Consider the following statements :
(1) Bridgehead halides are inert towards both $S {_{N}}^{1}$ and $S {_{N}}^{2}$ reactions (till one of the ring size is eight member ring)
(2) The first step in both $S {_{N}}^{1}$ and $E_{1}$
(3) $S {_{N}}^{2}$ reaction proceed with total retention of configuration
(4) $E_{2}$ elemnations are by use of asolvent of low polarity and high concentration of a strong base
Which og the above statements are correct?

1, 2, and 4

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1 and 3

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2, 3, and 4

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1, 2, 3 and 4

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Solution

The correct option is A 1, 2, and 4
Bridge head halide can't form a stable intermediate hence they can't undergo with both $S N^{1}$ and $S n^{2}$ . The $S N^{1}$ and $E^{1}$ reactions have some first step, leaving of leaving group and $E_{2}$ requires high concentration of bases to occur.

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Similar questions

Energy diagram of $S_{N} 1$ and $S_{N} 2$ reactions The order of hydrolysis of RX by $S N^{1}$ is $3^{\circ} > 2^{\circ} > 1^{\circ}$ RX and $S N^{2}$ path is $1^{\circ} > 2^{\circ} > 3^{\circ}$ RX.

Which of the following statements are correct for the above energy diagrams of $S_{N} 1$ and $S_{N} 2$ reactions?

Q. The first steps of

{S N}^{1}

and

{S N}^{2}

reactions are, respectively:

Consider the reactions

(i) $(C H_{3})_{2} C H - C H_{2} B r C_{2} H_{5} O H - ---- \to (C H_{3})_{2} C H - C H_{2} O C_{2} H_{5} + H B r$

(ii) $(C H_{3})_{2} C H - C H_{2} B r C_{2} H_{S} O^{-} - ---- \to (C H_{3})_{2} C H C H_{2} O C_{2} H_{5} + B r^{-}$

The mechanism of reactions (i) & (ii) are respectively

Q. Which of the following compound readily undergo both

S_{N} 1

and

S_{N} 2

reaction?

Which of the following halides will be most reactive in $S N^{2}$ reaction and $S N^{1}$ reaction, respectively?

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The correct option is A 1, 2, and 4Bridge head halide can't form a stable intermediate hence they can't undergo with both SN1 and Sn2. The SN1 and E1 reactions have some first step, leaving of leaving group and E2 requires high concentration of bases to occur.

The correct option is A 1, 2, and 4
Bridge head halide can't form a stable intermediate hence they can't undergo with both $S N^{1}$ and $S n^{2}$ . The $S N^{1}$ and $E^{1}$ reactions have some first step, leaving of leaving group and $E_{2}$ requires high concentration of bases to occur.