allylic and benzylic positions

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xoangelxo

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if you have a hydrogen directly connect to a benzene ring..is that the benzylic position, or would it be the hydrogen connected to a C that is off of a benzene ring?

What is the allylic position referring to?

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if you have a hydrogen directly connect to a benzene ring..is that the benzylic position, or would it be the hydrogen connected to a C that is off of a benzene ring?

What is the allylic position referring to?

Benzylic is the hydrogen attached to the carbon on the side chain.

Allylic is the first staurated carbon right after the double double bonds on wither side of the alkene.
 
well...it may be easier it you can imagine like a toluene - the methyl carbon will have the benzylic position.

if you have, lets say, a 2 pentene, then carbon 1 and carbon 4 can be considered allylic positions
 
okay, for the toluene example, is the carbon and the hydrogens attached to that carbon considered benzyllic? Or just the H's?

And for the allylic postions...is it just the carbons you both referred to, or also the H's attached to those C's.

Thanks!!
 
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i think it's to refer to anything/substituent on that carbon -
for example, 2-propenol - if it undergoes an E1 rxn...then the alcohol group will leave as water and there will be a carbo cation at the allylic position.
a halogen can be on that carbon and it'll be considered allyl bromide...

if you could provide a sample question with what you're confused on...that may be easier to clarify.
 
The best way to remember allylic and benzylic is to think of the word
adjacent ( which means next to). Both allylic and Benzylic "H" or "any Hallogen" or "any other groups such as -OH" are the ones on the adjacent "C molecule" that is adjacent to the Double Bond (allylic) and on the adjacent "C of the Benzen (benzylic)

Examples
C=C-C-Br allylic the " Br " here is on the adjacent carbon of the the Doble bond and similarly
Pretent the Smile face is a benzen ring...
🙂-C-I Benzylic - here " I " is on a carbon adjacent ( or next to) to the benzen ring

Hope it helps!
 
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