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  • (a) Out of t-butyl alcohol and n-butanol, which one will undergo acid catalyzed dehydration faster and why? Unit-7 2 (b) Carry out the following conversions (i) Phenol to Sa

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

Answers (1)

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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

  1. In the presence of strong base (NaOH), chloroform forms a reactive intermediate (dichlorocarbene :CCl?).

  2. This intermediate attacks the ortho position of the activated benzene ring (due to –OH group).

  3. 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

  1. Since t-Butyl chloride is a tertiary halide, the reaction does not proceed by S<sub>N</sub>2 (because of strong steric hindrance).

  2. 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

  1. 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?).

  2. Water molecule (H?O) attacks the carbocation, forming a protonated alcohol.

  3. 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.

Posted by

Saumya Singh

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