What is the major product obtained in the bromination of #(CH_3)_(2)CHCH_2CH_3#?

Answer 1

(CH₃)₃CBrCH₃CH₃ is the main product obtained from the bromination of (CH₃)₃CHCH₃CH₃.

Four products are obtained from the free-radical bromination of this alkane:

1-(bromo-3-methylbutane) 2-(bromo-3-methylbutane) 2-(bromo-2-methylbutane) (A, B, C, D)

Nevertheless, the regions that bromine targets are extremely selective.

The various forms of H atoms have relative attack rates of 3°:2°:1° = 1640:82:1.

We multiply the numbers of each type of atom by the relative reactivity to get the relative amounts of each product.

(1) A: (1°): 3 × 1 = 3; (2) B: (2°): 2 × 82 = 164; (3) C: (3°): 1 × 1640 = 1640; and (1°) D: (6 × 1 = 6

Each isomer's percentages are

A: #3/1813× 100 % = 0.1 %#
B: #164/1813 × 100 % = 9.0 %#
C: #1640/1813 × 100 % = 90.5 %#
D: #6/1813 × 100 % = 0.3 %#

Therefore, C, 2-bromo-2-methylbutane, is the main product.

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Answer 2

The major product obtained in the bromination of (CH₃)₂CHCH₂CH₃ is 2-bromo-2-methylbutane.

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Answer 3

The major product obtained in the bromination of ((CH_3)_2CHCH_2CH_3) is 2-bromo-2-methylbutane.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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