Q.3,3-Dimethylbutan-2-ol is dehydrated with hot concentrated $\text{H}_2\text{SO}_4$ and gives a product with a rearranged carbon skeleton. Explain, in mechanistic terms, why this rearrangement occurs.

Concentrated $H_2SO_4$ acts as a proton donor; the alcohol oxygen (nucleophilic) accepts a proton first, converting $-OH$ into a better leaving group $-OH_2^+$. The answer is (C).

The acid-catalysed dehydration mechanism always begins with the -OH oxygen picking up a proton from H₂SO₄ to become a good leaving group (water).

The accepted (E1-type) mechanism for acid-catalysed dehydration of an alcohol to an alkene has three steps, in this order:

1. Protonation of the alcohol: the lone pair on the -OH oxygen attacks a proton from conc. H₂SO₄, converting -OH into the much better leaving group -OH₂⁺ (a protonated/oxonium alcohol).
2. Formation of the carbocation: the C-O bond breaks heterolytically, water leaves, and a carbocation forms at that carbon. …

Ethanol undergoes acid-catalysed dehydration with concentrated $\text{H}_2\text{SO}_4$ at 443 K to give ethene.

Concentrated sulphuric acid protonates the –OH group of ethanol, which then leaves as water to form a carbocation; loss of a $\beta$-hydrogen (E1 elimination) gives the alkene. At the higher temperature of 443 K, elimination (dehydration) dominates over substitution. …

Concentrated sulphuric acid at 443 K dehydrates ethanol by an E1 elimination to give ethene.

Mechanism (brief)

Conc. $H_2SO_4$ first protonates the –OH oxygen of ethanol, converting it into a good leaving group ($-OH_2^+$). At 443 K, this leaving group departs to generate a carbocation ($CH_3-CH_2^+$), and a proton is then removed from the adjacent (β) carbon by $HSO_4^-$/water, forming the C=C double bond:

$CH_3CH_2OH \xrightarrow{H^+} CH_3CH_2\overset{+}{O}H_2 \xrightarrow{-H_2O} CH_3-\overset{+}{C}H_2 \xrightarrow{-H^+} CH_2=CH_2$

Overall:

$CH_3CH_2OH \xrightarrow[443\,K]{conc.\ H_2SO_4} CH_2=CH_2 + H_2O$

…

Ethanol undergoes acid-catalysed intermolecular dehydration at the lower of its two characteristic temperatures (413 K) to give diethyl ether; at the higher temperature (443 K) it instead dehydrates intramolecularly to ethene.

When ethanol is heated with excess concentrated sulphuric acid at a lower temperature (413 K, 140 °C), two ethanol molecules combine with loss of one water molecule (intermolecular dehydration), forming diethyl ether:

$2\,CH_3CH_2OH \xrightarrow[\;413\,K\;]{\text{excess conc. }H_2SO_4} CH_3CH_2\text{–O–}CH_2CH_3 + H_2O$

…

At 413 K, excess ethanol with conc. H2SO4 undergoes intermolecular dehydration (an SN2 Williamson-type step) to give diethyl ether; higher temperature (443 K) instead gives ethene by intramolecular (elimination) dehydration.

Concentrated $H_2SO_4$ protonates the -OH of ethanol, making it a good leaving group ($H_2O$). What happens next depends on temperature:

• At 413 K (lower temperature, excess alcohol present): a second, un-protonated ethanol molecule acts as a nucleophile and attacks the protonated ethanol via $S_N2$, displacing water — an intermolecular dehydration giving the ether: …

At the higher temperature of 440 K, concentrated sulphuric acid acts as a dehydrating agent, converting ethanol into ethene via an elimination (E1) pathway.

When ethanol is heated with excess concentrated sulphuric acid at 440 K (a relatively high temperature), the sulphuric acid acts as a strong dehydrating agent, protonating the $-OH$ group (making it a good leaving group, water) and then removing a $\beta$-hydrogen to eliminate water across the C–C bond:

$CH_3{-}CH_2{-}OH \xrightarrow[\ 440\ K\ ]{\text{conc. } H_2SO_4} CH_2{=}CH_2\ (\text{ethene}) + H_2O$

…