Question

The Maximum number of RBr producing 2-methylbutane by above sequence of reaction

• A. 3
• B. 5
• C. 4
• D. 6

Hint:

When working with Grignard reagents, consider all possible structural isomers that could lead to the desired product, taking into account the reaction conditions and the nature of the alkyl halides.

Answer 1

The Correct Option is A

Step 1: Possible alkyl groups with 4 carbon atoms. 
We can form different structural isomers of butyl groups (C₄H₉–) as the alkyl part \( R \). There are four distinct isomeric forms:

1. n-Butyl bromide \[ CH_3 - CH_2 - CH_2 - CH_2Br \]
2. sec-Butyl bromide \[ CH_3 - CHBr - CH_2 - CH_3 \]
3. iso-Butyl bromide \[ (CH_3)_2CH - CH_2Br \]
4. tert-Butyl bromide \[ (C H_3)_3CBr \]

Each of these has a unique arrangement of carbon atoms and bromine, leading to different structures and physical properties.

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Step 2: Total number of possible structures

\[ \boxed{\text{Total number of possible structures of } RBr = 4} \] 

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Final Answer:

\[ \boxed{4 \text{ isomeric structures of } RBr \text{ (C}_4H_9Br)} \]

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Quick Concept:

For alkyl halides \( R–X \), the number of possible structures corresponds to the number of structural isomers of the alkyl group \( R \). For butyl bromide, these are: n-butyl, sec-butyl, iso-butyl, and tert-butyl bromide.

Answer 2

The given reaction involves the alkylation of a halide (RBr) with magnesium in dry ether to form a Grignard reagent, which then reacts with water to produce 2-methylbutane. Considering the structural isomers of RBr, the maximum number of isomers that can produce 2-methylbutane in this reaction is 3.
The possible isomers of RBr that would produce 2-methylbutane are:
- 1-Bromo-3-methylbutane
- 2-Bromo-2-methylbutane
- 3-Bromo-2-methylbutane
Hence, the maximum number of RBr producing 2-methylbutane is 3.