Q.63. Shown below is the structure of methyl acetate with three different , and carbon-oxygen bond

Q: What is the correct answer for this exam Chemistry Q63?

The correct answer is B. Sol. The structure of methyl acetate is $H_3C-C(=O)-O-CH_3$. The $\alpha$ bond is $C=O$, $\beta$ bond is $C-O$ (ester linkage), and $\gamma$ bond is $O-CH_3$.Due to resonance, the $C=O$ bond ($\alpha$) has partial single bond character, and the $C-O$ bond ($\beta$) has partial double bond character. The $O-CH_3$ bond ($\gamma$) is a pure single bond.The resonance structures are: $H_3C-C(=O)-O-CH_3

Q: Q.63. Shown below is the structure of methyl acetate with three different , and carbon-oxygen bonds. [Figure: Structure of methyl acetate with bonds labeled , , .]The correct order of bond length…

The correct answer is B. Sol. The structure of methyl acetate is $H_3C-C(=O)-O-CH_3$. The $\alpha$ bond is $C=O$, $\beta$ bond is $C-O$ (ester linkage), and $\gamma$ bond is $O-CH_3$.Due to resonance, the $C=O$ bond ($\alpha$) has partial single bond character, and the $C-O$ bond ($\beta$) has partial double bond character. The $O-CH_3$ bond ($\gamma$) is a pure single bond.The resonance structures are: $H_3C-C(=O)-O-CH_3

Q: What are the options for this this exam Chemistry question?

A. $\alpha > \beta > \gamma$ · B. $\alpha < \beta < \gamma$ · C. $\alpha = \beta = \gamma$ · D. $\alpha < \beta = \gamma$

Question

Q.63. Shown below is the structure of methyl acetate with three different ,  and  carbon-oxygen bonds. [Figure: Structure of methyl acetate with bonds labeled , , .]The correct order of bond length of these bonds is :

FirstInTest · Accepted Answer

Sol. The structure of methyl acetate is $H_3C-C(=O)-O-CH_3$. The $\alpha$ bond is $C=O$, $\beta$ bond is $C-O$ (ester linkage), and $\gamma$ bond is $O-CH_3$.Due to resonance, the $C=O$ bond ($\alpha$) has partial single bond character, and the $C-O$ bond ($\beta$) has partial double bond character. The $O-CH_3$ bond ($\gamma$) is a pure single bond.The resonance structures are: $H_3C-C(=O)-O-CH_3 \leftrightarrow H_3C-C(O^-)=O^+-CH_3$.This means the $\alpha$ bond is shorter than a typical $C=O$ double bond (due to partial single bond character) and the $\beta$ bond is shorter than a typical $C-O$ single bond (due to partial double bond character). The $\gamma$ bond is a pure single bond.Therefore, the order of bond lengths is $\alpha < \beta < \gamma$.

Answer

$\alpha > \beta > \gamma$

Answer

$\alpha < \beta < \gamma$

Answer

$\alpha = \beta = \gamma$

Answer

$\alpha < \beta = \gamma$

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