Arrange the following compounds in increasing order of reactivity towards the addition of HBr
`RCH=CHR,overset(CH_(2))overset(||)(CH_(2)),R_(2)C=CHR,R_(2)C=CR_(2)`

Options

A. `underset(CH_(2))overset(CH_(2))(||)ltRCH=CHRltR_(2)C=CHRltR_(2)C=CR_(2)`
B. `R_(2)C=CHRltRCH=CHRltCH_(2)=CR_(2)ltR_(2)C=CR_(2)`
C. `R_(2)C=CR_(2)ltR_(2)C=CHRltRCH=CHRltCH_(2)=CH_(2)`
D. `R_(2)C=CR_(2)ltCH_(2)=CH_(3)ltRCH=CHRltR_(2)C=CHR`

Answer

To determine the increasing order of reactivity towards the addition of HBr for the given compounds, we need to analyze the structure of each compound and how the presence of alkyl groups affects their reactivity. The compounds provided are:

1. \( RCH=CHR \)
2. \( R_2C=CHR \)
3. \( R_2C=CR_2 \)
4. \( CH_2=CH_2 \)

### Step-by-Step Solution:

**Step 1: Understand the Mechanism of HBr Addition**
- The addition of HBr to alkenes involves the formation of a carbocation intermediate. The more stable the carbocation, the more reactive the alkene will be towards HBr addition.

**Hint for Step 1:** Recall that carbocation stability increases with the number of alkyl groups attached to the positively charged carbon.

**Step 2: Analyze Each Compound**
- **Compound 1: \( RCH=CHR \)**
- This has one alkyl group on one side of the double bond. It can form a secondary carbocation upon addition of HBr.

- **Compound 2: \( R_2C=CHR \)**
- This has two alkyl groups on one side of the double bond. It can form a more stable secondary carbocation.

- **Compound 3: \( R_2C=CR_2 \)**
- This has two alkyl groups on both sides of the double bond, allowing for the formation of a tertiary carbocation, which is the most stable.

- **Compound 4: \( CH_2=CH_2 \)**
- This is ethylene with no alkyl groups. It will form a primary carbocation, which is the least stable.

**Hint for Step 2:** Identify the type of carbocation that can form from each compound (primary, secondary, tertiary) based on the number of alkyl groups.

**Step 3: Determine the Reactivity Order**
- The stability of carbocations dictates the reactivity:
- \( R_2C=CR_2 \) (most stable, tertiary) > \( R_2C=CHR \) (secondary) > \( RCH=CHR \) (secondary) > \( CH_2=CH_2 \) (least stable, primary).

**Hint for Step 3:** Remember that tertiary carbocations are more stable than secondary, which are more stable than primary.

**Step 4: Write the Increasing Order of Reactivity**
- Based on the stability of the carbocations formed, the increasing order of reactivity towards the addition of HBr is:
1. \( CH_2=CH_2 \) (least reactive)
2. \( RCH=CHR \)
3. \( R_2C=CHR \)
4. \( R_2C=CR_2 \) (most reactive)

### Final Answer:
The increasing order of reactivity towards the addition of HBr is:
\[ CH_2=CH_2 < RCH=CHR < R_2C=CHR < R_2C=CR_2 \]