Which one of the following statements is not correct?

We start by testing each of the four statements one by one, because the question asks us to find the single statement that is not correct.

Statement A says: “Alcohols are weaker acids than water.”

To compare acidities we use the definition $$pK_a = -\log K_a.$$ A larger $$pK_a$$ means a smaller $$K_a$$ and therefore a weaker acid. Experimentally we have

$$pK_a(\text{H}_2\text{O}) \approx 15.7,$$ $$pK_a(\text{ROH})$$ (for a simple primary alcohol such as ethanol) $$\approx 16\text{--}18.$$

Because $$16-18 \gt 15.7,$$ the $$pK_a$$ of an alcohol is larger, so its $$K_a$$ is smaller, hence an alcohol is indeed a weaker acid than water. Therefore Statement A is correct.

Statement B claims: “Acid strength of alcohols decreases in the order $$\text{RCH}_2\text{OH} \gt \text{R}_2\text{CHOH} \gt \text{R}_3\text{COH}.$$”

Acidity depends on the stability of the conjugate base. After deprotonation an alkoxide ion $$\text{RO}^-$$ is formed. Alkyl groups are electron-donating by the inductive effect $$\left(+I\right).$$ More alkyl groups push electron density toward oxygen, destabilising the negative charge. Hence:

• A primary alkoxide $$(\text{RCH}_2\text{O}^-)$$ has one $$+I$$ group.
• A secondary alkoxide $$(\text{R}_2\text{CHO}^-)$$ has two $$+I$$ groups and is less stable.
• A tertiary alkoxide $$(\text{R}_3\text{CO}^-)$$ has three $$+I$$ groups and is the least stable.

Lower stability of the conjugate base means lower acidity of the conjugate acid, so the acid strength indeed falls in the sequence primary > secondary > tertiary. Hence Statement B is correct.

Statement C says: “Carbon-oxygen bond length in methanol, $$\text{CH}_3\text{OH},$$ is shorter than that of the C-O bond length in phenol.”

Let us recall typical experimental bond lengths:

• In methanol (a simple aliphatic alcohol) the $$\text{C-O}$$ bond is a pure single bond; its length is about $$143\ \text{pm}.$$
• In phenol, resonance causes partial double-bond character between the ring carbon and oxygen:

$$\text{Ar-O-H} \;\;\;\rightleftharpoons\;\;\; \text{Ar}=\text{O}^- -H^+$$

Because of this partial double-bond character the $$\text{C-O}$$ bond in phenol is shorter, about $$136\ \text{pm}.$$ Numerically, $$136\ \text{pm} \lt 143\ \text{pm},$$ so the bond in phenol is the shorter one. Statement C claims the opposite, therefore Statement C is not correct.

Statement D states: “The bond angle O-C-H in methanol is 108.9°.”

In methanol the carbon atom is $$sp^3$$-hybridised. An ideal $$sp^3$$ angle is $$109.5^\circ.$$ Slight differences arise because an $$\text{O-C}$$ bond and $$\text{C-H}$$ bonds are not identical in size or repulsion. Experimental gas-phase measurements give $$\angle(\text{O-C-H}) \approx 108.9^\circ,$$ well within normal error of the ideal tetrahedral value. Therefore Statement D is correct.

Summarising our findings:

• A is correct, • B is correct, • C is incorrect, • D is correct.

Hence, the only statement that is not correct is Statement C, which corresponds to Option 3.

Hence, the correct answer is Option C.