The mechanism, leaving group and stereochemistry, of the above-shown reaction are, respectively:

S_{N} 2, C l^{-}

, inversion

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S_{N} 2, B r^{-}

, inversion

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S_{N} 1, B r^{-}

, racemisation

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E 2, B r^{-}

, anti elimination

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Solution

The correct option is D $S_{N} 2, B r^{-}$ , inversion
NaSH provides $S H^{-}$ which acts as nucleophile and attacks at carbon attached with Br from the backside (as bromide is a better leaving group than chloride). So, the mechanism is $S_{N} 2$ and inversion will take place.

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Similar questions

S_{N} 2

S_{N} 1

S_{N} 2

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

Which reaction follows inversion of configuration SN1 or SN2?

S_{N} 1, S_{N} 2

E_{1}, E_{2}

Predict the order of reactivity of the following compounds in SN1 and SN2 reactions

C6H5CH2Br , C6H5CH(C6H5)Br , C6H5CH(CH3)Br , C6H5C(CH3)(C6H5)

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