MCQ
The $ IUPAC$ name of $C{H_3}C \equiv CCH\,{(C{H_3})_2}$ is
- ✓$4$ 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.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Above reaction is known as
| List-I Tetrahedral Complex | List-II Electronic configuration | ||
| (A) | $TiCl _4$ | (I) | $e ^2, t _2^0$ |
| (B) | $\left[ FeO _4\right]^{2-}$ | (II) | $e ^4, t _2^3$ |
| (C) | $\left[ FeCl _4\right]^{-}$ | (III) | $e^0, t_2^0$ |
| (D) | $\left[ CoCl _4\right]^{2-}$ | (IV) | $e ^2, t _2^3$ |