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Reaction of 1 molr of each $Br_2$ and phco$CH_2CH_3$ in basic solution yields 0.5 moles of ph$COCBr_2CH_3$ And 0.5 moles of unreacted ph$COCH_2CH_3$. It suggests:

$\begin{array}{1 1}a)\; Bromintion\; decreases\; the\; furthure\; rate \;of\; bromination\\b)\; Bromination\; increases\; the\; further\; rate\; of \;bromination\\c) \;Bromintion\; decreases\; the\; rate\; of\; tautomerization\\d) \; Bromination \;increases \;the\; reactivity\; of\; carbonyl\; carbon\; due\; to\; its\; bigger\; size\end{array}$

1 Answer

on initial bromination we get ph$COCBr_2CH_3$ which naturally decreases further bromination due to bulkiness of the bromine atom. So,further bromination is difficult.
Hence (a) is the correct answer.
answered Dec 5, 2013 by sreemathi.v
 
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Tag:MathPhyChemBioOther
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