How do you draw #(1)# 3-fluoro-4,5-dimethyl-1-heptyne and #(2)# 2-chloro-4-ethyl-1,3-cyclohexadiene?
Not sure why you have two questions.
1) 3-fluoro-4,5-dimethyl-1-heptyne
Hence, we have:
The main chain (a seven-membered straight-chained hydrocarbon) is highlighted. 2) 2-chloro-4-ethyl-1,3-cyclohexadiene
That's all you need to know to generate this:
The parent compound (a six-membered ring) is highlighted.
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To draw 3-fluoro-4,5-dimethyl-1-heptyne:
- Start with a straight chain of seven carbon atoms for the heptyne.
- Add a triple bond between the first and second carbon atoms.
- Add a fluorine atom on the third carbon atom.
- Add methyl groups on the fourth and fifth carbon atoms.
To draw 2-chloro-4-ethyl-1,3-cyclohexadiene:
- Draw a cyclohexadiene ring with double bonds between the first and second, and third and fourth carbon atoms.
- Add a chlorine atom on the second carbon atom.
- Add an ethyl group on the fourth carbon atom.
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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