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I am trying to measure the reactivity of the carbon (of C=O and C-OH bonds) as described below. Lets assume the groups R,R′

are identical in both molecules. Now I am trying here to to figure out the reactivity of the central carbon towards nucleophilic attack. Now from steric point of view, the ketonic carbon is less hindered, hence more reactive.

Now on charge separation (polarity) point of view, and this is where I am not clear, in case of alcohol, the O-H bond is polar, but to make the central carbon electron deficient the oxygen of O-H group has to pull electrons from σ
bond which is harder than π
bond of ketone.

Hence I think central carbon of ketone (all else identical) is more susceptible to nucleophilic attack.

Can someone kindly comment on my reasoning?

User Palsch
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Final Answer:

Yes, your reasoning is correct. The central carbon in a ketone is indeed more susceptible to nucleophilic attack compared to the central carbon in an alcohol, considering steric hindrance and charge separation factors.

Step-by-step explanation:

In terms of steric hindrance, the ketonic carbon is less hindered compared to the carbon in an alcohol due to the presence of the bulky -OH group in the latter. This makes the ketone more reactive towards nucleophilic attack.

Now, considering charge separation, the polarity of the O-H bond in an alcohol might suggest increased reactivity. However, when examining the mechanism of electron deficiency, it's crucial to note that pulling electrons from a σ bond (as in O-H) is energetically more demanding than from a π bond (as in the C=O of a ketone). The oxygen in the alcohol needs to engage in a more challenging process to make the central carbon electron deficient.

In conclusion, while the O-H bond in an alcohol is polar, the combination of steric hindrance and the energetically more favorable π bond in a ketone makes its central carbon more susceptible to nucleophilic attack. This dual consideration provides a comprehensive understanding of the reactivity difference between the two functional groups.

User OmniBug
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