Interpretation:
The reason has to be explained for formation of racemic mixture of 2-bromooctane when the reaction of NaBr with (R)-2-Bromooctane.
Concept introduction:
SN2 reaction:
The alcohol is reaction with acids like hydrochloric acid or hydrobromic acid, the bromine atom attacks back side of the carbon atoms in simultaneous manner and which is bearing alcohol group which yield the corresponding product.
Example:
The tosylated compound is reaction with sodium methoxide which undergoes again type of reaction, the methoxide ion attacks the carbon atom through the back side and provides Inverse configuration of methoxy compound. This is shown below,
Given information:
The given compound is shown below,
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Chapter 11 Solutions
Organic Chemistry
- Explain this observation: Ethyl 3-phenylpropanoate(C6H5CH2CH2CO2CH2CH3 ) reacts with electrophiles to afford ortho- and para-disubstituted arenes, but ethyl 3-phenylprop-2-enoate (C6H5CH=CHCO2CH2CH3) reacts with electrophiles to afford metadisubstituted arenes.arrow_forwardIn an aqueous solution containing sodium bicarbonate, aniline reacts quickly withbromine to give 2,4,6-tribromoaniline. Nitration of aniline requires very strong conditions,however, and the yields (mostly m-nitroaniline) are poor.(a) What conditions are used for nitration, and what form of aniline is present under theseconditions?arrow_forwardH 9. (a) Optically active 2-bromobutane undergoes racemization on treatment with a solution of KBr. Give a mechanism for this racemization. (b) In contrast, optically active butan-2-ol does not racemize on treatment with a solution of KOH. Explain why a reaction like that in part (a) does not occur. (c) Optically active butan-2-ol racemizes in dilute acid. Propose a mechanism for this racemization.arrow_forward
- Compounds X and Y are both C7H15Cl products formed in the radical chlorination of 2,4-dimethylpentane. Base-promoted E2 elimination of X and Y gives, in each case, a single C7H₁4 alkene. Both X and Y undergo an SN2 reaction with sodium iodide in acetone solution to give C7H15l products; in this reaction Y reacts faster than X. What is the structure of X? • Do not use stereobonds in your answer. • In cases where there is more than one possible structure for each molecule, just give one for each. . Draw one structure per sketcher. Add additional sketchers using the drop-down menu in the bottom right corner. Separate structures with + signs from the drop-down menu. наarrow_forwardThe reaction of 3,4-dimethyl-3-hexanol (3,4-dimethylhexan-3-ol) with HBr generates compound A as the major product. Treatment of compound A with a strong base gives two isomers of compound B as the major product, along with one isomer of compound C and one isomer of compound D as minor products, all of which have one double-bond equivalent. Identify compounds A, B, C, and D and give their names. By what mechanism does the reaction of 3,4-dimethyl-3-hexanol with HBr occur? By what mechanism does the reaction of A with strong base to form B occur? Propose reaction conditions for an alternative, one-step method for converting 3,4-dimethyl-3-hexanol directly to compound B.arrow_forward2-bromo-2-methylbutane undergoes hydrolysis reaction with water, H2O toform compound W. Compound X and compound Y are produced when 2-bromo-2-methylbutane undergoes elimination reaction with alcoholic ofsodium hydroxide, NaOH. (ii) What is the type of reaction involved in the formation of compound W? (iii) Identify the major product of the elimination reaction between compound Xand compound Y based on Zaitsev’s rule.arrow_forward
- please write out answers and explanations to make it easy to understandarrow_forwardFriedel–Crafts alkylation of benzene with (R)-2-chlorobutane and AlCl3 affords sec-butylbenzene. Would you expect the product to exhibit optical activity? Explain, with reference to the mechanism.arrow_forwardPredict the major monoalkylation products you would expect to obtain from reactionof the following substances with, chloromethane and AlCl3:(a) Bromobenzene(b) m-Bromophenol(c) p-Chloroaniline(d) 2, 4-Dichloronitrobenzene(e) 2, 4-Dichlorophenol(f) Benzoic acid(g) p-Methylbenzenesulfonic acid(h) 2, 5-Dibromotoluenearrow_forward
- Reaction of (CH3)3CCHO with (C6H5)3P=C(CH3)OCH3, followed by treatment with aqueous acid, affords R (C7H14O). R has a strong absorption in its IR spectrum at 1717 cm−1 and three singlets in its 1H NMR spectrum at 1.02 (9 H), 2.13 (3 H), and 2.33 (2 H) ppm. What is the structure of R? We will learn about this reaction in Chapter 18.arrow_forwardExplain the following observation. Ethyl 3-phenylpropanoate (C6H5CH2CH2CO2CH2CH3) reacts with electrophiles to afford ortho- and para-disubstituted arenes, but ethyl 3-phenylprop-2-enoate (C6H5CH= CHCO2CH2CH3) reacts with electrophiles to afford meta-disubstituted arenes.arrow_forwardIdentify products A and B from the given 1H NMR data. Treatment of acetone [(CH3)2C=O] with dilute aqueous base forms B. Compound B exhibits four singlets in its 1H NMR spectrum at 1.3 (6 H), 2.2 (3 H), 2.5 (2 H), and 3.8 (1H) ppm. What is the structure of B?arrow_forward
- Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning