Chemistry, asked by anantpushpraj2000, 9 hours ago

41. Write the name of following compounds according to E and Z system :
CH3
H
C2H5
H
С
11
С
II
C с
H н
C1
HO
C1​

Answers

Answered by singhtrisha386
0

Answer:

The traditional system for naming the geometric isomers of an alkene, in which the same groups are arranged differently, is to name them as cis or trans. However, it is easy to find examples where the cis-trans system is not easily applied. IUPAC has a more complete system for naming alkene isomers. The R-S system is based on a set of "priority rules", which allow you to rank any groups. The rigorous IUPAC system for naming alkene isomers, called the E-Z system, is based on the same priority rules.These priority rules are often called the Cahn-Ingold-Prelog (CIP) rules, after the chemists who developed the system

The general strategy of the E-Z system is to analyze the two groups at each end of the double bond. At each end, rank the two groups, using the CIP priority rules, discussed in Ch 15. Then, see whether the higher priority group at one end of the double bond and the higher priority group at the other end of the double bond are on the same side (Z, from German zusammen = together) or on opposite sides (E, from German entgegen = opposite) of the double bond.

The Figure below shows the two isomers of 2-butene. You should recognize them as cis and trans. Let's analyze them to see whether they are E or Z.

  • Start with the left hand structure (the cis isomer). On C2 (the left end of the double bond), the two atoms attached to the double bond are C and H. By the CIP priority rules, C is higher priority than H (higher atomic number). Now look at C3 (the right end of the double bond). Similarly, the atoms are C and H, with C being higher priority. We see that the higher priority group is "down" at C2 and "down" at C3. Since the two priority groups are both on the same side of the double bond ("down", in this case), they are zusammen = together. Therefore, this is (Z)-2-butene.

Now look at the right hand structure (the trans isomer). In this case, the priority group is "down" on the left end of the double bond and "up" on the right end of the double bond. Since the two priority groups are on opposite sides of the double bond, they are entgegen = opposite. Therefore, this is (E)-2-butene.

Explanation:

i hope u understand dear

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