Which of the following biphenyls is optically active?

Options

A.
B.
C.
D.

Answer

To determine which biphenyl is optically active, we need to follow these steps:

### Step 1: Understand Optical Activity
Optical activity requires the presence of a chiral center (a carbon atom bonded to four different groups) and the absence of a plane of symmetry.

**Hint:** Look for chiral centers and symmetry in the structures.

### Step 2: Analyze Each Option
We will analyze each biphenyl option provided to check for chirality and symmetry.

#### Option A:
- In this biphenyl, there are two methyl (CH₃) groups at the ortho positions on the same phenyl ring.
- Since both groups are identical, this biphenyl has a plane of symmetry and is optically inactive.

**Hint:** Identical groups at chiral centers lead to symmetry.

#### Option B:
- This biphenyl has no substitutions at the ortho positions.
- It remains planar, allowing for a plane of symmetry to be drawn through it. Thus, it is optically inactive.

**Hint:** A planar structure with no substitutions at chiral centers has symmetry.

#### Option C:
- Here, we have bromine (Br) and iodine (I) at the ortho positions on both phenyl rings.
- The presence of different groups at the ortho positions means that when the biphenyl rotates around the single bond, the two phenyl rings become perpendicular to each other.
- This configuration prevents a plane of symmetry from being drawn, making this biphenyl optically active.

**Hint:** Different groups at chiral centers can lead to optical activity if steric hindrance occurs.

#### Option D:
- In this biphenyl, there are iodine (I) substituents at the ortho positions on both phenyl rings.
- Since both substituents are identical and located on opposite sides, this biphenyl is planar and has a plane of symmetry, rendering it optically inactive.

**Hint:** Identical substituents on opposite sides lead to symmetry.

### Step 3: Conclusion
After analyzing all options, we find that **Option C** is the only biphenyl that is optically active due to the lack of a plane of symmetry and the presence of different groups at the ortho positions.

**Final Answer:** Option C is optically active.