a)
Interpretation:
The product of the Wolff-Kishner reduction reaction shown is to be given. The electron-pushing mechanism, beginning from the hydrazone intermediate is to be provided.
Concept introduction:
In Wolff-Kishner reduction
To give:
The product of the Wolff-Kishner reduction reaction shown.
To provide:
The electron-pushing mechanism, beginning from the hydrazone intermediate.
Answer to Problem 38MP
The product of the Wolff-Kishner reduction reaction shown is
The electron-pushing mechanism for the formation of the alkane, beginning from the hydrazone intermediate, is given below.
Explanation of Solution
The hydroxide ion from KOH abstracts a weakly acidic H from –NH2 of the hydrazone of isopropyl phenyl ketone to yield a carbanion which picks up a proton to yield a neutral intermediate. Deprotonation of the remaining hydrogen on N by the hydroxide ion occurs with the eliminartion of nitrogen to yield another carbanion which is protonated to give the alkane, ethyl benzene, as the product.
The product of the Wolff-Kishner reduction reaction shown is
The electron-pushing mechanism, beginning from the hydrazone intermediate is given below.
b)
Interpretation:
The product of the Wolff-Kishner reduction reaction shown is to be given. The electron-pushing mechanism, beginning from the hydrazone intermediate is to be provided.
Concept introduction:
In Wolff-Kishner reduction aldehydes and ketones react with hydrazine in the presence of a base to yield alkanes. First a hydrazone is formed which is then converted into an alkane.
To give:
The product of the Wolff-Kishner reduction reaction shown.
To provide:
The electron-pushing mechanism the formation of the alkane beginning from the hydrazone intermediate.
Answer to Problem 38MP
The product of the Wolff-Kishner reduction reaction shown is
The electron-pushing mechanism the formation of the alkane, beginning from the hydrazone intermediate, is given below.
Explanation of Solution
The hydroxide ion from KOH abstracts a weakly acidic H from –NH2 of the hydrazone of diethyl ketone to yield a carbanion which picks up a proton to yield a neutral intermediate. Deprotonation of the remaining hydrogen on N by the hydroxide ion occurs with the eliminartion of nitrogen to yield another carbanion which is protonated to give the alkane, n-pentane, as the product.
The product of the Wolff-Kishner reduction reaction shown is
The electron-pushing mechanism, beginning from the hydrazone intermediate is given below.
c)
Interpretation:
The product of the Wolff-Kishner reduction reaction shown is to be given. The electron-pushing mechanism, beginning from the hydrazone intermediate is to be provided.
Concept introduction:
In Wolff-Kishner reduction aldehydes and ketones react with hydrazine in the presence of a base to yield alkanes. First a hydrazone is formed which is then converted into an alkane.
To give:
The product of the Wolff-Kishner reduction reaction shown.
To provide:
The electron-pushing mechanism the formation of the alkane beginning from the hydrazone intermediate.
Answer to Problem 38MP
The product of the Wolff-Kishner reduction reaction shown is
The electron-pushing mechanism the formation of the alkane, beginning from the hydrazone intermediate, is given below.
Explanation of Solution
The hydroxide ion from KOH abstracts a weakly acidic H from –NH2 of the hydrazone of the ketone to yield a carbanion which picks up a proton to yield a neutral intermediate. Deprotonation of the remaining hydrogen on N by the hydroxide ion occurs with the eliminartion of nitrogen to yield another carbanion which is protonated to give the alkane as the product.
The product of the Wolff-Kishner reduction reaction shown is
The electron-pushing mechanism, beginning from the hydrazone intermediate is given below.
d)
Interpretation:
The product of the Wolff-Kishner reduction reaction shown is to be given. The electron-pushing mechanism, beginning from the hydrazone intermediate is to be provided.
Concept introduction:
In Wolff-Kishner reduction aldehydes and ketones react with hydrazine in the presence of a base to yield alkanes. First a hydrazone is formed which is then converted into an alkane.
To give:
The product of the Wolff-Kishner reduction reaction shown.
To provide:
The electron-pushing mechanism, beginning from the hydrazone intermediate.
Answer to Problem 38MP
The product of the Wolff-Kishner reduction reaction shown is
The electron-pushing mechanism for the formation of the alkane beginning from the hydrazone intermediate, is given below.
Explanation of Solution
The hydroxide ion from KOH abstracts a weakly acidic H from –NH2 of the hydrazone of ketone to yield a carbanion which picks up a proton to yield a neutral intermediate. Deprotonation of the remaining hydrogen on N by the hydroxide ion occurs with the eliminartion of nitrogen to yield another carbanion which is protonated to give the alkane as the product.
The product of the Wolff-Kishner reduction reaction shown is
The electron-pushing mechanism for the formation of the alkane beginning from the hydrazone intermediate, is given below.
Want to see more full solutions like this?
Chapter 19 Solutions
Student Value Bundle: Organic Chemistry, + OWLv2 with Student Solutions Manual eBook, 4 terms (24 months) Printed Access Card (NEW!!)
- Understanding the general acid-base properties of amino acids O Proteins Imagine each of the molecules shown below was found in an aqueous solution. Can you tell whether the solution is acidic, basic, or neutral? molecule The solution is... 010 H3N-CH-C-OH CH HO CH3 O acidic O basic neutral O (unknown) H3N HO 0 O acidic O basic neutral ○ (unknown) H3N-CH-C-O CH2 CH3-CH-CH3 O acidic O basic Oneutral ○ (unknown) O= X H2N-CH-C-O CH3 CH CH3 acidic O basic O neutral ○ (unknown) ? 000arrow_forwardImagine each of the molecules shown below was found in an aqueous solution. Can you tell whether the solution is acidic, basic, or neutral? molecule 0=0 H3N-CH-C-o HO CH2 OH The solution is... O acidic O basic O neutral O (unknown) H₂N acidic O basic O neutral ○ (unknown) + H3N O OH O acidic O basic O neutral O (unknown) H2N-CH-C-O CH3 O acidic O basic neutral ○ (unknown) X ? olo HEarrow_forwardRecognizing ampli Draw an a amino acid with a methyl (-CH3) side chain. Explanation Check Click and drag to start drawing a structure. X Carrow_forward
- Write the systematic name of each organic molecule: structure name × HO OH ☐ OH CI CI O CI OH OHarrow_forwardく Check the box under each a amino acid. If there are no a amino acids at all, check the "none of them" box under the table. Note for advanced students: don't assume every amino acid shown must be found in nature. COO H3N-C-H CH2 HO CH3 NH3 O CH3-CH CH2 OH Onone of them Explanation Check + H3N O 0. O OH + NH3 CH2 CH3-CH H2N C-COOH H O HIC + C=O H3N-C-O CH3- - CH CH2 OH Х 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center Accesarrow_forwardWrite the systematic name of each organic molecule: structure HO-C-CH2-CH3 O -OH CH3-CH2-CH2-CH2-CH2-C-OH CH3 CH3-CH-CH2-C-OH Explanation Check S namearrow_forward
- theres 2 productsarrow_forwardDraw the major product of this solvolysis reaction. Ignore any inorganic byproducts. + CH3CH2OH Drawing Q Atoms, Bonds and Rings OCH2CH3 || OEt Charges OH 00-> | Undo Reset | Br Remove Done Drag To Pan +arrow_forwardDraw the major product of this SN1 reaction. Ignore any inorganic byproducts. CH3CO2Na CH3CO2H Drawing + Br Q Atoms, Bonds and Rings OAC Charges OH ОАс Na ဂ Br Undo Reset Remove Done Drag To Pan +arrow_forward
- Organic Functional Groups entifying positions labeled with Greek letters in acids and derivatives 1/5 ssible, replace an H atom on the a carbon of the molecule in the drawing area with a ce an H atom on the ẞ carbon with a hydroxyl group substituent. ne of the substituents can't be added for any reason, just don't add it. If neither substi er the drawing area. O H OH Oneither substituent can be added. Check D 1 Accessibility ado na witharrow_forwardDifferentiate between electrophilic and nucleophilic groups. Give examples.arrow_forwardAn aldehyde/ketone plus an alcohol gives a hemiacetal, and an excess of alcohol gives an acetal. The reaction is an equilibrium; in aldehydes, it's shifted to the right and in ketones, to the left. Explain.arrow_forward
Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning
EBK A SMALL SCALE APPROACH TO ORGANIC LChemistryISBN:9781305446021Author:LampmanPublisher:CENGAGE LEARNING - CONSIGNMENT