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Which of the following compounds can be oxidised to the corresponding carbonyl compound with pyridininum chlorochromate (PCC)? (A)Propan-2-ol (B) Cyclohexanol (C) Acetaldehyde (D) 2-Methylpropan-2-ol

Short Answer

Expert verified
The compounds (A) Propan-2-ol and (B) Cyclohexanol can be oxidized to their corresponding carbonyl compounds with pyridinium chlorochromate (PCC).

Step by step solution

01

Identify the functional group of each compound

First, let's identify the functional group present in each of the given compounds: (A) Propan-2-ol: Secondary alcohol (B) Cyclohexanol: Primary alcohol (C) Acetaldehyde: Aldehyde (D) 2-Methylpropan-2-ol: Tertiary alcohol
02

Consider the effect of PCC on the functional groups

Now we evaluate the ability of PCC to oxidize the given functional groups: - PCC can oxidize primary alcohols to aldehydes. - PCC can oxidize secondary alcohols to ketones. - PCC does not have the ability to further oxidize aldehydes. - PCC does not oxidize tertiary alcohols since there is no hydrogen on the carbon that the PCC can remove.
03

Apply the understanding of PCC to the given compounds

Based on our understanding of the reactivity of PCC with different functional groups, let's evaluate each compound: (A) Propan-2-ol: This is a secondary alcohol, which means that PCC can oxidize it to a ketone (the corresponding carbonyl compound). (B) Cyclohexanol: This is a primary alcohol, which means that PCC can oxidize it to an aldehyde (the corresponding carbonyl compound). (C) Acetaldehyde: This is already an aldehyde, and as mentioned earlier, PCC does not have the ability to further oxidize aldehydes. (D) 2-Methylpropan-2-ol: This is a tertiary alcohol, which as mentioned earlier, PCC does not oxidize.
04

Determine the correct answer

Based on our analysis, compounds (A) Propan-2-ol and (B) Cyclohexanol can be oxidized to their corresponding carbonyl compounds by PCC. Thus, the correct answer is (A) and (B).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

PCC Oxidation
Pyridinium chlorochromate, or PCC, plays a specific role in organic chemistry as an oxidizing agent. It is particularly useful when working with alcohols. PCC is unique because it provides a method to selectively oxidize \( \text{primary} \) and \( \text{secondary alcohols} \) to their respective carbonyl compounds, which are aldehydes and ketones.

One key advantage of PCC is its ability to achieve this transformation without further oxidizing the resulting compound. For example, it can oxidize a primary alcohol to an aldehyde without pushing the reaction towards a carboxylic acid. Similarly, it oxidizes secondary alcohols to ketones without proceeding further.
  • PCC is not effective in oxidizing aldehydes to carboxylic acids, making it a mild and selective oxidizing agent.
  • Unlike other oxidizing agents, PCC is not capable of oxidizing tertiary alcohols due to the lack of a hydrogen atom on the carbon.
This selectivity makes PCC a valuable reagent in synthetic organic chemistry, particularly when precision is essential.
Functional Groups
In organic chemistry, functional groups are specific groups of atoms within molecules responsible for the characteristic chemical reactions of those molecules. Identifying the functional groups is crucial because it allows chemists to predict the reactivity and properties of compounds.

For example, alcohols contain the hydroxyl group \(-OH\) and can be further categorized based on the carbon atom to which this group is attached. In the context of PCC oxidation, we focus on primary, secondary, and tertiary alcohols.
  • Primary alcohols have the \(-OH\) group connected to a carbon atom that is bonded to only one other carbon atom.
  • Secondary alcohols have the \(-OH\) group connected to a carbon atom bonded to two other carbon atoms.
  • Tertiary alcohols have the \(-OH\) group connected to a carbon that is attached to three other carbon atoms.
Each type of alcohol undergoes oxidation differently, which is why identifying functional groups is a fundamental skill in organic chemistry.
Secondary Alcohols
Secondary alcohols are a type of alcohol where the hydroxyl group is attached to a carbon atom that is connected to two other carbon atoms. This configuration allows secondary alcohols to be oxidized to ketones.

With PCC as the oxidizing agent, the hydrogen atom from the hydroxyl group and one hydrogen atom from the carbon atom are removed. This removal results in the formation of a ketone. Unlike primary alcohols, secondary alcohols do not further oxidize beyond this point with PCC.
  • The oxidation process involves converting the \(-OH\) group to a carbonyl group \( (\text{C}=\text{O}) \).
  • The lack of further oxidation ensures that only a one-step transformation to the ketone occurs.
This precision is important for creating specific molecules in laboratory settings where secondary alcohols are involved.
Primary Alcohols
Primary alcohols have the hydroxyl group \(-OH\) attached to a carbon atom that is only bonded to one other carbon atom. When primary alcohols undergo oxidation with PCC, they form aldehydes.

The transformation takes place through the removal of two hydrogen atoms: one from the hydroxyl group and one from the carbon atom. This results in the creation of a carbonyl group \(\text{(C=O)}\), which is characteristic of the aldehyde functional group.
  • PCC allows primary alcohols to convert to aldehydes without further oxidation to carboxylic acids, which could occur using stronger oxidizing agents.
  • The reaction conditions for PCC are mild, suitable for preventing over-oxidation of sensitive alcohol structures.
This selective oxidation makes PCC particularly useful in maintaining the integrity of the desired product in synthetic routes.

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Most popular questions from this chapter

Ninhydrin when undergoes hydration, which carbonyl carbon is first attacked by \(\mathrm{H}_{2} \mathrm{O}\) ? (A) 1 (B) 2 (C) 3 (D) None

An organic compound \(\mathrm{B}\) is formed by the reaction of ethyl magnesium iodide \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{MgI}\right)\) with a substance \(\mathrm{A}\), followed by treatment with a dilute aqueous acid. Compound B doesn't react with PCC or PDC in dichloromethane. Which of the following is the most suitable as A? (A) \(\mathrm{CH}_{3} \mathrm{CHO}\) (B) \(\mathrm{HCHO}\) (C) \(\mathrm{H}_{2} \mathrm{C}-\mathrm{CH}_{2}\) (D) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CCH}_{3}\)

Reaction in which amino acid is not obtained as a product is (A) CC(=O)CCCC(N)=O \(\frac{\text { (i) } \mathrm{KOBr} \text { (excess) }}{\text { (ii) } \mathrm{H}_{3} \mathrm{O}^{\oplus}}\) (B) \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{NH} \frac{\text { (i) } \mathrm{KCN}}{\text { (ii) } \mathrm{H}_{2} \mathrm{O} / \mathrm{H}^{\mathrm{q}}}\) (C) (D) \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{O} \frac{\text { (i) } \mathrm{KCN}}{\text { (ii) } \mathrm{H}_{2} \mathrm{O} / \mathrm{H}^{\mathrm{F}}}{\longrightarrow}\)

Generally aldehydes are more reactive than ketones because of : (A) Less steric crowding (B) Hydrogen bonding (C) More electrophilic carbon (D) Both (A) and (C)

Match the following columns: For given reaction scheme If \((\mathrm{H})\) has molecular formula \(\mathrm{C}_{7} \mathrm{H}_{14} \mathrm{O}_{2}\) which on reaction with Red P/HI gives 1 -methyl cyclohexane \(12^{2}\) (J) does not give haloform test (All products are major product). Column-I Column - II (Compounds) (Can be distinguished by) (A) \((\mathrm{H})\) and \((\mathrm{N})\) (P) 2,4 -DNP (B) (I) and (Z) (Q) \(\mathrm{Br}_{2} / \mathrm{H}_{2} \mathrm{O}\) (C) (J) and (M) (R) Iodoform test (D) \((\mathrm{G})\) and \((\mathrm{O})\) (S) Ammoniacal \(\mathrm{AgNO}_{3}\) (T) \(\mathrm{NaHSO}_{3}\) test

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