Question

Which of the following is optically active?

(A question about a specific molecule structure would need to be provided for options)
A) CH3CH(CH(Cl)C(CH3)4
B) CH3CH2CH(NH2)COOH
C) (CH3)2C(NH2)COOH
D) NH2CH2COOH

Answer 1

Out of the given molecules, B) CH₃CH₂CH(NH₂)COOH is optically active due to its chiral center, which has four different substituents, allowing it to rotate the plane of plane-polarized light.

Step-by-step explanation:

The question asks which of the given molecules is optically active. A molecule is considered optically active if it can rotate the plane of plane-polarized light, a characteristic of chiral molecules.

A chiral molecule must have a carbon atom bonded to four different atoms or groups, creating a non-superimposable mirror image. Except for glycine, all amino acids are chiral due to the presence of a central carbon atom (called an alpha carbon) bonded to four different groups.

Let's examine the options given:

• B) CH₃CH₂CH(NH₂)COOH: This molecule possesses an alpha carbon with four different groups (hydrogen, CH₃, NH₂, and COOH), making it chiral and thus optically active.
• C) (CH3)₂C(NH₂)COOH: Although this has a carbon with different groups attached, the carbon is not chiral as two of the groups are the same (CH₃), meaning it is not optically active.

Therefore, the molecule B) CH₃CH₂CH(NH₂)COOH is optically active because it has a chiral center with four different substituents.