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Chapter 23: Problem 138

One of the constituents of the carnauba wax used in floor and furniture polish is an ester of a \(\mathrm{C}_{32}\) straight-chain alcohol with a \(\mathrm{C}_{20}\) straight-chain carboxylic acid. Draw the structure of this ester.

Short Answer

Expert verified
CH3-(CH2)30-CH2-O-C(=O)-(CH2)18-CH3 is the structure of the ester.

Step by step solution

01

Understand the chemical structure components

An ester is formed when a carboxylic acid reacts with an alcohol. In this exercise, we are combining a C32 straight-chain alcohol with a C20 straight-chain carboxylic acid. This means we have a 32-carbon alcohol molecule and a 20-carbon acid molecule.
02

Formulate the alcohol structure

The alcohol in this esterification process has 32 carbon atoms, noted chemically as C32. The general formula for a straight-chain alcohol is CnH2n+1OH. For a C32 alcohol, this becomes C32H65OH.
03

Formulate the carboxylic acid structure

The carboxylic acid has 20 carbon atoms, noted as C20. The general formula for a straight-chain carboxylic acid is CnH2n+1COOH. For a C20 acid, this becomes C19H39COOH, accounting for the carbon in the carboxyl group (-COOH).
04

Understand ester formation

Esterification is a chemical reaction between an alcohol and a carboxylic acid where water (\( H_2O \)) is removed and an ester bond forms. The hydroxyl group (-OH) from the alcohol and the hydrogen from the carboxyl group (-COOH) combine to create water and leave behind an ester linkage (-COO-).
05

Drafting the ester structure

The structure of the ester will combine the alcohol and acid structures. The ester linkage, -COO-, connects the C32 alcohol and the C20 acid. Start with the C32 alcohol, followed by the ester linkage (C=O-O), and end with the C19 alkyl chain from the carboxylic acid (as the carbon in the -COOH forms the ester linkage with the alcohol).
06

Draw the ester

The complete structure of the ester would look like this: CH3-(CH2)30-CH2-O-C(=O)-(CH2)18-CH3. Here, the CH3-(CH2)30-CH2 represents the C32 alcohol carbon chain and -(CH2)18-CH3 represents the carbon chain of the C20 acid, linked by the ester linkage -O-C(=O)-.

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Key Concepts

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

Carnauba wax
Carnauba wax is an all-natural wax derived from the leaves of the carnauba palm tree, native to Brazil. It is highly valued for its hardness and glossy finish, making it an ideal component in items like floor and furniture polishes.
The toughness of carnauba wax is because of its complex mixture of components, including esters, fatty acids, and alcohols. One important aspect of carnauba wax is esterification processes it undergoes, combining long-chain alcohols and acids to form esters that contribute to its remarkable characteristics.
  • Durability: Carnauba wax is exceptionally durable, providing a strong protective layer against water and dust.
  • Glossy Finish: The wax gives a shiny, polished surface without discoloring the material underneath.
  • Biodegradable: Unlike synthetic counterparts, carnauba wax is biodegradable, which makes it environmentally friendly.
C32 alcohol
C32 alcohol refers to a long-chain alcohol molecule with 32 carbon atoms. In organic chemistry, alcohols are a group of compounds characterized by the presence of one or more hydroxyl (-OH) groups.
The C32 alcohol used in making carnauba wax esters has a chemical formula of C32H65OH. This large structure enables the formation of a stable Ester linkage when combined with a substantial carbon-based acid.
Characteristics of C32 alcohol include:
  • Straight-Chain Structure: With 32 carbons in a linear arrangement, it is quite long compared to other more common alcohols.
  • Solubility: These alcohols are less soluble in water due to their large hydrophobic carbon chain.
  • Uses: Besides being used in wax production, such alcohols are critical in making high-grade industrial applications due to their efficacy as lubricants and surfactants.
C20 carboxylic acid
The C20 carboxylic acid in carnauba wax works alongside the C32 alcohol to form an ester through esterification. Carboxylic acids are vital in organic chemistry, identified by their carboxyl group (-COOH).
A C20 carboxylic acid specifically refers to a compound with 20 carbon atoms, making it a long-chain acid with the formula C19H39COOH. The -COOH group at the end of the chain readily reacts with alcohols to form esters.
Key features of C20 carboxylic acids:
  • Long Carbon Chain: The chain impacts the acid's physical and chemical properties, like melting point and solubility.
  • Reactivity: This acid is highly reactive, particularly due to its polar carboxyl group, making it essential in synthetic reactions, such as esterification.
  • Applications: Long-chain carboxylic acids are utilized not only in waxes but also in detergents, metalworking, and as components in complex chemical syntheses.
Organic chemistry
Organic chemistry is a branch of science focusing on the structure, properties, composition, reactions, and synthesis of organic compounds. Primarily, it deals with carbon-hydrogen bonds, which are fundamental in compounds such as carnauba wax.
Understanding organic chemistry is crucial as it lays the groundwork for fields like biochemistry, medicine, and materials science. It's concerned with numerous reactions and compounds, including the esterification process that creates esters from carboxylic acids and alcohols.
  • Esterification: A pivotal reaction in organic chemistry where an alcohol and a carboxylic acid react to form an ester, releasing water.
  • Complex Compounds: Organic chemistry encompasses a vast range of substances, from simple hydrocarbons to more complicated polymers and biopolymers.
  • Innovative Applications: Advances in organic chemistry have led to developments in pharmaceuticals, organic electronics, and sustainable energy solutions.

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Most popular questions from this chapter

Label the configuration around the double bond in the fatty acids as cis or trans. (a) Elaidic acid \(\left(\mathrm{C}_{18} \mathrm{H}_{34} \mathrm{O}_{2}\right)\) (b) Oleic acid \(\left(\mathrm{C}_{18} \mathrm{H}_{34} \mathrm{O}_{2}\right)\)

In the molecule below, the triple bond has one \(\pi\) bond that is in conjugation with the neighboring double bond, and it has one \(\pi\) bond that is not conjugated. Visualize a simplified orbital overlap picture and explain why one \(\pi\) bond of the triple bond has delocalized electrons, but the other does not. (The hybridization of each atom is indicated.)

Describe the structure of the functional group in each of the following families: (a) Alkene (b) Alcohol (c) Ester (d) Amine

The relative amount of unsaturation in a fat or oil is expressed as an iodine number. Olive oil, for instance, is highly unsaturated and has an iodine number of 172, while butter is much less unsaturated and has an iodine number of \(37 .\) Defined as the number of grams of \(\mathrm{I}_{2}\) absorbed per 100 grams of fat, the iodine number is based on the fact that the carbon-carbon double bonds in fats and oils undergo an addition reaction with \(\mathrm{I}_{2}\). The larger the number of double bonds, the larger the amount of \(\mathrm{I}_{2}\) that reacts. To determine an iodine number, a known amount of fat is treated with a known amount of \(\mathrm{I}_{2}\). When the addition reaction is complete, the amount of excess \(\mathrm{I}_{2}\) remaining is determined by titration with \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) according to the equation $$ 2 \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}(a q)+\mathrm{I}_{2}(a q) \rightarrow \mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}(a q)+2 \mathrm{NaI}(a q) $$ Knowing both the amount of \(\mathrm{I}_{2}\) originally added and the amount remaining after reaction, the iodine number can be calculated. Assume that \(0.500 \mathrm{~g}\) of human milk fat is allowed to react with \(25.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{I}_{2}\) solution and that \(81.99 \mathrm{~mL}\) of \(0.100\) \(\mathrm{M} \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) is required for complete reaction with the excess \(\mathrm{I}_{2}\). (a) What amount (in grams) of \(\mathrm{I}_{2}\) was added initially? (b) How many grams of \(\mathrm{I}_{2}\) reacted with the milk fat, and how many grams were in excess? (c) What is the iodine number of human milk fat? (d) Assuming a molecular weight of 800 , how many double bonds does an average molecule of milk fat contain?

Use the three-letter shorthand notations to name the two isomeric dipeptides that can be made from valine and cysteine. Draw both structures.
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