Predict the major products, `P_1` and `P_2` in the following two reactions
`Me_2CHBroverset(CH_3COO^(-))rarrP_1`
`CH_3(CH_2)_15CH_2CH_2Broverset((CH_3)_3CO^(-))rarrP_2`

Options

A. `P_1" " is" " Me_2CHOCOCH_3,P_2" "is" " CH_3(CH_2)_15CH_2CH_2OCMe_3`
B. `P_1" " is" " Me_2CHOCOCH_3,P_2" "is" " CH_3(CH_2)_15CH=CH_2`
C. `P_1" "is" "CH_3CH=CH_2,P_2" "is" "CH_3(CH_2)_15CH_2CH_2OCMe_3`
D. `P_1 " " is" "CH_3CH=CH_2,P_2" "is" "CH_3(CH_2)_15CH=CH_2`

Explanation

To predict the major products \( P_1 \) and \( P_2 \) from the given reactions, we need to analyze each reaction step by step.

### Reaction 1: \( \text{Me}_2\text{CHBr} + \text{CH}_3\text{COO}^- \rightarrow P_1 \)

1. **Identify the Reactants**: We have \( \text{Me}_2\text{CHBr} \) (which is 2-bromopropane) and \( \text{CH}_3\text{COO}^- \) (acetate ion).

2. **Determine the Type of Reaction**: The acetate ion is a weak base and will act as a nucleophile. In this case, it will perform a nucleophilic substitution reaction.

3. **Mechanism of Reaction**: The acetate ion will attack the carbon atom bonded to the bromine atom (the leaving group).

4. **Formation of Product**: The bromine will leave, and the acetate will bond to the carbon, resulting in the formation of:
\[ P_1 = \text{Me}_2\text{C}(\text{OOCCH}_3)\text{H} \]
This product can be represented as:
\[ P_1 = \text{Me}_2\text{C}(\text{OOCCH}_3)\text{H} \quad \text{(2-acetoxypropane)} \]

### Reaction 2: \( \text{CH}_3(\text{CH}_2)_{15}\text{CH}_2\text{Br} + (\text{CH}_3)_3\text{CO}^- \rightarrow P_2 \)

1. **Identify the Reactants**: We have a long-chain alkyl bromide \( \text{CH}_3(\text{CH}_2)_{15}\text{CH}_2\text{Br} \) and a strong base \( (\text{CH}_3)_3\text{CO}^- \) (tert-butoxide ion).

2. **Determine the Type of Reaction**: The tert-butoxide ion is a strong base and will favor elimination over substitution due to steric hindrance around the carbon bonded to bromine.

3. **Mechanism of Reaction**: The strong base will abstract a proton from the carbon adjacent to the carbon bonded to bromine, leading to the formation of a double bond (alkene) and the elimination of bromine.

4. **Formation of Product**: The elimination will result in the formation of:
\[ P_2 = \text{CH}_3(\text{CH}_2)_{15}\text{CH}=\text{CH}_2 \]
This product can be represented as:
\[ P_2 = \text{1-Hexadecene} \]

### Summary of Products
- \( P_1 = \text{2-acetoxypropane} \)
- \( P_2 = \text{1-Hexadecene} \)