Q.A) Write the mechanism of dehydration of ethanol to form ethene. B) Give the name of Enzyme used to convert glucose into ethanol. C) Write chemical equation of phenol with chloroform and KOH. OR A) How to convert methanol into ethanol? Write chemical equations only. B) The mixture of alcohol and Ether, used in place of petrol, is known as? C) How will you obtain p-hydroxybenzaldehyde from phenol? Give chemical equation.

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:

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

…