$2 {C H}_{3} C H O A l {(O E t)}_{3} - ---- \to {C H}_{3} C O O {C H}_{2} {C H}_{3}$
The above reaction is disproportionation reaction. If true enter 1, else enter 0.

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Solution

${C H}_{3} C O O C_{6} H_{5}$ (ester) is formed from ${C H}_{3} C O O H$ (oxidation product of ${C H}_{3} C H O$ ) and ${C H}_{3} {C H}_{2} O H$ (reduction product ${C H}_{3} C H O$ ).
Hence, ${C H}_{3} C H O$ disproportionates in presence of aluminium ethoxide (Tischenko reaction).

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