MCQ
$IUPAC$ name of $C{H_3}CH(OH)C{H_2}C{H_2}COOH$ is
- ✓$4-$ hydroxy pentanoic acid
- B$1-$ carboxy $-3-$ butanoic acid
- C$1-$ carboxy $-4-$ butanol
- D$4-$ carboxy $-2-$ butanol
Therefore, the IUPAC name of given compound is $4$-Hydroxypentanoic acid.
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(Mass of proton $= 1.67 \times 10^{-27}\, kg$ and $h = 6.63 \times 10^{-34}\, Js$)
How can this reaction is made to proceed in forward direction?
$CH_4(g) +2O_2(g) \rightarrow CO_2(g)+2H_2O(l)\,;$ $\Delta H= -890 \,kJ$
$CO_2(g) \rightarrow C($graphite$) + O_2(g) \,;$ $\Delta H= 393 \,kJ$
$2H_2O(l) \rightarrow 2H_2(g) + O_2(g) \,;$ $\Delta H= 571 \,kJ$
$2H_2(g) \rightarrow 4H(g)\, ;$ $\Delta H= 871\, kJ$
$C($graphite$) \rightarrow C(g)\, ;$ $\Delta H= 716\, kJ$