O-.  Perhaps the resulting product would be dipropyl peroxide?  (Though I'm not sure what part the water would play in this reaction, or if the peroxide could even form in the first place.  I need to go through my old o-chem books again to try and remember this stuff).  

[movies]

What's the problem with an S_N1 reaction (or possible an S_N2' reaction) with HBr to make allyl bromide?  Then the allyl Grignard could be formed and then quenched with water to make propene.

[Blitz]

I am suggesting another route.
Step 1: Protonation of the hydroxyl group.
Step 2: Formation of Cyclopropanyl cation (SNI reaction under elimination of water, double bond acts as nucleophile)
Step 3: Formation of Bromo-cyclopropane
Step 4: Formation of Grignard Bromo-Mg-Cyclopropane
Step 5: Last step (hydrolysis) is leading to the formation of CYCLOPROPANE.

 

[miaskows]

Under certain conditions HBr would be able to attack both double bond and -OH group of Allylalcohol with 1,2dibromopropane formation :
CH2=CH-CH2OH +2HBr=CH3-CHBr-CH2Br
following treatment of the dibromoalkyl with Mg will give number of Grignard compounds.
 They will be hydrolysed with water to make Propane:
CH3-CHBr-CH2Br +Mg= CH3-CH(MgBr)-CH2(MgBr) (one from a wide scope of possible products)

CH3-CH(MgBr)-CH2(MgBr) + 2H2O= CH3-CH2-CH3 +
2Mg(OH)Br

[movies]

That's an interesting idea too.  However, if you did make 1,2-dibromopropane then whichever bromine formed the Grignard first would almost immediated eliminate the other bromine to form propene.  Remember that you can use 1,2-dibromoethane to activate the Mg turnings when you make a Grignard reagent and in that case, essentially the same process occurs (producing ethylene in that case).  I have some experience with making multiply charged Grignard reagents (that is, molecules with more than one R-Mg-Br) and it is actually very difficult to get the second one to form while the first forms relatively rapidly.

[miaskows]

Thank's for your replay. I actually have not such experience in making multiplay charged Gringard reagents. But i do think that Allylalcohol can react with two molecules of HBr (gaseous , bubled into liquid Allylalcohol or very concentrated water solution >=48%) to give dibromopropane and water.