MCQ (Practice) - Isomerism (Level 2) - Class 11 MCQ

specific angle of rotation

1,3    1,4    2,3    2,4 are Diastereomers.
Ans. 2,4

Both K and L meso structure.  

Plane not present so optically active ⇒ Resolvable
Trans-1,3-dimethyl cyclo hexane.

Identicle compounds are = II & IV

I & III are diastereomers because they are not mirror image.

So compound (A) is meso compound

III and IV compound represents pair of enantiomers.

They are conformers as well as enantiomers.

asymmetric T.S.I = 2ⁿ = 2² = 4
n = no. of double bond + no. of chiral center = 1 + 1 = 2

A single C₂ rotational axis but no mirror plane.

double Bond for G.I. = 1
n = 3 + 1 = 4

Asymmetric 2³ = 8

Ind compound is meso compound so θ = 0°

% of L-enantiomer = 100 – 80 = 20%

Total 12 structure isomer possible

Two pure chiral center must be optically active.

Physical properties like M.P., B.P., solubility are different for diastereomers.

The isomerism exhibited by acetic acid and methyl formate is functional isomerism. These two compounds have the same molecular formula but belong to different chemical classes; acetic acid is a carboxylic acid and methyl formate is an ester. Each has different functional groups, which leads to different properties. Thus, the correct answer is: Functional

Cis-trans (geometrical) isomerism occurs in compounds with restricted rotation around a double bond and differing groups attached to the carbons involved in the double bond.

1. but-2-ene: This compound contains a double bond between the second and third carbons and has different groups on either side of the double bond, making it capable of cis-trans isomerism.
2. Butanol: This is an alcohol with no double bonds to allow for geometrical isomerism.
3. but-2-yne: This compound has a triple bond, which does not allow for geometrical isomerism since the linear arrangement of a triple bond prevents the existence of cis and trans forms.
4. buten-2-ol: While this compound has a double bond, the presence of a hydroxyl group (OH) on one of the carbons of the double bond restricts it from exhibiting cis-trans isomerism effectively due to the impact of hydrogen bonding and the polar nature of the OH group on the geometry.

Hence, the correct answer is: but-2-ene

C₃H₆Br₂

Two gem dibromide.