Initiators that can be used in anionic polymerisation is/are
(a) Potassium amide (b) n-butyl lithium (c) `AlCl_(3)` (d) `H_(2)SO_(4)`
The correct answer is

Options

A. only a
B. only a and b
C. only c and d
D. all

Answer

To determine which initiators can be used in anionic polymerization from the given options, we will analyze each option based on their chemical nature.

### Step-by-Step Solution:

1. **Understanding Anionic Polymerization**:
Anionic polymerization is a type of chain-growth polymerization that involves the reaction of monomers with an anion. The initiators used in this process must be able to generate an anion or provide a reactive site that can initiate the polymerization.

2. **Analyzing Each Option**:
- **(a) Potassium amide (KMH2)**:
- Potassium amide is a strong base and can generate an anion (amide ion) when it reacts with water or other proton sources. Therefore, it can act as an initiator in anionic polymerization.
- **(b) n-butyl lithium (n-BuLi)**:
- n-Butyl lithium is also a strong base and a powerful nucleophile. It can easily generate an anion and is commonly used as an initiator in anionic polymerization.
- **(c) AlCl3**:
- Aluminum chloride (AlCl3) is a Lewis acid and does not generate an anion. Instead, it accepts electrons, which makes it unsuitable as an initiator for anionic polymerization.
- **(d) H2SO4**:
- Sulfuric acid (H2SO4) is a strong acid that donates protons (H+ ions) rather than generating anions. Therefore, it cannot act as an initiator in anionic polymerization.

3. **Conclusion**:
Based on the analysis:
- **Potassium amide (a)** and **n-butyl lithium (b)** can be used as initiators in anionic polymerization.
- **AlCl3 (c)** and **H2SO4 (d)** cannot be used as initiators in this type of polymerization.

Thus, the correct answer is **(a) Potassium amide and (b) n-butyl lithium**.

### Final Answer:
The correct answer is **(a) and (b)**.

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