Q1. Out of t-butyl alcohol and n-butanol, which one will undergo acid catalyzed dehydration faster and why?
Q2. Carry out the following conversions:
(i) Phenol to Salicylaldehyde.
(ii) t-butylchloride to t-butyl ethyl ether.
(iii) Propene to Propanol
Q1. Ans
t-Butyl alcohol will undergo acid-catalyzed dehydration faster than n-butanol because:
Acid-catalyzed dehydration of alcohols proceeds via the formation of a carbocation intermediate (E1 mechanism).
In t-butyl alcohol, the dehydration gives a tertiary carbocation (CH3)3C+(CH_3)_3C^+(CH3?)3?C+, which is highly stable due to +I effect of three methyl groups and hyperconjugation.
In n-butanol, dehydration gives a primary carbocation, which is very unstable.
Therefore, t-Butyl alcohol undergoes dehydration faster than n-butanol because it forms a highly stable tertiary carbocation during the reaction, whereas n-butanol forms a less stable primary carbocation.
Q2. Ans
(i) Phenol → Salicylaldehyde
Step 1: Phenol (C?H?OH) is treated with chloroform (CHCl?) and aqueous sodium hydroxide (NaOH).
Step 2: This is the Reimer–Tiemann Reaction.
Step 3: Mechanism
In the presence of strong base (NaOH), chloroform forms a reactive intermediate (dichlorocarbene :CCl?).
This intermediate attacks the ortho position of the activated benzene ring (due to –OH group).
Hydrolysis of the resulting intermediate introduces an –CHO group at the ortho position.
Step 4: The final product is o-Hydroxybenzaldehyde (Salicylaldehyde).
Final Answer: Phenol, when treated with chloroform (CHCl?) and sodium hydroxide (NaOH), gives Salicylaldehyde via the Reimer–Tiemann reaction, where a –CHO group is introduced at the ortho position of phenol.
ii) t-Butyl chloride → t-Butyl ethyl ether
Step 1: Take t-Butyl chloride (CH3)3C–Cl(CH?)?C–Cl(CH3?)3?C–Cl and sodium ethoxide (C2H5ONa)(C?H?ONa)(C2?H5?ONa) in ethanol.
Step 2: This reaction is an example of Williamson’s Ether Synthesis.
Step 3: Mechanism
Since t-Butyl chloride is a tertiary halide, the reaction does not proceed by S<sub>N</sub>2 (because of strong steric hindrance).
Instead, it goes by the SN1 mechanism:
(i) First, the C–Cl bond breaks, forming a stable t-butyl carbocation (CH3)3C+
(ii) Then, the ethoxide ion (C?H?O?) attacks this carbocation.
Step 4: The final product formed is t-Butyl ethyl ether (CH3)3C–OC2H5(CH?)?C–OC?H?(CH3?)3?C–OC2?H5?, along with NaCl.
Final Answer: t-Butyl chloride reacts with sodium ethoxide in ethanol to give t-butyl ethyl ether by Williamson’s ether synthesis, following the SN1 pathway.
iii) Propene → Propanol
Step 1: Take propene (CH?–CH=CH?) and react it with water (H?O) in the presence of dilute sulphuric acid (H?SO?) as a catalyst.
Step 2: This is an example of Electrophilic Addition Reaction (Hydration of alkene).
Step 3: Mechanism
The double bond in propene gets protonated by H? from H?SO?, forming a carbocation. By Markovnikov’s rule, the H? attaches to the carbon with more hydrogens, giving a secondary carbocation (CH?–CH?–CH?).
Water molecule (H?O) attacks the carbocation, forming a protonated alcohol.
Loss of a proton (H?) gives the neutral alcohol.
Step 4: The final product is Propan-2-ol (CH?–CHOH–CH?), i.e., isopropanol.
Final Answer: Propene reacts with water in the presence of dilute H?SO? to give Propan-2-ol (propanol) by acid-catalyzed hydration following Markovnikov’s rule.