Acidic dehydration of alcohol involves

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(D) Acidic dehydration of alcohols typically proceeds via the $E_1$ mechanism. $1$. The hydroxyl group is protonated to form a good leaving group $(-O

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(D) Acidic dehydration of alcohols typically proceeds via the $E_1$ mechanism. $1$. The hydroxyl group is protonated to form a good leaving group $(-OH_2^+)$. $2$. The leaving group departs to form a carbocation intermediate. $3$. If the carbocation can rearrange to a more stable form (e.g.,from secondary to tertiary),it will do so. $4$. Finally,a proton is removed to form an alkene,where the most substituted alkene (Saytzeff's product) is generally the major product due to its higher stability. Therefore,all the given statements are correct.

Similar Questions

What are $X$ and $Y$ in the following reactions?
$R-CH_2OH \xrightarrow{Cu/573 \ K} X$
$(CH_3)_3C-OH \xrightarrow{Cu/573 \ K} Y$

Consider the following alcohols:
$(I)$ $C_6H_5CH_2OH$
$(II)$ $O_2N-C_6H_4-CH_2OH$ (para-nitro)
$(III)$ $CH_3O-C_6H_4-CH_2OH$ (para-methoxy)
$(IV)$ $Br-C_6H_4-CH_2OH$ (para-bromo)
The order of decreasing reactivities of these alcohols towards nucleophilic substitution with $HBr$ is:

$R-OH + HX \to R-X + H_2O$. In the above reaction,the reactivity of different alcohols is:

$CH_3-C(CH_3)_2-Cl$ $\xrightarrow{NaCN} A$ $\xrightarrow{dil. H_2SO_4} B$. Compound $B$ is:

Arrange the following alcohols in the decreasing order of their rate of dehydration via the $S_{N}1$ mechanism:

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