Question
Circle the stronger BASE in each pair. No explanation, no incomplete answers. I have 4 of these correct so I will know if you’re getting them wrong.
3. Circle the stronger base in each pair. (0.2 pts ea = 2.0 pts) a. (CH3) 3COK vs. CH3OK b. HCECNa vs. CH3CH2ONa C. NaNH2 vs.
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Answer #1

Sol :-

(a). (CH3)3COK is stronger base than CH3COK, because in (CH3)3COK has three electron donating groups that is +I -CH3 groups which increase the electron density on oxygen atom and make it a strong base. While CH3COK only contains one -CH3 group.

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(b). HC≡CNa is stronger base than CH3CH2ONa, because in HC≡CH is weaker acid than ethanol C2H5OH, therefore its conjugate base HC≡C- must be stronger base than conjugate base C2H5O- of C2H5OH.

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(c). NaNH2 is stronger base than HC≡CNa, because in NH3 is weaker acid than HC≡CH, therefore its conjugate base NH2- must be stronger base than conjugate base HC≡C- of HC≡CH.

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(d). NaCN is stronger base than NaI, because in HCN is weaker acid than HI, therefore its conjugate base CN- must be stronger base than conjugate base I- of HI.

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(e). NaOH is stronger base than NaHCO3, because in NaHCO3 is the salt of strong base NaOH and weak acid H2CO3. On the other hand NaOH is the stronger base.

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(f). NaOH is stronger base than C6H5COONa, because in C6H5COONa is the salt of strong base NaOH and weak acid C6H5​​​​​​​COOH. On the other hand NaOH is the stronger base.

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(g). C6H5ONa​​​​​​​ is stronger base than C6H5COONa, because in C6H5OH is weaker acid than C6H5​​​​​​​COOH, therefore its conjugate base C6H5O- must be stronger base than conjugate base C6H5​​​​​​​COO- of C6H5​​​​​​​COOH.

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(h). CH3ONa​​​​​​​ is stronger base than CH3COONa, because in CH3OH is weaker acid than CH3COOH, therefore its conjugate base CH3O- must be stronger base than conjugate base CH3COO- of CH3COOH.

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(i). NaH​​​​​​​ is stronger base than CH3ONa, because NaH gives strong base NaOH when dissolve in water.

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(j). NH3 is stronger base than CH3COONa. Because nitrogen in NH3​​​​​​​ good electron donar than oxygen in CH3COONa.

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