a)
Interpretation:
The product of the reaction shown is to be predicted.
Concept introduction:
Alcohols gets converted into the corresponding bromides on treatment with HBr in ether solution.
To predict:
The product of the reaction shown.
Answer to Problem 26AP
The product of the reaction shown is
Explanation of Solution
When 1-methylcyclohexanol is treated with HBr, the OH group in it is replaced by the bromine to yield 1-chloro-1-methylcyclohexane and water as products.
The product of the reaction shown is
b)
Interpretation:
The product of the reaction shown is to be predicted.
Concept introduction:
Alcohols when treated with SOCl2 yield the corresponding alkyl chlorides along with SO2 and HCl.
To predict:
The product of the reaction in which 1-butanol reacts with SOCl2 .
Answer to Problem 26AP
The product of the reaction in which 1-butanol reacts with SOCl2 is 1-chlorobutane.
Explanation of Solution
When 1-butanol reacts with SOCl2, the OH group in is replaced by chlorine to yield 1-chlorobutane. SO2 and HCl are obtained as other products.
The product of the reaction in which 1-butanol reacts with SOCl2 is 1-chlorobutane.
c)
Interpretation:
The product of the reaction shown is to be predicted.
Concept introduction:
Cycloalkenes on treatment with NBS in CCl4 in the presence of light undergo bromination at the allyl position.
To predict:
The product of the reaction shown.
Answer to Problem 26AP
The products of the reaction shown are
Explanation of Solution
NBS is normally used for allylic bromination. The reaction occurs through a radical mechanism. The radical produced by the homolytic cleavage of the C-H bond in the allyl position gets delocalized with the π electrons of the double bond to yield another radical with an odd electron at the junction of the rings. The attack of bromine radical hence leads to two different products as shown.
The products of the reaction shown are
d)
Interpretation:
The product of the reaction shown is to be predicted.
Concept introduction:
Alcohols give the corresponding bromides as products on treatment with PBr3 in ether solution.
To predict:
The product of the reaction in which cyclohexanol reacts with PBr3.
Answer to Problem 26AP
The product of the reaction in which cyclohexanol reacts with PBr3is
Explanation of Solution
When cyclohexanol is treated with PBr3, the OH group in it is replaced by the bromine to yield bromocyclohexane.
The product of the reaction in which cyclohexanol reacts with PBr3 is
e)
Interpretation:
The products of the reactions shown are to be predicted.
Concept introduction:
Alkyl bromides react with Mg in ether to give Grignard reagents. Grignard reagents when treated with water yield the corresponding
To predict:
The products of the reaction shown.
Answer to Problem 26AP
The products of the reaction shown are sec-butyl magnesium bromide (A) and n-butane (B).
Explanation of Solution
sec-butyl bromide when reacted with Mg in ether yields sec-butyl magnesium bromide, a Grignard reagent. The sec-butyl magnesium bromide reacts with water to yield n-butane.
The products of the reaction shown are sec-butyl magnesium bromide (A) and n-butane (B).
f)
Interpretation:
The products of the reaction shown are to be predicted.
Concept introduction:
To predict:
The product of the reaction shown.
Answer to Problem 26AP
The products of the reaction shown are N-butyllithium (A) and Lithium n-dibutyl copper (B).
Explanation of Solution
n-Butyl bromide reacts with Li to yield n-butyl lithium and lithium bromide. The n-butyl lithium produced when reacted with CuI yields Lithium n-dibutyl copper as the product.
The products of the reaction shown are N-butyllithium (A) and Lithium n-dibutyl
g)
Interpretation:
The product of the reaction shown is to be predicted.
Concept introduction:
When alkyl halides are treated with lithium dimethyl copper, the halogens are displaced by an alkyl group and a higher alkane is produced as the product.
To predict:
The product of the reaction in which n-butyl bromide reacts with lithium dimethylcopper.
Answer to Problem 26AP
The product of the reaction in which n-butyl bromide reacts with lithium dimethylcopper is n-pentane.
Explanation of Solution
When n-butyl bromide is treated with lithium dimethylcopper, the bromine is replaced by a methyl group to yield n-pentane as the product.
The product of the reaction in which n-butyl bromide reacts with lithium dimethylcopper is n-pentane.
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Chapter 10 Solutions
OWLv2 with Student Solutions Manual eBook, 4 terms (24 months) Printed Access Card for McMurry's Organic Chemistry, 9th
- Hi!! Please provide a solution that is handwritten. Ensure all figures, reaction mechanisms (with arrows and lone pairs please!!), and structures are clearly drawn to illustrate the synthesis of the product as per the standards of a third year organic chemistry course. ****the solution must include all steps, mechanisms, and intermediate structures as required. Please hand-draw the mechanisms and structures to support your explanation. Don’t give me AI-generated diagrams or text-based explanations, no wordy explanations on how to draw the structures I need help with the exact mechanism hand drawn by you!!! I am reposting this—ensure all parts of the question are straightforward and clear or please let another expert handle it thanks!!arrow_forwardHi!! Please provide a solution that is handwritten. Ensure all figures, reaction mechanisms (with arrows and lone pairs please!!), and structures are clearly drawn to illustrate the synthesis of the product as per the standards of a third year organic chemistry course. ****the solution must include all steps, mechanisms, and intermediate structures as required. Please hand-draw the mechanisms and structures to support your explanation. Don’t give me AI-generated diagrams or text-based explanations, no wordy explanations on how to draw the structures I need help with the exact mechanism hand drawn by you!!! I am reposting this—ensure all parts of the question are straightforward and clear or please let another expert handle it thanks!!arrow_forward. (11pts total) Consider the arrows pointing at three different carbon-carbon bonds in the molecule depicted below. Bond B 2°C. +2°C. < cleavage Bond A • CH3 + 26. t cleavage 2°C• +3°C• Bond C Cleavage CH3 ZC '2°C. 26. E Strongest 3°C. 2C. Gund Largest BDE weakest bond In that molecule a. (2pts) Which bond between A-C is weakest? Which is strongest? Place answers in appropriate boxes. Weakest C bond Produces A Weakest Bond Most Strongest Bond Stable radical Strongest Gund produces least stable radicals b. (4pts) Consider the relative stability of all cleavage products that form when bonds A, B, AND C are homolytically cleaved/broken. Hint: cleavage products of bonds A, B, and C are all carbon radicals. i. Which ONE cleavage product is the most stable? A condensed or bond line representation is fine. 人 8°C. formed in bound C cleavage ii. Which ONE cleavage product is the least stable? A condensed or bond line representation is fine. methyl radical •CH3 formed in bund A Cleavagearrow_forward
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Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning