CH3 - CH2 - CH2 - CH3-CH - CH3 product- at room temperature-Choose the correct option

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Byju's Answer

Standard XII

Chemistry

Free Radical Halogenation

CH3 - CH2 - C...

Question

$C H_{3} - C H_{2} - C H_{2} - C H | C H_{3} - C H_{3} C l_{2} - - \to h v$ product, at room temperature.
Choose the correct option

The major product of the reaction is

3^{o}

halide because

3^{o}

free radial is more stable

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The major product is

2^{o}

halide

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The major product is

1^{o}

halide

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all ,

1^{o}, 2^{o}

and

3^{o}

are formed in equal proportions

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Solution

The correct option is A The major product of the reaction is $3^{o}$ halide because $3^{o}$ free radial is more stable
The above reaction takes place in presence of sunlight. So, free radical takes place. The stability of free radical follows the order
$3^{o} > 2^{o} > 1^{o}$
The major product forms will be $3^{o}$ - radical.

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Similar questions

(e)

C H_{3} - C H_{2} - C H_{2} - N H - C H_{3}

N-Methylpropanamine

(2^{o})

(f)

C H_{3} - C H_{2} N H - C H_{2} C H_{3}

N-Ethylethanamine

(2^{o})

(g)

C H_{3} - C H_{3} | C H - N H - C H_{3}

N-Methylpropan-2-amine

(2^{o})

(h)

C H_{3} - C H_{2} - C H_{3} | N - C H_{3}

N, N-Dimethylethanamine

(3^{o})

In the given compound E,F and F,G are :

Q. In the given example B,C and B,D are :

(a)

C H_{3} - C H_{2} - C H_{2} - C H_{2} N H_{2}

Butanamine

(1^{o})

(b)

C H_{3} - C H_{2} - N H_{2} | C H - C H_{3}

Butan-2-amine

(1^{o})

(c)

C H_{3} - C H_{3} | C H - C H_{2} - N H_{2}

2-Methylpropanamine

(1^{o})

(d)

C H_{3} - C H_{3} | C | - C H_{3} N H_{2}

2-methylpropan-2-amine

(1^{o})

(e)

C H_{3} - C H_{2} - C H_{2} - N H - C H_{3}

N-methylpropanamine

(2^{o})

{C H}_{3} - C H = {C H}_{2} + H B r ⟶ ?

The major product formed is:

Q. In the given compound A,C and A,D are exbhits which of the following isomerism :

(a)

C H_{3} - C H_{2} - C H_{2} - C H_{2} N H_{2}

Butanamine

(1^{o})

(b)

C H_{3} - C H_{2} - N H_{2} | C H - C H_{3}

Butan-2-amine

(1^{o})

(c)

C H_{3} - C H_{3} | C H - C H_{2} - N H_{2}

2-Methylpropanamine

(1^{o})

(d)

C H_{3} - C H_{3} | C | - C H_{3} N H_{2}

2-methylpropan-2-amine

(1^{o})

(e)

C H_{3} - C H_{2} - C H_{2} - N H - C H_{3}

N-methylpropanamine

(2^{o})

C H_{3} - C H_{3} | C H - C H_{2} - C H_{3} + B r_{2} h v - \to

P (Major product)

(on monobromination)

The product P is :

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CH3−CH2−CH2−CH|CH3−CH3Cl2−−→hv product, at room temperature.Choose the correct option

The correct option is A The major product of the reaction is 3o halide because 3o free radial is more stableThe above reaction takes place in presence of sunlight. So, free radical takes place. The stability of free radical follows the order3o>2o>1oThe major product forms will be 3o- radical.

$C H_{3} - C H_{2} - C H_{2} - C H | C H_{3} - C H_{3} C l_{2} - - \to h v$ product, at room temperature.
Choose the correct option