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Basics of Organic Chemistry question

2009 · Shift 0 · Q3
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Basics of Organic Chemistry question

2009 · Shift 0 · Q3

JEE MainChemistryBasics of Organic ChemistryMCQ+4 / −1
The alkene that exhibits geometrical isomerism is :
  1. A
    propene
  2. B
    2-methyl propene
  3. C
    2-butene
  4. D
    2- methyl -2- butene
View written solutionFree

Correct answer: C

  1. Condition for geometrical isomerism in alkenes

For an alkene to show geometrical (cis-trans / E-Z) isomerism, each carbon atom of the double bond must have two different substituents.

If any one of the double-bonded carbon atoms has two identical groups, geometrical isomerism is not possible.

  1. Check each option

A. Propene

Structure: CH3−CH=CH2\mathrm{CH_3-CH=CH_2}CH3​−CH=CH2​

The terminal carbon of the double bond is CH2\mathrm{CH_2}CH2​, which has two identical H atoms.

So, geometrical isomerism is not possible.

B. 2-methyl propene

Structure: (CH3)2C=CH2\mathrm{(CH_3)_2C=CH_2}(CH3​)2​C=CH2​

One double-bonded carbon has two identical CH3\mathrm{CH_3}CH3​ groups, and the other has two identical H atoms.

So, geometrical isomerism is not possible.

C. 2-butene

Structure: CH3−CH=CH−CH3\mathrm{CH_3-CH=CH-CH_3}CH3​−CH=CH−CH3​

Each carbon of the double bond has:

  • one CH3\mathrm{CH_3}CH3​ group
  • one H atom

Thus, each double-bonded carbon has two different groups.

So, geometrical isomerism is possible.

It exists as:

  • cis-2-butene
  • trans-2-butene

D. 2-methyl-2-butene

Structure: CH3−C(CH3)=CH−CH3\mathrm{CH_3-C(CH_3)=CH-CH_3}CH3​−C(CH3​)=CH−CH3​

The left double-bonded carbon has two identical CH3\mathrm{CH_3}CH3​ groups.

So, geometrical isomerism is not possible.

  1. Conclusion

Only 2-butene satisfies the condition for geometrical isomerism.

Therefore, the correct option is: C: 2-butene\boxed{\text{C: 2-butene}}C: 2-butene​

  1. Comparison with stored correct answer

Stored correct answer: C

Derived answer: C

They match.

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