Which of the following sequence of reagents c | vedclass

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(A) The conversion of propene $(CH_3-CH=CH_2)$ to $1$-chloropropane $(CH_3-CH_2-CH_2-Cl)$ requires an anti-Markovnikov hydration followed by the conve

Vedclass · Accepted Answer

$(i) (BH_3)_2, (ii) H_2O_2 / OH^- ; HCl, ZnCl_2$

Vedclass · Answer

(A) The conversion of propene $(CH_3-CH=CH_2)$ to $1$-chloropropane $(CH_3-CH_2-CH_2-Cl)$ requires an anti-Markovnikov hydration followed by the conversion of the primary alcohol to an alkyl chloride.Step $1$: Hydroboration-oxidation of propene using $(i) (BH_3)_2$ and $(ii) H_2O_2 / OH^-$ yields propan-$1$-ol $(CH_3-CH_2-CH_2-OH)$.Step $2$: The conversion of propan-$1$-ol to $1$-chloropropane is typically achieved using $HCl$ in the presence of a catalyst like $ZnCl_2$ (Lucas reagent) or other chlorinating agents. Although primary alcohols react slowly with Lucas reagent,the sequence provided in option $A$ represents the standard synthetic route for this transformation.

Which of the following sequence of reagents converts propene to $1$-chloropropane?

Similar Questions

In the following sequence of reactions,the product $C$ is: $CH_3-CH_2-CH_2-Br$ $\xrightarrow{KOH(alc)} (A)$ $\xrightarrow{HBr} (B)$ $\xrightarrow{KOH(aq.)} (C)$

$CH \equiv CH$ $\xrightarrow{Hg^{2+}/H_2SO_4} B$ $\xrightarrow{CH_3MgX/H_2O} C$ $\xrightarrow{[O]} D$. What is the final product $D$?

Which of the following alcohols will yield alkyl chloride at once on reaction with concentrated $HCl$ and $ZnCl_2$ at room temperature?

$CH_3MgBr + \text{Cyclopentanone}$ $\xrightarrow{H_3O^{+}} A$ $\xrightarrow{HBr} B$ $\xrightarrow{Mg/\text{ether}} C$ $\xrightarrow{HCHO}_{H_3O^{+}} D$ $\xrightarrow{HI} E$
$E$ is

The reaction shown below is an example of:
$CH_2=CH-COOCH_3 + CH_3CH_2CH_2OH \xrightarrow{\text{Conc. } HCl} CH_2=CH-COOCH_2CH_2CH_3 + CH_3OH$

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