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Answers
5) 2-bromo-3,3-dimethylbutanoic acid
- OR 2-bromo-3,3-dimethylbutan-1-oic acid
6) Propan-2-oic acid
7) 2,3-dimethyl-hexanoic acid
8) 3-hydroxy-butanoic acid
9) 2-methyl-butanoic acid
10) 4,4-dimethyl-hexanoic acid
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Was too lazy for maintaining the explanation,if you want, inform me I'll provide it for sure. Thanks!
Answer:
1: carbon chain has five carbon
2: Numbering starts from where alkene has the lowest number
3: bromine at number 4position use prefix as bromo.
4: methyl at number 3 carbon position use prefix is 3 methyl.
5: combined all. 4- Bromo-3-methyl,pent 2-ene.
6 H3−C2H=C3H−C4H=C5H−C6≡C7−C8H3
The ene functional group is preferred over yne-group if both are present at same distance from their respective terminal carbons. Here from right side of the chain yne-group is present at 2nd carbon and ene-group is also present at 2nd carbon from left side of the chain. Hence, the numbering is done from the left side.
Hence, the IUPAC name of the compound is
Octa−2,4−diene−6−yne
7 The photo is the answer 0f no 7
8 wo functional groups are present in the given compound. Priority will be given to −COOH group. The numbering will start from carboxylic acid carbon. Double bond will get number 3 and hydroxy group will get number 4.
Hence, the IUPAC name is 4-hydroxy-3-pentenoic acid.
9 Carboxylic carbon is numbered as 1 as per the priority rule.Consider the longest carbon chain, at C-2 and C-4, methyl group is present and total 6 carbon atoms are present in longest carbon chain, hence name is 2,4-dimethyl hexanoic acid
