Question

When ethanol is treated with sulphuric acid at \(413\ K\), the reaction involved in it is:

• A. \(S_N2\)
• B. \(S_N1\)
• C. Addition
• D. Elimination

Hint:

Ethanol with concentrated \(H_2SO_4\): \[ 413\ K \rightarrow \text{Diethyl ether} \quad (S_N2) \] \[ 443\ K \rightarrow \text{Ethene} \quad (\text{Elimination}) \] Temperature determines the major product.

Answer 1

The Correct Option is A

Step 1: Recall the temperature-dependent behavior of ethanol with concentrated $H_2SO_4$.
At 413 K (about 140 degrees C): major product is diethyl ether (intermolecular dehydration). At 443 K (about 170 degrees C): major product is ethene (intramolecular elimination). The question specifies 413 K, so ether formation is favored.
Step 2: Write the overall reaction at 413 K.
\[ 2C_2H_5OH \xrightarrow[413\ K]{\text{Conc. }H_2SO_4} C_2H_5{-}O{-}C_2H_5 + H_2O \]
Step 3: First step of the mechanism: protonation of ethanol.
\[ C_2H_5OH + H^+ \rightarrow C_2H_5\overset{+}{O}H_2 \] The $-OH$ is converted into a much better leaving group ($H_2O$).
Step 4: Second step: nucleophilic attack by a second ethanol molecule.
A second ethanol molecule attacks the carbon of the protonated ethanol from the back (backside attack). Water is displaced as the leaving group in a single concerted step. This is the $S_N2$ mechanism.
Step 5: Why not $S_N1$?
An $S_N1$ mechanism would require formation of a primary carbocation ($C_2H_5^+$), which is extremely unstable and not formed under normal conditions. $S_N1$ is ruled out.
Step 6: Why not elimination at 413 K?
Elimination (producing ethene) requires breaking a $C{-}H$ bond and is thermodynamically disfavored at the lower temperature of 413 K. At 443 K, elimination becomes predominant. At 413 K, the $S_N2$ pathway to ether dominates.
Step 7: State the final answer.
\[ \boxed{S_N2} \]