Which of the following reagents/reactions will convert ‘A’ to ‘B’?

Question

3,3-Dimethyl-2-butanol cannot be prepared by:

(A) Reaction of the given aldehyde with MeMgBr followed by H3O+
(B) Acid-catalysed hydration of the given alkene
(C) Ozonolysis followed by reduction using NaBH4/MeOH
(D) Reduction using LiAlH4/H3O+
(E) Acid-catalysed hydration of the given alkyne using Hg2+/H+
Choose the correct answer from the options given below:

Answer 1

The question asks us to identify which reagent or reaction will convert compound 'A' to compound 'B'. Here, compound 'A' is a terminal alkene, and compound 'B' is an aldehyde. The transformation involves converting the terminal double bond into an aldehyde group.

To achieve this transformation, we can use the following steps:

Hydroboration-Oxidation: This involves the use of \(BH_3\) followed by \(H_2O_2\) and \(–OH\) to convert the terminal alkene to the corresponding alcohol. The mechanism adds the OH group to the less substituted carbon of the alkene (anti-Markovnikov addition), resulting in a primary alcohol.
PCC Oxidation: Pyridinium Chlorochromate (PCC) is used to oxidize the primary alcohol to an aldehyde without further oxidation to a carboxylic acid.

Therefore, the correct reagent/reaction to convert ‘A’ to ‘B’ is:

\(BH_3\), \(H_2O_2\)/\(–OH\) followed by PCC oxidation

Explanation of Incorrect Options:

PCC oxidation alone would not work because it only oxidizes alcohols, not alkenes.
Ozonolysis cleaves the double bond completely, leading to the formation of different products, namely carboxylic acids or ketones, not aldehydes.
HBr, hydrolysis followed by oxidation by \(K_2Cr_2O_7\) would result in the formation of a carboxylic acid from a primary alcohol due to the strong oxidation strength of \(K_2Cr_2O_7\).

Thus, the correct answer is the hydroboration-oxidation sequence, followed by PCC oxidation to get the aldehyde.

Answer 2

The question asks us to identify which reagent or reaction will convert compound 'A' to compound 'B'. Here, compound 'A' is a terminal alkene, and compound 'B' is an aldehyde. The transformation involves converting the terminal double bond into an aldehyde group.

To achieve this transformation, we can use the following steps:

Hydroboration-Oxidation: This involves the use of \(BH_3\) followed by \(H_2O_2\) and \(–OH\) to convert the terminal alkene to the corresponding alcohol. The mechanism adds the OH group to the less substituted carbon of the alkene (anti-Markovnikov addition), resulting in a primary alcohol.
PCC Oxidation: Pyridinium Chlorochromate (PCC) is used to oxidize the primary alcohol to an aldehyde without further oxidation to a carboxylic acid.

Therefore, the correct reagent/reaction to convert ‘A’ to ‘B’ is:

\(BH_3\), \(H_2O_2\)/\(–OH\) followed by PCC oxidation

Explanation of Incorrect Options:

PCC oxidation alone would not work because it only oxidizes alcohols, not alkenes.
Ozonolysis cleaves the double bond completely, leading to the formation of different products, namely carboxylic acids or ketones, not aldehydes.
HBr, hydrolysis followed by oxidation by \(K_2Cr_2O_7\) would result in the formation of a carboxylic acid from a primary alcohol due to the strong oxidation strength of \(K_2Cr_2O_7\).

Thus, the correct answer is the hydroboration-oxidation sequence, followed by PCC oxidation to get the aldehyde.

Which of the following reagents/reactions will convert ‘A’ to ‘B’?

The Correct Option is C

Solution and Explanation

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