Question

State True (T) or False (F) and select the correct option:

• [(a)] Alcohols can react both as electrophile and nucleophile.

• [(b)] Ethanol is stronger than methanol with respect to acidic character.

• [(c)] The C--O bond length in phenol is slightly less than that in methanol.

• [(d)] The \(\angle C-O-C\) bond angle in methoxyethane is slightly smaller than normal tetrahedral angle.

• A. (a)-F, (b)-T, (c)-F, (d)-T
• B. (a)-T, (b)-F, (c)-T, (d)-F
• C. (a)-T, (b)-F, (c)-F, (d)-T
• D. (a)-F, (b)-T, (c)-T, (d)-F

Hint:

• Methanol is more acidic than ethanol.

• Phenol has shorter C--O bond due to resonance.

• Alcohols may behave as both nucleophiles and electrophiles.

• In ethers, \(\angle C-O-C\) is generally greater than \(109.5^\circ\).

Answer 1

The Correct Option is B

Concept: The behaviour of alcohols and ethers can be understood from bond polarity, resonance and inductive effects.

Step 1: Analyze Statement (a). Alcohols contain lone pairs on oxygen and can act as nucleophiles. The protonated alcohol can also behave as an electrophile. \[\begin{aligned} \text{Statement (a) is True} \end{aligned}\]

Step 2: Analyze Statement (b). Alkyl groups exhibit \(+I\) effect, which decreases acidity. Since ethanol has a larger electron-releasing group than methanol, \[\begin{aligned} CH_3OH \gt C_2H_5OH \end{aligned}\] in acidic strength. Therefore, \[\begin{aligned} \text{Statement (b) is False} \end{aligned}\]

Step 3: Analyze Statement (c). In phenol, resonance gives partial double-bond character to the C--O bond. Hence the bond length becomes shorter than that in methanol. \[\begin{aligned} \text{Statement (c) is True} \end{aligned}\]

Step 4: Analyze Statement (d). The C--O--C bond angle in ethers is generally larger than the tetrahedral angle due to repulsion between bulky alkyl groups. Thus, \[\begin{aligned} \text{Statement (d) is False} \end{aligned}\]

Step 5: Prepare the truth table. \[ \begin{aligned} (a) &:\quad T \\ (b) &:\quad F \\ (c) &:\quad T \\ (d) &:\quad F \end{aligned} \] \[\begin{aligned} \boxed{ (a)-T,\; (b)-F,\; (c)-T,\; (d)-F } \end{aligned}\] Hence, option \(\mathbf{(B)}\) is correct.