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1. The S_N2 reaction of 1-chloro-3-methylbutane with sodium methoxide is relatively slow, but can be accelerated by the addition of a small amount of NaI. How is this catalysis best explained?

The NaI changes the mechanism to SN1

SN2 reaction of iodide ion converts the alkyl chloride to the more reactive alkyl iodide

The iodide anion activates the methoxide nucleophile

The sodium cation helps pull off the chloride anion

The addition of NaI in the reaction converts the alkyl chloride to a more reactive alkyl iodide. This is because the iodide ion is a better leaving group than the chloride ion, making the alkyl iodide more susceptible to nucleophilic attack. Therefore, the SN2 reaction is accelerated by the presence of NaI.

Explanation

The addition of NaI in the reaction converts the alkyl chloride to a more reactive alkyl iodide. This is because the iodide ion is a better leaving group than the chloride ion, making the alkyl iodide more susceptible to nucleophilic attack. Therefore, the SN2 reaction is accelerated by the presence of NaI.

2. Which reagent(s) would best accomplish the following transformation?

(i) HOBr (ii) Mg in ether

(i) B2H6 in ether (ii) H2O2 and base

(i) HgSO4 in H2O (ii) NaBH4

H3O+ & heat

The given transformation involves the addition of a hydroxyl group (-OH) to a double bond. This is a typical reaction of hydroboration-oxidation, where the first step is the addition of borane (B2H6) in ether to the double bond, forming a boron intermediate. In the second step, the boron intermediate is oxidized by H2O2 and base, resulting in the formation of the hydroxyl group. Therefore, the reagents B2H6 in ether and H2O2 with base would best accomplish this transformation.

Explanation

The given transformation involves the addition of a hydroxyl group (-OH) to a double bond. This is a typical reaction of hydroboration-oxidation, where the first step is the addition of borane (B2H6) in ether to the double bond, forming a boron intermediate. In the second step, the boron intermediate is oxidized by H2O2 and base, resulting in the formation of the hydroxyl group. Therefore, the reagents B2H6 in ether and H2O2 with base would best accomplish this transformation.

3. Addition of HBr to 2,3-dimethyl-1,3-cyclohexadiene may occur in the absence or presence of peroxides. In each case two isomeric C₈H₁₃Br products are obtained. Which of the following is a common product from both reactions?

B

D

A

C

The common product obtained from both reactions is B.

Explanation

The common product obtained from both reactions is B.

4. Which of the following will be the kinetically favored product from the depicted reaction ?

B

D

A

C

5. A C₈H₁₂ chiral hydrocarbon, X, is reduced by catalytic hydrogenation to an achiral C₈H₁₄ compound. Ozonolysis of X (H₂O₂ workup) gave a chiral C₈H₁₂O₄ dicarboxylic acid. Which of the following could be X ?

A

D

B

C

6. A C₈H₁₄ hydrocarbon (X) is reduced by sodium in liquid ammonia to a single C₈H₁₆ product (Y). Both of these compounds undergo hydrogenation (Pt catalyst) to give 2,5-dimethylhexane. Ozonolysis of Y with an oxidative workup produces a single C₄H₈O₂ carboxylic acid. Reaction of Y with perbenzoic acid (C₆H₅CO₃H) gives a chiral C₈H₁₄O product, but reaction with bromine gives an achiral C₈H₁₄Br₂ product What are X and Y?

X is 2,5-dimethyl-2,4-hexadiene ; Y is cis-2,5-dimethyl-3-hexene.

X is 2.5-dimethyl-1,5-hexadiene ; Y is 2,5-dimethyl-3-hexyne.

X is 2,5-dimethyl-3-hexyne ; Y is trans-2,5-dimethyl-3-hexene.

X is 2,5-dimethyl-3-hexyne ; Y is cis-2,5-dimethyl-3-hexene.

The given information suggests that X is a C8H14 hydrocarbon that undergoes reduction to form Y, a C8H16 hydrocarbon. Y is then subjected to ozonolysis, which produces a single C4H8O2 carboxylic acid. This implies that Y must contain a double bond, which is cleaved during ozonolysis to form the carboxylic acid. Additionally, Y reacts with perbenzoic acid to form a chiral product, indicating the presence of a stereocenter. However, Y reacts with bromine to form an achiral product, suggesting that the bromine addition occurs in a way that does not generate a stereocenter. The only option that satisfies all of these conditions is X being 2,5-dimethyl-3-hexyne and Y being trans-2,5-dimethyl-3-hexene.

Explanation

The given information suggests that X is a C8H14 hydrocarbon that undergoes reduction to form Y, a C8H16 hydrocarbon. Y is then subjected to ozonolysis, which produces a single C4H8O2 carboxylic acid. This implies that Y must contain a double bond, which is cleaved during ozonolysis to form the carboxylic acid. Additionally, Y reacts with perbenzoic acid to form a chiral product, indicating the presence of a stereocenter. However, Y reacts with bromine to form an achiral product, suggesting that the bromine addition occurs in a way that does not generate a stereocenter. The only option that satisfies all of these conditions is X being 2,5-dimethyl-3-hexyne and Y being trans-2,5-dimethyl-3-hexene.

7. The structures of two 3º-bicyclic chlorides (I and II) are shown below.

II is more reactive than I for both substitution and eleimination

I is more reactive than II for both substitution and eleimination

On treatment with KOH in ethanol, I undergoes substitution and II undergoes eleimination.

On treatment with KOH in ethanol, both compounds undergo E2 elimination.

8. Which of the following reagents would not effect the following transformation?

PCl3

SOCl2

HCl & ZnCl2

KCl (5 molar solution)

KCl (5 molar solution) would not affect the given transformation because it is a neutral salt and does not possess any reactive or transforming properties. The other reagents, PCl3, SOCl2, and HCl & ZnCl2, are all known to be reactive and can bring about various chemical transformations. However, KCl (5 molar solution) does not possess any such reactivity and would not cause any change in the given transformation.

Explanation

KCl (5 molar solution) would not affect the given transformation because it is a neutral salt and does not possess any reactive or transforming properties. The other reagents, PCl3, SOCl2, and HCl & ZnCl2, are all known to be reactive and can bring about various chemical transformations. However, KCl (5 molar solution) does not possess any such reactivity and would not cause any change in the given transformation.

9. Which of the following represents the most stable conformation of all cis-1,2,4-trimethylcyclohexane?

D

B

A

C

The most stable conformation of cis-1,2,4-trimethylcyclohexane is represented by option A. This is because in this conformation, the bulky methyl groups are in equatorial positions, which reduces steric hindrance. The axial positions are occupied by the smaller hydrogen atoms, resulting in a more stable and energetically favorable conformation.

Explanation

The most stable conformation of cis-1,2,4-trimethylcyclohexane is represented by option A. This is because in this conformation, the bulky methyl groups are in equatorial positions, which reduces steric hindrance. The axial positions are occupied by the smaller hydrogen atoms, resulting in a more stable and energetically favorable conformation.

10. Identify "AC" product in following chart from following answers ?

A

B

C

D

The correct answer is A because it is the only option that matches the given "AC" product in the chart.

Explanation

The correct answer is A because it is the only option that matches the given "AC" product in the chart.

11. Cyclooctatetraene, C₈H₈, appears to be a conjugated annulene. In addition to the planar configuration, three non-planar structures may be considered, all of which are shown below. Which of these is most likely to be the favored conformation? note that D is planar.

A

B

D

C

The most likely favored conformation is C. This is because in order to have conjugation in a molecule, it is favorable for the p-orbitals to overlap and align in a planar configuration. The other structures shown are non-planar, which would disrupt the conjugation and make it less stable. Therefore, C, which is also planar, is the most likely favored conformation for cyclooctatetraene.

Explanation

The most likely favored conformation is C. This is because in order to have conjugation in a molecule, it is favorable for the p-orbitals to overlap and align in a planar configuration. The other structures shown are non-planar, which would disrupt the conjugation and make it less stable. Therefore, C, which is also planar, is the most likely favored conformation for cyclooctatetraene.

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