The enantiopure bromide shown reacts by S_N2 with various nucleophiles in aprotic solvents. Select all correct statements.

With one equivalent of NaI in acetone, the first-formed product is (R)-2-iodobutane

With a large excess of NaI and a long reaction time, the iodide product becomes racemic

With one equivalent of NaI in acetone, the first-formed product is (R)-2-iodobutane

With a large excess of NaI and a long reaction time, the iodide product becomes racemic

With NaSH, the product is (S)-butane-2-thiol

Solvolysis in hot aqueous ethanol gives the alcohol with more than 95 percent retention

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The enantiopure bromide shown reacts by $S X_{N} 2$ with various nucleophiles in aprotic solvents. Select all correct statements.

AWith one equivalent of $NaI$ in acetone, the first-formed product is (R)-2-iodobutane
BWith a large excess of $NaI$ and a long reaction time, the iodide product becomes racemic
CWith $NaSH$ , the product is (S)-butane-2-thiol
DSolvolysis in hot aqueous ethanol gives the alcohol with more than 95 percent retention

Answer: A, B

The compound is (S)-2-bromobutane. Iodide gives (R)-2-iodobutane (inversion, labels checked), but iodide also displaces iodide, so with excess

NaI

and long reaction time the product is racemised.

NaSH

gives the thiol with inverted arrangement, which is R (not S), and solvolysis in hot aqueous ethanol is not 100 percent retention. Correct: first two.

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The enantiopure bromide shown reacts by SXN​2 with various nucleophiles in aprotic solvents. Select all correct statements.

Answer: A, B

More questions on Stereochemical outcome of substitution on drawn chiral substrates

The enantiopure bromide shown reacts by $S X_{N} 2$ with various nucleophiles in aprotic solvents. Select all correct statements.