Question
The product formed on reaction of $\mathrm{n}$ -butanol with $\mathrm{SOCl}_{2}$ in presence of pyridine is(a) chlorobutanol(b) 1 -chlorobutane(c) chlorobutanone(d) 2 -chlorobutane
Step 1
Step 1: The reaction starts with n-butanol, which has the formula $\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{OH}$. Show more…
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The product formed on reaction of $\mathrm{n}$ -butanol with $\mathrm{SOCl}_{2}$ in presence of pyridine is (a) chlorobutanol (b) 1 -chlorobutane (c) chlorobutanone (d) 2 -chlorobutane
When (R)-2-butanol is treated with TsCl in Pyridine and then reacted with NaCl, the product is (A) (S)-2-chlorobutane (B) (R)-2-chlorobutane (C) meso-2-chlorobutane (D) a racemic mixture of (A) and (B)
If the reaction of an alcohol with SOCl$_2$ and pyridine follows an S$_N$2 mechanism, what is the stereochemistry of the alkyl chloride formed from ($2R$)-2-butanol?
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Conversion of Alcohols to Alkyl Halides with SOCl2 and PBr3
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