- ✓$4$ methyl$-2 $ pentyne
- B$4, 4-$dimethyl$-2-$butyne
- CMethyl isopropyl acetylene
- D$2-$methyl$-4-$pentyne
Now, we start numbering from the leftmost carbon as by doing that, the triple bond will be on $2^{nd}$ carbon.
Hence, the name of the compound will be $4$methyl-$2$-pentyne as there are $5$ carbons on the parent chain.
Hence, (a) is correct answer.
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$(A)$ $E, F$ and $\mathrm{G}$ are resonance structures
$(B)$ $E, F$ and $E, G$ are tautomers
$(C)$ $\mathrm{F}$ and $\mathrm{G}$ are geometrical isomers
$(D)$ $\mathrm{F}$ and $\mathrm{G}$ are diasteromers
$2{C_4}{H_{10(g)}}\, + \,13{O_{2(g)}}\, \to \,8C{O_{2(g)}}\, + \,10{H_2}{O_{(l)}}\,;$ $\Delta {H^o}\, = \, - \,5756\,KJ$
$ \mathrm{N}_2=3.0 \times 10^{-3} \mathrm{M}, \mathrm{O}_2=4.2 \times 10^{-3} \mathrm{M} \text { and } \mathrm{NO}=2.8 \times 10^{-3} \mathrm{M} . $
$ 2 \mathrm{NO}_{(\mathrm{g})} \rightleftharpoons \mathrm{N}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})}$
If $0.1 \mathrm{~mol} \mathrm{~L} \mathrm{~L}^{-1}$ of $\mathrm{NO}_{(\mathrm{g})}$ is taken in a closed vessel, what will be degree of dissociation ( $\alpha$ ) of $\mathrm{NO}_{(\mathrm{g})}$ at equilibrium?