Comparison of Bond Lengths in Methyl Acetate

To determine the correct order of the carbon-oxygen bond lengths in methyl acetate, we need to analyze the bond orders of the three specified bonds ($\alpha$, $\beta$, and $\gamma$) by considering their resonance structures:

1. Bond $\alpha$ ($\text{C}=\text{O}$ carbonyl bond): In the main contributing canonical structure, the $\alpha$ bond is a double bond. It participates in resonance with the adjacent oxygen atom: $\text{H}_3\text{C}-\overset{\overset{\text{O}}{||}}{\text{C}}-\text{O}-\text{CH}_3 \longleftrightarrow \text{H}_3\text{C}-\overset{\overset{\text{O}^-}{|}}{\text{C}}=\overset{+}{\text{O}}-\text{CH}_3$ Due to this delocalization, the $\alpha$ bond has a major double-bond character with a small fraction of single-bond character. Thus, its bond order is close to $2$ (approximately $1.8 - 1.9$). High bond order corresponds to the shortest bond length.

2. Bond $\beta$ ($\text{C}-\text{O}$ ester acyl bond): In the primary structure, the $\beta$ bond is a single bond. However, the lone pairs on the ester oxygen atom undergo resonance with the carbonyl group, imparting partial double-bond character to the $\beta$ bond ($\text{C}=\text{O}^+$). Thus, its bond order is strictly greater than $1$ (approximately $1.1 - 1.25$). Consequently, its length is intermediate—shorter than a standard single bond, but longer than the double bond $\alpha$.

3. Bond $\gamma$ ($\text{O}-\text{CH}_3$ ester alkyl bond): The $\gamma$ bond connects the ester oxygen to an $\text{sp}^3$-hybridized methyl carbon. This bond does not participate in resonance and retains a pure single-bond character with a bond order of exactly $1$. Therefore, it is the longest bond among the three.

Comparing the bond orders: $\text{Bond Order: } \alpha > \beta > \gamma$

Since bond length is inversely proportional to bond order, the order of bond lengths is: $\alpha < \beta < \gamma$

Thus, the correct option is B.