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Predict the principal products to be expected in each of the following reactions; give your reasoning: a. \(\mathrm{CH}_{3} \mathrm{CHO}+\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO} \stackrel{\mathrm{NaOH}}{\longrightarrow}\) b. \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}(\mathrm{OH}) \mathrm{CHCOCH}_{3} \stackrel{\mathrm{NaOH}}{\longrightarrow}\) c. \(\mathrm{CH}_{2} \mathrm{O}+\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCHO} \stackrel{\mathrm{NaOH}}{\longrightarrow}\) d. \(\mathrm{CH}_{2} \mathrm{O}+\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCHO} \stackrel{\mathrm{Ca}(\mathrm{OH})_{2}}{\longrightarrow}\)

Short Answer

Expert verified
a. Aldol product; b. α,β-unsaturated ketone; c. Aldol product; d. Aldol product.

Step by step solution

01

Analyzing Reaction a

In reaction a, we have acetaldehyde, \( \mathrm{CH}_{3} \mathrm{CHO} \), and acetone, \( (\mathrm{CH}_{3})_{2} \mathrm{CO} \), under basic conditions with \( \mathrm{NaOH} \). The reaction is likely an Aldol reaction. The acetaldehyde, as it has an alpha hydrogen, will undergo deprotonation and form an enolate ion, which can attack the carbonyl carbon of acetone, providing an aldol addition product, \( \mathrm{CH}_{3} \mathrm{CH} (\mathrm{OH})\mathrm{CH}_{2} \mathrm{C(O)CH}_{3} \).
02

Analyzing Reaction b

Reaction b involves a β-hydroxy ketone, \( (\mathrm{CH}_{3})_{2} \mathrm{C(OH)CHCOCH}_{3} \), and \( \mathrm{NaOH} \). The identification is a dehydration step following the aldol addition (if previous steps involved aldol), leading to the formation of an α,β-unsaturated ketone. The elimination of water from the β-hydroxy ketone gives \( (\mathrm{CH}_{3})_{2} \mathrm{C} \text{=} \mathrm{CHCOCH}_{3} \).
03

Analyzing Reaction c

Reaction c involves formaldehyde, \( \mathrm{CH}_{2} \mathrm{O} \), reacting with tert-butyl aldehyde, \( \left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCHO} \), in the presence of \( \mathrm{NaOH} \). The formaldehyde can act as the nucleophile due to the basic environment forming an enolate-like species, which then attacks the carbonyl carbon of the tert-butyl aldehyde. The major product is likely \( \left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH(OH)CH}_{2} \mathrm{OH} \) due to an aldol type reaction.
04

Analyzing Reaction d

Reaction d combines formaldehyde with isobutyraldehyde \( \left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCHO} \) under basic conditions with \( \mathrm{Ca(OH)}_{2} \). The formation of an aldol product from the attack on isobutyraldehyde by the enolate form of formaldehyde could result in \( \left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHC(OH)CH}_{2} \mathrm{OH} \), after protonation.

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Key Concepts

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

Alpha Hydrogen
An alpha hydrogen is a hydrogen atom attached to the carbon adjacent to a carbonyl group. This specific positioning makes it more acidic than other hydrogen atoms in an organic molecule. The acidity is typically due to the ability of the enolate formation. In the case of an Aldol reaction, the alpha hydrogen plays a crucial role.
  • The presence of an alpha hydrogen allows for deprotonation under basic conditions, leading to enolate formation.
  • Acetaldehydes, like the one in the exercise, readily lose their alpha hydrogen to create a negatively charged enolate ion.
The enolate ion is not only stable but highly reactive, ready to further participate in addition reactions. This makes understanding the concept of alpha hydrogens fundamental in predicting the course of Aldol reactions.
Enolate Ion
The enolate ion is formed when the alpha hydrogen of a carbonyl compound is abstracted by a base. This ion is a resonance-stabilized intermediate, which can exist in two forms, namely, the keto form and the enol form. In Aldol reactions, enolate ions are key players, participating actively in the formation of new bonds.
  • The enolate ion acts as a nucleophile due to the negative charge on the oxygen, allowing it to attack electrophilic species like carbonyl carbons.
  • Its ability for resonance provides additional stability, making it a strong intermediate for reactions under basic conditions.
Recognizing the formation and role of enolate ions helps in understanding why certain products are favored and how they are formed in Aldol reactions.
Beta-Hydroxy Ketone
A beta-hydroxy ketone is a result of an Aldol addition. It's characterized by a hydroxyl group (–OH) on the beta carbon relative to a carbonyl group in a key step before any potential dehydration occurs to form α,β-unsaturated carbonyl compounds.
  • In the exercise, after the enolate ion attacks the carbonyl carbon, it leads to the formation of a new carbon-carbon bond.
  • This intermediate product is a beta-hydroxy ketone, named from the placements of the -OH group and the keto group.
Understanding beta-hydroxy ketones is essential for seeing how these intermediates can easily dehydrate, given their structural setup, leading to the formation of more complex products.
Carbonyl Carbon Attack
In an Aldol reaction, one of the quintessential maneuvers is the attack on the carbonyl carbon, which is typically the electrophilic center. Due to the partial positive charge on this carbon, it becomes a prime site for nucleophilic attack by the enolate ion.
  • The electrophilicity of the carbonyl carbon is heightened by resonance and the polar nature of the C=O double bond.
  • The enolate ion, having a negatively charged oxygen, is attracted to it, leading to the formation of a new C-C bond.
This attack is key to the Aldol reaction, setting the stage for the creation of intricate organic molecules, thus showing the importance of understanding how and why carbonyl carbon sites are typically chosen for these reactions.

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Most popular questions from this chapter

If you wished to prepare the methyl ether of 4 -hydroxy-3-penten- 2 -one by \(\mathrm{O}\) -alkylation, what base and which fo the methylating agents listed would you choose? \(\mathrm{CH}_{3} \mathrm{Cl}, \mathrm{CH}_{3} \mathrm{I}, \mathrm{CH}_{3} \mathrm{OCO}_{2} \mathrm{OCH}_{3},\left(\mathrm{CH}_{3}\right)_{3} \mathrm{OBF}_{4}\), or \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{O} .\) Give your reasoning.

If methyl iodide gives mainly \(\mathrm{C}\) -alkylation with the enolate anion of 2 -propanone, which of the following halides would you expect to be candidates to give \(\mathrm{O}\) -alkylation: tert-butyl chloride, phenylmethyl chloride, 3 -chloropropene, neopentyl chloride?

A detailed study of the rate of bromination of 2 -propanone in water, in the presence of ethanoic acid and ethanoate \(\quad\) ions, \(\quad\) has \(\quad\) shown \(\quad\) that \(v=\left\\{6 \times 10^{-9}+5.6 \times 10^{-4}\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]+1.3 \times 10^{-7}\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]+7\left[\mathrm{OH}^{-}\right]+3.3 \times 10^{-6}\left[\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}\right]+3.5 \quad\right.\) in which \(\left.\times 10^{-6}\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]\left[\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}\right]\right\\}\left[\mathrm{CH}_{3} \mathrm{COCH}_{3}\right]\) the rate \(v\) is expressed in \(\mathrm{mol} \mathrm{L}^{-1} \mathrm{sec}^{-1}\) when the concentrations are in \(\mathrm{mol} \mathrm{L}^{-1}\). a. Calculate the rate of the reaction for a 1 M solution of 2 -propanone in water at \(\mathrm{pH} 7\) in the absence of \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\) and \(\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}\) b. Calculate the rate of the reaction for \(1 \mathrm{M} 2\) -propanone in a solution made by neutralizing \(1 \mathrm{M}\) ethanoic acid with sufficient sodium hydroxide to give \(\mathrm{pH} 5.0\) ( \(K_{a}\) of ethanoic acid \(=1.75 \times 10^{-5}\) ). c. Explain how the numerical values of the coefficients for the rate equation may be obtained from observations of the reaction at various \(\mathrm{pH}\) values and ethanoate ion concentrations. d. The equilibrium concentration of enol in 2 -propanone is estimated to be \(\sim 1.5 \times 10^{-4} \% .\) If the rate of conversion of \(1 \mathrm{M}\) 2-propanone to enol at \(\mathrm{pH} 7\) (no \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\) or \(\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}\) present) is as calculated in Part a, calculate the rate of the reverse reaction from enol to ketone at \(\mathrm{pH} 7\) if the enol were present in \(1 \mathrm{M}\) concentration. e. Suggest a mechanistic explanation for the term \(3.5 \times 10^{-6}\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]\left[\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}\right]\) in the rate expression.

Devise a reasonable synthesis of each of the following compounds from the indicated starting materials. Assume that other needed reagents are available. (Not all of the syntheses involve aldol-addition reactions, but all involve at some stage or the other carbonyl-addition reactions.) a. propenenitrile from ethanal b. 1 -(trichloromethyl) cyclohexanol from cyclohexanone c. 2,2 -dimethyl- 1,3 -propanediol from 2 -methylpropanal d. 2 -(phenylmethylidene)cyclohexanone form cyclohexanone e. 2,3 -diphenylpropenenitrile from phenylethanenitrile f. \(\quad\) OH \(\quad 0\) from a compound with only one cyclohexane ring g. 3-methyl-2-cyclopentenone from an open-chain compound

Explain why the \(D\) or \(L\) enantiomer of a chiral ketone such as 4 -phenyl-3-methyl-2-butanone racemizes in the presence of dilute acid or dilute base.

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