Question

$C{H}_2-\stackrel{OH}{\stackrel{|}{C}}=\underset{Cl}{\underset{|}{CH}}-\stackrel{O-Me}{\stackrel{|}{CH}}-C=C-COOH$

Correct IUPAC name of the compound :

• A. 4-methoxy-5-chloro-6-hydroxy heptanoic acid
• B. 5-chloro-4-methoxy-6-hydroxy heptan-2-en-3-ynoic acid
• C. 5-chloro-6-hydroxy-4-methoxy heptan-5-en-2-ynoic acid
• D. 4-chloro-5-ethoxy-6-hydroxy heptan-5-yn-2-enoic acid

Answer 1

The correct option is C 5-chloro-6-hydroxy-4-methoxy heptan-5-en-2-ynoic acid

$\underset{7}{C}{H}_3-\stackrel{\stackrel{OH}{|}}{\underset{6}{C}}=\stackrel{5}{\underset{Cl}{\underset{|}{CH}}}-\stackrel{O-Me}{\stackrel{|}{\underset{4}{CH}}}-\underset{3}{C}\equiv \stackrel{2}{C}-\stackrel{1}{C}OOH$

Rules for Naming Hydrocarbons:-

Rule-I Selection of the longest carbon chain.

Rule-II Numbering the longest chain: Longest chain is numbered starting from a terminal carbon. Numbering is done is such a way that lowest possible number is given to the side cahin or a functional group.

Rule-III Writing names of branched, saturated hydrocarbons, locants are written first in alphabetical order followed by name of parent chain.

Rule-IV Some common names are retained in $IUPAC$ system for unsubstituted hydrocarbon only.

Correct $IUPAC$ name of the compound is $5-chloro-6-hydroxy-4-methoxy$ $heptan-5-en-2-ynoic$ $acid$