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According to Blanc Rule for Dibasic Carboxylic acids, A cyclic Ketone is formed unless a five or six membered cyclic acid anhydride can be formed. This rule is valid for 1,4; 1,5; 1,6. dicarboxylic acids

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On rotating the compound B by 180°, we get the two -OH groups on the left side and the two H atoms on the right. In other words, B would then be an enantiomer of A. Note: on rotating a Fischer projection by 180°, the compound remains the same; it's just that all the atoms have to be rotated, i.e. the atoms on C3 in this case would go to C2 and vice versa.

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I am directly going to address the source of your confusion: $\text{+M}$ of -$\ce{OCH3}$ is more than -$\ce{OH}$. You see, you are basically saying that the $\text{+I}$ effect of $\ce{CH3}$ in -$\ce{OCH3}$ is going to push more electron density towards the ring, hence, -$\ce{OCH3}$ is more activating than -$\ce{OH}$. But look at the bigger picture, and let'...

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Both -$\ce{OCH3}$ and -$\ce{OH}$ groups have exhibited two effects on the aromatic ring: (1) Electron donating resonance or mesomeric effect (+M) and (2) Electron withdrawing inductive effect (-I). For both acids in hand, the electron withdrawing inductive effect (-I) is almost same since both -$\ce{OCH3}$ and -$\ce{OH}$ groups are 4 carbons away from the ...

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