\HCHO + \CH_3\CHO \stackrel{\\Delta, \dil. \NaOH }{––––––––🢖} X \stackrel{\HCN}{––––––––🢖}_{\H_3\O^{+}}Y

Correct Answer is : H−C|H=C|H−C|OHH−COOH

\H - \C _{|}_{\H}= \C _{|}_{\H} - \C _{|}_{\OH}\H - \COOH

\C_2\H_5 - \CH_{|}_{\OH} - \COOH

\CH_3 - \CH_{|}↙\stackrel{|}{\OH}^{\CN} - \COOH

AP and TS EAMCET Engineering Model Paper 20

Question : 159 of 160

Marks: +1, -0

HCHO+CH3‌CHO‌ ‌

Δ,dil.NaOH

––––––––🢖

HCN

––––––––🢖

H3O+

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