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Match the columns. Column I (a) PCC (Pyridiniumchlorochromate) (b) NBS (N-Bromosuccinimide) (c) \(\mathrm{AlPO}_{4}\) (d) Li/Liquid \(\mathrm{NH}_{3}\) Column II (p) C1=CCc2ccccc2C1 BrC1C=CCc2ccccc21 (q) \(\mathrm{MeC} \equiv \mathrm{CMe} \longrightarrow\) (r) (s) \(\mathrm{CH}_{3} \mathrm{COOH} \longrightarrow \mathrm{CH}_{2}=\mathrm{C}=\mathrm{O}\)

Short Answer

Expert verified
(a) matches (s), (b) matches (p) & (r), (d) matches (q). \\(\mathrm{AlPO}_{4}\\) is not directly matched.

Step by step solution

01

Understand the Reactants in Column I

Each compound in Column I has a specific chemical function:- PCC (Pyridiniumchlorochromate) is an oxidizing agent, commonly used to oxidize alcohols to aldehydes and ketones.- NBS (N-Bromosuccinimide) is typically used for bromination of allylic and benzylic positions.- \(\mathrm{AlPO}_{4}\) is a catalyst used in various organic reactions.- Li/Liquid \(\mathrm{NH}_{3}\) is used in dissolving metal reductions, typically to reduce alkynes to trans-alkenes.
02

Match Reactants from Column I with Products/Transformations in Column II

Analyze the products and transformations given in Column II:- Reaction (q), \(\mathrm{MeC} \equiv \mathrm{CMe} \longrightarrow\), indicates a reduction of an alkyne to an alkene, most likely using Li/Liquid \(\mathrm{NH}_{3}\).- Reaction (s), \(\mathrm{CH}_{3} \mathrm{COOH} \longrightarrow \mathrm{CH}_{2}=\mathrm{C}=\mathrm{O}\), suggests the oxidation of acetic acid to ketene, which corresponds to PCA's oxidation properties.- The SMILES structure \(C1=CCc2ccccc2C1\) is a cyclic hydrocarbon that can undergo allylic bromination to form the second SMILES structure \(BrC1C=CCc2ccccc21\), usually achieved using NBS.So:- (a) PCC matches with reaction \(\mathrm{CH}_{3} \mathrm{COOH} \longrightarrow \mathrm{CH}_{2}=\mathrm{C}=\mathrm{O}\) (s).- (b) NBS matches with the SMILES structures (p) and (r).- (c) \(\mathrm{AlPO}_{4}\) needs an unspecified reaction, so it's not directly matched.- (d) Li/Liquid \(\mathrm{NH}_{3}\) matches with alkyne reduction \(\mathrm{MeC} \equiv \mathrm{CMe} \longrightarrow\) (q).
03

Confirm Reaction Matches

Verify that all compounds from Column I have appropriate chemical reactions that their typical roles suggest:- PCC is correctly matched with an oxidation reaction (s).- NBS is correctly used for allylic bromination between SMILES structures (p) and (r).- Li/Liquid \(\mathrm{NH}_{3}\) is used for alkyne reduction (q).After matching, column \( \mathrm{AlPO}_{4} \) remains unused, suggesting it does not have a reaction or it acts as a catalyst possibly in unspecified parts of the exercise.

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Key Concepts

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

Oxidizing Agents
Oxidizing agents are chemical compounds that facilitate the oxidation of other substances by accepting electrons from them. This process causes the oxidizing agent itself to get reduced. One of the classic examples of an oxidizing agent in organic chemistry is Pyridinium chlorochromate (PCC).

PCC is commonly utilized for the specific transformation of alcohols into aldehydes and ketones without further oxidation to carboxylic acids. This makes it especially valuable when you need to stop the oxidation process at an intermediate step. When acetic acid (\(\mathrm{CH}_{3}\mathrm{COOH}\)) is oxidized by PCC, for example, it forms ketene (\(\mathrm{CH}_{2} = \mathrm{C} = \mathrm{O}\)), an important intermediate in many chemical syntheses.

Understanding the role of oxidizing agents like PCC is crucial:
  • They are integral in transforming functional groups.
  • They often require careful handling due to their reactivity.
  • Choice of oxidizing agent impacts the selectivity and outcome of a reaction.
Allylic Bromination
Allylic bromination is a unique chemical reaction where a hydrogen atom adjacent to a double bond (an allylic position) is replaced by a bromine atom. This reaction is effectively carried out using N-Bromosuccinimide (NBS). The utility of NBS is attributed to its ability to maintain a low concentration of bromine, which helps in selectively attacking the allylic position.

In the given problem, NBS is used to brominate a cyclic hydrocarbon, converting it into an allylic bromide. This transformation indicates the preference for NBS to target the less crowded allylic position, avoiding direct attack on the double bonds themselves. Follow these key points to grasp allylic bromination better:
  • Allylic bromination is typically a radical substitution process.
  • An allylic hydrogen being substituted is key to increasing reactivity in subsequent reactions.
  • NBS ensures selectivity, minimizing unwanted reactions at other parts of the molecule.
  • It allows for transformations in complex organic structures by introducing bromine atoms at strategic positions.
NBS and allylic bromination are pivotal in organic synthesis, enabling the strategic modification of molecular frameworks necessary for further reactions.
Dissolving Metal Reductions
Dissolving metal reductions involve the use of metals like lithium in a liquid ammonia solution to reduce organic compounds. This type of reaction is notably used to convert alkynes into trans-alkenes. The process involves the transfer of electrons from the metal to the substrate, leading to the breaking of \(C\equiv C\) triple bonds and the formation of \(C=C\) double bonds.

For example, Li in liquid \(\mathrm{NH}_3\) is a notable representation of a dissolving metal reduction, turning an alkyne like \(\mathrm{MeC}\equiv\mathrm{CMe}\) into a trans-alkene. This reduces the alkyne while adding hydrogen atoms across the former triple bond.

Understanding dissolving metal reductions involves recognizing:
  • The mechanism usually includes single electron transfer processes, which differs from typical ionic or covalent bond forming reactions.
  • Stereochemistry of the reduction favoring anti-addition, leading predominantly to trans-alkenes.
  • The selectivity that allows specific transformations, useful in multi-step synthetic pathways.
Dissolving metal reductions offer unique opportunities in synthetic organic chemistry, particularly in modifying carbon skeletons to introduce specific stereochemical configurations.

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Most popular questions from this chapter

When 1-phenylpropyne reacts with \(\mathrm{H}_{2} \mathrm{O} / \mathrm{HgSO}_{4} / \mathrm{H}_{2} \mathrm{SO}_{4^{\prime}}\) the major product is propiophenone shown below. This is because (a) Alkyl groups are weak electron donors due to inductive effects and hyperconjugation. (b) Carbonyl groups are electron withdrawing groups due to resonance. (c) Phenyl groups can stabilise positive charge by resonance. (d) The reaction is controlled by steric factors.

What will be the \(\mathrm{pH}\) of an acetate-acetic acid solution when the ratio of \(\left[\mathrm{CH}_{3} \mathrm{CO}_{2}\right] /$$\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]\) is \(10 ?\) (A table of \(\mathrm{pK}\) data is given below.) $$ \begin{array}{|lc|} \hline \text { Some useful } & \mathrm{pK}_{\mathrm{a}} \text { values } \\ \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H} & 4.76 \\ \mathrm{H}_{3} \mathrm{PO}_{4} & 2.2 \\ \mathrm{H}_{2} \mathrm{PO}^{\rho} & 7.2 \\ \mathrm{HPO}_{4}^{2} & 12.4 \\ \hline \end{array} $$ (a) \(5.76\) (b) \(4.76\) (c) \(3.76\) (d) \(1.76\)

Match the reactions in column I with their reagents in column II. Column I Column II (a) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CHO} \rightarrow \mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCOOH}\) (p) LiAIH \(_{4}\) (b) \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}-\mathrm{CHO} \rightarrow \mathrm{CH}_{3} \mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{2} \mathrm{OH}\) (q) \(\mathrm{NaBH}_{4}\) (c) Ph-CH=CH-CHO \(\rightarrow\) Ph-CH \(_{2}-\mathrm{CH}_{2}-\mathrm{CHO}\) (r) \(\mathrm{Pd}-\mathrm{C} / \mathrm{H}_{2}\) (d) \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2} \mathrm{CHO} \rightarrow \mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) (s) \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{\oplus}\)

If only one equivalent of \(\mathrm{NH}_{3}\) (one mole of \(\mathrm{NH}_{3}\) reacts with one mole of cyclohexyl chloride) was used, the reaction may not go to completion. Why? (a) \(\mathrm{NH}_{3}\) is not a strong nucleophile. Therefore, more \(\mathrm{NH}_{3}\) is needed. (b) \(\mathrm{Cl}\) is not a good leaving. Therefore, more \(\mathrm{NH}_{3}\) is needed. (c) Cyclohexyl group presents large steric hindrance. Therefore, more \(\mathrm{NH}_{3}\) is needed. (d) \(\mathrm{NH}_{3}\) will be protonated by \(\mathrm{HCl}\) as reaction proceeds. Therefore, the reaction will be incomplete.

Which of the following compound has the highest boiling point? (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}\) (b) \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\) (c) \(\mathrm{CH}_{3} \mathrm{COCH}_{3}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\)

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