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Chapter 17: Q-17-73P (page 850)

Show how the following compound can be prepared from starting materials that have no more than five carbons:

Short Answer

Expert verified

If an aldol addition uses two different carbonyl compounds, such reactions are crossed aldol addition.

A single product can be obtained in this reaction if one of the aldehydes does not comprise alpha hydrogen and cannot form an enolate ion.

Step by step solution

01

Crossed Aldol addition

If an aldol addition uses two different carbonyl compounds, such reactions are crossed aldol addition.

A single product can be obtained in this reaction if one of the aldehydes does not comprise alpha hydrogen and cannot form an enolate ion.

02

Identifying how the given compound can be prepared from starting materials that have no more than five carbons

The given compound is prepared from the starting material with no more than five carbons through a series of steps. An - unsaturated carbonyl compound is reacted with a base leading to the formation of an enolate ion.

This ion attacks another cyclic unsaturated compound. An intramolecular cyclization happens in the last transformation, and the formation of the required compound takes place. The reaction is given below.

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Most popular questions from this chapter

Explain why alkylation of anα α -carbon works best if the alkyl halide used in the reaction is a primary alkyl halide, and why alkylation does not work at all if a tertiary alkyl halide is used.

Show how the following compound can be synthesized from the given starting material.

Compound A with molecular formula C6H10 has two peaks in its 1H NMR spectrum, both of which are singlets (with ratio 9 : 1). Compound Areacts

with an acidic aqueous solution containing mercuric sulfate to form compound B,which gives a positive iodoform test (Problem 58) and has an 1H NMR

spectrum that shows two singlets (with ratio 3 : 1). Identify A and B.

What is the product of each of the following reactions?

In the presence of excess base and excess halogen, a methyl ketone is converted to a carboxylate ion. The reaction is known as the haloform reaction because one of the products is haloform (chloroform, bromoform, or iodoform). Before spectroscopy became a routine analytical tool, the haloform reaction served as a test for methyl ketones: the formation of iodoform, a bright yellow compound, signaled that a methyl ketone was present. Why do only methyl ketones form a haloform?
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