Addition of OH to but-1,3-diene

[Function]

Hi everyone

 

An exercise that's driving me nuts. Completely nuts.

 

The question: Which products are being formed when but-1,3-diene undergoes 1 addition reaction?

 

The correct answer:

• R-but-3-ene-2-ol
• S-but-3-ene-2-ol
• But-2-ene-1-ol

Another possible answer would be:

• But-2-ene-1-ol
• But-3-ene-1-ol

However, this answer is incorrect. Why is but-3-ene-1-ol incorrect? I've already thought of Zaitsev's rule, but that one won't apply to addition reactions.

 

Can someone please help me?

 

Thanks.

 

Function

[AndresKiani]

This is very simple..

 

The carbocation would most likely occur on the second carbon, thus the end carbons would be protonated initially, which explains the R-but-3-ene-2-ol and S-but-3-ene-2-ol. Keep in mind that this is an allylic addition which is highly favorable.

 

As far as the superconfiguration, which occurs because it depends on which face the nucleophilic attack takes place Si or Re face. Which makes it just as likely for either of them too occur (however, this only makes sense for the depth of organic chemistry your in right now, further along the line you will learn conjugation, because the double bonds are 1 bond apart, this makes them extremely stabile and the interaction between the two double bonds is similar to any other covalent bond interaction). However, since your not there yet I wouldn't worry about it.. but yes this is what would occur a SI or Re face attack will give you both enantiomers.

 

In the But-2-ene-1-ol occurrence, this is due to carbocation rearrangement which is very likely with these addition reactions.. specially if your using a strong acid instead of Mercury, HOAc as catalyst.

[Function]

In the But-2-ene-1-ol occurrence, this is due to carbocation rearrangement which is very likely with these addition reactions.. specially if your using a strong acid instead of Mercury, HOAc as catalyst.

 

Could it be due to Markovnikov's rule?

Black numbers are the numbers of the C-atoms, red are the numbers of H-atoms bonded to the C-atoms.

Markovnikov would dictate us that when addition occurs, H-atoms are very much more likely to bind with C4 or C1, which occurs in the lower structure, but not in the right one: here, one H-atom has been added to C3. Contradictory to Markovnikov's rule.

Is this the reason?

[AndresKiani]

• Yes I didn't realize you were using a strong base. Strong bases in addition reactions lean towards anti-Markovnikov additions.

 

• Strong acids on the other hand follow markovnikov but are susceptible to various undesirable rearrangement.

 

• When Hg(OAc) I used however as you know allows us to evade these rearrangements and still follow a markovnikov pattern.