Show how you would accomplish the following multistep syntheses using the indicated starting material and any necessary reagents.

(a) hept-6-en-1-ol caprolactone

(b) methoxybenzenep-methoxybenzamide

(c)

(d)

Gallic acid Mescaline

Short Answer

Expert verified

(a)

Formation of caprolactone

(b)

Formation of p-methoxybenzamide

(c)

(bromomethyl) benzene 2-phenylacetonitrile 2-phenylethan-1-amine

Formation of2-phenylethan-1-amine

(d)

Formation of mescaline

Step by step solution

01

(a) Preparation of caprolactone

Treatment of hept-6-en-1-ol with ozone in the presence of dimethyl sulfide gives 6-hydroxyhexanal, which reacts with Tolle’s reagent and forms a carboxylate ion in the product.

Now, the former compound undergoes intramolecular cyclization in the presence of an acid that gives the desired compound.

02

Formation of the desired product

The overall synthesis for the preparation of caprolactone is as follows:

Formation of caprolactone

03

(b) Preparation of p-methoxybenzamide

Treatment of methoxybenzene with bromine molecule gives 1-Bromo-4-methoxybenzene, which further reacts with Mg in the presence of ether gives a Grignard product (4-methoxyphenyl) magnesium bromide).

Now, the formed compound (4-methoxyphenyl) magnesium bromide reacts with carbon dioxide gives 4-methoxy benzoic acid, which reacts with thionyl chloride gives 4-methoxybenzoyl chloride, which reacts with ammonia to form the desired compound.

04

Formation of p-methoxybenzamide

The formation of p-methoxybenzamide is represented as follows:

Formation of p-methoxybenzamide

05

(c) Preparation of the desired compound

Treatment of (bromomethyl) benzene with a cyanide ion gives 2-phenyl acetonitrile. Now, the cyanide group undergoes reduction with lithium aluminum hydride and forms the desired compound.

06

Formation of the desired compound

The desired compound is formed by the following reaction as follows:

(bromomethyl) benzene 2-phenylacetonitrile 2-phenylethan-1-amine

Formation of 2-phenylethan-1-amine

07

(d) Preparation of mescaline

Treatment of gallic acid with excess sodium hydroxide followed by the reaction with excess methyl iodide gives 3,4,5-tri methoxy benzoic acid.

Now, the carboxylic acid group in 3,4,5-tri methoxy benzoic acid undergoes reduction with lithium aluminum hydride gives (3,4,5-tri methoxyphenyl) methanol, which reacts with tosyl chloride in the presence of pyridine gives a tosylate in the product.

Now, the tosyl group reacts with cyanide ion and forms a new C-C bond in the product, which further reacts with lithium aluminum hydride and yields the desired compound.

08

Formation of the desired compound

The desired compound mescaline is formed by the following reactions as follows:

Formation of mescaline

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Most popular questions from this chapter

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