$Ph-CH=CH-CHO + CH_3-CH=CH-CHO \xrightarrow[EtOH, \Delta]{\text{base}} (A) (87\%)$; The product of this reaction is:

  • A
    $Ph-(CH=CH)_2-CHO$
  • B
    $Ph-(CH=CH)_3-CHO$
  • C
    $Ph-(CH=CH)_4-CHO$
  • D
    $Ph-CH=CH-CH=CH-CH_3$

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