
Concept explainers
Name the following

a)
Interpretation:
The name of the alkene shown to be given. The products of its reaction with 1) meta-chloroperoxybenzoic acid, 2) KMnO4 in aqueous acid and 3) O3, followed by Zn in acetic acid, are to be given.
Concept introduction:
When alkenes react with meta-chloroperoxybenzoic acid, oxygen atom adds to the double bond to give epoxides.
Upon treatment with KMnO4 in aqueous acid, if the alkene the double bonded carbon is di-substituted and has no hydrogen it is converted in to a ketone and if the double bonded carbon is mono-substituted and has one hydrogen it is oxidized to a carboxylic acid.
Upon treatment with O3, followed by Zn in acetic acid the double bond is cleaved and each carbon in the double bond gets attached to an oxygen atom to give carbonyl compounds as products. If in the alkene double bonded carbon is di-substituted and has no hydrogen it is converted in to a ketone and if the double bonded carbon is mono-substituted and has one hydrogen it is oxidized to an aldehyde.
To give:
The name of the alkene shown and the products of its reaction with 1) meta-chloroperoxybenzoic acid, 2) KMnO4 in aqueous acid and 3) O3, followed by Zn in acetic acid.
Answer to Problem 22VC
The name of the alkene shown is 2,5-dimethyl-2-heptene.
The product formed when it reacts with meta-chloroperoxybenzoic acid is
The products formed when it reacts with KMnO4 in aqueous acid are
The products formed when it reacts with O3, followed by Zn in acetic acid are
Explanation of Solution
2,5-dimethyl-2-heptene has a double bond between C2&C3. When treated with meta-chloroperoxybenzoic acid, oxygen atom adds to both C2&C3 to give an epoxide.
In 2,5-dimethyl-2-heptene, C2 has no hydrogens attached. So it is oxidized to a ketone.C3 has one hydrogen and is oxidized to a carboxylic acid.
Upon treatment with O3, followed by Zn in acetic acid the double bond between C2&C3 in 2,5-dimethyl-2-heptene is cleaved and each carbon gets attached to an oxygen atom to give a ketone and aldehyde as products.
The name of the alkene shown is 2,5-dimethyl-2-heptene is
The product formed when it reacts with meta-chloroperoxybenzoic acid is
The products formed when it reacts with KMnO4 in aqueous acid are
The products formed when it reacts with O3, followed by Zn in acetic acid are

b)
Interpretation:
The name of the alkene shown to be given. The products of its reaction with 1) meta-chloroperoxybenzoic acid, 2) KMnO4 in aqueous acid and 3) O3, followed by Zn in acetic acid, are to be given.
Concept introduction:
When alkenes react with meta-chloroperoxybenzoic acid, oxygen atom adds to the double bond to give epoxides.
Upon treatment with KMnO4 in aqueous acid, if in the alkene double bonded carbon is di-substituted and has no hydrogen it is converted in to a ketone and if the double bonded carbon is mono-substituted and has one hydrogen it is oxidized to a carboxylic acid.
Upon treatment with O3, followed by Zn in acetic acid the double bonded is cleaved and each carbon in the double bond gets attached to an oxygen atom to give carbonyl compounds as products. If in the alkene double bonded carbon is di-substituted and has no hydrogen it is converted in to a ketone and if the double bonded carbon is mono-substituted and has one hydrogen it is oxidized to an aldehyde.
To give:
The name of the alkene shown and the products of its reaction with 1) meta-chloroperoxybenzoic acid, 2) KMnO4 in aqueous acid and 3) O3, followed by Zn in acetic acid.
Answer to Problem 22VC
The name of the alkene shown is 3,3-dimethylcyclopentene.
The product formed when it reacts with meta-chloroperoxybenzoic acid is
The products formed when it reacts with KMnO4 in aqueous acid are
The products formed when it reacts with O3, followed by Zn in acetic acid are
Explanation of Solution
3,3-dimethylcyclopentene has a double bond between C1&C2. When treated with meta-chloroperoxybenzoic acid, oxygen atom adds to both C1&C2 to give an epoxide.
In 3,3-dimethylcyclopentene both C1&C2 have one hydrogen attached to them. So they are oxidized to carboxylic acids when treated with KMnO4.
Upon treatment with O3, followed by Zn in acetic acid the double bond between C1&C2 in 3,3-dimethylcyclopentene is cleaved and each carbon with one hydrogen gets attached to an oxygen atom to yield a dialdehyde as product.
The name of the alkene shown is 3,3-dimethylcyclopentene.
The products formed when it reacts with meta-chloroperoxybenzoic acid are
The products formed when it reacts with KMnO4 in aqueous acid are
The products formed when it reacts with O3, followed by Zn in acetic acid are
Want to see more full solutions like this?
Chapter 8 Solutions
Organic Chemistry
- Predict the major organic product(s) of the following reactions. Indicate which of the following mechanisms is in operation: SN1, SN2, E1, or E2.arrow_forward(c) (4pts) Mechanism: heat (E1) CH3OH + 1.5pts each _E1 _ (1pt) Br CH3OH (d) (4pts) Mechanism: SN1 (1pt) (e) (3pts) 1111 I H 10 Ill!! H LDA THF (solvent) Mechanism: E2 (1pt) NC (f) Bri!!!!! CH3 NaCN (3pts) acetone Mechanism: SN2 (1pt) (SN1) -OCH3 OCH3 1.5pts each 2pts for either product 1pt if incorrect stereochemistry H Br (g) “,、 (3pts) H CH3OH +21 Mechanism: SN2 (1pt) H CH3 2pts 1pt if incorrect stereochemistry H 2pts 1pt if incorrect stereochemistryarrow_forwardA mixture of butyl acrylate and 4'-chloropropiophenone has been taken for proton NMR analysis. Based on this proton NMR, determine the relative percentage of each compound in the mixturearrow_forward
- Q5: Label each chiral carbon in the following molecules as R or S. Make sure the stereocenter to which each of your R/S assignments belong is perfectly clear to the grader. (8pts) R OCH 3 CI H S 2pts for each R/S HO R H !!! I OH CI HN CI R Harrow_forwardCalculate the proton and carbon chemical shifts for this structurearrow_forwardA. B. b. Now consider the two bicyclic molecules A. and B. Note that A. is a dianion and B. is a neutral molecule. One of these molecules is a highly reactive compound first characterized in frozen noble gas matrices, that self-reacts rapidly at temperatures above liquid nitrogen temperature. The other compound was isolated at room temperature in the early 1960s, and is a stable ligand used in organometallic chemistry. Which molecule is the more stable molecule, and why?arrow_forward
- Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning

