When $CH_3-CH_2-COOH$ is reduced with $LiAlH_4$,the compound obtained will be

  • A
    $CH_3CH_2COOH$
  • B
    $CH_3CH_2CHO$
  • C
    $CH_3CH_2CH_2OH$
  • D
    $CH_3-CH=CH-OH$

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Arrange the products $I$,$II$,$III$ from the following reactions in decreasing order of their acid strength.
$A) \text{Propylbenzene} \xrightarrow[\text{ (ii) } H_3O^{+}]{\text{ (i) } KMnO_4 / OH^{-}} I$
$B) CH_3COOH \xrightarrow[\text{ (ii) } H_2O]{\text{ (i) } Br_2 / \text{red } P} II$
$C) \text{Bromobenzene}$ $\xrightarrow[\substack{\text{ (ii) } CO_2 / \text{dry ether} \\ \text{ (iii) } H_3O^{+}}]{\text{ (i) } Mg / \text{dry ether}} III$

The dehydration of malonic acid $CH_2(COOH)_2$ with $P_4O_{10}$ gives:

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