Question:
The compound(s) showing optical activity is/are
The compound(s) showing optical activity is/are
Show Hint
A molecule is optically active only if it is chiral: it cannot be superimposed on its mirror image. A carbon with four different groups is a chiral centre, but a meso compound has an internal plane of symmetry and is inactive.
Updated On: Nov 5, 2026
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The Correct Option is B, C
Solution and Explanation
Step 1: Understanding the Concept:
A compound shows optical activity only when it is chiral. Chiral means it cannot be placed over its mirror image. A common cause is a carbon bonded to four different groups.
Step 2: Key Approach:
For each drawn compound we look for a carbon with four different groups. We also check for a plane of symmetry, which would cancel the activity.
Step 3: Check option (A), glycine:
The central carbon holds two H atoms, an \( NH_2 \) group and a \( COOH \) group. Two groups are the same, so the carbon is not a chiral centre. Glycine is optically inactive. (A) is FALSE.
Step 4: Check option (B), lactic acid:
The central carbon holds H, OH, \( CH_3 \) and \( COOH \). All four are different, so it is a chiral centre. Lactic acid is optically active. (B) is TRUE.
Step 5: Check option (C), glyceraldehyde:
The central carbon holds H, OH, CHO and \( CH_2OH \). All four are different, so it is a chiral centre. Glyceraldehyde is optically active. (C) is TRUE.
Step 6: Check option (D), tartaric acid:
This drawing has two chiral carbons, but the two halves are mirror images of each other. There is a plane of symmetry through the middle of the C-C bond. The effects of the two centres cancel inside the molecule. This is the meso form and is optically inactive. (D) is FALSE.
Final Answer:
Only (B) and (C) show optical activity. \[ \boxed{\text{(B) and (C)}} \]
A compound shows optical activity only when it is chiral. Chiral means it cannot be placed over its mirror image. A common cause is a carbon bonded to four different groups.
Step 2: Key Approach:
For each drawn compound we look for a carbon with four different groups. We also check for a plane of symmetry, which would cancel the activity.
Step 3: Check option (A), glycine:
The central carbon holds two H atoms, an \( NH_2 \) group and a \( COOH \) group. Two groups are the same, so the carbon is not a chiral centre. Glycine is optically inactive. (A) is FALSE.
Step 4: Check option (B), lactic acid:
The central carbon holds H, OH, \( CH_3 \) and \( COOH \). All four are different, so it is a chiral centre. Lactic acid is optically active. (B) is TRUE.
Step 5: Check option (C), glyceraldehyde:
The central carbon holds H, OH, CHO and \( CH_2OH \). All four are different, so it is a chiral centre. Glyceraldehyde is optically active. (C) is TRUE.
Step 6: Check option (D), tartaric acid:
This drawing has two chiral carbons, but the two halves are mirror images of each other. There is a plane of symmetry through the middle of the C-C bond. The effects of the two centres cancel inside the molecule. This is the meso form and is optically inactive. (D) is FALSE.
Final Answer:
Only (B) and (C) show optical activity. \[ \boxed{\text{(B) and (C)}} \]
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