Addition of HBr to a double bond with an ether...

[skyisblue]

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Addition of HBr to a cyclohexene with an ether like OCH3 ether group bonded to the number one position should lead to only one product right according to Markovnikov addition?

OR should there be 2 products because the double bond is bonded to an Oxygen and not a Carbon??

 

[BodybldgDoc]

Addition of HBr to a cyclohexene with an ether like OCH3 ether group bonded to the number one position should lead to only one product right according to Markovnikov addition?

OR should there be 2 products because the double bond is bonded to an Oxygen and not a Carbon??

The proton will add to the less substituted carbon atom(the carbon with the most protons), since alkyl subtituents stabilize carbocations.
If it is a symmetrical alkene then two products will be formed.
If it is asymmetrical alkene then only the major product is formed.
hope this helps!

 

[poc91nc]

The proton will add to the less substituted carbon atom(the carbon with the most protons), since alkyl subtituents stabilize carbocations.
If it is a symmetrical alkene then two products will be formed.
If it is asymmetrical alkene then only the major product is formed.
hope this helps!

Good stuff, some other things to keep in mind are methyl shifts and hydride shifts whenever a carbocation is involved. Practice a few problems that keep those points in mind. If you're not used to looking out for methyl/hydride shifts you probably won't "see" it.

Whenever you have HBr in the presence of peroxide use anti-markovnikov. Remember in the presence of peroxides you go by free radical mechanism. Know the general mechanism for the free radical reaction. There are problems out there that expose your understanding of the HBr/peroxide mechanism. If no peroxide is present then you go by default rules.

Another anti markovnikov addition reaction is hydroboration/peroxide. Notice a theme with peroxide? Be wary of stereochemistry with this problem.

Also, halohydrin reaction....halogen to carbon with more H's.

 

[skyisblue]

Good stuff, some other things to keep in mind are methyl shifts and hydride shifts whenever a carbocation is involved. Practice a few problems that keep those points in mind. If you're not used to looking out for methyl/hydride shifts you probably won't "see" it.

Whenever you have HBr in the presence of peroxide use anti-markovnikov. Remember in the presence of peroxides you go by free radical mechanism. Know the general mechanism for the free radical reaction. There are problems out there that expose your understanding of the HBr/peroxide mechanism. If no peroxide is present then you go by default rules.

Another anti markovnikov addition reaction is hydroboration/peroxide. Notice a theme with peroxide? Be wary of stereochemistry with this problem.

Also, halohydrin reaction....halogen to carbon with more H's.

GOOD STUFF thanks Poc!

 

[skyisblue]

The proton will add to the less substituted carbon atom(the carbon with the most protons), since alkyl subtituents stabilize carbocations.
If it is a symmetrical alkene then two products will be formed.
If it is asymmetrical alkene then only the major product is formed.
hope this helps!

Hey Body!

It did help!

 

[BBB111]

Can NaBH4 reduce a carboxylic acid?

 

[Ibraiz]

Can NaBH4 reduce a carboxylic acid?

From my understanding, LAH can reduce a carboxylic acid to a primary alcohol.

NaBH4 only reduces Aldehydes and Ketones since its a mild reducing agent.
PCC oxides primary alcohol to aldehyde and secondary alcohol to ketones (opposite of NaBH4)

 

[RCT PC CRN]

Can NaBH4 reduce a carboxylic acid?

Based on my ochem 2 knowledge. No it wont. NaBH4 - mild reducing agent. LAH - strong one, will reduce carboxylic acid.

 

[Ibraiz]

Good stuff, some other things to keep in mind are methyl shifts and hydride shifts whenever a carbocation is involved. Practice a few problems that keep those points in mind. If you're not used to looking out for methyl/hydride shifts you probably won't "see" it.

Whenever you have HBr in the presence of peroxide use anti-markovnikov. Remember in the presence of peroxides you go by free radical mechanism. Know the general mechanism for the free radical reaction. There are problems out there that expose your understanding of the HBr/peroxide mechanism. If no peroxide is present then you go by default rules.

Another anti markovnikov addition reaction is hydroboration/peroxide. Notice a theme with peroxide? Be wary of stereochemistry with this problem.

Also, halohydrin reaction....halogen to carbon with more H's.

Poc,
is that mean that if we know which ones are anti-markovnikov then we can look out for the anti-markovnikov reagents like ROOR and peroxides and light and everything else is markovnikov?

Is that a good assumption to go by?
If HBR was with a base adding to an alkene, would that be mean its an anti-markovnikov since HRB with an acid adding to an alkene is markovnikov?

Thanks

 

[FredDurst]

Poc,
is that mean that if we know which ones are anti-markovnikov then we can look out for the anti-markovnikov reagents like ROOR and peroxides and light and everything else is markovnikov?

Is that a good assumption to go by?
If HBR was with a base adding to an alkene, would that be mean its an anti-markovnikov since HRB with an acid adding to an alkene is markovnikov?

Thanks

No. The reason for the anti-markovnikov showing up is because how the mechanism works. In the hydroboration, it isn't the peroxide that's causing the anti-markovnikov, it's the fact that the boron pairs with the least substituted carbon due to sterics. All the peroxide does it substitute an OH group for the boron group. For the addition of HBr in the presence of a peroxide, it isn't the peroxide that determines the anti-markovnikov, it's the fact that the radical formed from the addition of the Br radical is most stabilized on the higher-substituted carbon. All the peroxide does is help initially form the Br radical.

Look at the mechanisms for these reactions. It may help you to write them all down and group them, you will see many similarities in different reactions that add to alkenes.

 

[Ibraiz]

Thanks for the explanation Fred Durst.

My understanding:
Peroxide forms the Br Radical
Br Radical attacks the alkene
Carbon radical is formed - the most substituted, most stable radical
Proton is added to the most substituted radical

 

[Troyvdg]

^ yessir!