In 3,3-dimethylhex-1-ene-4-yne, there are _____ $$sp^3$$, _____ $$sp^2$$ and _____ $$sp$$ hybridised carbon atoms respectively :

The compound is named $$3,3\text{-dimethylhex}\!-\!1\text{-ene-4-yne}$$.

Step 1 : Draw the parent (hex-) chain and locate the multiple bonds
Hex means six carbon atoms in the parent chain. Numbering starts from the end that gives the lowest locants to the multiple bonds.
  • Double bond at C-1 ⇒ $$C_1 = C_2$$.
  • Triple bond between C-4 and C-5 ⇒ $$C_4 \equiv C_5$$.
Thus the skeleton is
$$C_1{=}C_2{-}C_3{-}C_4{\equiv}C_5{-}C_6$$.

Step 2 : Add the substituents
“3,3-dimethyl” means two methyl groups (-$$CH_3$$) attached to C-3:
$$C_1{=}C_2{-}C_3(-CH_3)(-CH_3){-}C_4{\equiv}C_5{-}C_6$$.

Step 3 : Assign the hybridisation of each carbon

Step 4 : Count the hybridised carbons
$$sp^3$$ carbons : C-3, C-6, and two methyl carbons ⇒ $$4$$.
$$sp^2$$ carbons : C-1, C-2 ⇒ $$2$$.
$$sp$$ carbons : C-4, C-5 ⇒ $$2$$.

Therefore the molecule contains $$4$$ $$sp^3$$, $$2$$ $$sp^2$$ and $$2$$ $$sp$$ hybridised carbon atoms.

The correct option is Option A $$4,\;2,\;2$$.