Shown below is the structure of methyl acetate with three different \(\alpha\), \(\beta\) and \(\gamma\) carbon–oxygen bonds. The correct order of bond lengths of these bonds is :}
Step 1: Understanding the Question:
We need to compare the lengths of three different carbon-oxygen bonds in the methyl acetate molecule (\(CH_{3}-CO-OCH_{3}\)). Step 2: Key Formula or Approach:
1. Bond Length Order: Triple Bond<Double Bond<Single Bond.
2. Resonance effect: Resonance can give partial double bond character to a single bond, making it shorter.
3. Electronegativity/Hybridization: Bonds involving higher 's' character (like \(sp^{2}\) vs \(sp^{3}\)) are generally shorter. Step 3: Detailed Explanation:
The structure of Methyl Acetate is: \(CH_{3}-C(=O_{\alpha})-O_{\beta}-CH_{3(\gamma)}\).
1. Bond \(\alpha\): This is a C=O double bond. Double bonds are shorter than single bonds. Thus, \(\alpha\) is the shortest.
2. Bond \(\beta\): This is a \(C-O\) bond where the Oxygen is attached to the carbonyl Carbon. In esters, there is resonance:
\[ CH_{3}-C(=O)-O-CH_{3} \longleftrightarrow CH_{3}-C(O^{-})=O^{+}-CH_{3} \]
Due to this resonance, the \(\beta\) bond has partial double bond character. This makes it shorter than a pure single bond.
3. Bond \(\gamma\): This is a pure \(O-CH_{3}\) single bond. There is no resonance involving this bond that gives it double bond character. It involves an \(sp^{3}\) hybridized carbon.
Comparing \(\beta\) and \(\gamma\): \(\beta\) is shorter than \(\gamma\) because \(\beta\) has partial double bond character and involves an \(sp^{2}\) carbon. Step 4: Final Answer:
The order is Double Bond (\(\alpha\))<Partial Double Bond (\(\beta\))<Pure Single Bond (\(\gamma\)).
Order: \(\alpha<\beta<\gamma\).
