The IUPAC name of the following compound is :

The given structure is $$HC \equiv C - CH_2 - CH(OH) - CH_2 - CH = CH_2$$.

Step 1: Select the parent chain
The longest continuous chain that contains the hydroxyl group as well as both the double and the triple bonds has seven carbon atoms, so the parent hydrocarbon is $$hept$$.

Step 2: Decide the principal characteristic group
According to IUPAC priority order, the $${-}OH$$ group (alcohol) has higher suffix priority than double or triple bonds. Hence the compound will be named with the suffix $$-ol$$, not with the prefix “hydroxy”.

Step 3: Number the chain
We need the lowest possible set of locants for the principal group, the multiple bonds, and any substituents.
Case A (left → right):
  C-1≡C-2-C-3-C-4(OH)-C-5-C-6= C-7
Locants: 1-yne, 4-ol, 6-ene ⇒ set $$(1,4,6)$$.

Case B: (right → left):
  C-1= C-2-C-3-C-4(OH)-C-5-C-6≡C-7
Locants: 1-ene, 4-ol, 6-yne ⇒ set $$(1,4,6)$$.

Both directions give the same numerical set $$(1,4,6)$$. When a tie exists, IUPAC rules state that the lower locant is assigned to the double bond (-ene) rather than to the triple bond (-yne). Therefore Case B is preferred, keeping the double bond at $$C_1$$.

Step 4: Write the complete name
• Parent chain: $$hept$$ (7 carbons)
• Double bond between C-1 and C-2 ⇒ $$1-en$$
• Triple bond between C-6 and C-7 ⇒ $$6-yn$$
• Hydroxyl group on C-4 ⇒ $$4-ol$$ (suffix of highest priority)

Combining these parts in the correct order (ene, yne, ol):
$$\text{Hept-1-en-6-yn-4-ol}$$

Step 5: Match with the options
Option D states “Hept-1-en-6-yn-4-ol”, which is exactly the name obtained.

Hence, the correct answer is Option D (Hept-1-en-6-yn-4-ol).