The Question is from neet-chemistry hydrocarbons , and it was created by the author THIRUPATHI REDDY

Question

The major product obtained on the monobromination (with $Br_2$ and UV light) of neopentane is:

THIRUPATHI REDDY · Accepted Answer

## CORE INFORMATION
✓ **Answer**: B
## CONCEPT FRAMEWORK
	**Primary Concept**: Free radical halogenation of alkanes
	**Related Topics**:  Reaction mechanisms, stability of free radicals, selectivity
	**Prerequisites**: Understanding of free radical substitution reactions, tertiary, secondary, and primary carbon atoms.
## SOLUTION PATHWAY
**Given Information**:
	Neopentane reacts with $Br_2$ in the presence of UV light (monobromination).
	Neopentane's structure:  $CH_3C(CH_3)_3$
**Method & Steps**:
	1. Initiation: UV light breaks the $Br-Br$ bond to form two bromine radicals ($Br·$).
	2. Propagation:
		(a) A bromine radical abstracts a hydrogen atom from neopentane, forming HBr and a neopentyl radical.  The most stable radical is formed preferentially.  In neopentane, all hydrogens are equivalent and bonded to a tertiary carbon. 
		(b) The neopentyl radical reacts with another $Br_2$ molecule to form the product and a new bromine radical.
	3. Termination:  Radical-radical combinations end the chain reaction.
**Formula/Rule**: The stability of free radicals follows the order: tertiary > secondary > primary > methyl.  The reaction will favor the formation of the most stable radical intermediate.
		The reaction produces 1-bromo-2,2-dimethylpropane.
		Neopentane has only one type of hydrogen atom, which is bonded to a tertiary carbon. Therefore, only one product is possible.

3. Final Step: The major product is 1-bromo-2,2-dimethylpropane because all hydrogens are equivalent and bonded to a tertiary carbon, leading to only one possible brominated product.
		Result: 1-bromo-2,2-dimethylpropane
		Verification: Check the structure of the product and verify that it is indeed a monobrominated derivative of neopentane.
## OPTION ANALYSIS
A :x: 2-Bromo-2-methylbutane is not a possible product from the monobromination of neopentane.
B ✓: 1-Bromo-2,2-dimethylpropane is the correct product. This is because all the hydrogen atoms in neopentane are equivalent and attached to a tertiary carbon atom.  Bromination occurs at one of these tertiary carbon atoms.
C :x: 2-Bromo-3-methylbutane is not a possible product from the monobromination of neopentane.
D :x: 1-Bromo-3-methylbutane is not a possible product from the monobromination of neopentane.
## LEARNING AIDS
	**Common Mistakes**:
		1. Incorrectly identifying the most stable radical:  Understanding the relative stability of free radicals (tertiary > secondary > primary) is crucial for predicting the major product.
		2. Assuming multiple products:  In this specific case, only one product is formed because all hydrogens are equivalent.
	**Quick Tips**:
		1. Draw the structure of the reactant carefully and identify all possible sites of hydrogen abstraction.
		2. Remember that free radical halogenation favors the formation of the most stable free radical intermediate.

THIRUPATHI REDDY · Answer

2-Bromo-2-methylbutane

THIRUPATHI REDDY · Answer

The correct Option is "1-Bromo-2,2-dimethylpropane"

THIRUPATHI REDDY · Answer

2-Bromo-3-methylbutane

THIRUPATHI REDDY · Answer

1-Bromo-3-methylbutane

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