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HALOALKANES AND HALOARENES

91101 The major product of the following reaction is
original image

1

{CH}_{3} - CHBr - {CH}_{2} Br

2

{CH}_{2} = CH - {CH}_{2} Br

3

{CH}_{3} - CHBr - {CH}_{2} - OH

4

{CH}_{3} - CHOH - {CH}_{2} OH

Explanation:

Alkene react with excess of hydrogen bromide $(HBr)$ to form major product 1,2-dibromo propane.
In the reaction $- Br$ added in the place of double bond and $- OH$ is removed for $- Br$ group.
original image

**[Kerala CEE-29.08.2021]**

HALOALKANES AND HALOARENES

91103 For the following reactions :
original image
Where

1

μ_{B} > μ_{A}

and

k_{e} (B) > k_{e}

(A)

2

μ_{B} > μ_{A}

and

k_{e} (A) > k_{e}

(B)

3

μ_{A} > μ_{B}

and

k_{e} (A) > k_{e}

(B)

4

μ_{A} > μ_{B}

and

k_{e} (B) > k_{e} (A)

Explanation:

original image
is bulky group, therefore it cannot attack
$∴$ it will act like base.
$k_{s} << k_{e}$
$μ_{B}$ will be very small
(A) ${CH}_{3} {CH}_{2} O^{-}$ will act like nucleophile as
$\begin{aligned} ∴ μ_{A} will be large \\ ∴ μ_{A} > μ_{B} and k_{e} (B) > k_{e} (A) \end{aligned}$

7 Jan Shift-I]**

HALOALKANES AND HALOARENES

91098 Identify the major product of the following reaction
original image

1

2

3

4

Shift-I]**

HALOALKANES AND HALOARENES

91100 The major product (A) formed in the reaction given below is:
original image

1

2

3

4

**[JEE MAINS 2021

HALOALKANES AND HALOARENES

91104 The major product of the following reaction is:
original image

1

2

3

4

9 Jan Shift-I]**

HALOALKANES AND HALOARENES

91101 The major product of the following reaction is
original image

1

{CH}_{3} - CHBr - {CH}_{2} Br

2

{CH}_{2} = CH - {CH}_{2} Br

3

{CH}_{3} - CHBr - {CH}_{2} - OH

4

{CH}_{3} - CHOH - {CH}_{2} OH

Explanation:

Alkene react with excess of hydrogen bromide $(HBr)$ to form major product 1,2-dibromo propane.
In the reaction $- Br$ added in the place of double bond and $- OH$ is removed for $- Br$ group.
original image

**[Kerala CEE-29.08.2021]**

HALOALKANES AND HALOARENES

91103 For the following reactions :
original image
Where

1

μ_{B} > μ_{A}

and

k_{e} (B) > k_{e}

(A)

2

μ_{B} > μ_{A}

and

k_{e} (A) > k_{e}

(B)

3

μ_{A} > μ_{B}

and

k_{e} (A) > k_{e}

(B)

4

μ_{A} > μ_{B}

and

k_{e} (B) > k_{e} (A)

Explanation:

original image
is bulky group, therefore it cannot attack
$∴$ it will act like base.
$k_{s} << k_{e}$
$μ_{B}$ will be very small
(A) ${CH}_{3} {CH}_{2} O^{-}$ will act like nucleophile as
$\begin{aligned} ∴ μ_{A} will be large \\ ∴ μ_{A} > μ_{B} and k_{e} (B) > k_{e} (A) \end{aligned}$

7 Jan Shift-I]**

HALOALKANES AND HALOARENES

91098 Identify the major product of the following reaction
original image

1

2

3

4

Shift-I]**

HALOALKANES AND HALOARENES

91100 The major product (A) formed in the reaction given below is:
original image

1

2

3

4

**[JEE MAINS 2021

HALOALKANES AND HALOARENES

91104 The major product of the following reaction is:
original image

1

2

3

4

9 Jan Shift-I]**

HALOALKANES AND HALOARENES

91101 The major product of the following reaction is
original image

1

{CH}_{3} - CHBr - {CH}_{2} Br

2

{CH}_{2} = CH - {CH}_{2} Br

3

{CH}_{3} - CHBr - {CH}_{2} - OH

4

{CH}_{3} - CHOH - {CH}_{2} OH

Explanation:

Alkene react with excess of hydrogen bromide $(HBr)$ to form major product 1,2-dibromo propane.
In the reaction $- Br$ added in the place of double bond and $- OH$ is removed for $- Br$ group.
original image

**[Kerala CEE-29.08.2021]**

HALOALKANES AND HALOARENES

91103 For the following reactions :
original image
Where

1

μ_{B} > μ_{A}

and

k_{e} (B) > k_{e}

(A)

2

μ_{B} > μ_{A}

and

k_{e} (A) > k_{e}

(B)

3

μ_{A} > μ_{B}

and

k_{e} (A) > k_{e}

(B)

4

μ_{A} > μ_{B}

and

k_{e} (B) > k_{e} (A)

Explanation:

original image
is bulky group, therefore it cannot attack
$∴$ it will act like base.
$k_{s} << k_{e}$
$μ_{B}$ will be very small
(A) ${CH}_{3} {CH}_{2} O^{-}$ will act like nucleophile as
$\begin{aligned} ∴ μ_{A} will be large \\ ∴ μ_{A} > μ_{B} and k_{e} (B) > k_{e} (A) \end{aligned}$

7 Jan Shift-I]**

HALOALKANES AND HALOARENES

91098 Identify the major product of the following reaction
original image

1

2

3

4

Shift-I]**

HALOALKANES AND HALOARENES

91100 The major product (A) formed in the reaction given below is:
original image

1

2

3

4

**[JEE MAINS 2021

HALOALKANES AND HALOARENES

91104 The major product of the following reaction is:
original image

1

2

3

4

9 Jan Shift-I]**

HALOALKANES AND HALOARENES

91101 The major product of the following reaction is
original image

1

{CH}_{3} - CHBr - {CH}_{2} Br

2

{CH}_{2} = CH - {CH}_{2} Br

3

{CH}_{3} - CHBr - {CH}_{2} - OH

4

{CH}_{3} - CHOH - {CH}_{2} OH

Explanation:

Alkene react with excess of hydrogen bromide $(HBr)$ to form major product 1,2-dibromo propane.
In the reaction $- Br$ added in the place of double bond and $- OH$ is removed for $- Br$ group.
original image

**[Kerala CEE-29.08.2021]**

HALOALKANES AND HALOARENES

91103 For the following reactions :
original image
Where

1

μ_{B} > μ_{A}

and

k_{e} (B) > k_{e}

(A)

2

μ_{B} > μ_{A}

and

k_{e} (A) > k_{e}

(B)

3

μ_{A} > μ_{B}

and

k_{e} (A) > k_{e}

(B)

4

μ_{A} > μ_{B}

and

k_{e} (B) > k_{e} (A)

Explanation:

original image
is bulky group, therefore it cannot attack
$∴$ it will act like base.
$k_{s} << k_{e}$
$μ_{B}$ will be very small
(A) ${CH}_{3} {CH}_{2} O^{-}$ will act like nucleophile as
$\begin{aligned} ∴ μ_{A} will be large \\ ∴ μ_{A} > μ_{B} and k_{e} (B) > k_{e} (A) \end{aligned}$

7 Jan Shift-I]**

HALOALKANES AND HALOARENES

91098 Identify the major product of the following reaction
original image

1

2

3

4

Shift-I]**

HALOALKANES AND HALOARENES

91100 The major product (A) formed in the reaction given below is:
original image

1

2

3

4

**[JEE MAINS 2021

HALOALKANES AND HALOARENES

91104 The major product of the following reaction is:
original image

1

2

3

4

9 Jan Shift-I]**

HALOALKANES AND HALOARENES

91101 The major product of the following reaction is
original image

1

{CH}_{3} - CHBr - {CH}_{2} Br

2

{CH}_{2} = CH - {CH}_{2} Br

3

{CH}_{3} - CHBr - {CH}_{2} - OH

4

{CH}_{3} - CHOH - {CH}_{2} OH

Explanation:

Alkene react with excess of hydrogen bromide $(HBr)$ to form major product 1,2-dibromo propane.
In the reaction $- Br$ added in the place of double bond and $- OH$ is removed for $- Br$ group.
original image

**[Kerala CEE-29.08.2021]**

HALOALKANES AND HALOARENES

91103 For the following reactions :
original image
Where

1

μ_{B} > μ_{A}

and

k_{e} (B) > k_{e}

(A)

2

μ_{B} > μ_{A}

and

k_{e} (A) > k_{e}

(B)

3

μ_{A} > μ_{B}

and

k_{e} (A) > k_{e}

(B)

4

μ_{A} > μ_{B}

and

k_{e} (B) > k_{e} (A)

Explanation:

original image
is bulky group, therefore it cannot attack
$∴$ it will act like base.
$k_{s} << k_{e}$
$μ_{B}$ will be very small
(A) ${CH}_{3} {CH}_{2} O^{-}$ will act like nucleophile as
$\begin{aligned} ∴ μ_{A} will be large \\ ∴ μ_{A} > μ_{B} and k_{e} (B) > k_{e} (A) \end{aligned}$

7 Jan Shift-I]**

HALOALKANES AND HALOARENES

91098 Identify the major product of the following reaction
original image

1

2

3

4

Shift-I]**

HALOALKANES AND HALOARENES

91100 The major product (A) formed in the reaction given below is:
original image

1

2

3

4

**[JEE MAINS 2021

HALOALKANES AND HALOARENES

91104 The major product of the following reaction is:
original image

1

2

3

4

9 Jan Shift-I]**

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