---
title: "E or Z configuration of given alkenes"
subject: "Engineering Chemistry"
chapter: 9
key: "semester_2/engineering_chemistry:ez-config"
source: https://www.bctengineer.com/q/semester_2/engineering_chemistry/ez-config
site: BCT Engineering
---

# E or Z configuration of given alkenes

Engineering Chemistry · Chapter 9: Stereochemistry

- Asked in: 2074 Bhadra, 2 marks
- Video: [E/Z Absolute Configuration of Alkenes](https://www.youtube.com/watch?v=DWyRymumUJ4) — Professor Dave Explains
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/engineering_chemistry/ez-config
- Whole chapter: https://www.bctengineer.com/md/semester_2/engineering_chemistry/chapter-9

## Question

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]

## 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.

## Answer

**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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is CIP priority decided?
   A. By atomic number at the first point of difference
   B. By size of the group
   C. By number of hydrogens
   D. By alphabetical order
   Answer: A. Higher atomic number gets higher priority. (Key point: Method)

2. In case (i), which groups are higher priority?
   A. Br and F
   B. Br and Cl
   C. I on the left carbon and Cl on the right carbon
   D. I and F
   Answer: C. I (53) > Br (35) and Cl (17) > F (9). (Key point: (i) Br/I and F/Cl)

3. In case (ii), which has higher priority, F or C2H5?
   A. F, since Z = 9 > 6
   B. C2H5, since it is larger
   C. Equal priority
   D. C2H5, since it has more atoms
   Answer: A. Priority depends on atomic number of the attached atom. (Key point: F vs ethyl)

4. In case (iii), why does C6H5 outrank CH3?
   A. Phenyl has fewer atoms
   B. Methyl has higher atomic number
   C. Phenyl carbon has (C,C,C) while methyl has (H,H,H)
   D. They are equal
   Answer: C. At the first carbon they tie, so substituents decide. (Key point: Phenyl vs methyl)

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