Alkene Geometrical Isomerism

8 MCQs9-step worked example
Source: NCERT HydrocarbonsOfficial key: NTA-verifiedLast updated: 8 Oct 2026

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Why can an alkene such as but-2-ene exist as two different geometrical isomers?
  1. A.The two carbon atoms of C=C rotate freely, so the groups can swap sides
  2. B.Rotation about the C=C bond is restricted, so the groups stay fixed in their arrangement
  3. C.The double bond is weaker than a single bond
  4. D.A hydrogen atom can migrate from one carbon to the other
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Answer: B. The restricted rotation of atoms or groups around the doubly bonded carbon atoms gives rise to different geometries; the stereoisomers of this type are called geometrical isomers (NCERT Class 11 Chemistry, Chapter 9, page 308).

A is wrong: A is wrong because rotation about C=C is not free. If it were free, the two arrangements would have the same geometry and there would be only one compound (NCERT Class 11 Chemistry, Chapter 9, page 308).

C is wrong: C is wrong because the strength of the bond does not create two spatial arrangements; the cause named by NCERT is restricted rotation (NCERT Class 11 Chemistry, Chapter 9, page 308).

D is wrong: D is wrong because the two forms have the same structure and differ only in the arrangement of groups in space; nothing moves from one carbon to the other (NCERT Class 11 Chemistry, Chapter 9, page 308).

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Alkene Geometrical Isomerism, explained for NEET

The slip is to assume that every alkene has a cis form and a trans form. It does not. Each of the two doubly bonded carbons must carry two different atoms or groups.

Each carbon of C=C satisfies its remaining two valences with two atoms or groups. Rotation about the C=C bond is not free; it is restricted. So when each carbon carries two different groups, there are two arrangements in space that cannot change into each other. These stereoisomers are called geometrical isomers (NCERT Class 11 Chemistry, Chapter 9, page 308).

In the cis isomer the two identical atoms or groups lie on the same side of the double bond; in the trans isomer they lie on opposite sides. The two have the same structure but a different configuration, so they differ in melting point, boiling point, dipole moment and solubility (NCERT Class 11 Chemistry, Chapter 9, page 308).

The cis form is the more polar one. The dipole moment of cis-but-2-ene is 0.33 Debye, while that of the trans form is almost zero (NCERT Class 11 Chemistry, Chapter 9, page 308). In trans-but-2-ene the two methyl groups point in opposite directions, so the C–CH₃ bond dipoles cancel and the molecule is non-polar (NCERT Class 11 Chemistry, Chapter 9, page 309). For solids, the trans isomer has the higher melting point than the cis form (NCERT Class 11 Chemistry, Chapter 9, page 309).

Cis-trans isomerism is also shown by alkenes of the types XYC=CXZ and XYC=CZW. If one doubly bonded carbon holds two identical groups, as in CH₂=CBr₂, there is no cis-trans isomerism (NCERT Class 11 Chemistry, Chapter 9, page 309).

The preparation of alkenes from alkynes decides which isomer forms: partial reduction with Lindlar's catalyst gives the cis alkene, while sodium in liquid ammonia gives the trans alkene (NCERT Class 11 Chemistry, Chapter 9, page 309).

Watch out: check both carbons before you draw cis and trans. One carbon with two identical groups is enough to rule isomerism out.

How do you solve a Alkene Geometrical Isomerism question? A worked example

  1. 1

    Given

    The compound pent-2-ene, CH₃–CH=CH–CH₂–CH₃.

  2. 2

    Required

    Whether pent-2-ene shows geometrical isomerism, and how the cis and trans forms differ.

  3. 3

    Concept

    Geometrical isomerism needs restricted rotation about C=C and two different atoms or groups on each doubly bonded carbon (NCERT Class 11 Chemistry, Chapter 9, pages 308 and 309).

  4. 4

    Formula

    Condition: the alkene must be of the type XYC=CXZ or XYC=CZW (NCERT Class 11 Chemistry, Chapter 9, page 309).

  5. 5

    Substitution

    Carbon 2 carries H and CH₃. Carbon 3 carries H and CH₂CH₃. Both carbons carry two different groups, and the two hydrogen atoms are the identical pair.

  6. 6

    Calculation

    Place the two hydrogens on the same side of the double bond: that is the cis isomer. Place them on opposite sides: that is the trans isomer. Two arrangements exist, and rotation about C=C is restricted, so they do not interconvert. No measured quantity is used here, so no significant-figure question arises.

  7. 7

    Final answer

    Pent-2-ene shows geometrical isomerism and has two isomers, cis-pent-2-ene and trans-pent-2-ene. They share the same structure but differ in configuration, so their physical properties differ.

  8. 8

    Common trap

    Pent-1-ene, CH₂=CH–CH₂–CH₂–CH₃, has the same chain but a CH₂ end with two identical hydrogens, so it has no cis-trans isomerism. Moving the double bond changes the answer.

  9. 9

    Similar NEET-style question

    Does 2-methylbut-2-ene, (CH₃)₂C=CH–CH₃, show geometrical isomerism? (Answer: no. One doubly bonded carbon holds two identical methyl groups, so only one arrangement exists.)

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Can you answer these Alkene Geometrical Isomerism MCQs?

Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.

MCQ 1Easy RecallPractice

Why can an alkene such as but-2-ene exist as two different geometrical isomers?

Show answer and why every option is right or wrong

Answer: B. The restricted rotation of atoms or groups around the doubly bonded carbon atoms gives rise to different geometries; the stereoisomers of this type are called geometrical isomers (NCERT Class 11 Chemistry, Chapter 9, page 308).

Why A is wrong: A is wrong because rotation about C=C is not free. If it were free, the two arrangements would have the same geometry and there would be only one compound (NCERT Class 11 Chemistry, Chapter 9, page 308).

Why C is wrong: C is wrong because the strength of the bond does not create two spatial arrangements; the cause named by NCERT is restricted rotation (NCERT Class 11 Chemistry, Chapter 9, page 308).

Why D is wrong: D is wrong because the two forms have the same structure and differ only in the arrangement of groups in space; nothing moves from one carbon to the other (NCERT Class 11 Chemistry, Chapter 9, page 308).

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MCQ 2Easy RecallPractice

In which of the following descriptions is the isomer correctly called the cis isomer?

Show answer and why every option is right or wrong

Answer: B. The isomer in which two identical atoms or groups lie on the same side of the double bond is called the cis isomer (NCERT Class 11 Chemistry, Chapter 9, page 308).

Why A is wrong: A is wrong because identical groups on opposite sides of the double bond describe the trans isomer (NCERT Class 11 Chemistry, Chapter 9, page 308).

Why C is wrong: C is wrong because cis and trans refer to where the identical groups sit across the bond, not to both carbons being the same (NCERT Class 11 Chemistry, Chapter 9, page 308).

Why D is wrong: D is wrong because cis and trans isomers have the same structure and differ in configuration, the arrangement of atoms or groups in space (NCERT Class 11 Chemistry, Chapter 9, page 308).

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MCQ 3Easy RecallPractice

Partial reduction of an alkyne with sodium in liquid ammonia gives mainly

Show answer and why every option is right or wrong

Answer: C. Alkynes on reduction with sodium in liquid ammonia form trans alkenes (NCERT Class 11 Chemistry, Chapter 9, page 309).

Why A is wrong: A is wrong because the cis alkene comes from partial reduction with dihydrogen over Lindlar's catalyst, not from sodium in liquid ammonia (NCERT Class 11 Chemistry, Chapter 9, page 309).

Why B is wrong: B is wrong because the reduction stops at the alkene stage under these conditions; NCERT states that trans alkenes are formed (NCERT Class 11 Chemistry, Chapter 9, page 309).

Why D is wrong: D is wrong because NCERT assigns a definite geometry to each method: cis with Lindlar's catalyst and trans with sodium in liquid ammonia (NCERT Class 11 Chemistry, Chapter 9, page 309).

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MCQ 4Direct ApplicationPractice

Which of the following compounds shows cis-trans isomerism?

Show answer and why every option is right or wrong

Answer: C. In CH₃–CH=CH–Cl each doubly bonded carbon carries two different groups (CH₃ and H on one, Cl and H on the other), the XYC=CXZ type, so cis and trans forms exist (NCERT Class 11 Chemistry, Chapter 9, page 309).

Why A is wrong: A is wrong because one doubly bonded carbon (CH₂) holds two identical hydrogen atoms, so there is no cis-trans isomerism (NCERT Class 11 Chemistry, Chapter 9, page 309).

Why B is wrong: B is wrong because one doubly bonded carbon holds two identical methyl groups, so there is no cis-trans isomerism (NCERT Class 11 Chemistry, Chapter 9, page 309).

Why D is wrong: D is wrong because one doubly bonded carbon holds two identical chlorine atoms, so there is no cis-trans isomerism (NCERT Class 11 Chemistry, Chapter 9, page 309).

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MCQ 5Direct ApplicationPractice

The cis and trans isomers of a solid alkene are compared. Which statement is correct?

Show answer and why every option is right or wrong

Answer: A. The cis form of an alkene is found to be more polar than the trans form, and for solids the trans isomer has the higher melting point than the cis form (NCERT Class 11 Chemistry, Chapter 9, pages 308 and 309).

Why B is wrong: B is wrong because it swaps both properties: the cis form is the more polar and the trans form melts higher (NCERT Class 11 Chemistry, Chapter 9, pages 308 and 309).

Why C is wrong: C is wrong because cis and trans isomers differ in melting point, boiling point, dipole moment and solubility (NCERT Class 11 Chemistry, Chapter 9, page 308).

Why D is wrong: D is wrong because the cis form is the more polar one, but for solids it is the trans isomer that has the higher melting point (NCERT Class 11 Chemistry, Chapter 9, page 309).

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MCQ 6Direct ApplicationPractice

How many geometrical isomers (counting both cis and trans) does C₂H₅–C(CH₃)=C(CH₃)–C₂H₅ have?

Show answer and why every option is right or wrong

Answer: B. Each doubly bonded carbon carries two different groups (CH₃ and C₂H₅), so the cis and trans arrangements both exist and there are two geometrical isomers (NCERT Class 11 Chemistry, Chapter 9, page 309).

Why A is wrong: A is wrong because symmetry of the formula does not remove the isomerism; each doubly bonded carbon has two different groups, so cis and trans forms exist (NCERT Class 11 Chemistry, Chapter 9, page 309).

Why C is wrong: C is wrong because only two arrangements are possible about a single C=C bond: identical groups on the same side or on opposite sides (NCERT Class 11 Chemistry, Chapter 9, page 308).

Why D is wrong: D is wrong because the double bond gives just the cis and the trans arrangement; two more forms would need another source of isomerism (NCERT Class 11 Chemistry, Chapter 9, page 308).

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MCQ 7CalculationPractice

But-2-yne is reduced in two separate experiments. In experiment X it is treated with dihydrogen over palladised charcoal partly deactivated with quinoline. In experiment Y it is treated with sodium in liquid ammonia. Which statement about the products is correct?

Show answer and why every option is right or wrong

Answer: C. Deactivated palladised charcoal is Lindlar's catalyst and gives the cis alkene, which is polar (cis-but-2-ene, 0.33 Debye). Sodium in liquid ammonia gives the trans alkene, whose dipole moment is almost zero (NCERT Class 11 Chemistry, Chapter 9, pages 308 and 309).

Why A is wrong: A is wrong because it swaps the stereochemistry: Lindlar's catalyst gives the cis alkene and sodium in liquid ammonia gives the trans alkene (NCERT Class 11 Chemistry, Chapter 9, page 309).

Why B is wrong: B is wrong because the two methods give different geometries, cis and trans, which are different compounds with different properties (NCERT Class 11 Chemistry, Chapter 9, pages 308 and 309).

Why D is wrong: D is wrong because only the cis isomer has the dipole moment 0.33 Debye; trans-but-2-ene is almost non-polar since its C–CH₃ bond dipoles cancel (NCERT Class 11 Chemistry, Chapter 9, page 309).

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MCQ 8CalculationPractice

Each of four alkynes is partially reduced to the alkene. Which alkyne gives a product that can exist as cis and trans isomers?

Show answer and why every option is right or wrong

Answer: D. Partial reduction of CH₃–C≡C–CH₃ gives but-2-ene, CH₃–CH=CH–CH₃, where each doubly bonded carbon carries H and CH₃, so cis and trans forms exist (NCERT Class 11 Chemistry, Chapter 9, page 308). Ethyne gives ethene and propyne gives propene, and each has a CH₂ end with two identical hydrogens (NCERT Class 11 Chemistry, Chapter 9, page 309).

Why A is wrong: A is wrong because ethyne gives ethene, CH₂=CH₂, whose carbons each hold two identical hydrogen atoms, so there is no cis-trans isomerism (NCERT Class 11 Chemistry, Chapter 9, page 309).

Why B is wrong: B is wrong because propyne gives propene, CH₃–CH=CH₂, and the CH₂ carbon holds two identical hydrogen atoms, so propene shows no geometrical isomerism (NCERT Class 11 Chemistry, Chapter 9, page 309).

Why C is wrong: C is wrong because neither product has two different groups on both doubly bonded carbons (NCERT Class 11 Chemistry, Chapter 9, page 309).

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What to remember before solving Alkene Geometrical Isomerism questions

9 NCERT lines
Definition

Geometrical isomerism: restricted rotation about C=C gives stereoisomers called geometrical isomers

Geometrical isomerism: Doubly bonded carbon atoms have to satisfy the remaining two valences by joining with two atoms or groups. If the two atoms or groups attached to each carbon atom are different, they can be represented by YX C = C XY like structure. YX C = C XY can be represented in space in the following two ways : In (a), the two identical atoms i.e., both the X or both the Y lie on the same side of the double bond but in (b) the two X or two Y lie across the double bond or on the opposite sides of the double bond. This results in different geometry of (a) and (b) i.e. disposition of atoms or groups in space in the two arrangements is different. Therefore, they are stereoisomers. They would have the same geometry if atoms or groups around C=C bond can be rotated but rotation around C=C bond is not free. It is restricted. For understanding this concept, take two pieces of strong cardboards and join them with the help of two nails. Hold one cardboard in your one hand and try to rotate the other. Can you really rotate the other cardboard ? The answer is no. The rotation is restricted. This illustrates that the restricted rotation of atoms or groups around the doubly bonded carbon atoms gives rise to different geometries of such compounds. The stereoisomers of this type are called geometrical isomers.

-- NCERT Class 11 Chemistry, Ch. 9, p. 308
Definition

Cis isomer: identical groups on the same side of the double bond; trans: on opposite sides

The isomer of the type (a), in which two identical atoms or groups lie on the same side of the double bond is called cis isomer and the other isomer of the type (b), in which identical atoms or groups lie on the opposite sides of the double bond is called trans isomer . Thus cis and trans isomers have the same structure but have different configuration (arrangement of atoms or groups in space).

-- NCERT Class 11 Chemistry, Ch. 9, p. 308
Key Fact

trans-But-2-ene is non-polar: the C-CH3 bond dipoles cancel

trans-but-2-ene is non-polar. This can be understood by drawing geometries of the two forms as given below from which it is clear that in the trans-but-2-ene, the two methyl groups are in opposite directions, Threfore, dipole moments of C-CH3 bonds cancel, thus making the trans form non-polar.

-- NCERT Class 11 Chemistry, Ch. 9, p. 309

Alkene Geometrical Isomerism: NEET previous year questions (PYQs) with answers

11 questions in Hydrocarbons

No question in our NEET 2020–2025 set targets this topic directly.

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More in Hydrocarbons: 4 exam traps and mistakes · 1 formula · 2 question patterns from its other lessons.

Sources

NCERT refs: Class 11 Chemistry Chapter 9, p.308 | Class 11 Chemistry Chapter 9, p.309

Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.

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