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Chapter 5: Problem 57

O=C1C=CCCC1 O=C1CCCCC1 (ii) \(\mathrm{H}_{3} \mathrm{O}^{+} /\). O=C1C=CCCC1=C1CCCCC1 (i)(X) The reagent (X) required is (a) \(\mathrm{NaOH}\) (b) \(\mathrm{I}_{2} / \mathrm{NaOH}\) (c) KCN (d) \(\mathrm{CF}_{3} \mathrm{CO}_{3} \mathrm{H}\)

Short Answer

Expert verified
Answer: The correct reagent (X) needed for the second reaction is (b) I2/NaOH. However, the first reaction is performed using reagent (d) CF_3CO_3H.

Step by step solution

01

Identify structural changes between compounds

We will first identify the structural changes between the compounds. We can see that compound (i) has an alpha, beta-unsaturated carbonyl group (C=C-C=O), while compound (ii) has a simple carbonyl group (C=O) and an additional carbon is added between the double bond and carbonyl group. In compound (iii), another benzene ring is attached to the alpha position.
02

Determine the reagent for the first reaction

To convert compound (i) into compound (ii), we need to add a carbon between the double bond and the carbonyl group, specifically via a nucleophilic addition. From the available options, the reagent which can achieve this is (d) CF_3CO_3H or Trifluoroacetic acid. It adds a carboxylic acid group to the alpha, beta-unsaturated carbonyl group and then decarboxylation takes place to form compound (ii).
03

Determine the reagent for the second reaction

For the conversion of compound (ii) into compound (iii), we need to introduce another benzene ring at the alpha position. To add a benzene ring, we need an electrophilic aromatic substitution to occur. From the available options, the best reagent to perform this transformation is (b) I2/NaOH, which first reacts to form Iodonium ion. The Iodonium ion is a good electrophile and reacts with compound (ii) to form compound (iii).
04

Conclude the answer

After analyzing the given compounds and the structural changes required to form products, the correct reagent (X) needed is (b) I2/NaOH for the second reaction. However, the first reaction is performed using reagent (d) CF_3CO_3H.

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Most popular questions from this chapter

p-cresol \(\stackrel{\mathrm{CHCl}_{3} \mathrm{NaOH}}{\longrightarrow} \mathrm{A} \stackrel{\mathrm{HCN}}{\longrightarrow} \mathrm{B} \stackrel{\mathrm{H}, \mathrm{O}^{+}}{\longrightarrow} \mathrm{C}\) Which of the following statements is not true about the product (C) in the above sequence of reactions? (a) It gives violet colour with neutral ferric chloride solution. (b) It is optically active. (c) There are six \(\mathrm{sp}^{2}\) hybridised carbon atoms in one molecule of it. (d) It liberates \(\mathrm{CO}_{2}\) when treated with \(\mathrm{NaHCO}_{3}\) solution.

Explain why o-hydroxybenzaldehyde is a liquid at room temperature while p-hydroxybenzaldehyde is a solid?

When acetone reacts with chloroform in the presence of alkali like \(\mathrm{NaOH}\), the product formed is (a) Chloropionium (b) acetic acid (c) Chloretone (d) mesityloxide

Aldol condensation cannot occur between (a) two aldehydes (b) an aldehyde and a ketone (c) two ketones (d) an aldehyde and an ester

The acylating species in the Friedel-Crafts acylation of benzene to give acetophenone is (a) \(\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{O}^{\odot}\) (b) \(\mathrm{AlCl}_{4}\) (c) \(\mathrm{CH}_{3}-\stackrel{\oplus}{\mathrm{C}}=\ddot{\mathrm{O}}\) (d) \(\mathrm{CH}_{3}^{-\mathrm{C} \mathrm{C} \mathrm{Cl}}-\mathrm{O}-\mathrm{AlCl}\)
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