Questions 1. Could 1-pentanol and 2-pentanol be converted to the bromide merely
ID: 590295 • Letter: Q
Question
Questions 1. Could 1-pentanol and 2-pentanol be converted to the bromide merely by shaking with concentrated hydrobromic acid in the manner the 2-methyl-2-butanol did? (Hint: What are the experimental conditions for the Lucas test?) 2. How would you expect the results of the alkoxide-promoted eliminations to change if the alcohols used as solvents contained water? 3. Would an amine be a good solvent for reaction one? Why or why not? 4. What differences in product distribution would be expected for the elimination reactions of 2- and 2-bromo-2,3-dimethylbutane with excess potassium hydroxide inExplanation / Answer
1. Primary and secondary alcohols are less reactive towards Sn1 substitution reaction condition and therefore 1-pentanol and 2-pentanol would not form the bromo product merely by shaling with HBr. the carbocation intermediate formed in the two are not very stable and thus forward reaction is not favoured. On the other hand a tertiary carbocation formed in case of 2-methyl-2-butanol is very stable and thus this reaction is favoured.
2. If alcohols had water in it, the alkoxide would react with water to form alcohol and thus quenching of alkoxide would occur. Reagent is destroyed by this reaction.
3. Amine would react with HBr and form salt and thus Hbr will not be available for the reaction. Thus amine sare not a good solvent for this reaction.
4. More substituted the alkene is, more stable and higher chances of its formation in the reaction. So when 2-bromo-2,3-dimethylbutane is subjected to elimination reaction, we get trisubstituted alkene. Along with this we would also get terminal alkene 2-methyl-1-butene product. However we get exclusively only one product formation from elimination reaction of 2-bromo-2-methylbutane.
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