- 2074 Bhadra · 2 marks
E or Z configuration of given alkenes
Short answer
Using CIP priority (higher atomic number gets higher priority), if the two higher-priority groups are on same side of double bond it is Z, opposite side E. (i) Z, (ii) E, (iii) Z, (iv) Z.
Find out E or Z configuration in the following molecules. (i) [Figure: C=C with Br and I on the left carbon; F and Cl on the right carbon] (ii) [Figure: C=C with CH₃ and H on the left carbon; C₂H₅ and F on the right carbon] (iii) [Figure: C=C with I and Br on the left carbon; C₆H₅ and CH₃ on the right carbon] (iv) [Figure: C=C with C₂H₅ and CH₃ on the left carbon; I and OH on the right carbon]
Method: CIP priority is decided by atomic number at the first point of difference. If the two higher-priority groups are on the same side of C=C, the compound is Z; if on opposite sides, it is E.
Assumption (figure not shown): for each alkene, the first-named group is drawn at the top and the second-named group at the bottom of its carbon, as in the question. If the figure is the other way round, the Z/E label of that case reverses.
| No. | Left carbon (higher) | Right carbon (higher) | Positions | Configuration |
|---|---|---|---|---|
| (i) | I > Br → I (bottom) | Cl > F → Cl (bottom) | same side | Z |
| (ii) | CH₃ > H → CH₃ (top) | F > C₂H₅ → F (bottom) | opposite | E |
| (iii) | I > Br → I (top) | C₆H₅ > CH₃ → C₆H₅ (top) | same side | Z |
| (iv) | C₂H₅ > CH₃ → C₂H₅ (top) | I > OH → I (top) | same side | Z |
Notes: in (ii), F (Z = 9) outranks C₂H₅ (carbon, Z = 6). In (iii), phenyl (C,C,C) outranks methyl (H,H,H). In (iv), I (Z = 53) outranks O.
✓ Answer checked by Claude Opus 5.5 (high effort)
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