
Tertiary butyl alcohol gives tertiary butyl chloride on treatment with:
(a) conc. \[HCl\] /anhydrous \[ZnC{l_2}\]
(b) \[KCN\]
(c) \[NaOCl\]
(d) \[C{l_2}\]
Answer
543.6k+ views
Hint: First we have to know the various reagents that can be used to change alcohol into alkyl halides and then we go for further specification of the reagent which doesn't change the degree of the carbon and also the number of carbons before and after the reaction.
Complete answer:
So, the various reagents which change the alcohol into alkyl halides are:
(a) Using Zinc chloride \[ZnC{l_2}\] in presence of $HCl$ :
$R - OH + HCl\xrightarrow{{ZnC{l_2}}}R - Cl + {H_2}O$ (where, $R - $ is alkyl group)
For this reaction, conc. $HCl$ requires anhydrous \[ZnC{l_2}\] as the catalyst.
(b) Using $NaBr$ and ${H_2}S{O_4}$ :
$R - OH + NaBr + {H_2}S{O_4} \to R - Br + NaHS{O_4}$ (where, $R - $ is alkyl group)
This process is used for preparation of alkyl bromide.
(c) Using $P{X_3}$ :
$3R - OH + P{X_3} \to 3R - X + {H_3}P{O_3}$ (where, $R - $ is alkyl group and $X = Cl,Br$ )
(d) Using Red phosphorus in presence of ${X_2}$
$R - OH\xrightarrow[{{X_2} = B{r_2},{I_2}}]{{redP/{X_2}}}R - X$ (where, $R - $ is alkyl group and $X = Br,I$ )
(e) Using $PC{l_5}$ ,
$R - OH + PC{l_5} \to R - Cl + POC{l_3} + HCl$ (where, $R - $ is alkyl group)
(f) Using $SOC{l_2}$ ,
$R - OH + SOC{l_2} \to R - Cl + S{O_2} + HCl$ (where, $R - $ is alkyl group)
In all the above reagents $SOC{l_2}$ is most preferable for making alkyl chloride, because the by-products of this reaction are $S{O_2}$ and $HCl$ which are released in the form of gases. Hence, easy to separable.
So, from the above all discussions,
We are trying to find out the correct option which converts tertiary butyl alcohol to tertiary butyl chloride.
In all the above options, only conc. \[HCl\] /anhydrous \[ZnC{l_2}\] can be used for converting tertiary butyl alcohol to tertiary butyl chloride.
Hence, the correct option is (a) conc. \[HCl\] /anhydrous \[ZnC{l_2}\]
Note:
Degree of the carbon is identified by the number of carbons attached with that carbon for which the degree of carbon is calculated. If the carbon is attached with one carbon, it is called Primary carbon, if with two, Secondary carbon and if with three tertiary carbon.
Complete answer:
So, the various reagents which change the alcohol into alkyl halides are:
(a) Using Zinc chloride \[ZnC{l_2}\] in presence of $HCl$ :
$R - OH + HCl\xrightarrow{{ZnC{l_2}}}R - Cl + {H_2}O$ (where, $R - $ is alkyl group)
For this reaction, conc. $HCl$ requires anhydrous \[ZnC{l_2}\] as the catalyst.
(b) Using $NaBr$ and ${H_2}S{O_4}$ :
$R - OH + NaBr + {H_2}S{O_4} \to R - Br + NaHS{O_4}$ (where, $R - $ is alkyl group)
This process is used for preparation of alkyl bromide.
(c) Using $P{X_3}$ :
$3R - OH + P{X_3} \to 3R - X + {H_3}P{O_3}$ (where, $R - $ is alkyl group and $X = Cl,Br$ )
(d) Using Red phosphorus in presence of ${X_2}$
$R - OH\xrightarrow[{{X_2} = B{r_2},{I_2}}]{{redP/{X_2}}}R - X$ (where, $R - $ is alkyl group and $X = Br,I$ )
(e) Using $PC{l_5}$ ,
$R - OH + PC{l_5} \to R - Cl + POC{l_3} + HCl$ (where, $R - $ is alkyl group)
(f) Using $SOC{l_2}$ ,
$R - OH + SOC{l_2} \to R - Cl + S{O_2} + HCl$ (where, $R - $ is alkyl group)
In all the above reagents $SOC{l_2}$ is most preferable for making alkyl chloride, because the by-products of this reaction are $S{O_2}$ and $HCl$ which are released in the form of gases. Hence, easy to separable.
So, from the above all discussions,
We are trying to find out the correct option which converts tertiary butyl alcohol to tertiary butyl chloride.
In all the above options, only conc. \[HCl\] /anhydrous \[ZnC{l_2}\] can be used for converting tertiary butyl alcohol to tertiary butyl chloride.
Hence, the correct option is (a) conc. \[HCl\] /anhydrous \[ZnC{l_2}\]
Note:
Degree of the carbon is identified by the number of carbons attached with that carbon for which the degree of carbon is calculated. If the carbon is attached with one carbon, it is called Primary carbon, if with two, Secondary carbon and if with three tertiary carbon.
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