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Chapter 4: Problem 102

A dextrorotatory alkyl halide is hydrolysed under \(\mathrm{S}_{\mathrm{N}} 2\) conditions to form the corresponding alcohol. The resulting alcohol (a) will be levorotary (b) will be dextrorotatory (c) will be optically inactive due to racemization (d) may be dextro or laevorotatory

Short Answer

Expert verified
Answer: Levorotary

Step by step solution

01

Recall the SN2 mechanism

The SN2 (Substitution Nucleophilic Bimolecular) reaction involves the direct attack of a nucleophile (such as water or hydroxide) on a chiral center, leading to the inversion of its stereochemistry. The nucleophile attacks from the opposite side of the leaving group (halide) and the leaving group departs, resulting in a 180-degree flip of the configuration.
02

Determine the stereochemistry of the initial dextrorotatory alkyl halide

A dextrorotatory compound has a positive optical rotation, meaning it rotates plane-polarized light in a clockwise direction. The initial alkyl halide has a chiral center that results in this dextrorotatory orientation.
03

Predict the stereochemistry of the resulting alcohol after the SN2 reaction

Since the SN2 reaction involves the attack of the nucleophile from the opposite side of the leaving group and results in the inversion of the stereochemistry, the resulting alcohol will have the opposite optical rotation from the initial alkyl halide. Since the initial alkyl halide was dextrorotatory (positive optical rotation), the resulting alcohol will be levorotatory (negative optical rotation). Answer: (a) The alcohol resulting from the hydrolysis of a dextrorotatory alkyl halide under SN2 conditions will be levorotary.

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Most popular questions from this chapter

Each question in this section has four suggested answers of which ONE OR MORE answers will be correct. Which of the following will give yellow precipitate when heated with powdered iodine and aqueous \(\mathrm{NaOH}\) ? (a) 2 -Butanol (b) 2-Methyl-2- propanol (c) 2-Pentanone (d) 3 -pentanol

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\(\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{CH}_{3}+\mathrm{Cl}_{2} \stackrel{800 \mathrm{~K}}{\longrightarrow} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{Cl}+\mathrm{HCl}\). This reaction takes place mainly by (a) \(\mathrm{S}_{\mathrm{N}} 1\) mechanism (b) \(\mathrm{S}_{\mathrm{N}} 2\) mechanism (c) free radical mechanism (d) E1 mechanism

Five alcohols (A) \(-(\mathrm{E})\) are given below (A) \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{4} \mathrm{CH}_{2} \mathrm{OH}\) (B) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{CH}(\mathrm{OH}) \mathrm{C}_{3} \mathrm{H}_{7}\) (C) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}(\mathrm{OH}) \mathrm{C}_{3} \mathrm{H}_{7}\) (D) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}(\mathrm{OH}) \mathrm{CH}\left(\mathrm{CH}_{3}\right)_{2}\) (E) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}\left(\mathrm{CH}_{3}\right)_{2}\) The ease of dehydration in the decreasing order will be (a) \(\mathrm{D}>\mathrm{C}>\mathrm{E}>\mathrm{B}>\mathrm{A}\) (b) \(C>D>B>A>E\) (c) \(\mathrm{D}>\mathrm{C}>\mathrm{A}>\mathrm{B}>\mathrm{E}\) (d) \(\mathrm{D}>\mathrm{E}>\mathrm{B}>\mathrm{A}>\mathrm{C}\)
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