(a)
Interpretation:
The major product has to be identified.
Concept introduction:
SN1 reaction:
The reaction of alcohols with acids like hydrochloric acid or hydrobromic which yield the corresponding carbocation intermediate, this carbocation intermediate undergoes substitution reaction which yields the corresponding substitution product.
Tertiary alcohols undergo substitution very fast than the secondary alcohols because tertiary carbocation is more stable than the secondary carbocation than the primary carbocation.
Primary alcohol is less stable therefore it won’t undergoes SN1 substitution reaction.
(b)
Interpretation:
The major product has to be identified.
Concept introduction:
SN2 reaction:
The alcohols are reaction with acids like hydrochloric acid or hydrobromic which yield the corresponding substitution product. Primary alcohol undergoes SN2 substitution reaction than secondary alcohol than tertiary alcohol because SN2 reaction is simultaneous reaction.
(c)
Interpretation:
The major product has to be identified.
Concept introduction:
Dehydration reaction:
Removal of water molecule from the reaction, the alcohol is treated with strong acid like sulfuric acid.
The stability of carbocation is given below,
Tertiary carbocation is more stable than the secondary and primary.
(d)
Interpretation:
The major product should be identified.
Concept introduction:
SN1 reaction:
The alcohols is reaction with acids like hydrochloric acid or hydrobromic which yield the corresponding carbocation intermediate, this carbocation intermediate undergoes substitution reaction which yields the corresponding substitution product.
Tertiary alcohols undergo substitution very fast than the secondary alcohols because tertiary carbocation is more stable than the secondary carbocation than the primary carbocation.
Primary alcohol is less stable therefore it won’t undergoes SN1 substitution reaction.
(e)
Interpretation:
The major product should be identified.
Concept introduction:
In the presence of acid catalyst, this reaction takes place through partial SN1 and partial SN2 pathway.
Epoxides are reactive, methoxide ion attacks the Epoxides in a less sterically hindered position which forms the alkoxide ion, and then it gets proton from alcohol which form the product.
(f)
Interpretation:
The major product should be identified.
Concept introduction:
In the presence of acid catalyst, this reaction takes place through partial SN1 and partial SN2 pathway. It is not a pure SN1 reaction because a carbocation is not formed fully and not a pure SN2 reaction because the leaving group begins to depart before the compound is attacked by the nucleophile. Epoxides are reactive; Epoxides get protonated followed by alcohol attacks to the stable carbocation and form the product.
Epoxides are reactive, methoxide ion attacks the Epoxides in a less sterically hindered position which forms the alkoxide ion, and then it gets proton from alcohol which form the product. When a nucleophile attacks an unprotonated epoxide, the reaction is a pure SN2 reaction.
Note: Under acidic conditions, the nucleophile preferentially attacks the more substuituted ring carbon. Under Basic conditions, the nucleophile preferentially attacks the less substuituted ring carbon.

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Chapter 9 Solutions
EBK ESSENTIAL ORGANIC CHEMISTRY
- How do I explain this? Thank you!arrow_forwardWhen an unknown amine reacts with an unknown acid chloride, an amide with a molecular mass of 163 g/mol (M* = 163 m/z) is formed. In the infrared spectrum, important absorptions appear at 1661, 750 and 690 cm. The 13C NMR and DEPT spectra are provided. Draw the structure of the product as the resonance contributor lacking any formal charges. 13C NMR DEPT 90 200 160 120 80 40 0 200 160 120 80 40 0 DEPT 135 T 200 160 120 80 40 0 Draw the unknown amide. Select Dow Templates More Fragearrow_forwardIdentify the unknown compound from its IR and proton NMR spectra. C4H6O: 'H NMR: 82.43 (1H, t, J = 2 Hz); 8 3.41 (3H, s); 8 4.10 (2H, d, J = 2 Hz) IR: 2125, 3300 cm¹ The C4H6O compound liberates a gas when treated with C2H5 MgBr. Draw the unknown compound. Select Draw с H Templates Morearrow_forward
- Please help with number 6 I got a negative number could that be right?arrow_forward1,4-Dimethyl-1,3-cyclohexadiene can undergo 1,2- or 1,4-addition with hydrogen halides. (a) 1,2-Addition i. Draw the carbocation intermediate(s) formed during the 1,2-addition of hydrobromic acid to 1,4-dimethyl-1,3-cyclohexadiene. ii. What is the major 1,2-addition product formed during the reaction in (i)? (b) 1,4-Addition i. Draw the carbocation intermediate(s) formed during the 1,4-addition of hydrobromic acid to 1,4-dimethyl-1,3-cyclohexadiene. ii. What is the major 1,4-addition product formed from the reaction in (i)? (c) What is the kinetic product from the reaction of one mole of hydrobromic acid with 1,4-dimethyl-1,3-cyclohexadiene? Explain your reasoning. (d) What is the thermodynamic product from the reaction of one mole of hydrobro-mic acid with 1,4-dimethyl-1,3-cyclohexadiene? Explain your reasoning. (e) What major product will result when 1,4-dimethyl-1,3-cyclohexadiene is treated with one mole of hydrobromic acid at - 78 deg * C ? Explain your reasoning.arrow_forwardGive the product of the bimolecular elimination from each of the isomeric halogenated compounds. Reaction A Reaction B. КОВ CH₂ HotBu +B+ ко HOIBU +Br+ Templates More QQQ Select Cv Templates More Cras QQQ One of these compounds undergoes elimination 50x faster than the other. Which one and why? Reaction A because the conformation needed for elimination places the phenyl groups and to each other Reaction A because the conformation needed for elimination places the phenyl groups gauche to each other. ◇ Reaction B because the conformation needed for elimination places the phenyl groups gach to each other. Reaction B because the conformation needed for elimination places the phenyl groups anti to each other.arrow_forward
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- Complete the clean-pushing mechanism for the given ether synthesia from propanol in concentrated sulfurica140°C by adding any mining aloms, bands, charges, nonbonding electron pairs, and curved arrows. Draw hydrogen bonded to cayan, when applicable. ore 11,0 HPC Step 1: Draw curved arrows Step 2: Complete the intend carved Q2Q 56 QQQ Step 3: Complete the intermediate and add curved Step 4: Modify the structures to draw the QQQ QQQarrow_forward6. In an experiment the following replicate set of volume measurements (cm3) was recorded: (25.35, 25.80, 25.28, 25.50, 25.45, 25.43) A. Calculate the mean of the raw data. B. Using the rejection quotient (Q-test) reject any questionable results. C. Recalculate the mean and compare it with the value obtained in 2(a).arrow_forwardA student proposes the transformation below in one step of an organic synthesis. There may be one or more reactants missing from the left-hand side, but there are no products missing from the right-hand side. There may also be catalysts, small inorganic reagents, and other important reaction conditions missing from the arrow. • Is the student's transformation possible? If not, check the box under the drawing area. • If the student's transformation is possible, then complete the reaction by adding any missing reactants to the left-hand side, and adding required catalysts, inorganic reagents, or other important reaction conditions above and below the arrow. • You do not need to balance the reaction, but be sure every important organic reactant or product is shown. + T G OH де OH This transformation can't be done in one step.arrow_forward
