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CH(3)-CH(2)-CH(2)-underset(CH(3))underse...

`CH_(3)-CH_(2)-CH_(2)-underset(CH_(3))underset(|)overset(Br)overset(|)(C)-CH_(2)-CH_(3)`
Total number of `SN_(1) + E_(1)` products obtained will be -

A

5

B

6

C

7

D

8

Text Solution

AI Generated Solution

The correct Answer is:
To determine the total number of \( \text{SN}_1 + \text{E}_1 \) products obtained from the given compound, we will analyze both the \( \text{E}_1 \) and \( \text{SN}_1 \) mechanisms step by step. ### Step 1: Identify the Structure The compound given is: \[ \text{CH}_3-\text{CH}_2-\text{CH}_2-\text{C}(\text{Br})(\text{CH}_3)-\text{CH}_2-\text{CH}_3 \] This structure indicates a tertiary carbon (due to the presence of a bromine atom) which can undergo both \( \text{E}_1 \) and \( \text{SN}_1 \) reactions. ### Step 2: Analyze \( \text{E}_1 \) Mechanism 1. **Formation of Carbocation**: The bromine atom leaves, forming a carbocation: \[ \text{C}^+ \] 2. **Hydrogen Abstraction**: The next step involves the removal of a hydrogen atom from an adjacent carbon to form a double bond (alkene). 3. **Identify Possible Alkenes**: - If we remove a hydrogen from the first \( \text{CH}_2 \) (next to the carbocation), we can form: \[ \text{CH}_3-\text{C}=\text{CH}-\text{CH}_2-\text{CH}_3 \] This can exist as both cis and trans isomers. - If we remove a hydrogen from the second \( \text{CH}_2 \) (next to the carbocation), we can form: \[ \text{CH}_3-\text{CH}_2-\text{C}=\text{CH}-\text{CH}_3 \] This also can exist as both cis and trans isomers. - Removing a hydrogen from the \( \text{CH}_3 \) group adjacent to the carbocation will yield a different alkene: \[ \text{CH}_3-\text{C}=\text{CH}_2-\text{CH}_2-\text{CH}_3 \] This alkene does not have cis/trans isomers due to symmetry. ### Step 3: Count \( \text{E}_1 \) Products - From the first \( \text{CH}_2 \): 2 products (cis and trans) - From the second \( \text{CH}_2 \): 2 products (cis and trans) - From the \( \text{CH}_3 \): 1 product (no cis/trans) Thus, total \( \text{E}_1 \) products = \( 2 + 2 + 1 = 5 \). ### Step 4: Analyze \( \text{SN}_1 \) Mechanism 1. **Formation of Carbocation**: Similar to \( \text{E}_1 \), the bromine leaves, forming a carbocation. 2. **Nucleophilic Attack**: A nucleophile can attack the carbocation from either side, leading to the formation of different stereoisomers. 3. **Chirality**: The carbocation is chiral, which means it can form two stereoisomers (R and S configurations). ### Step 5: Count \( \text{SN}_1 \) Products - The \( \text{SN}_1 \) reaction will yield 2 products (due to the chirality of the carbocation). ### Step 6: Total Products Now, we combine the total products from both mechanisms: \[ \text{Total products} = \text{E}_1 \text{ products} + \text{SN}_1 \text{ products} = 5 + 2 = 7 \] ### Final Answer The total number of \( \text{SN}_1 + \text{E}_1 \) products obtained will be **7**. ---
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