Oxidation numbers can be assigned to determine if an atom is reduced or oxidized during the course of a reaction. An organic compound that has a carbon atom that has been reduced has generally obtained more hydrogen atoms. Hydrogen is less electronegative than carbon, so bonds to additional hydrogen atoms increase the electron density around a carbon atom. In contrast, when oxidation occurs in an organic molecule, a carbon atom loses H atoms, or forms additional bonds to more electronegative elements such as oxygen or nitrogen. One of the most commonly used methods for the preparation of 1° and 2° alcohols is via the reduction of aldehydes or ketones. In the generic reactions shown below, [H] represents the reducing agent, the reactant that causes reduction of the organic molecule. The conversion of a carbonyl compound to an aldehyde can be recognized as a reduction reaction because the carbonyl carbon forms an additional bond to a less electronegative H atom, while at the same time losing one of the bonds to the more electronegative O atom. Many different reducing agents may be employed to reduce an aldehyde or ketone to an alcohol. The two most commonly used are lithium aluminum hydride (LiAlH4) and sodium borohydride (NaBH4). Both of these reagents act as sources of the strongly basic hydride anion (H–). LiAlH4 is highly reactive and can react explosively with water or alcoholic solvents (the H– combines with H+ from the solvent to form the flammable gas H2). Because of this danger, use of LiAlH4 is typically restricted to situations where a very strong reducing agent is required – for instance the reduction of a carboxylic acid, ester, or amide, which are somewhat less reactive than aldehydes or ketones.Since NaBH4 is a milder reducing agent than LiAlH4, it is much safer to use in the laboratory. Freshly prepared solutions of NaBH4 in water or methanol are stable at room temperature for hours (decomposition will take place at higher temperatures, however). NaBH4 is effective for the reduction of an aldehyde or ketone, but it will not react with other carbonyl compounds. In this experiment, you will reduce the aldehyde functional group in vanillin (4-hydroxy-3-methoxybenzaldehyde) using NaBH4. The product of this reaction is vanillyl alcohol. The overall balanced equation is shown below. The NaBH4 will be dissolved in aqueous NaOH. It is necessary to use a basic solution because vanillin also contains an acidic phenol group. If neutral reaction conditions were used, the H– released by the NaBH4 would preferentially react with the phenolic proton rather than with the carbonyl C atom of the aldehyde.The hydroxide present under the reaction conditions will serve to deprotonate the phenol group, allowing the hydride to react selectively at the carbonyl C atom. The mechanism of the reduction is outlined below.Notice that one molecule of borohydride (BH4–) contains 4 reactive B–H bonds. In theory, 1 BH4– molecule should be able to reduce 4 alcohol molecules. At the end of the reaction, boric acid (H3BO3) will be formed as a byproduct that will remain in solution. TABLE OF PROPERTIES Compound MM (g/mol) mp (oC) Solubility (g/100 mL H2O) vanillin 152.2 81-83 0.1 vanillyl alcohol 154.2 113 0.2 sodium borohydride 37.8 400 miscible, reacts with H2O 6 M hydrochloric acid 36.5 miscible If you were to follow the progress of this reaction using IR spectroscopy, you would see Disappearance of a broad peak above 3000cm-1 Disappearance of a peak around 1700cm-1 Disappearance of a broad peak above 3000cm-1 and appearance of peak around 1700cm-1 Disappearance of a peak above 1700-cm1 and appearance of a broad peak around 3000cm-1 QUESTION 2 If you were following this reaction using Mass spec, a strong peak at M/z of 28 would indicate: Complete conversion of the reactant to product possible side reactions presence of unreacted aldehyde a high yielding reaction QUESTION 3 If the proton NMR showed a singlet and 2 doublets clustered between 7-8ppm. You can infer that: The conversion from aldehyde to alcohol was incomplete there was no reaction does not provide any information regarding the success of the reaction. The reaction was high yielding.

Oxidation numbers can be assigned to determine if an atom is reduced or oxidized during the course of a reaction. An organic compound that has a carbon atom that has been reduced has generally obtained more hydrogen atoms. Hydrogen is less electronegative than carbon, so bonds to additional hydrogen atoms increase the electron density around a carbon atom. In contrast, when oxidation occurs in an organic molecule, a carbon atom loses H atoms, or forms additional bonds to more electronegative elements such as oxygen or nitrogen. One of the most commonly used methods for the preparation of 1° and 2° alcohols is via the reduction of aldehydes or ketones. In the generic reactions shown below, [H] represents the reducing agent, the reactant that causes reduction of the organic molecule. The conversion of a carbonyl compound to an aldehyde can be recognized as a reduction reaction because the carbonyl carbon forms an additional bond to a less electronegative H atom, while at the same time losing one of the bonds to the more electronegative O atom. Many different reducing agents may be employed to reduce an aldehyde or ketone to an alcohol. The two most commonly used are lithium aluminum hydride (LiAlH4) and sodium borohydride (NaBH4). Both of these reagents act as sources of the strongly basic hydride anion (H–). LiAlH4 is highly reactive and can react explosively with water or alcoholic solvents (the H– combines with H+ from the solvent to form the flammable gas H2). Because of this danger, use of LiAlH4 is typically restricted to situations where a very strong reducing agent is required – for instance the reduction of a carboxylic acid, ester, or amide, which are somewhat less reactive than aldehydes or ketones.Since NaBH4 is a milder reducing agent than LiAlH4, it is much safer to use in the laboratory. Freshly prepared solutions of NaBH4 in water or methanol are stable at room temperature for hours (decomposition will take place at higher temperatures, however). NaBH4 is effective for the reduction of an aldehyde or ketone, but it will not react with other carbonyl compounds. In this experiment, you will reduce the aldehyde functional group in vanillin (4-hydroxy-3-methoxybenzaldehyde) using NaBH4. The product of this reaction is vanillyl alcohol. The overall balanced equation is shown below. The NaBH4 will be dissolved in aqueous NaOH. It is necessary to use a basic solution because vanillin also contains an acidic phenol group. If neutral reaction conditions were used, the H– released by the NaBH4 would preferentially react with the phenolic proton rather than with the carbonyl C atom of the aldehyde.The hydroxide present under the reaction conditions will serve to deprotonate the phenol group, allowing the hydride to react selectively at the carbonyl C atom. The mechanism of the reduction is outlined below.Notice that one molecule of borohydride (BH4–) contains 4 reactive B–H bonds. In theory, 1 BH4– molecule should be able to reduce 4 alcohol molecules. At the end of the reaction, boric acid (H3BO3) will be formed as a byproduct that will remain in solution. TABLE OF PROPERTIES Compound MM (g/mol) mp (oC) Solubility (g/100 mL H2O) vanillin 152.2 81-83 0.1 vanillyl alcohol 154.2 113 0.2 sodium borohydride 37.8 400 miscible, reacts with H2O 6 M hydrochloric acid 36.5 miscible If you were to follow the progress of this reaction using IR spectroscopy, you would see Disappearance of a broad peak above 3000cm-1 Disappearance of a peak around 1700cm-1 Disappearance of a broad peak above 3000cm-1 and appearance of peak around 1700cm-1 Disappearance of a peak above 1700-cm1 and appearance of a broad peak around 3000cm-1 QUESTION 2 If you were following this reaction using Mass spec, a strong peak at M/z of 28 would indicate: Complete conversion of the reactant to product possible side reactions presence of unreacted aldehyde a high yielding reaction QUESTION 3 If the proton NMR showed a singlet and 2 doublets clustered between 7-8ppm. You can infer that: The conversion from aldehyde to alcohol was incomplete there was no reaction does not provide any information regarding the success of the reaction. The reaction was high yielding.

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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BACKGROUND

Oxidation numbers can be assigned to determine if an atom is reduced or oxidized during the course of a reaction.

An organic compound that has a carbon atom that has been reduced has generally obtained more hydrogen atoms. Hydrogen is less electronegative than carbon, so bonds to additional hydrogen atoms increase the electron density around a carbon atom. In contrast, when oxidation occurs in an organic molecule, a carbon atom loses H atoms, or forms additional bonds to more electronegative elements such as oxygen or nitrogen.

One of the most commonly used methods for the preparation of 1° and 2° alcohols is via the reduction of aldehydes or ketones. In the generic reactions shown below, [H] represents the reducing agent, the reactant that causes reduction of the organic molecule. The conversion of a carbonyl compound to an aldehyde can be recognized as a reduction reaction because the carbonyl carbon forms an additional bond to a less electronegative H atom, while at the same time losing one of the bonds to the more electronegative O atom.

Many different reducing agents may be employed to reduce an aldehyde or ketone to an alcohol. The two most commonly used are lithium aluminum hydride (LiAlH4) and sodium borohydride (NaBH4). Both of these reagents act as sources of the strongly basic hydride anion (H–).

LiAlH4 is highly reactive and can react explosively with water or alcoholic solvents (the H– combines with H+ from the solvent to form the flammable gas H2). Because of this danger, use of LiAlH4 is typically restricted to situations where a very strong reducing agent is required – for instance the reduction of a carboxylic acid, ester, or amide, which are somewhat less reactive than aldehydes or ketones.Since NaBH4 is a milder reducing agent than LiAlH4, it is much safer to use in the laboratory. Freshly prepared solutions of NaBH4 in water or methanol are stable at room temperature for hours (decomposition will take place at higher temperatures, however). NaBH4 is effective for the reduction of an aldehyde or ketone, but it will not react with other carbonyl compounds. In this experiment, you will reduce the aldehyde functional group in vanillin (4-hydroxy-3-methoxybenzaldehyde) using NaBH4. The product of this reaction is vanillyl alcohol. The overall balanced equation is shown below.

The NaBH4 will be dissolved in aqueous NaOH. It is necessary to use a basic solution because vanillin also contains an acidic phenol group. If neutral reaction conditions were used, the H– released by the NaBH4 would preferentially react with the phenolic proton rather than with the carbonyl C atom of the aldehyde.The hydroxide present under the reaction conditions will serve to deprotonate the phenol group, allowing the hydride to react selectively at the carbonyl C atom. The mechanism of the reduction is outlined below.Notice that one molecule of borohydride (BH4–) contains 4 reactive B–H bonds. In theory, 1 BH4– molecule should be able to reduce 4 alcohol molecules. At the end of the reaction, boric acid (H3BO3) will be formed as a byproduct that will remain in solution.

TABLE OF PROPERTIES

Compound	MM (g/mol)	mp (oC)	Solubility (g/100 mL H2O)
vanillin	152.2	81-83	0.1
vanillyl alcohol	154.2	113	0.2
sodium borohydride	37.8	400	miscible, reacts with H2O
6 M hydrochloric acid	36.5		miscible

If you were to follow the progress of this reaction using IR spectroscopy, you would see

		Disappearance of a broad peak above 3000cm^-1
		Disappearance of a peak around 1700cm^-1
		Disappearance of a broad peak above 3000cm^-1 and appearance of peak around 1700cm^-1
		Disappearance of a peak above 1700^-cm¹ and appearance of a broad peak around 3000cm^-1

QUESTION 2

If you were following this reaction using Mass spec, a strong peak at M/z of 28 would indicate:

Complete conversion of the reactant to product

possible side reactions

presence of unreacted aldehyde

a high yielding reaction

QUESTION 3

If the proton NMR showed a singlet and 2 doublets clustered between 7-8ppm. You can infer that:

		The conversion from aldehyde to alcohol was incomplete
		there was no reaction
		does not provide any information regarding the success of the reaction.
		The reaction was high yielding.

QUESTION 4

The limiting reagent in this experiment is

vanilla

vanillin

sodium borohydride

sodium hydroxide

QUESTION 5

The gas evolved in this reaction is

carbon dioxide

nitrogen

oxygen

Hydrogen

QUESTION 6

Reduction reaction involves

Addition of hydrogens

removal of hydrogens

addition of oxygens

removal of H⁺

QUESTION 7

Theoretically, one mole of sodium borohydride can react with _______moles of vanilin

4

3

2

5

QUESTION 8

Care has to be taken to keep the reaction mixture at _________during the addition of HCl.

55 deg

25 deg

0 deg

15 deg

QUESTION 9

Sodium borohydride reduces___________ and ____________ but not _____________

aldehydes and ketones not esters

aldehydes and esters not ketones

alkanes and ketones not aldehydes

aldehydes and ketones

Definition Definition Organic compounds that have a carbonyl group, C = O, as their functional group. The carbonyl group in ketones is placed somewhere in the middle of the molecular structure, which means, the C=O is attached to two alkyl groups or benzene rings. Just like all other homologous series, the naming of ketones is done using a common suffix “-one”. The general molecular formula is C n H 2n O. Functional group of a ketone

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		Complete conversion of the reactant to product
		possible side reactions
		presence of unreacted aldehyde
		a high yielding reaction

		Addition of hydrogens
		removal of hydrogens
		addition of oxygens
		removal of H⁺

		aldehydes and ketones not esters
		aldehydes and esters not ketones
		alkanes and ketones not aldehydes
		aldehydes and ketones

		55 deg
		25 deg
		0 deg
		15 deg