- ✓$4$ 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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[Use mass of electron $\left.=9.1 \times 10^{-31}\, \mathrm{~kg}, \mathrm{~h}=6.63 \times 10^{-34}\, \mathrm{~J} \mathrm{~s}, \pi=3.14\right]$
$(I)$ Only compounds with chiral centers can be optically active
$(II)$ Absence of elements of symmetry is the reason for a molecule's optical activity
$(II)$ All organic compounds contain carbon & hydrogen
$(IV)$ Different cannonical forms of a molecule represent the actual structures of a molecule which has resonace in it
