Identify the correct structure of the compounds whose IUPAC names are respectively as follows:
(i) 4-Chloro-3-ethylbutan-1-ol and (ii) Cyclopent-3-en-1-ol
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → the pair of structures shown in Option (1)
To draw a structure from a name, read the name from the end: the parent chain or ring and its suffix first, then the numbered prefixes. The locants count from the carbon carrying the principal group (here –OH, which gets C1).
(i) 4-Chloro-3-ethylbutan-1-ol: butan-1-ol is a four-carbon chain with –OH on C1. An ethyl group is on C3 and a chlorine atom on C4:
\( \displaystyle \ce{HO-CH2-CH2-CH(C2H5)-CH2-Cl} \)
(ii) Cyclopent-3-en-1-ol: a five-membered ring with –OH on C1 and the double bond between C3 and C4, that is, on the side of the ring opposite the C–OH carbon, not next to it.
Option (1) shows both: \( \ce{Cl-CH2} \) on the carbon bearing the ethyl group, two carbons away from the \( \ce{CH2OH} \) end, and a cyclopentene ring with the double bond away from the –OH carbon.
Why the other options are incorrect
Option (2) — Incorrect. In (i) the chlorine and the ethyl group sit on the same carbon (4-chlorohexan-1-ol), and in (ii) the double bond is next to the C–OH carbon (cyclopent-2-en-1-ol).
Option (3) — Incorrect. In (i) the –OH is on a carbon carrying a methyl group, a secondary alcohol (4-chlorohexan-2-ol), and in (ii) the –OH is on a doubly bonded carbon (an enol).
Option (4) — Incorrect. In (i) the –OH and Cl are on the same carbon (1-chloro-2-ethylbutan-1-ol), and in (ii) the double bond is again next to the C–OH carbon.
Note: It seems that there is a mistake in the original NTA paper, which prints 4-chloro-3-ethylbutan-1-ol when it should read 3-(chloromethyl)pentan-1-ol (the longest chain through the C–OH carbon has five carbons).