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    In the nitration of 1,3-dimethylbenzene, why doesn't the electrophile get substituted to the carbon in between the 2 methyl groups because that's the 2nd carbon...

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    (Original post by coconut64)
    In the nitration of 1,3-dimethylbenzene, why doesn't the electrophile get substituted to the carbon in between the 2 methyl groups because that's the 2nd carbon...

    Thanks
    A methyl group on a benzene ring directs electrophillic attack to both the ortho and para positions (these are positions 2 and 4).So when nitrating 1,3-dimethylbenzene, it would be sensible to expect attack at the 2 position (ortho to both methyl groups), however you should also note that the 4 and 6 positions (which are equivalent to one another) are both ortho to one methyl and para to the other, so these two positions are also favoured by both methyl groups.

    In reality, some of both product will be formed, however the product with attack at the 4 (or 6) position dominates for two main reasons.

    Firstly, there are two of these sites and only 1 between the methyl's so even if it were random chance you would expect a 2:1 ratio of products.

    On top of that, methyl groups are quite big and so the electrophile's approach to the 2 position is blocked by two methyl groups whilst attack at 4 (or 6) positions is only hindered by 1 methyl. This effect is known as steric hindrance and plays an important role in organic compounds reactivity.
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    (Original post by MexicanKeith)
    A methyl group on a benzene ring directs electrophillic attack to both the ortho and para positions (these are positions 2 and 4).So when nitrating 1,3-dimethylbenzene, it would be sensible to expect attack at the 2 position (ortho to both methyl groups), however you should also note that the 4 and 6 positions (which are equivalent to one another) are both ortho to one methyl and para to the other, so these two positions are also favoured by both methyl groups.

    In reality, some of both product will be formed, however the product with attack at the 4 (or 6) position dominates for two main reasons.

    Firstly, there are two of these sites and only 1 between the methyl's so even if it were random chance you would expect a 2:1 ratio of products.

    On top of that, methyl groups are quite big and so the electrophile's approach to the 2 position is blocked by two methyl groups whilst attack at 4 (or 6) positions is only hindered by 1 methyl. This effect is known as steric hindrance and plays an important role in organic compounds reactivity.
    I get where you are coming from now. So you will never have something like this?
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    (Original post by coconut64)
    I get where you are coming from now. So you will never have something like this?
    Name:  lolll;l.png
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Size:  18.3 KB thanks for helping
    You might see something like that because as you said the 2 position is favourable electronically, but it isnt the major product in this reaction because of the steric restrictions of the methyl groups! (it would probably account for less than 10% of the product). it's quite common for organic reactions to give multiple products but at A-level they virtually always want the major product (often some marks are given for minor products too)
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    (Original post by MexicanKeith)
    You might see something like that because as you said the 2 position is favourable electronically, but it isnt the major product in this reaction because of the steric restrictions of the methyl groups! (it would probably account for less than 10% of the product). it's quite common for organic reactions to give multiple products but at A-level they virtually always want the major product (often some marks are given for minor products too)
    I can't rep you for some reason, but thanks for helping out so late
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    (Original post by coconut64)
    I can't rep you for some reason, but thanks for helping out so late
    No worries, happy to help
 
 
 
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